Transcript bachelor girl n 1

ELEMENTARY
STATISTICS
Introduction to Geostatistics
Dept of Energy & Mineral Resources Eng.,
Sejong University
Myung Chae JUNG
ELEMENTARY
MARIO F. TRIOLA
STATISTICS
EIGHTH
EDITION
Handling Data and Figures of Merit
Data come in different formats
time
Histograms
Lists
But….
Can contain the same information about quality
What is meant by quality?
(figures of merit)
Precision, separation (selectivity), limits of detection,
Linear range
My weight
day
weight
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140.1
139.8
140.6
140
139.8
139.6
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140.8
139.7
140.2
141.7
141.9
141.4
142.3
142.3
141.9
142.1
142.5
142.3
142.1
142.5
143.5
143
143.2
143
143.4
143.5
142.7
143.7
day
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weight
day
143.9
144
142.5
142.9
142.8
143.9
144
144.8
143.9
144.5
143.9
144
144.2
143.8
143.5
143.8
143.2
143.5
143.6
143.4
143.9
143.6
144
143.8
143.6
143.8
144
144.2
144
143.9
weight
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Plot as a function of time data was acquired:
144
144.2
144.5
144.2
143.9
144.2
144.5
144.3
144.2
144.9
144
143.8
144
143.8
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144.5
143.7
143.9
144
144.2
144
144.4
143.8
144.1
day
Comments:
background is white (less ink);
Font size is larger than Excel
default (use 14 or 16)
146
145
144
weight (lbs)
weight
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140
140.1
139.8
140.6
140
139.8
139.6
140
140.8
139.7
140.2
141.7
141.9
141.4
142.3
142.3
141.9
142.1
142.5
142.3
142.1
142.5
143.5
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143.2
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143.4
143.5
142.7
143.7
day
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weight
day
143.9
144
142.5
142.9
142.8
143.9
144
144.8
143.9
144.5
143.9
144
144.2
143.8
143.5
143.8
143.2
143.5
143.6
143.4
143.9
143.6
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143.8
143.6
143.8
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144.2
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143.9
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142
Do not use curved lines to connect data
points
– that assumes you know more about the
relationship of the data than you really do
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Day
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weight
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144
144.2
144.5
144.2
143.9
144.2
144.5
144.3
144.2
144.9
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143.8
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143.8
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144.5
143.7
143.9
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144.2
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144.4
143.8
144.1
Bin refers to what groups of
weight to cluster. Like
A grade curve which lists
number of students who got
between 95 and 100 pts
95-100 would be a bin
Assume my weight is a single, random, set of similar data
25 Make a frequency chart (histogram) of the data
146
145
# of Observations
144
weight (lbs)
20
143
142
141
15
140
139
0
10
20
30
40
50
60
Day
10
5
0
Weight (lbs)
Create a “model” of my weight and determine average
Weight and how consistent my weight is
25
average
143.11
# of Observations
20
15
10
Inflection pt
s = 1.4 lbs
5
0
Weight (lbs)
s = standard deviation
= measure of the consistency, or similarity, of weights
Characteristics of the Model Population
(Random, Normal)
A
f  x 
e
s 2
1 x m 
 

2 s 
2
Peak height, A
Peak location (mean or average), m
Peak width, W, at baseline
Peak width at half height, W1/2
Standard deviation, s, estimates the variation in an infinite population, s
Related concepts
0.45
0.4
0.35
Amplitude
Width is measured
At inflection point =
s
0.3
0.25
0.2
W1/2
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
0
1
2
3
4
s
Triangulated peak: Base width is 2s < W < 4s
5
0.45
0.4
Pp = peak to peak – or
– largest separation of
measurements
0.35
+/- 1s Area = 68.3%
Amplitude
0.3
pp ~ 6s
0.25
0.2
0.15
0.1
Area +/- 2s = 95.4%
0.05
0
-5
-4
-3
-2
Area +/- 3s = 99.74 %
-1
0
1
2
3
4
5
s
Peak to peak is sometimes
Easier to “see” on the data vs time plot
pp ~ 6s
(Calculated s= 1.4)
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144.9
145
Peak to
peak
143
25
142
20
# of Observations
weight (lbs)
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141
15
10
5
140
139.5
0
Weight (lbs)
139
0
10
20
30
Day
s~ pp/6 = (144.9-139.5)/6~0.9
40
50
60
There are some other important characteristics of a normal (random)
population
0.45
0.4
0.35
0.3
Amplitude
0.25
1st derivative
0.2
2nd derivative
0.15
0.1
0.05
0
-5
-4
-3
-2
-1
-0.05
0
1
2
3
4
s
Scale up the first derivative and second derivative to see better
5
Population, 0th derivative
0.6
0.4
Amplitude
0.2
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
-0.2
-0.4
-0.6
2nd derivative
Peak is at the inflection
Of first derivative – should
Be symmetrical for normal
Population; goes to zero at
Std. dev.
-0.8
-1
s
1st derivative,
Peak is at the inflection
Determines the std. dev.
Asymmetry can be determined from principle component analysis
A. F. (≠Alanah Fitch) = asymmetric factor
Comparing TWO populations of measurements
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School Begins
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Baseline
Vacation
weight (lbs)
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30
Day
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60
Is there a difference between my “baseline” weight and school weight?
Can you “detect” a difference? Can you “quantitate” a difference?
Exact same information displayed differently, but now we divide
The data into different measurement populations
25
school
# of Observations
20
15
10
baseline
5
0
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139
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Weight (lbs)
Model of the data as two normal populations
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weight (lbs)
# of Observations
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15
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139
0
10
20
30
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50
60
Standard deviation
Of the school weigh
Day
10
Standard deviation
Of baseline weight
5
0
138
139
140
141
Average
Baseline weight
142
143
Weight (lbs)
144
145
146
Average school
weight
147
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20
20
# of Observations
15
10
15
10
5
5
0
138
0
139
140
141
142
Weight (lbs)
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147
Weight (lbs)
We have two models to describe the population of measurements
Of my weight.
In one we assume that all measurements fall into a single population.
In the second we assume that the measurements
Have sampled two different populations.
25
20
Which is the better model?
How to we quantify “better”?
# of Observations
# of Observations
25
15
10
5
0
138
139
140
141
142
143
Weight (lbs)
144
145
146
147
25
The red bars represent the difference
Between the two population model and
The data
# of Observations
20
15
10
5
Compare how close
The measured data
Fits the model
The purple lines represent
The difference between
The single population
Model and the data
Which model
Has less summed
differences?
0
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Weight (lbs)
Did I gain weight?
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147
Normally sum the square of the difference in order to account for
Both positive and negative differences.
This process (summing of the squares of the differences)
Is essentially what occurs in an ANOVA
Analysis of variance
In the bad old days you had to work out all the sums of squares.
In the good new days you can ask Excel program to do it for you.
Anova: Single Factor
5% certainty
SUMMARY
Groups
Count
Column 1
12
Column 2
12
ANOVA
Source of Variation
Between Groups
Within Groups
SS
194.4273
167.2408
Total
361.6682
Sum
Average
277.41 23.1175
345.72
28.81
df
MS
1 194.4273
22 7.601856
Variance
8.70360227
6.50010909
F
P-value
F crit
25.5762995 4.59E-05 4.300949
Source of Variation
23
Test: is F<Fcritical? If true = hypothesis true, single population
if false = hypothesis false, can not be explained
by a single population at the
5% certainty level
0.3
0.35
Red, N=12, Sum sq diff=0.11, stdev=3.27 White, N=12, Sum sq diff=0.037, stdev=2.55
Red, N=40, Sum sq diff=0.017, stdev-2.67 White, N=38, Sum sq diff=0.028, stdev=2.15
N=24 Sum sq diff=0.0449, stdev=3.96
N=78, sum sq diff=0.108, stdev=4.05
0.3
0.25
0.25
Frequency
Frequency
0.2
0.2
0.15
0.15
0.1
0.1
0.05
0.05
0
0
14
19
24
29
Length (cm)
34
39
14
19
24
29
34
39
Length, cm
In an Analysis of Variance you test the hypothesis that the sample is
Best described as a single population.
1. Create the expected frequency (Gaussian from normal error curve)
2. Measure the deviation between the histogram point and the expected
frequency
3. Square to remove signs
4. SS = sum squares
5. Compare to expected SS which scales with population size
6. If larger than expected then can not explain deviations assuming a
single population
0.3
0.35
Red, N=12, Sum sq diff=0.11, stdev=3.27 White, N=12, Sum sq diff=0.037, stdev=2.55
Red, N=40, Sum sq diff=0.017, stdev-2.67 White, N=38, Sum sq diff=0.028, stdev=2.15
N=24 Sum sq diff=0.0449, stdev=3.96
N=78, sum sq diff=0.108, stdev=4.05
0.3
0.25
0.25
Frequency
0.2
0.15
0.15
0.1
0.1
0.05
0.05
0
0
14
19
24
29
34
14
39
19
24
29
34
39
Length, cm
Length (cm)
0.04
0.035
Square Difference Expected Measured
Frequency
0.2
0.03
0.025
0.02
0.015
0.01
0.005
0
15
17
19
21
23
25
Length (cm)
27
29
31
33
35
The square differences
For an assumption of
A single population
Is larger than for
The assumption of
Two individual
populations
There are other measurements which describe the two populations
Resolution of two peaks
xa  xb
R
Wa Wb

2
2
Mean or average
Baseline width
x a  xb
4.5
xa
xb
4
3.5
Signal
Wa Wb
3
2 2.5 2
2
1.5
Wa
2
1
Wb
2
0.5
0
1
1.5
In this example
R  1:
Wa Wb
xa  xb 

2
2
2
2.5
3
3.5
4
x
Peaks are baseline resolved when R > 1
x a  xb
1.8
xa
xb
1.6
1.4
Signal
Wa Wb
1.2
2 12
0.8
0.6
Wa
2
0.4
Wb
2
0.2
0
1
1.5
In this example
R  1:
2
Wa Wb
xa  xb 

2
2
2.5
x
3
3.5
Peaks are just baseline
resolved when R = 1
4
x a  xb
1.6
xa
xb
1.4
1.2
Signal
Wa Wb
1
2
2
0.8
0.6
Wa
2
0.4
0.2
Wb
2
0
1
1.5
In this example
R  1:
2
Wa Wb
xa  xb 

2
2
2.5
3
3.5
x
Peaks are not baseline resolved
when R < 1
4
2008 Data
0.35
White, N=12, Sum sq diff=0.037
Red, N=12, Sum sq diff=0.11
0.3
xp 
Frequency
0.25
0.2
1
2
W
R
 WW 
R1
0.15
0.1
0.05
0
14
19
24
29
Length (cm)
What is the R for this data?
34
39
Comparison of 1978 Low Lead to 1978 High Lead
25
20
Comparison of 1978 Low Lead to 1979 High Lead
25
% Measured
20
15
15
10
10
5
5
0
0
0
0
20
40
60
80
100
120
140
20
160
40
60
80
Series2
100
120
140
Series3
IQ Verbal
Visually less resolved
Anonymous 2009 student analysis of Needleman data
Wa
 ~ 112   ~ 70  42
2
Wb
 ~ 130   ~ 95  35
2
R
xa  xb
Wa Wb

2
2
Visually better resolved
160
Comparison of 1978 Low Lead to 1978 High Lead
25
20
Comparison of 1978 Low Lead to 1979 High Lead
25
% Measured
20
15
15
10
10
5
5
0
0
0
0
20
40
60
80
100
120
140
20
40
60
80
Series2
160
100
120
140
Series3
IQ Verbal
Visually less resolved
Visually better resolved
Anonymous 2009 student analysis of Needleman data
Wa
 ~ 112   ~ 70  42
2
Wb
 ~ 130   ~ 95  35
2
x a  x b  ~ 112  ~ 95  17
R
xa  xb
17
~
 0.22
Wa Wb
42  35

2
2
160
Other measures of the quality of separation of the
Peaks
1. Limit of detection
2. Limit of quantification
3. Signal to noise (S/N)
X blank
X limit of detection
0.45
99.74%
Of the observations
Of the blank will lie
below the mean of the
First detectable signal
(LOD)
0.4
0.35
Amplitude
0.3
0.25
0.2
0.15
0.1
0.05
0
-6
-4
-2
0
x LOD  xblank  3sblank
2
3s 4
s
6
8
10
12
Two peaks are visible when all the data is summed together
0.45
0.4
0.35
Amplitude
0.3
0.25
0.2
0.15
0.1
0.05
0
-6
-4
-2
0
2
3s
s
4
6
8
10
12
146
25
# of Observations
20
145
15
10
5
144
0
138
139
140
141
142
143
144
145
146
147
weight (lbs)
Weight (lbs)
143
142
141
140
139
0
10
20
30
Day
Estimate the LOD (signal) of this
data
40
50
60
Other measures of the quality of separation of the
Peaks
1. Limit of detection
2. Limit of quantification
3. Signal to noise (S/N)
x LOQ  xblank  9sblank
Your book suggests 10
0.45
0.4
0.35
Amplitude
0.3
0.25
0.2
0.15
0.1
0.05
0
-6
-4
-2
0
2
4
6
8
9s
10
12
Limit of squantification requires absolute
Certainty that no blank is part of the
146
25
# of Observations
20
145
15
10
5
144
0
138
139
140
141
142
143
144
145
146
147
weight (lbs)
Weight (lbs)
143
142
141
140
139
0
10
20
30
Day
Estimate the LOQ (signal) of this
data
40
50
60
Other measures of the quality of separation of the
Peaks
1. Limit of detection
2. Limit of quantification
3. Signal to noise (S/N)
Signal = xsample - xblank
Noise = N = standard deviation, s
x sample  xblank x sample  xblank
S


N
s
 pp 
 
 6
(This assumes pp school ~ pp baseline)
146
25
# of Observations
20
145
School Begins
15
Baseline
Vacation
10
5
144
0
138
139
140
141
142
143
144
145
146
147
weight (lbs)
Weight (lbs)
143
Peak to peak variation within
mean school
~ 6s where s = N for Noise
142
Signal
141
140
139
0
10
20
30
Estimate the S/N of this data Day
40
50
60
35
30
length (cm)
25
20
15
Can you “tell” where the switch between
Red and white potatoes begins?
10
What is the signal (length of white)?
What is the background (length of red)?
What is the S/N ?
5
0
0
5
10
15
Sample number
20
25
30
Effect of sample size on the measurement
Error curve
Peak height grows with # of measurements.
+ - 1 s always has same proportion of total number of measurements
However, the actual value of s decreases as population grows
ssample 
s population
nsample
2008 Data
27
5
4.5
4
26
3
25
2.5
24.5
2
24
Red Running Stdev
3.5
25.5
ssample 
1.5
23.5
1
23
0.5
22.5
s population
nsample
0
0
2
4
6
8
10
12
14
Sample number
4.1
3.9
3.7
stdev red length cm
Red Running Length Average
26.5
3.5
3.3
3.1
y = -0.8807x + 5.9303
R2 = 0.9491
2.9
2.7
2.5
1.5
2
2.5
3
sqrt number of samples
3.5
4
0.35
Red, N=12, Sum sq diff=0.11, stdev=3.27 White, N=12, Sum sq diff=0.037, stdev=2.55
Red, N=40, Sum sq diff=0.017, stdev-2.67 White, N=38, Sum sq diff=0.028, stdev=2.15
0.3
Frequency
0.25
0.2
0.15
0.1
0.05
0
14
19
24
29
Length (cm)
34
39
Calibration Curve
A calibration curve is based on a selected measurement as linear
In response to the concentration of the analyte.
y  a  bx
Or… a prediction of measurement due to some change
Can we predict my weight change if I had spent a longer time on
Vacation?
fitch lbs  a  bdays on vacation
25
# of Observations
20
15
10
5
0
138
139
140
141
142
143
Weight (lbs)
144
145
146
147
5 days
fitch lbs  a  bdays on vacation
The calibration curve contains information about the sampling
Of the population
143
Can get this by using “trend line”
142.5
Fitch Weight, lbs
142
y = 0.3542x + 140.04
2
R = 0.7425
141.5
141
140.5
140
139.5
139
0
1
2
3
Days on Vacation
4
5
6
This is just a trendline
From “format” data
4.1
3.9
stdev red length cm
3.7
3.5
3.3
3.1
y = -0.8807x + 5.9303
R2 = 0.9491
2.9
2.7
Sample
1
2
3
4
5
6
7
8
9
10
11
12
sqrt(#samples)
1
1.414213562
1.732050808
2
2.236067977
2.449489743
2.645751311
2.828427125
3
3.16227766
3.31662479
3.464101615
stdev
#DIV/0!
2.036468
4.475727
4.31441
3.844045
3.844604
3.735124
3.458414
3.235055
3.093053
2.935944
2.950187
2.5
1.5
2
2.5
3
3.5
4
sqrt number of samples
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.296113395
R Square
0.087683143
Adjusted R Square
-0.013685397
Standard Error
0.703143388
Observations
11
Using the analysis
Data pack
ANOVA
df
Regression
Residual
Total
Intercept
X Variable 1
1
9
10
Coefficients
3.884015711
-0.06235252
SS
MS
F
Significance F
0.427662048 0.427662 0.864994 0.376617
4.449695616 0.494411
4.877357664
Standard Error
t Stat
P-value Lower 95%
0.514960076 7.542363 3.53E-05 2.719094
0.067042092 -0.93005 0.376617 -0.21401
Get an error
Associated with
The intercept
In the best of all worlds you should have a series of blanks
That determine you’re the “noise” associated with the background
x LOD  xblank  3sblank
Sometimes you forget, so to fall back and punt, estimate
The standard deviation of the “blank” from the linear regression
But remember, in doing this you are acknowledging
A failure to plan ahead in your analysis
x LOD  x blank  b[conc. LOD]
Signal LOD
Sensitivity (slope)
x LOD  xblank  3sblank
x blank  3sblank  x blank  b[conc. LOD]
3sblank
[conc. LOD] 
b
Extrapolation of the associated error
Can be obtained from the Linear
Regression data
The concentration LOD depends on BOTH
Stdev of blank and sensitivity
!!Note!!
Signal LOD ≠ Conc LOD
We want Conc. LOD
Selectivity
pHpM
or pM
pH or
0
0
12 12
10 10
8
8
6
6
4
4
2
2
0
0
Difference in slope is one measure selectivity
-50 -50
Pb2+
y = -31.143x - 74.333
R2 = 0.9994
mV
-150
+
-150
H
-200-200
-250-250
y = -41x - 118.5
R2 = 0.9872
In a perfect method the sensing device would have zero
Slope for the interfering species
-300-300
-350-350
mV
-100-100
Limit of linearity
5% deviation
Summary: Figures of Merit Thus far
R = resolution
S/N
LOD = both signal and concentration
Can be expressed in terms of signal, but better
LOQ
Expression is in terms of concentration
LOL
Sensitivity (calibration curve slope)
Selectivity (essentially difference in slopes)
Tests: Anova
Why is the limit of detection important?
Why has the limit of detection changed so much in the
Last 20 years?
The End
25
20
20
% of Measurements
% of Measurements
25
15
10
15
10
5
5
0
0
40
60
80
100
120
Verbal IQ
140
160
40
60
80
100
120
140
Verbal IQ
Which of these two data sets would be likely
To have better numerical value for the
Ability to distinguish between two different
Populations?
Needleman’s data
160
Height for normalized
Bell curve <1
2008 Data
0.35
White, N=12, Sum sq diff=0.037
Red, N=12, Sum sq diff=0.11
0.3
Frequency
0.25
0.2
0.15
0.1
0.05
0
14
19
24
29
Length (cm)
Which population is more variable?
How can you tell?
34
39
0.35
Red, N=12, Sum sq diff=0.11, stdev=3.27 White, N=12, Sum sq diff=0.037, stdev=2.55
Red, N=40, Sum sq diff=0.017, stdev-2.67 White, N=38, Sum sq diff=0.028, stdev=2.15
0.3
Frequency
0.25
0.2
0.15
0.1
0.05
0
14
19
24
29
34
39
Length (cm)
Increasing the sample size decreases the std dev and increases separation
Of the populations, notice that the means also change, will do so until
We have a reasonable sample of the population
25
% of Measurements
20
15
10
5
0
40
60
80
100
Verbal IQ
120
140
160
25
% of Measurements
20
15
10
5
0
40
60
80
100
Verbal IQ
120
140
160
ELEMENTARY
MARIO F. TRIOLA
STATISTICS
EIGHTH
EDITION
Chapter 1 Introduction to Statistics
1-1 Overview
1-2 The Nature of Data
1-3 Uses and Abuses of Statistics
1-4 Design of Experiments
Chapter 2 Describing, Exploring, and Comparing Data
2-1 Overview
2-2 Summarizing Data with Frequency Tables
2-3 Pictures of Data
2-4 Measures of Center
2-5 Measures of Variation
2-6 Measures of Position
2-7 Exploratory Data Analysis (EDA)
Chapter 3 Probability
3-1 Overview
3-2 Fundamentals
3-3 Addition Rule
3-4 Multiplication Rule: Basics
3-5 Multiplication Rule: Complements and Conditional Probability
3-6 Counting
Chapter 4 Probability Distributions
4-1 Overview
4-2 Random Variables
4-3 Binomial Probability Distributions
4-4 Mean, Variance, and Standard Deviation for the Binomial Distribution
Chapter 5 Normal Probability Distributions
5-1 Overview
5-2 The Standard Normal Distribution
5-3 Nonstandard Normal Distributions: Finding Probabilities
5-4 Nonstandard Normal Distributions: Finding Values
5-5 The Central Limit Theorem
5-6 Normal Distribution as Approximation to Binomial Distribution
Chapter 6 Estimates and Sample Sizes
6-1 Overview
6-2 Estimating a Population Mean: Large Samples
6-3 Estimating a Population Mean: Small Samples
6-4 Determining Sample Size
6-5 Estimating a Population Proportion
6-6 Estimating a Population Variance
Chapter 7 Hypothesis Testing
7-1 Overview
7-2 Fundamentals of Hypothesis Testing
7-3 Testing a Claim about a Mean: Large Samples
7-4 Testing a Claim about a Mean: Small Samples
7-5 Testing a Claim about a Proportion
7-6 Testing a Claim about a Standard Deviation or Variance
Chapter 8 Inferences from Two Samples
8-1 Overview
8-2 Inferences about Two Means: Independent and Large Samples
8-3 Inferences about Two Means: Matched Pairs
8-4 Inferences about Two Proportions
Chapter 9 Correlation and Regression
9-1 Overview
9-2 Correlation
9-3 Regression
9-4 Variation and Prediction Intervals
9-5 Rank Correlation
Chapter 10 Multinomial Experiments and Contingency Tables
10-1 Overview
10-2 Multinomial Experiments: Goodness-0f-Fit
10-3 Contingency Tables: Independence and Homogeneity
10-4 One-Way ANOVA
ELEMENTARY
Chapter 1
STATISTICS
Introduction to Statistics
MARIO F. TRIOLA
EIGHTH
EDITION
Chapter 1
Introduction to Statistics
1-1 Overview
1-2 The Nature of Data
1-3 Uses and Abuses of Statistics
1-4 Design of Experiments
1-1
Overview
Statistics
Two Meanings
 Specific numbers
 Method of analysis
Statistics
 Specific number
numerical measurement determined by a
set of data
Example: Twenty-three percent of people
polled believed that there are
too many polls.
Statistics
 Method of analysis
a collection of methods for planning
experiments, obtaining data, and then
then organizing, summarizing, presenting,
analyzing, interpreting, and drawing
conclusions based on the data
Definitions
Population
the complete collection of all
elements (scores, people,
measurements, and so on) to be
studied. The collection is complete
in the sense that it includes all
subjects to be studied.
Definitions
Census
the collection of data from every
element in a population
Sample
a subcollection of elements drawn
from a population
1-2
The Nature of Data
Definitions
 Parameter
a numerical measurement describing
some characteristic of a population
Definitions
 Parameter
a numerical measurement describing
some characteristic of a population
population
parameter
Definitions
Statistic
a numerical measurement describing
some characteristic of a sample
Definitions
Statistic
a numerical measurement describing
some characteristic of a sample
sample
statistic
Definitions
Quantitative data
numbers representing counts or
measurements
Definitions
Quantitative data
numbers representing counts or
measurements
 Qualitative (or categorical or
attribute) data
can be separated into different categories
that are distinguished by some nonnumeric
characteristics
Definitions
Quantitative data
the incomes of college graduates
Definitions
Quantitative data
the incomes of college graduates
 Qualitative (or categorical or
attribute) data
the genders (male/female) of college
graduates
Definitions
Discrete
data result when the number of possible values
is
either a finite number or a ‘countable’
number of
possible values
0, 1, 2, 3, . . .
Definitions
Discrete
data result when the number of possible values
is
either a finite number or a ‘countable’
number of
possible values
0, 1, 2, 3, . . .
 Continuous
(numerical) data result from infinitely many
possible values that correspond to some
continuous scale
that 2covers a range
of
3
values without gaps,
interruptions,
Definitions
Discrete
The number of eggs that hens lay; for
example, 3 eggs a day.
Definitions
Discrete
The number of eggs that hens lay; for
example, 3 eggs a day.
 Continuous
The amounts of milk that cows produce;
for example, 2.343115 gallons a day.
Definitions
 nominal level of measurement
characterized by data that consist of names, labels, or
categories only. The data cannot be arranged in an
ordering scheme (such as low to high)
Example: survey responses yes, no, undecided
Definitions
 ordinal level of measurement
involves data that may be arranged in some order, but
differences between data values either cannot be
determined or are meaningless
Example: Course grades A, B, C, D, or F
Definitions
 interval level of measurement
like the ordinal level, with the additional property that the
difference between any two data values is meaningful.
However, there is no natural zero starting point (where
none of the quantity is present)
Example: Years 1000, 2000, 1776, and 1492
Definitions
 ratio level of measurement
the interval level modified to include the natural zero
starting point (where zero indicates that none of the
quantity is present). For values at this level, differences
and ratios are meaningful.
Example: Prices of college textbooks
Levels of Measurement
 Nominal - categories only
 Ordinal - categories with some order
 Interval - differences but no natural
starting point
 Ratio - differences and a natural starting
point
Levels of Measurement
 Nominal - categories only
 Ordinal - categories with some order
 Interval - differences but no natural
starting point
 Ratio - differences and a natural starting
point
ELEMENTARY
STATISTICS
Section 1-3 Uses and Abuses of Statistics
MARIO F. TRIOLA
EIGHTH
EDITION
Uses of Statistics
 Almost all fields of
study benefit from the
application of statistical methods
Abuses of Statistics
 Bad Samples
Definitions
self-selected survey
(or voluntary response sample)
one in which the respondents themselves decide
whether to be included
Abuses of Statistics




Bad Samples
Small Samples
Loaded Questions
Misleading Graphs
Figure 1-1 Salaries of People with Bachelor’s Degrees and with High
School Diplomas
$40,500
$40,500
$40,000
$40,000
35,000
30,000
30,000
20,000
25,000
$24,400
20,000
$24,400
10,000
0
Bachelor High School
Degree Diploma
(a)
Bachelor High Schoo
Degree(b) Diploma
We should analyze the numerical
information given in the graph
instead of being mislead by its
general shape.
Abuses of Statistics





Bad Samples
Small Samples
Loaded Questions
Misleading Graphs
Pictographs
Double the length, width, and height of a cube, and the
volume increases by a factor of eight
Figure 1-2
Abuses of Statistics
 Bad Samples
 Small Samples
 Loaded Questions
 Misleading Graphs
 Pictographs
 Precise Numbers
 Distorted Percentages
 Partial Pictures
“Ninety percent of all our cars sold in
this country in the last 10 years are still
on the road.”
Abuses of Statistics









Bad Samples
Small Samples
Loaded Questions
Misleading Graphs
Pictographs
Precise Numbers
Distorted Percentages
Partial Pictures
Deliberate Distortions
ELEMENTARY
Section 1-4
MARIO F. TRIOLA
STATISTICS
Design of Experiments
EIGHTH
EDITION
Definitions
 Observational Study
observing and measuring specific
characteristics without attempting to modify the
subjects being studied
Definitions
 Experiment
apply some treatment and then observe its
the subjects
effects on
Designing an Experiment
 Identify your objective
 Collect sample data
 Use a random procedure that
avoids bias
 Analyze the data and
form conclusions
Definitions
 Confounding
occurs in an experiment when the effects
from two or more variables cannot be
distinguished from each other
Definitions
 Replication
used when an experiment is repeated on a
sample of subjects that is large enough so that
we
can see the true nature of any effects
(instead of being
misled by erratic behavior of samples that are too small)
Definitions
 Random Sample
members of the population are selected in
such
a way that each has an equal chance of
being selected
Definitions
 Random Sample
members of the population are selected in
such
a way that each has an equal chance of
being selected
 Simple Random Sample (of size n)
subjects selected in such a way that every
possible sample of size n has the same
chance of being chosen
Random Sampling - selection so that each has
an equal chance of being selected
Systematic Sampling - Select some starting
point and then select every K th element in the population
Convenience Sampling - use results that are
readily available
Hey!
Do you believe
in the death
penalty?
Stratified Sampling - subdivide the population
into subgroups that share the same characteristic, then draw a
sample from each stratum
Cluster Sampling - divide the population into sections
(or clusters); randomly select some of those clusters; choose all
members from selected clusters
Methods of Sampling
 Random
 Systematic
 Convenience
 Stratified
 Cluster
Definitions

Sampling Error
the difference between a sample result and the true
population result; such an error results from chance
sample fluctuations.

Nonsampling Error
sample data that are incorrectly collected, recorded, or
analyzed (such as by selecting a biased sample, using
a defective instrument, or copying the data
incorrectly).
STATISTICS
ELEMENTARY
Chapter 2
Descriptive Statistics
MARIO F. TRIOLA
EIGHTH
EDITION
Chapter 2
Descriptive Statistics
2-1 Overview
2-2 Summarizing Data with Frequency Tables
2-3 Pictures of Data
2-4 Measures of Center
2-5 Measures of Variation
2-6 Measures of Position
2-7 Exploratory Data Analysis (EDA)
2 -1
Overview
 Descriptive Statistics
summarize or describe the important
characteristics of a known set of
population data
 Inferential Statistics
use sample data to make inferences (or
generalizations) about a population
Important Characteristics of Data
1. Center: A representative or average value that indicates
where the middle of the data set is located
2. Variation: A measure of the amount that
among themselves
the values vary
3. Distribution: The nature or shape of the distribution of data
(such as bell-shaped, uniform, or skewed)
4. Outliers: Sample values that lie very far away from the vast
majority of other sample values
5. Time: Changing characteristics of the data over time
2-2 Summarizing Data With
Frequency Tables
 Frequency Table
lists classes (or categories) of values,
along
with frequencies (or counts) of the
number of
values that fall into each class
Table 2-1
Qwertry Keyboard Word Ratings
2
2
5
1
2
6
3
3
4
2
4
0
5
7
7
5
6
6
8
10
7
2
2
10
5
8
2
5
4
2
6
2
6
1
7
2
7
2
3
8
1
5
2
5
2
14
2
2
6
3
1
7
Table 2-3
Frequency Table of Qwerty Word Ratings
Rating
Frequency
0-2
20
3-5
14
6-8
15
9 - 11
2
12 - 14
1
Frequency Table
Definitions
Lower Class Limits
are the smallest numbers that can actually
belong to different classes
Lower Class Limits
are the smallest numbers that can actually
belong to different classes
Rating
Frequency
0-2
20
3-5
14
6-8
15
9 - 11
2
12 - 14
1
Lower Class Limits
are the smallest numbers that can actually
belong to different classes
Rating
Lower Class
Limits
Frequency
0-2
20
3-5
14
6-8
15
9 - 11
2
12 - 14
1
Upper Class Limits
are the largest numbers that can actually belong
to different classes
Upper Class Limits
are the largest numbers that can actually belong
to different classes
Rating
Upper Class
Limits
Frequency
0-2
20
3-5
14
6-8
15
9 - 11
2
12 - 14
1
Class Boundaries
are the numbers used to separate classes, but without the
gaps created by class limits
Class Boundaries
number separating classes
Rating
Frequency
- 0.5
0-2
20
2.5
3-5
5.5
14
6-8
15
8.5
9 - 11
2
12
11.5- 14
1
14.5
Class Boundaries
number separating classes
Rating
Frequency
- 0.5
0-2
20
2.5
Class
Boundaries
3-5
5.5
14
6-8
15
8.5
9 - 11
2
12
11.5- 14
1
14.5
Class Midpoints
midpoints of the classes
Class Midpoints
midpoints of the classes
Rating
Class
Midpoints
Frequency
0- 1 2
20
3- 4
5
14
6- 7
8
15
9 - 10 11
2
12 - 13 14
1
Class Width
is the difference between two consecutive lower class limits
or two consecutive class boundaries
Class Width
is the difference between two consecutive lower class limits
or two consecutive class boundaries
Rating
Class Width
Frequency
3 0-2
20
3 3-5
14
3 6-8
15
3 9 - 11
2
3 12 - 14
1
Guidelines For Frequency Tables
1. Be sure that the classes are mutually exclusive.
2. Include all classes, even if the frequency is zero.
3. Try to use the same width for all classes.
4. Select convenient numbers for class limits.
5. Use between 5 and 20 classes.
6. The sum of the class frequencies must equal the
original data values.
number of
Constructing A Frequency Table
1.
Decide on the number of classes .
2. Determine the class width by dividing the range by the number
classes (range = highest score - lowest score) and round up.
class width
3.

round up of
of
range
number of classes
Select for the first lower limit either the lowest score or a
convenient value slightly less than the lowest score.
4.
Add the class width to the starting point to get the second lower
class
limit, add the width to the second lower limit to get the
third, and
so on.
5.
List the lower class limits in a vertical column and enter the
upper class limits.
6.
Represent each score by a tally mark in the appropriate class.
Total tally marks to find the total frequency for each class.
Figure 2-1
Relative Frequency Table
class frequency
relative frequency =
sum of all frequencies
Relative Frequency Table
Rating
Frequency
Relative
Rating Frequency
0-2
20
0-2
38.5%
3-5
14
3-5
26.9%
6-8
15
6-8
28.8%
9 - 11
2
9 - 11
3.8%
12 - 14
1
12 - 14
1.9%
Total frequency = 52
Table 2-5
20/52 = 38.5%
14/52 = 26.9%
etc.
Cumulative Frequency Table
Rating
Frequency
Cumulative
Frequency
Rating
0-2
20
Less than 3
20
3-5
14
Less than 6
34
6-8
15
Less than 9
49
9 - 11
2
Less than 12
51
12 - 14
1
Less than 15
52
Table 2-6
Cumulative
Frequencies
Frequency Tables
Rating Frequency
Rating
Relative
Frequency
Cumulative
Frequency
Rating
0-2
20
0-2
38.5%
Less than 3
20
3-5
14
3-5
26.9%
Less than 6
34
6-8
15
6-8
28.8%
Less than 9
49
9 - 11
2
9 - 11
3.8%
Less than 12
51
12 - 14
1
12 - 14
1.9%
Less than 15
52
Table 2-3
Table 2-5
Table 2-6
STATISTICS
ELEMENTARY
Section 2-3
Pictures of Data
MARIO F. TRIOLA
EIGHTH
EDITION
Pictures of Data
depict the nature or shape of the data
distribution
Histogram
a bar graph in which the horizontal scale
represents classes and the vertical scale
represents frequencies
Table 2-3
Frequency Table of Qwerty Word Ratings
Rating
Frequency
0-2
20
3-5
14
6-8
15
9 - 11
2
12 - 14
1
Histogram of Qwerty Word Ratings
Figure 2-2
Rating
Frequency
0-2
20
3-5
14
6-8
15
9 - 11
2
12 - 14
1
Relative Frequency Histogram
of Qwerty Word Ratings
Rating
Relative
Frequency
0-2
38.5%
3-5
26.9%
6-8
28.8%
9 - 11
3.8%
12 - 14
1.9%
Figure 2-3
Histogram
and
Relative Frequency Histogram
Figure 2-2
Figure 2-3
Frequency Polygon
Figure 2-4
Ogive
Figure 2-5
Dot Plot
Figure 2-6
Stem-and Leaf Plot
Stem
Raw Data (Test Grades)
67 72 85 75
89
89
88 90 99 100
6
7
8
9
10
Leaves
7
25
5899
09
0
Pareto Chart
45,000
40,000
35,000
30,000
Accidental Deaths by Type
Frequency
25,000
20,000
15,000
10,000
5,000
Firearms
Ingestion of food
or object
Fire
Poison
Falls
Drowning
Figure 2-7
Motor Vehicle
0
Pie Chart
Firearms
(1400. 1.9%)
Ingestion of food or object
(2900. 3.9%
Fire
(4200. 5.6%)
Motor vehicle
(43,500. 57.8%)
Drowning
(4600. 6.1%)
Poison
(6400. 8.5%)
Figure 2-8
Accidental Deaths by Type
Falls
(12,200. 16.2%)
Scatter Diagram
20
TAR
•
10
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
0
0.0
0.5
1.0
NICOTINE
1.5
Deaths in British Military Hospitals During the Crimean War
Figure 2-9
Other Graphs
 Boxplots (textbook section 2-7)
 Pictographs
 Pattern of data over time
STATISTICS
ELEMENTARY
Section 2-4
Measures of Center
MARIO F. TRIOLA
EIGHTH
EDITION
Measures of Center
a value at the
center or middle
of a data set
Definitions
Mean
(Arithmetic Mean)
AVERAGE
the number obtained by adding the values and
dividing the total by the number of values
Notation

denotes the addition of a set of values
is the variable usually used to represent the
individual
data values
x
n
represents the number of data values in a sample
N represents the number of data values in a
population
Notation
pronounced ‘x-bar’ and denotes the mean of a set of
x issample
values
x
x =
n
Notation
pronounced ‘x-bar’ and denotes the mean of a set of
x issample
values
x
x =
n
µ is pronounced ‘mu’ and denotes the mean of all
values
in a population
µ =
x
N
Calculators can calculate the mean of data
Definitions
 Median
the middle value when the original
data values are arranged in order of
increasing (or decreasing) magnitude
Definitions
 Median
the middle value when the original
data values are arranged in order of
increasing (or decreasing) magnitude
 often denoted by x
~
(pronounced
‘x-tilde’)
Definitions
 Median
the middle value when the original
data values are arranged in order of
increasing (or decreasing) magnitude
 often denoted by x
~
(pronounced
‘x-tilde’)
 is not affected by an extreme value
6.72
3.46
3.60
6.44
3.46
3.60
6.44
6.72
(even number of values)
no exact middle -- shared by two numbers
3.60 + 6.44
2
MEDIAN is 5.02
6.72
3.46
3.60
6.44
3.46
3.60
6.44
6.72
(even number of values)
no exact middle -- shared by two numbers
3.60 + 6.44
MEDIAN is 5.02
2
6.72
3.46
3.60
6.44
26.70
3.46
3.60
6.44
6.72
26.70
(in order exact middle
odd number of values)
MEDIAN is 6.44
Definitions
 Mode
the score that occurs most frequently
Bimodal
Multimodal
No Mode
denoted by M
the only measure of central tendency that can be used with
nominal data
Examples
a. 5 5 5 3 1 5 1 4 3 5
Mode is 5
b. 1 2 2 2 3 4 5 6 6 6 7 9
Bimodal -
c. 1 2 3 6 7 8 9 10
and 6
No Mode
2
Definitions
 Midrange
the value midway between the highest
lowest values in the original data set
and
Definitions
 Midrange
the value midway between the highest
lowest values in the original data set
highest score + lowest score
Midrange =
2
and
Round-off Rule for
Measures of Center
Carry one more decimal place than is present in
the original set of values
Mean from a Frequency Table
use class midpoint of classes for variable x
Mean from a Frequency Table
use class midpoint of classes for variable x
 (f • x)
x =
f
Formula 2-2
Mean from a Frequency Table
use class midpoint of classes for variable x
 (f • x)
x =
f
Formula 2-2
x = class midpoint
f = frequency
f=n
Weighted Mean
 (w • x)
x =
w
Best Measure of Center
Advantages - Disadvantages
Table 2-13
Definitions
 Symmetric
Data is symmetric if the left half of its
histogram is roughly a mirror of its
right half.
 Skewed
Data is skewed if it is not symmetric
and if it extends more to one side than the
other.
Skewness
Figure
2-13 (b)
Mode = Mean
= Median
SYMMETRIC
Skewness
Figure
2-13 (b)
Mode = Mean
= Median
SYMMETRIC
Mean
Mode
Median
Figure
2-13 (a)
SKEWED LEFT
(negatively)
Skewness
Figure
2-13 (b)
Mode = Mean
= Median
SYMMETRIC
Mean
Mode
Median
Figure
2-13 (a)
SKEWED LEFT
(negatively)
Mean
Mode
Median
SKEWED RIGHT
(positively)
Figure
2-13 (c)
STATISTICS
ELEMENTARY
Section 2-5
Measures of Variation
MARIO F. TRIOLA
EIGHTH
EDITION
Waiting Times of Bank Customers
at Different Banks
in minutes
Jefferson Valley Bank
6.5
6.6
6.7
6.8
7.1
7.3
7.4
7.7
7.7
7.7
Bank of Providence
4.2
5.4
5.8
6.2
6.7
7.7
7.7
8.5
9.3
10.0
Waiting Times of Bank Customers
at Different Banks
in minutes
Jefferson Valley Bank
6.5
6.6
6.7
6.8
7.1
7.3
7.4
7.7
7.7
7.7
Bank of Providence
4.2
5.4
5.8
6.2
6.7
7.7
7.7
8.5
9.3
10.0
Jefferson Valley Bank
Bank of Providence
Mean
7.15
7.15
Median
7.20
7.20
Mode
7.7
7.7
Midrange
7.10
7.10
Dotplots of Waiting Times
Figure 2-14
Measures of Variation
Measures of Variation
Range
highest
value
lowest
value
Measures of Variation
Standard Deviation
a measure of variation of the scores about
the mean
(average deviation from the mean)
Sample Standard Deviation Formula
Sample Standard Deviation Formula
S=
 (x - x)
n-1
2
Formula 2-4
calculators can compute the sample
standard deviation of data
Sample Standard Deviation Shortcut
Formula
n (x ) - (x)
n (n - 1)
2
s=
Formula 2-5
calculators can compute the sample
standard deviation of data
2
Mean Deviation Formula
(absolute deviation)
Mean Absolute Deviation Formula
 x-x
n
Population Standard Deviation
s =
 (x - µ)
2
N
calculators can compute the
population standard deviation of data
Symbols
for Standard Deviation
Symbols
for Standard Deviation
Sample
Symbols
for Standard Deviation
Sample
s
Symbols
for Standard Deviation
Sample
Textbook
s
Symbols
for Standard Deviation
Sample
Textbook
s
Sx
Symbols
for Standard Deviation
Sample
Textbook
Some graphics
calculators
s
Sx
Symbols
for Standard Deviation
Sample
Textbook
Some graphics
calculators
s
Sx
xsn-1
Symbols
for Standard Deviation
Sample
Textbook
Some graphics
calculators
Some
non-graphics
calculators
s
Sx
xsn-1
Symbols
for Standard Deviation
Sample
Textbook
Some graphics
calculators
Some
non-graphics
calculators
s
Sx
xsn-1
Population
Symbols
for Standard Deviation
Sample
Textbook
Some graphics
calculators
Some
non-graphics
calculators
Population
s
s
Sx
sx
xsn
xsn-1
Symbols
for Standard Deviation
Sample
Textbook
Some graphics
calculators
Some
non-graphics
calculators
Population
s
s
Sx
sx
xsn
xsn-1
Book
Some graphics
calculators
Some
non-graphics
calculators
Symbols
for Standard Deviation
Sample
Textbook
Some graphics
calculators
Some
non-graphics
calculators
Population
s
s
Sx
sx
xsn
xsn-1
Book
Some graphics
calculators
Some
non-graphics
calculators
Articles in professional journals and reports often use SD for
standard deviation and VAR for variance.
Measures of Variation
Variance
Measures of Variation
Variance
standard deviation squared
Measures of Variation
Variance
standard deviation squared
Notation
}
s
s
2
2
use square key
on calculator
Variance
2
s =
s
2
=
 (x - x )
2
n-1
 (x - µ)
N
2
Sample
Variance
Population
Variance
Round-off Rule
for measures of variation
Carry one more decimal place than is
present in the original set of values.
Round only the final answer, never in the middle of a
calculation.
Standard Deviation from a Frequency
Table
Formula 2-6
n [(f • x 2)] -[(f • x)]2
S=
n (n - 1)
Use the class midpoints as the x values
Calculators can compute the standard deviation for frequency
table
Estimation of Standard Deviation
Range Rule of Thumb
x - 2s
(minimum
usual value)
x + 2s
x
Range 
or
(maximum usual
value)
4s
Estimation of Standard Deviation
Range Rule of Thumb
x - 2s
(minimum
usual value)
Range 
or
s
x + 2s
x
Range
4
(maximum usual
value)
4s
Estimation of Standard Deviation
Range Rule of Thumb
x - 2s
x + 2s
x
(minimum
usual value)
Range 
(maximum usual
value)
4s
or
s
Range
4
=
highest value - lowest value
4
Usual Sample Values
Usual Sample Values
minimum ‘usual’ value  (mean) - 2 (standard
deviation)
minimum  x - 2(s)
Usual Sample Values
minimum ‘usual’ value  (mean) - 2 (standard
deviation)
minimum  x - 2(s)
maximum ‘usual’ value  (mean) + 2 (standard
deviation)
maximum  x + 2(s)
FIGURE 2-15
The Empirical Rule
(applies to bell-shaped distributions)
x
FIGURE 2-15
The Empirical Rule
(applies to bell-shaped distributions)
68% within
1 standard deviation
34%
x-s
34%
x
x+s
The Empirical Rule
FIGURE 2-15
(applies to bell-shaped distributions)
95% within
2 standard deviations
68% within
1 standard deviation
34%
34%
13.5%
x - 2s
13.5%
x-s
x
x+s
x + 2s
The Empirical Rule
FIGURE 2-15
(applies to bell-shaped distributions)
99.7% of data are within 3 standard deviations of the mean
95% within
2 standard deviations
68% within
1 standard deviation
34%
34%
2.4%
2.4%
0.1%
0.1%
13.5%
x - 3s
x - 2s
13.5%
x-s
x
x+s
x + 2s
x + 3s
Chebyshev’s Theorem
 applies to distributions of any shape.
 the proportion (or fraction) of any set of data lying
within K standard deviations of the mean is always at
2
least 1 - 1/K , where K is any positive number greater
than 1.
 at least 3/4 (75%) of all values lie within 2 standard
deviations of the mean.
 at least 8/9 (89%) of all values lie within 3 standard
deviations of the mean.
Measures of Variation Summary
For typical data sets, it is unusual for a score to
differ from the mean by more than 2 or 3 standard
deviations.
STATISTICS
ELEMENTARY
Section 2-6
Measures of Position
MARIO F. TRIOLA
EIGHTH
EDITION
Measures of Position
Measures of Position
 z Score
(or standard score)
the number of standard deviations
that a given value x is above or below
the mean
Measures of Position
z score
Sample
x
x
z= s
Measures of Position
z score
Sample
x
x
z= s
Population
x
µ
z=
s
Measures of Position
z score
Sample
x
x
z= s
Population
x
µ
z=
s
Round to 2 decimal places
FIGURE 2-16
Interpreting Z Scores
Unusual
Values
Ordinary
Values
-3
-2
-1
0
Z
Unusual
Values
1
2
3
Measures of Position
Quartiles, Deciles,
Percentiles
Quartiles
Quartiles
Q1, Q2, Q3
Quartiles
Q1, Q2, Q3
divides ranked scores into four equal parts
Quartiles
Q1, Q2, Q3
divides ranked scores into four equal parts
25%
25%
25% 25%
Q1 Q2 Q3
Quartiles
Q1, Q2, Q3
divides ranked scores into four equal parts
25%
(minimum)
25%
25% 25%
Q1 Q2 Q3
(median)
(maximum)
Deciles
D1, D2, D3, D4, D5, D6, D7, D8, D9
divides ranked data into ten equal parts
Deciles
D1, D2, D3, D4, D5, D6, D7, D8, D9
divides ranked data into ten equal parts
10% 10% 10%
D1
D2
D3
10% 10% 10%
D4
D5
D6
10% 10% 10% 10%
D7
D8
D9
Percentiles
99 Percentiles
Quartiles, Deciles, Percentiles
Fractiles
Quartiles, Deciles, Percentiles
Fractiles
(Quantiles)
partitions data into approximately equal parts
Finding the Percentile of a
Given Score
Finding the Percentile of a
Given Score
Percentile of score x =
number of scores less than x
• 100
total number of scores
Finding the Score
Given a Percentile
Finding the Score
Given a Percentile
L=
k
100
•n
n total number of values in the data set
k
percentile being used
L locator that gives the position of a value
Pk kth percentile
Start
Finding the Value
of the
kth Percentile
Sort the data.
(Arrange the data in
order of lowest to
highest.)
Compute
k n
L=
100
(
)
where
n = number of values
k = percentile in question
Is
L a whole
number
?
No
Yes
The value of the kth percentile
is midway between the Lth value
and the next value in the
sorted set of data. Find Pk by
adding the L th value and the
next value and dividing the
total by 2.
Change L by rounding
it up to the next
larger whole number.
The value of Pk is the
Lth value, counting
from the lowest
Figure 2-17
Quartiles
Q1 = P25
Q2 = P50
Q3 = P75
Quartiles
Q1 = P25
Q2 = P50
Q3 = P75
Deciles
D1 = P10
D2 = P20
D3 = P30
•
•
•
D9 = P90
Interquartile Range (or IQR): Q3 - Q1
Interquartile Range (or IQR): Q3 - Q1
Semi-interquartile Range:
Q3 - Q1
2
Interquartile Range (or IQR): Q3 - Q1
Semi-interquartile Range:
Q3 - Q1
2
Midquartile:
Q 1 + Q3
2
Interquartile Range (or IQR): Q3 - Q1
Semi-interquartile Range:
Q3 - Q1
2
Midquartile:
Q 1 + Q3
2
10 - 90 Percentile Range: P90 - P10
ELEMENTARY
Section 2-7
STATISTICS
Exploratory Data Analysis (EDA)
MARIO F. TRIOLA
EIGHTH
EDITION
Exploratory Data Analysis
the process of using statistical tools (such as
graphs, measures of center, and measures of
variation) to investigate the data sets in order to
understand their important characteristics
Outliers
 a value located very far away from
of the other values
almost all
 an extreme value
 can have a dramatic effect on the
mean, standard deviation, and on the scale
of the histogram so that the true
nature of
the distribution is totally
obscured
Boxplots
(Box-and-Whisker Diagram)
Reveals the:
 center of the data
 spread of the data
 distribution of the data
 presence of outliers
Excellent for comparing two or
more data sets
Boxplots
5 - number summary
 Minimum
 first quartile Q1
 Median (Q2)
 third quartile Q3
 Maximum
Boxplots
2
4
6
14
0
0
Figure 2-18
2
4
6
8
10
12
14
Boxplot of Qwerty Word Ratings
Figure 2-19
Boxplots
Bell-Shaped
Figure 2-19
Boxplots
Bell-Shaped
Uniform
Figure 2-19
Boxplots
Bell-Shaped
Uniform
Skewed
Exploring
 Measures of center: mean, median, and mode
 Measures of variation: Standard deviation and
range
 Measures of spread and relative
location: minimum values, maximum value, and
quartiles
 Unusual values: outliers
 Distribution: histograms, stem-leaf plots, and
ELEMENTARY
Chapter 3
MARIO F. TRIOLA
STATISTICS
Probability
EIGHTH
EDITION
Chapter 3
Probability
3-1
Overview
3-2
Fundamentals
3-3
Addition Rule
3-4
Multiplication Rule: Basics
3-5
Multiplication Rule: Complements and
Conditional Probability
3-6
Probabilities Through Simulations
3-7
Counting
3-1
Overview
Objectives
 develop sound understanding of
probability values used in subsequent
chapters
 develop basic skills necessary to solve
simple probability problems
Rare Event Rule for Inferential Statistics:
If, under a given assumption, the probability of a
particular observed event is extremely small, we
conclude that the assumption is probably not correct.
3-2
Fundamentals
Definitions
Event - any collection of results or
outcomes from some procedure
 Simple event - any outcome or event that
cannot be broken down into
simpler components
 Sample space - all possible simple events
Notation
P - denotes a probability
A, B, ... - denote specific events
P (A) -
denotes the probability of
event A occurring
Basic Rules for
Computing Probability
Rule 1: Relative Frequency Approximation
Conduct (or observe) an experiment a large
number of times, and count the number of
times event A actually occurs, then an
estimate of P(A) is
Basic Rules for
Computing Probability
Rule 1: Relative Frequency Approximation
Conduct (or observe) an experiment a large
number of times, and count the number of
times event A actually occurs, then an
estimate of P(A) is
P(A) =
number of times A occurred
number of times trial was repeated
Basic Rules for
Computing Probability
Rule 2: Classical approach
(requires equally likely outcomes)
If a procedure has n different simple events,
each with an equal chance of occurring, and s is
the number of ways event A can occur, then
Basic Rules for
Computing Probability
Rule 2: Classical approach
(requires equally likely outcomes)
If a procedure has n different simple events,
each with an equal chance of occurring, and s is
the number of ways event A can occur, then
s
=
P(A) =
n
number of ways A can occur
number of different simple
events
Basic Rules for
Computing Probability
Rule 3: Subjective Probabilities
P(A), the probability of A, is found by simply
guessing or estimating its value based on
knowledge of the relevant circumstances.
Rule 1
The relative frequency approach is an
approximation.
Rule 1
The relative frequency approach is an
approximation.
Rule 2
The classical approach is the
actual probability.
Law of Large Numbers
As a procedure is repeated again and
again, the relative frequency probability
(from Rule 1) of an event tends to
approach the actual probability.
Illustration of
Law of Large Numbers
Figure 3-2
Example:
Find the probability that a randomly
selected person will be struck by lightning this year.
The sample space consists of two simple events: the
person is struck by lightning or is not. Because
these simple events are not equally likely, we can
use the relative frequency approximation (Rule 1) or
subjectively estimate the probability (Rule 3). Using
Rule 1, we can research past events to determine
that in a recent year 377 people were struck by
lightning in the US, which has a population of about
274,037,295. Therefore,
P(struck by lightning in a year)

377 / 274,037,295  1/727,000
Example: On an ACT or SAT test, a typical multiple-choice
question has 5 possible answers. If you make a random guess on
one such question, what is the probability that your response is
wrong?
There are 5 possible outcomes or
answers, and there are 4 ways to answer
incorrectly. Random guessing implies that
the outcomes in the sample space are
equally likely, so we apply the classical
approach (Rule 2) to get:
P(wrong answer) = 4 / 5 = 0.8
Probability Limits
 The probability of an impossible event is 0.
 The probability of an event that is certain
to occur is 1.
Probability Limits
 The probability of an impossible event is 0.
 The probability of an event that is certain
to occur is 1.
0  P(A)  1
Probability Limits
 The probability of an impossible event is 0.
 The probability of an event that is certain
to occur is 1.
0  P(A)  1
Impossible
to occur
Probability Limits
 The probability of an impossible event is 0.
 The probability of an event that is certain
to occur is 1.
0  P(A)  1
Impossible
to occur
Certain
to occur
Possible Values for Probabilities
1
Certain
Likely
0.5
50-50 Chance
Unlikely
Figure 3-3
0
Impossible
Complementary Events
Complementary Events
The complement of event A, denoted
by A, consists of all outcomes in
which event A does not occur.
Complementary Events
The complement of event A, denoted
by A, consists of all outcomes in
which event A does not occur.
P(A)
P(A)
(read “not A”)
Example:
Testing Corvettes
The General Motors Corporation wants to conduct a test of a new
model of Corvette. A pool of 50 drivers has been recruited, 20 or
whom are men. When the first person is selected from this pool,
what is the probability of not getting a male driver?
Example:
Testing Corvettes
The General Motors Corporation wants to conduct a test of a new
model of Corvette. A pool of 50 drivers has been recruited, 20 or
whom are men. When the first person is selected from this pool,
what is the probability of not getting a male driver?
Because 20 of the 50 subjects are men,
it follows that 30 of the 50 subjects are
women so,
P(not selecting a man) = P(man)
= P(woman)
= 30 = 0.6
50
Rounding Off Probabilities
give the exact fraction or decimal
or
Rounding Off Probabilities
give the exact fraction or decimal
or
round off the final result to three
significant
digits
Odds
Odds
actual odds against event A occurring
are the ratio P(A) P(A), usually
expressed in the form of a:b
(or ‘a to b’), where a and b are
integers with no common factors
actual odds in favor of event A are the
reciprocal of the odds against that
event, b:a (or ‘b to a’)
Odds
 The payoff odds against event A
represent
the ratio of net profit (if you win) to the amount
of the bet.
Payoff odds against event A = (net
profit):(amount bet)
ELEMENTARY
Section 3-3
STATISTICS
Addition Rule
MARIO F. TRIOLA
EIGHTH
EDITION
Definition
 Compound Event
•
Any event combining 2 or more
simple events
Definition
 Compound Event
•
Any event combining 2 or more
simple events
 Notation
•
P(A or B) = P (event A occurs or event B
occurs or they both occur)
Compound Event
• General Rule
•
When finding the probability that event A
occurs or event B occurs, find the total number
of ways A can occur and the number of ways B
can occur, but find the total in such a way that
no outcome is counted more than once.
Compound Event
• Formal Addition Rule
•
•
P(A or B) = P(A) + P(B) - P(A and B)
where P(A and B) denotes the probability that A and B
both occur at the same time.
Compound Event
• Formal Addition Rule
•
•
P(A or B) = P(A) + P(B) - P(A and B)
where P(A and B) denotes the probability that A and B
both occur at the same time.
• Intuitive Addition Rule
•
To find P(A or B), find the sum of the number of ways
event A can occur and the number of ways event B can occur,
adding in such a way that every outcome is counted only once.
P(A or B) is equal to that sum, divided by the total number of
outcomes.
Definition
• Events A and B are mutually exclusive if they
cannot occur simultaneously.
Definition
• Events A and B are mutually exclusive if they
cannot occur simultaneously.
Total Area = 1
P(A)
P(B)
P(A and B)
Overlapping Events
Figures 3-5
Definition
• Events A and B are mutually exclusive if they
cannot occur simultaneously.
Total Area = 1
P(A)
P(B)
Total Area = 1
P(A)
P(B)
P(A and B)
Overlapping Events
Figures 3-5 and 3-6
Non-overlapping Events
Figure 3-7
Applying the Addition Rule
P(A or B)
Addition Rule
Are
A and B
mutually
exclusive
?
Yes
No
P(A or B) = P(A)+ P(B) - P(A and B)
P(A or B) = P(A) + P(B)
Contingency Table
Survived
Died
Total
Men
Women
332
318 29
27
706
1360
104 35
18
1517
1692
422
Boys
64
Girls Totals
56
2223
• Find the probability of randomly selecting a man or a boy.
Contingency Table
Survived
Died
Total
Men
Women
332
318 29
27
706
1360
104 35
18
1517
1692
422
Boys
64
Girls Totals
56
2223
• Find the probability of randomly selecting a man or a boy.
Contingency Table
Survived
Died
Total
Men
Women
332
318 29
27
706
1360
104 35
18
1517
1692
422
Boys
64
Girls Totals
56
2223
• Find the probability of randomly selecting a man or a boy.
• P(man or boy) = 1692 + 64 = 1756 = 0.790
•
2223 2223 2223
Contingency Table
Survived
Died
Total
Men
Women
332
318 29
27
706
1360
104 35
18
1517
1692
422
Boys
64
Girls Totals
56
2223
• Find the probability of randomly selecting a man or a boy.
• P(man or boy) = 1692 + 64 = 1756 = 0.790
•
2223 2223 2223
* Mutually Exclusive *
Contingency Table
Survived
Died
Total
Men
Women
332
318 29
27
706
1360
104 35
18
1517
1692
422
Boys
64
Girls Totals
56
2223
• Find the probability of randomly selecting a man or someone
who survived.
Contingency Table
Survived
Died
Total
Men
Women
332
318 29
27
706
1360
104 35
18
1517
1692
422
Boys
64
Girls Totals
56
2223
• Find the probability of randomly selecting a man or someone
who survived.
Contingency Table
Survived
Died
Total
Men
Women
332
318 29
27
706
1360
104 35
18
1517
1692
422
Boys
64
Girls Totals
56
2223
• Find the probability of randomly selecting a man or someone
who survived.
• P(man or survivor) = 1692 + 706 - 332 = 1756
•
2223 2223 2223 2223
= 0.929
Contingency Table
Survived
Died
Total
Men
Women
332
318 29
27
706
1360
104 35
18
1517
1692
422
Boys
64
Girls Totals
56
2223
• Find the probability of randomly selecting a man or someone
who survived.
• P(man or survivor) = 1692 + 706 - 332 = 1756
•
2223 2223 2223 2223
= 0.929
* NOT Mutually Exclusive *
Complementary Events
Complementary Events
• P(A) and P(A)
• are
• mutually exclusive
Complementary Events
• P(A) and P(A)
• are
• mutually exclusive
• All simple events are either in A or A.
Complementary Events
• P(A) and P(A)
• are
• mutually exclusive
• All simple events are either in A or A.
• P(A) + P(A) = 1
Rules of Complementary Events
P(A) + P(A) = 1
Rules of Complementary Events
P(A) + P(A) = 1
P(A)
= 1 - P(A)
Rules of Complementary Events
P(A) + P(A) = 1
P(A)
= 1 - P(A)
P(A) = 1 - P(A)
Figure 3-8
Venn Diagram for the
Complement of Event A
Total Area = 1
P (A)
P (A) = 1 - P (A)
ELEMENTARY
Section 3-4
STATISTICS
Multiplication Rule: Basics
MARIO F. TRIOLA
EIGHTH
EDITIO
Finding the Probability of Two or
More Selections
 Multiple selections
 Multiplication Rule
Notation
P(A and B) =
P(event A occurs in a first trial and
event B occurs in a second trial)
FIGURE 3-9
Tree Diagram of Test Answers
FIGURE 3-9
Tree Diagram of Test Answers
T
F
a
b
c
d
e
a
b
c
d
e
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
FIGURE 3-9
Tree Diagram of Test Answers
T
F
a
b
c
d
e
a
b
c
d
e
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
FIGURE 3-9
Tree Diagram of Test Answers
T
F
P(T) =
1
2
a
b
c
d
e
a
b
c
d
e
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
FIGURE 3-9
Tree Diagram of Test Answers
a
b
c
d
T
P(T) =
F
e
a
b
1
2
c
d
e 1
5
P(c) =
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
FIGURE 3-9
Tree Diagram of Test Answers
a
b
c
d
T
P(T) =
F
e
a
b
1
2
c
d
e 1
5
P(c) =
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
P(T and c) =
1
10
P (both correct)
P (both correct) = P (T and c)
P (both correct) = P (T and c)
1
10
1
2
1
5
P (both correct) = P (T and c)
1 =
10
1
1
•
2
5
Multiplication
Rule
P (both correct) = P (T and c)
1 =
10
1
1
•
2
5
Multiplication
Rule
INDEPENDENT EVENTS
Notation for Conditional
Probability
P(B A) represents the probability of event B occurring
after it is assumed that event A has already occurred
(read B A as “B given A”).
Definitions
 Independent Events
Two events A and B are independent if the
occurrence of one does not affect the probability
of the occurrence of the other.
 Dependent Events
If A and B are not independent, they are said to
be dependent.
Formal Multiplication Rule
P(A and B) = P(A) • P(B A)
If A and B are independent events,
P(B A) is really the same as P(B)
Figure 3-10
Applying the Multiplication Rule
P(A or B)
Multiplication Rule
Are
A and B
independent
?
Yes
No
P(A and B) = P(A) • P(B A)
P(A and B) = P(A) • P(B)
Intuitive Multiplication
When finding the probability that event A occurs in one
trial and B occurs in the next trial, multiply the
probability of event A by the probability of event B, but
be sure that the probability of event B takes into
account the previous occurrence of event A.
Small Samples
from
Large Populations
If a sample size is no more than 5% of the size
of the population, treat the selections as being
independent (even if the selections are made
without replacement, so they are technically
dependent).
ELEMENTARY
Section 3-5
STATISTICS
Multiplication Rule: Complements and
Conditional Probability
MARIO F. TRIOLA
EIGHTH
EDITION
Probability of ‘At Least One’
Probability of ‘At Least One’
 ‘At least one’ is equivalent to ‘one or
more’.
Probability of ‘At Least One’
 ‘At least one’ is equivalent to ‘one or
more’.
 The complement of getting at least one item
of a particular type is that you get no items of
that type.
Probability of ‘At Least One’
 ‘At least one’ is equivalent to ‘one or
more’.
 The complement of getting at least one item
of a particular type is that you get no items of
that type.
If P(A) = P(getting at least one), then
Probability of ‘At Least One’
 ‘At least one’ is equivalent to ‘one or
more’.
 The complement of getting at least one item
of a particular type is that you get no items of
that type.
If P(A) = P(getting at least one), then
P(A) = 1 - P(A)
Probability of ‘At Least One’
 ‘At least one’ is equivalent to ‘one or
more’.
 The complement of getting at least one item
of a particular type is that you get no items of
that type.
If P(A) = P(getting at least one), then
P(A) = 1 - P(A)
where P(A) is P(getting none)
Probability of ‘At Least One’
 Find the probablility of a couple have at
least 1 girl among 3 children.
Probability of ‘At Least One’
 Find the probablility of a couple have at
least 1 girl among 3 children.
If P(A) = P(getting at least 1 girl), then
P(A) = 1 - P(A)
where P(A) is P(getting no girls)
Probability of ‘At Least One’
 Find the probablility of a couple have at
least 1 girl among 3 children.
If P(A) = P(getting at least 1 girl), then
P(A) = 1 - P(A)
where P(A) is P(getting no girls)
P(A) = (0.5)(0.5)(0.5) = 0.125
Probability of ‘At Least One’
 Find the probablility of a couple have at
least 1 girl among 3 children.
If P(A) = P(getting at least 1 girl), then
P(A) = 1 - P(A)
where P(A) is P(getting no girls)
P(A) = (0.5)(0.5)(0.5) = 0.125
P(A) = 1 - 0.125 = 0.875
Conditional Probability
Definition
The conditional probability of event B
occurring, given that A has already occurred, can
be found by dividing the probability of events A
and B both occurring by the probability of event A.
Conditional Probability
P(A and B) = P(A) • P(B|A)
Conditional Probability
P(A and B) = P(A) • P(B|A)
Formal
Conditional Probability
P(A and B) = P(A) • P(B|A)
Formal
P(B|A) =
P(A and B)
P(A)
Conditional Probability
P(A and B) = P(A) • P(B|A)
Formal
P(B|A) =
Intuitive
P(A and B)
P(A)
Conditional Probability
P(A and B) = P(A) • P(B|A)
Formal
P(B|A) =
Intuitive
P(A and B)
P(A)
The conditional probability of B given A can be
found by assuming the event A has occurred and,
operating under that assumption, calculating the
probability that event B will occur.
Testing for Independence
Testing for Independence
If P(B|A) = P(B)
then the occurrence of A has no effect on the
probability of event B; that is, A and B are
independent events.
Testing for Independence
If P(B|A) = P(B)
then the occurrence of A has no effect on the
probability of event B; that is, A and B are
independent events.
or
If P(A and B) = P(A) • P(B)
then A and B are independent events.
ELEMENTARY
STATISTICS
Chapter 4
Probability Distributions
MARIO F. TRIOLA
EIGHTH
EDITION
Chapter 4
Probability Distributions
4-1 Overview
4-2 Random Variables
4-3 Binomial Probability Distributions
4-4 Mean, Variance, Standard Deviation
for the Binomial Distribution
4-5 The Poisson Distribution
4-1
Overview
This chapter will deal with the
construction of
probability distributions
by combining the methods of Chapter 2
with the those of Chapter 3.
Probability Distributions will describe
what will probably happen instead of
what actually did happen.
Combining Descriptive Statistics Methods and
Probabilities to Form a Theoretical Model of
Behavior
Figure 4-1
4-2
Random Variables
Definitions
 Random Variable
a variable (typically represented by x) that has a
single numerical value, determined by chance,
for each outcome of a procedure
Probability Distribution
a graph, table, or formula that gives the
probability for each value of the random variable
Table 4-1
Probability Distribution
Number of Girls Among Fourteen Newborn Babies
x
P(x)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0.000
0.001
0.006
0.022
0.061
0.122
0.183
0.209
0.183
0.122
0.061
0.022
0.006
0.001
0.000
Definitions
Discrete random variable
has either a finite number of values or countable
number of values, where ‘countable’ refers to the
fact that there might be infinitely many values,
but they result from a counting process.
Continuous random variable
has infinitely many values, and those values can
be associated with measurements on a
continuous scale with no gaps or interruptions.
Probability Histogram
Figure 4-3
Requirements for
Probability Distribution
Requirements for
Probability Distribution
 P(x) = 1
where x assumes all possible values
Requirements for
Probability Distribution
 P(x) = 1
where x assumes all possible values
0  P(x)  1
for every value of x
Mean, Variance and Standard Deviation
of a Probability Distribution
Formula 4-1
µ =  [x • P(x)]
Formula 4-2
2
2
s =  [(x - µ) • P(x)]
Formula 4-3
2
2
2
s = [ x • P(x)] - µ (shortcut)
Mean, Variance and Standard Deviation
of a Probability Distribution
Formula 4-1
µ =  [x • P(x)]
Formula 4-2
2
2
s =  [(x - µ) • P(x)]
Formula 4-3
2
2
2
s = [ x • P(x)] - µ (shortcut)
Formula 4-4
s = [ x 2 • P(x)] - µ 2
Mean, Variance and Standard Deviation
of a Probability Distribution
Formula 4-1
µ =  [x • P(x)]
Formula 4-2
2
2
s =  [(x - µ) • P(x)]
Formula 4-3
2
2
2
s = [ x • P(x)] - µ (shortcut)
Formula 4-4
s = [ x 2 • P(x)] - µ 2
Roundoff Rule for µ, s , and s
2
Round results by carrying one more decimal
place than the number of decimal places used
for the random variable x. If the values of x
are integers, round µ, s2, and s to one
decimal place.
Definition
Expected Value
The average value of outcomes
E =  [x • P(x)]
E =  [x • P(x)]
Event
Win
Lose
E =  [x • P(x)]
Event
x
Win
$499
Lose
- $1
E =  [x • P(x)]
Event
x
P(x)
Win
$499
0.001
Lose
- $1
0.999
E =  [x • P(x)]
Event
x
P(x)
x • P(x)
Win
$499
0.001
0.499
Lose
- $1
0.999
- 0.999
E =  [x • P(x)]
Event
x
P(x)
x • P(x)
Win
$499
0.001
0.499
Lose
- $1
0.999
- 0.999
E = -$.50
ELEMENTARY
Section 4-3
STATISTICS
Binomial Probability Distributions
MARIO F. TRIOLA
EIGHTH
EDITION
Definitions
Binomial Probability Distribution
1. The experiment must have a fixed number of trials.
2. The trials must be independent. (The outcome of
any individual trial doesn’t affect the probabilities
in the other trials.)
3. Each trial must have all outcomes classified into
two categories.
4. The probabilities must remain constant for each
trial.
Notation for Binomial Probability
Distributions
n = fixed number of trials
x = specific number of successes in n trials
p = probability of success in one of n trials
q = probability of failure in one of n trials
(q = 1 - p )
P(x) = probability of getting exactly x
success among n trials
Be sure that x and p both refer to the same category
being called a success.
Method 1
Binomial Probability
Formula
Method 1
Binomial Probability
Formula
 P(x) =
n!
•
(n - x )! x!
px •
n-x
q
Method 1
Binomial Probability
Formula
 P(x) =
n!
•
(n - x )! x!
 P(x) = nCx • px
px •
•
n-x
q
qn-x
for calculators with nCr key, where r = x
Example: Find the probability of getting exactly
3 correct responses among 5 different requests
from AT&T directory assistance. Assume in
general, AT&T is correct 90% of the time.
This is a binomial experiment where:
n=5
x=3
p = 0.90
q = 0.10
Example: Find the probability of getting exactly
3 correct responses among 5 different requests
from AT&T directory assistance. Assume in
general, AT&T is correct 90% of the time.
This is a binomial experiment where:
n=5
x=3
p = 0.90
q = 0.10
Using the binomial probability formula to solve:
P(3) =
5C3
3
2
• 0.9 • 01 = 0.0.0729
Method 2
Table A-1 in Appendix A
For n = 15 and p = 0.10
Table A-1
n
x
P(x)
15
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.206
0.343
0.267
0.129
0.043
0.010
0.002
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
For n = 15 and p = 0.10
Table A-1
n
x
P(x)
15
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.206
0.343
0.267
0.129
0.043
0.010
0.002
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
For n = 15 and p = 0.10
Table A-1
Binomial Probability Distribution
n
x
P(x)
x
P(x)
15
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.206
0.343
0.267
0.129
0.043
0.010
0.002
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0.0+
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.206
0.343
0.267
0.129
0.043
0.010
0.002
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
Example: Using Table A-1 for n = 5 and p = 0.90,
find the following:
a) The probability of exactly 3 successes
b) The probability of at least 3 successes
a) P(3) = 0.073
b) P(at least 3) = P(3 or 4 or 5)
= P(3) or P(4) or P(5)
= 0.073 + 0.328 + 0.590
= 0.991
Method 3
Using Technology
 STATDISK
 Minitab
Excel
 TI-83 Plus
Binomial Probability
Formula
P(x) =
n!
•
(n - x )! x!
Number of
outcomes with
exactly x
successes
among n trials
px •
n-x
q
Binomial Probability
Formula
P(x) =
n!
•
(n - x )! x!
Number of
outcomes with
exactly x
successes
among n trials
px •
n-x
q
Probability of x
successes
among n trials
for any one
particular order
ELEMENTARY
Section 4-4
STATISTICS
Mean, Variance, and Standard Deviation
for the Binomial Distribution
MARIO F. TRIOLA
EIGHTH
EDITION
For Any Discrete Probability
Distribution:
• Formula 4-1 µ = [x • P(x)]
• Formula 4-3 s
2
= [ x • P(x) ] - µ
2
2
For Any Discrete Probability
Distribution:
• Formula 4-1 µ = [x • P(x)]
• Formula 4-3 s
2
Formula 4-4 s =
= [ x • P(x) ] - µ
2
[ x • P(x) ] - µ
2
2
2
For Binomial Distributions:
• Formula 4-6 µ
=n•p
• Formula 4-7 s = n • p • q
2
For Binomial Distributions:
• Formula 4-6 µ
=n•p
• Formula 4-7 s = n • p • q
2
Formula 4-8 s= n • p • q
Example:
Find the mean and standard
deviation for the number of girls in
groups of 14 births.
• We previously discovered that this scenario could be
considered a binomial experiment where:
• n = 14
• p = 0.5
• q = 0.5
• Using the binomial distribution formulas:
Example:
Find the mean and standard
deviation for the number of girls in
groups of 14 births.
• We previously discovered that this scenario could be
considered a binomial experiment where:
• n = 14
• p = 0.5
• q = 0.5
• Using the binomial distribution formulas:
• µ = (14)(0.5) = 7 girls
 s=
(14)(0.5)(0.5) = 1.9 girls (rounded)
Reminder
• Maximum usual values = µ + 2 s
• Minimum usual values = µ - 2 s
Example:
Determine whether 68 girls among 100
babies could easily occur by chance.
• For this binomial distribution,
• µ = 50 girls
 s= 5 girls
• µ + 2 s = 50 + 2(5) = 60
• µ - 2 s = 50 - 2(5) = 40
•
The usual number girls among 100 births would be from
40 to 60. So 68 girls in 100 births is an unusual result.
ELEMENTARY
Chapter 5
STATISTICS
Normal Probability Distributions
MARIO F. TRIOLA
EIGHTH
EDITION
Chapter 5
Normal Probability Distributions
5-1
Overview
5-2
The Standard Normal Distribution
5-3
Normal Distributions: Finding Probabilities
5-4
Normal Distributions: Finding Values
5-5
The Central Limit Theorem
5-6
Normal Distribution as Approximation to
Binomial Distribution
5-7
Determining Normality
5-1
Overview
 Continuous random variable
 Normal distribution
5-1
Overview
 Continuous random variable
 Normal distribution
Curve is bell shaped
and symmetric
Figure 5-1
µ
Score
5-1
Overview
 Continuous random variable
 Normal distribution
Curve is bell shaped
and symmetric
Figure 5-1
µ
Score
Formula 5-1
y=
e
1
2
s
( x - µ)
s
2
2
5-2
The Standard Normal
Distribution
Definitions
 Uniform Distribution
a probability distribution in which the
continuous random variable values are
spread evenly over the range of
possibilities; the graph results in a
rectangular shape.
Definitions
 Density Curve (or probability
density
function)
the graph of a continuous
probability distribution
Definitions
 Density Curve (or probability
density
function)
the graph of a continuous
probability distribution
1. The total area under the curve must equal 1.
2. Every point on the curve must have a vertical height
that is 0 or greater.
Because the total area under
the density curve is equal to 1,
there is a correspondence
between area and probability.
Times in First or Last Half Hours
Figure 5-3
Heights of Adult Men and Women
Women:
µ = 63.6
s = 2.5
Figure 5-4
Men:
µ = 69.0
s = 2.8
63.6
69.0
Height (inches)
Definition
Standard Normal Deviation
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
Definition
Standard Normal Deviation
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
Area found in
Table A-2
Area = 0.3413
0.4429
-3
-2
-1
0
1
2
3
0
z = 1.58
Score (z )
Figure 5-5
Figure 5-6
Table A-2
 Back left cover of text book
 Formulas and Tables card
 Appendix
Table A-2
Standard Normal Distribution
s=1
µ=0
0
x
z
Table A-2 Standard Normal (z) Distribution
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
.0000
.0398
.0793
.1179
.1554
.1915
.2257
.2580
.2881
.3159
.3413
.3643
.3849
.4032
.4192
.4332
.4452
.4554
.4641
.4713
.4772
.4821
.4861
.4893
.4918
.4938
.4953
.4965
.4974
.4981
.4987
.0040
.0438
.0832
.1217
.1591
.1950
.2291
.2611
.2910
.3186
.3438
.3665
.3869
.4049
.4207
.4345
.4463
.4564
.4649
.4719
.4778
.4826
.4864
.4896
.4920
.4940
.4955
.4966
.4975
.4982
.4987
.0080
.0478
.0871
.1255
.1628
.1985
.2324
.2642
.2939
.3212
.3461
.3686
.3888
.4066
.4222
.4357
.4474
.4573
.4656
.4726
.4783
.4830
.4868
.4898
.4922
.4941
.4956
.4967
.4976
.4982
.4987
.0120
.0517
.0910
.1293
.1664
.2019
.2357
.2673
.2967
.3238
.3485
.3708
.3907
.4082
.4236
.4370
.4484
.4582
.4664
.4732
.4788
.4834
.4871
.4901
.4925
.4943
.4957
.4968
.4977
.4983
.4988
.0160
.0557
.0948
.1331
.1700
.2054
.2389
.2704
.2995
.3264
.3508
.3729
.3925
.4099
.4251
.4382
.4495
.4591
.4671
.4738
.4793
.4838
.4875
.4904
.4927
.4945
.4959
.4969
.4977
.4984
.4988
.0199
.0596
.0987
.1368
.1736
.2088
.2422
.2734
.3023
.3289
.3531
.3749
.3944
.4115
.4265
.4394
.4505
.4599
.4678
.4744
.4798
.4842
.4878
.4906
.4929
.4946
.4960
.4970
.4978
.4984
.4989
.0239
.0636
.1026
.1406
.1772
.2123
.2454
.2764
.3051
.3315
.3554
.3770
.3962
.4131
.4279
.4406
.4515
.4608
.4686
.4750
.4803
.4846
.4881
.4909
.4931
.4948
.4961
.4971
.4979
.4985
.4989
.0279
.0675
.1064
.1443
.1808
.2157
.2486
.2794
.3078
.3340
.3577
.3790
.3980
.4147
.4292
.4418
.4525
.4616
.4693
.4756
.4808
.4850
.4884
.4911
.4932
.4949
.4962
.4972
.4979
.4985
.4989
.0319
.0714
.1103
.1480
.1844
.2190
.2517
.2823
.3106
.3365
.3599
.3810
.3997
.4162
.4306
.4429
.4535
.4625
.4699
.4761
.4812
.4854
.4887
.4913
.4934
.4951
.4963
.4973
.4980
.4986
.4990
.0359
.0753
.1141
.1517
.1879
.2224
.2549
.2852
.3133
.3389
.3621
.3830
.4015
.4177
.4319
.4441
.4545
.4633
.4706
.4767
.4817
.4857
.4890
.4916
.4936
.4952
.4964
.4974
.4981
.4986
.4990
*
*
To find:
z Score
the distance along horizontal scale of the
standard normal distribution; refer to the
leftmost column and top row of Table A-2
Area
the region under the curve; refer to the
values in the body of Table A-2
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
P ( 0 < x < 1.58 ) =
0
1.58
Table A-2 Standard Normal (z) Distribution
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
.0000
.0398
.0793
.1179
.1554
.1915
.2257
.2580
.2881
.3159
.3413
.3643
.3849
.4032
.4192
.4332
.4452
.4554
.4641
.4713
.4772
.4821
.4861
.4893
.4918
.4938
.4953
.4965
.4974
.4981
.4987
.0040
.0438
.0832
.1217
.1591
.1950
.2291
.2611
.2910
.3186
.3438
.3665
.3869
.4049
.4207
.4345
.4463
.4564
.4649
.4719
.4778
.4826
.4864
.4896
.4920
.4940
.4955
.4966
.4975
.4982
.4987
.0080
.0478
.0871
.1255
.1628
.1985
.2324
.2642
.2939
.3212
.3461
.3686
.3888
.4066
.4222
.4357
.4474
.4573
.4656
.4726
.4783
.4830
.4868
.4898
.4922
.4941
.4956
.4967
.4976
.4982
.4987
.0120
.0517
.0910
.1293
.1664
.2019
.2357
.2673
.2967
.3238
.3485
.3708
.3907
.4082
.4236
.4370
.4484
.4582
.4664
.4732
.4788
.4834
.4871
.4901
.4925
.4943
.4957
.4968
.4977
.4983
.4988
.0160
.0557
.0948
.1331
.1700
.2054
.2389
.2704
.2995
.3264
.3508
.3729
.3925
.4099
.4251
.4382
.4495
.4591
.4671
.4738
.4793
.4838
.4875
.4904
.4927
.4945
.4959
.4969
.4977
.4984
.4988
.0199
.0596
.0987
.1368
.1736
.2088
.2422
.2734
.3023
.3289
.3531
.3749
.3944
.4115
.4265
.4394
.4505
.4599
.4678
.4744
.4798
.4842
.4878
.4906
.4929
.4946
.4960
.4970
.4978
.4984
.4989
.0239
.0636
.1026
.1406
.1772
.2123
.2454
.2764
.3051
.3315
.3554
.3770
.3962
.4131
.4279
.4406
.4515
.4608
.4686
.4750
.4803
.4846
.4881
.4909
.4931
.4948
.4961
.4971
.4979
.4985
.4989
.0279
.0675
.1064
.1443
.1808
.2157
.2486
.2794
.3078
.3340
.3577
.3790
.3980
.4147
.4292
.4418
.4525
.4616
.4693
.4756
.4808
.4850
.4884
.4911
.4932
.4949
.4962
.4972
.4979
.4985
.4989
.0319
.0714
.1103
.1480
.1844
.2190
.2517
.2823
.3106
.3365
.3599
.3810
.3997
.4162
.4306
.4429
.4535
.4625
.4699
.4761
.4812
.4854
.4887
.4913
.4934
.4951
.4963
.4973
.4980
.4986
.4990
.0359
.0753
.1141
.1517
.1879
.2224
.2549
.2852
.3133
.3389
.3621
.3830
.4015
.4177
.4319
.4441
.4545
.4633
.4706
.4767
.4817
.4857
.4890
.4916
.4936
.4952
.4964
.4974
.4981
.4986
.4990
*
*
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
Area = 0.4429
P ( 0 < x < 1.58 ) = 0.4429
0
1.58
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
Area = 0.4429
P ( 0 < x < 1.58 ) = 0.4429
0
1.58
The probability that the chosen
thermometer will measure freezing water
between 0 and 1.58 degrees is 0.4429.
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
Area = 0.4429
P ( 0 < x < 1.58 ) = 0.4429
0
1.58
There is 44.29% of the thermometers with
readings between 0 and 1.58 degrees.
Using Symmetry to Find the Area
to the Left of the Mean
Because of symmetry, these areas are equal.
Figure 5-7
(a)
(b)
0.4925
0.4925
0
z = - 2.43
0
Equal distance away from 0
z = 2.43
NOTE: Although a z score can be negative, the area
under the curve (or the corresponding probability)
can never be negative.
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water, and if one thermometer is randomly
selected, find the probability that it reads freezing water
between -2.43 degrees and 0 degrees.
Area = 0.4925
P ( -2.43 < x < 0 ) = 0.4925
-2.43
0
The probability that the chosen thermometer
will measure freezing water between -2.43
and 0 degrees is 0.4925.
The Empirical Rule
Standard Normal Distribution: µ = 0 and s = 1
The Empirical Rule
Standard Normal Distribution: µ = 0 and s = 1
68% within
1 standard deviation
34%
x-s
34%
x
x+s
The Empirical Rule
Standard Normal Distribution: µ = 0 and s = 1
95% within
2 standard deviations
68% within
1 standard deviation
34%
34%
13.5%
x - 2s
13.5%
x-s
x
x+s
x + 2s
The Empirical Rule
Standard Normal Distribution: µ = 0 and s = 1
99.7% of data are within 3 standard deviations of the mean
95% within
2 standard deviations
68% within
1 standard deviation
34%
34%
2.4%
2.4%
0.1%
0.1%
13.5%
x - 3s
x - 2s
13.5%
x-s
x
x+s
x + 2s
x + 3s
Probability of Half of a Distribution
0.5
0
Finding the Area to the Right of z = 1.27
Value found
in Table A-2
0.3980
0
Figure 5-8
This area is
0.5 - 0.3980 = 0.1020
z = 1.27
Finding the Area Between z = 1.20 and z = 2.30
0.4893 (from Table A-2 with z = 2.30)
Area A is 0.4893 - 0.3849 =
0.1044
0.3849
A
0
Figure 5-9
z = 1.20
z = 2.30
Notation
P(a < z < b)
denotes the probability that the z score is
between a and b
P(z > a)
denotes the probability that the z score is
greater than a
P (z < a)
denotes the probability that the z score is
less than a
Figure 5-10
Interpreting Area Correctly
Interpreting Area Correctly
Figure 5-10
‘greater than
‘at least
x’
x’
‘more than
Subtract
from
0.5
Add to
0.5
x’
‘not less than
x’
0.5
x
x
Interpreting Area Correctly
Figure 5-10
‘greater than
‘at least
x’
Add to
0.5
x’
‘more than
Subtract
from
0.5
x’
‘not less than
x’
0.5
x
Add to
0.5
x
‘less than
‘at most
x’
x’
‘no more than
x’
‘not greater than
Subtract
from
0.5
x’
0.5
x
x
Interpreting Area Correctly
Figure 5-10
‘greater than
‘at least
x’
Add to
0.5
x’
‘more than
Subtract
from
0.5
x’
‘not less than
x’
0.5
x
Add to
0.5
x
‘less than
‘at most
x’
x’
‘no more than
x’
‘not greater than
Subtract
from
0.5
x’
0.5
x
x
Add
C
‘between
x1
and
Use
A=C-B
x2’
A
x1
x2
x1 x2
B
Finding a z - score when given a probability
Using Table A-2
1. Draw a bell-shaped curve, draw the centerline, and
identify the region under the curve that corresponds to
the given probability. If that region is not bounded by
the centerline, work with a known region that is
bounded by the centerline.
2. Using the probability representing the area bounded by
the centerline, locate the closest probability in the body
of Table A-2 and identify the corresponding z score.
3. If the z score is positioned to the left of the centerline,
make it a negative.
Finding z Scores when Given Probabilities
95%
5%
5% or 0.05
0.45
0.50
z
0
( z score will be positive )
FIGURE 5-11
Finding the 95th Percentile
Finding z Scores when Given Probabilities
95%
5%
5% or 0.05
0.45
0.50
0
1.645
(z score will be positive)
FIGURE 5-11
Finding the 95th Percentile
Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
0.10
0.40
z
0
(z score will be negative)
FIGURE 5-12
Finding the 10th Percentile
Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
0.10
0.40
-1.28
0
(z score will be negative)
FIGURE 5-12
Finding the 10th Percentile
ELEMENTARY
Section 5-3
STATISTICS
Normal Distributions: Finding Probabilities
MARIO F. TRIOLA
EIGHTH
EDITION
Other Normal Distributions
m
s
If
 0 or  1 (or both), we will
convert values to standard scores using
Formula 5-2, then procedures for working
with all normal distributions are the same
as those for the standard normal
distribution.
Other Normal Distributions
m
s
If
 0 or  1 (or both), we will
convert values to standard scores using
Formula 5-2, then procedures for working
with all normal distributions are the same
as those for the standard normal
distribution.
Formula 5-2
z=
x-µ
s
Converting to Standard Normal
Distribution
P
(a)
Figure 5-13
m
x
Converting to Standard Normal
Distribution
x-m
z=
s
P
P
(a)
Figure 5-13
m
x
(b)
0
z
Probability of Weight between 143
pounds and 201 pounds
z=
x = 143
s = 29
143
Figure 5-14
0
201
2.00
201 - 143
29
= 2.00
Weight
z
Probability of Weight between 143
pounds and 201 pounds
Value found
in Table A-2
x = 143
s = 29
143
Figure 5-14
0
201
2.00
Weight
z
Probability of Weight between 143
pounds and 201 pounds
0.4772
x = 143
s = 29
143
Figure 5-14
0
201
2.00
Weight
z
Probability of Weight between 143
pounds and 201 pounds
There is a 0.4772 probability of
randomly selecting a woman with
a weight between 143 and 201
lbs.
x = 143
s = 29
143
Figure 5-14
0
201
2.00
Weight
z
Probability of Weight between 143
pounds and 201 pounds
OR - 47.72% of women have
weights between 143 lb and 201 lb.
x = 143
s = 29
143
Figure 5-14
0
201
2.00
Weight
z
ELEMENTARY
Section 5-4
STATISTICS
Normal Distributions: Finding Values
MARIO F. TRIOLA
EIGHTH
EDITION
Cautions to keep in mind
1. Don’t confuse z scores and areas.
Z scores are distances along the
horizontal scale, but areas are regions
under the normal curve. Table A-2 lists z
scores in the left column and across the
top row, but areas are found in the body of
the table.
2. Choose the correct (right/left) side of
the
graph.
3. A z score must be negative whenever it
is
located to the left of the centerline of 0.
Finding z Scores when Given Probabilities
95%
5%
5% or 0.05
0.45
0.50
0
1.645
(z score will be positive)
FIGURE 5-11
Finding the 95th Percentile
Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
0.10
0.40
-1.28
0
(z score will be negative)
FIGURE 5-12
Finding the 10th Percentile
Procedure for Finding Values
Using Table A-2 and Formula 5-2
1. Sketch a normal distribution curve, enter the given probability or
percentage in the appropriate region of the graph, and identify the
value(s) being sought.
x
2. Use Table A-2 to find the z score corresponding to the region bounded
by x and the centerline of 0. Cautions:


Refer to the BODY of Table A-2 to find the closest area, then
identify the corresponding z score.
Make the z score negative if it is located to the left of the centerline.
3. Using Formula 5-2, enter the values for µ, s, and the z score found in
step 2, then solve for x.
x = µ + (z • s)
(Another form of Formula 5-2)
4. Refer to the sketch of the curve to verify that the solution makes sense
in the context of the graph and the context of the problem.
Finding P10 for Weights of Women
10%
90%
40%
x=?
FIGURE 5-17
143
50%
Weight
Finding P10 for Weights of Women
0.10
0.40
FIGURE 5-17
0.50
x=?
143
-1.28
0
Weight
Finding P10 for Weights of Women
x = 143 + (-1.28 • 29) = 105.88
0.10
0.40
FIGURE 5-17
0.50
x=?
143
-1.28
0
Weight
Finding P10 for Weights of Women
The weight of 106 lb (rounded) separates
the lowest 10% from the highest 90%.
0.10
0.40
x = 106
FIGURE 5-17
-1.28
0.50
143
0
Weight
Forgot to make z score negative???
x = 143 + (1.28 • 29) = 180
0.10
0.40
FIGURE 5-17
0.50
x=?
143
1.28
0
Weight
Forgot to make z score negative???
x = 143 + (1.28 • 29) = 180
0.10
0.40
x = 180
FIGURE 5-17
1.28
0.50
143
0
Weight
Forgot to make z score negative???
UNREASONABLE ANSWER!
0.10
0.40
x = 180
FIGURE 5-17
1.28
0.50
143
0
Weight
REMEMBER!
Make the z score negative if the
value is located to the left (below)
the mean. Otherwise, the z score
will be positive.
ELEMENTARY
Section 5-5
S
TATISTICS
The Central Limit Theorem
MARIO F. TRIOLA
EIGHTH
EDITION
Definition
Sampling Distribution of the mean
the probability distribution of
sample means, with all
samples having the same sample
size n.
Central Limit Theorem
Given:
1. The random variable x has a distribution (which
may or may not be normal) with mean µ and
standard deviation s.
2. Samples all of the same size n are randomly
selected from the population of x values.
Central Limit Theorem
Conclusions:
Central Limit Theorem
Conclusions:
1. The distribution of sample x will, as the sample
size increases, approach a normal distribution.
Central Limit Theorem
Conclusions:
1. The distribution of sample x will, as the sample
size increases, approach a normal distribution.
2. The mean of the sample means will be the
population mean µ.
Central Limit Theorem
Conclusions:
1. The distribution of sample x will, as the sample
size increases, approach a normal distribution.
2. The mean of the sample means will be the
population mean µ.
3. The standard deviation of the sample means
will approach s
n
Practical Rules Commonly Used:
1. For samples of size n larger than 30, the distribution of
the sample means can be approximated reasonably well
by a normal distribution. The approximation gets better
as the sample size n becomes larger.
2. If the original population is itself normally distributed,
then the sample means will be normally distributed for
any sample size n (not just the values of n larger than 30).
Notation
Notation
the mean of the sample means
µx = µ
Notation
the mean of the sample means
µx = µ
the standard deviation of sample mean
s
sx = n
Notation
the mean of the sample means
µx = µ
the standard deviation of sample mean
s
sx = n
(often called standard error of the mean)
Distribution of 200 digits from
Social Security Numbers
Frequency
(Last 4 digits from 50 students)
20
10
0
0
1
2
3
4
5
6
7
Distribution of 200 digits
Figure 5-19
8
9
Table 5-2
x
SSN digits
1
5
9
5
9
4
7
9
5
7
2
6
2
2
5
0
2
7
8
5
8
3
8
1
3
2
7
1
3
3
7
7
3
4
4
4
5
1
3
6
6
3
8
2
3
6
1
5
3
4
6
7
3
7
3
3
8
3
7
6
4
6
8
5
5
2
6
4
9
4.75
4.25
8.25
3.25
5.00
3.50
5.25
4.75
5.00
2
6
1
9
5
7
8
6
4
0
7
4.00
5.25
4.25
4.50
4.75
3.75
5.25
3.75
4.50
6.00
Frequency
Distribution of 50 Sample Means for
50 Students
15
10
5
0
0
Figure 5-20
1
2
3
4
5
6
7
8
9
As the sample size increases,
the sampling distribution of
sample means approaches a
normal distribution.
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the probability
that her weight is greater than 150 lb.
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the probability
that her weight is greater than 150 lb.
z = 150-143 = 0.24
29
0.5 - 0.0948 = 0.4052
0.0948
m = 143
s= 29
0
150
0.24
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, the probability that
her weight is greater than 150 lb. is 0.4052.
0.5 - 0.0948 = 0.4052
0.0948
m = 143
s= 29
0
150
0.24
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
mx = 143
150
sx = 29 = 4.83333
36
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
z = 150-143 = 1.45
29
36
0.4265
mx = 143
sx = 4.83333
0
150
1.45
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
z = 150-143 = 1.45
29
36
0.5 - 0.4265 = 0.0735
0.4265
mx = 143
sx = 4.83333
0
150
1.45
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, the
probability that their mean weight is greater than 150 lb is
0.0735.
z = 150-143 = 1.45
29
36
0.5 - 0.4265 = 0.0735
0.4265
mx = 143
sx = 4.83333
0
150
1.45
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the
probability that her weight is greater than 150 lb.
P(x > 150) = 0.4052
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the
probability that her weight is greater than 150 lb.
P(x > 150) = 0.4052
b.) if 36 different women are randomly selected, their
mean weight is greater than 150 lb.
P(x > 150) = 0.0735
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the
probability that her weight is greater than 150 lb.
P(x > 150) = 0.4052
b.) if 36 different women are randomly selected, their
mean weight is greater than 150 lb.
P(x > 150) = 0.0735
It is much easier for an individual to deviate from the
mean than it is for a group of 36 to deviate from the mean.
Sampling Without Replacement
If n > 0.05 N
Sampling Without Replacement
If n > 0.05 N
sx =
s
n
N-n
N-1
Sampling Without Replacement
If n > 0.05 N
sx =
s
n
N-n
N-1
finite population
correction factor
ELEMENTARY
Section 5-6
STATISTICS
Normal Distribution as Approximation to
Binomial Distribution
MARIO F. TRIOLA
EIGHTH
EDITION
Review
Binomial Probability Distribution
1. The procedure must have fixed number of trials.
2. The trials must be independent.
3. Each trial must have all outcomes classified into
two categories.
4. The probabilities must remain constant for each
trial.
Solve by binomial probability formula, Table A-1,
or technology
Approximate a Binomial Distribution
with a Normal Distribution if:
np  5
nq  5
Approximate a Binomial Distribution
with a Normal Distribution if:
np  5
nq  5
then µ = np and s = npq
and the random variable has
a
distribution.
(normal)
Solving Binomial Probability
Problems
Using a Normal
Approximation
Figure 5-24
Solving Binomial Probability
Problems
Using a Normal
Approximation
Figure 5-24
Start
1
2
3
4
First try to solve the binomial probability
problem by using
1. Software or a calculator
2. Table A-1
3. The binomial probability formula
Are np  5 and
nq  5
both true ?
No
Use binomial probability formula
P(x) =
n!
• px • qn-x
(n - x)!x!
Yes
Compute µ = np and s = npq
Draw the normal curve, and identify the region representing the probability to be found.
Be sure to include the continuity correction. (Remember, the discrete value x is adjusted for
continuity by adding and subtracting 0.5)
Solving Binomial Probability
Problems
Using a Normal
Approximation
Figure 5-24
Draw the normal curve, and identify the region representing the probability to be
found. Be sure to include the continuity correction. (Remember, the discrete value x is
adjusted for continuity by adding and subtracting 0.5)
4
5
6
Use Table A-2 for the
standard normal
distribution
Calculate z  x  m
s
where µ and sare the values
already found and x has been
adjusted for continuity
Refer to Table A-2 to find the area between
µ and the value of x adjusted for continuity.
Use that area to find the probability being sought.
Use a TI-83
calculator
Press 2nd, VARS, 2 (for
normalcdf), enter (lower score,
upper, µ, s).
Procedure for Using a Normal Distribution to
Approximate a Binomial Distribution
1. Establish that the normal distribution is a suitable
approximation to the binomial distribution by verifying
np  5 and nq  5.
2. Find the values of the parameters µ and s by
calculating µ = np and s = npq.
3. Identify the discrete value of x (the number of
successes). Change the discrete value x by replacing
it with the interval from x - 0.5 to x + 0.5. Draw a normal
curve and enter the values of µ , s, and either x - 0.5 or
x + 0.5, as appropriate.
continued
Procedure for Using a Normal Distribution to
Approximate a Binomial Distribution
continued
4. Change x by replacing it with x - 0.5 or x + 0.5, as
appropriate.
5. Find the area corresponding to the desired probability.
Finding the Probability of “At Least” 520 Men Among
1000 Accepted Applicants
Figure 5-25
Definition
When we use the normal distribution
(which is continuous) as an
approximation to the binomial
distribution (which is discrete), a
continuity correction is made to a
discrete whole number x in the binomial
distribution by representing the single
value x by the interval from
x - 0.5 to x + 0.5.
Procedure for Continuity Corrections
1. When using the normal distribution as an approximation to the
binomial distribution, always use the continuity correction.
2. In using the continuity correction, first identify the discrete whole
number x that is relevant to the binomial probability problem.
3. Draw a normal distribution centered about µ, then draw a vertical
strip area centered over x . Mark the left side of the strip with the
number x - 0.5, and mark the right side with x + 0.5. For x =520,
draw a strip from 519.5 to 520.5. Consider the area of the strip to
represent the probability of discrete number x.
continued
Procedure for Continuity Corrections
continued
4. Now determine whether the value of x itself should be included in the
probability you want. Next, determine whether you want the
probability of at least x, at most x, more than x, fewer than
x, or
exactly x. Shade the area to the right or left of the strip, as
appropriate; also shade the interior of the strip itself if and only if
itself is to be included. The total shaded region corresponds to
probability being sought.
x
x = at least 520
= 520, 521, 522, . . .
.
520
519.5
Figure 5-26
x = at least 520
= 520, 521, 522, . . .
.
520
519.5
x = more than 520
= 521, 522, 523, . . .
521
520.5
Figure 5-26
x = at least 520
= 520, 521, 522, . . .
.
520
519.5
x = more than 520
= 521, 522, 523, . . .
521
520.5
x = at most 520
= 0, 1, . . . 518, 519, 520
520
520.5
Figure 5-26
x = at least 520
= 520, 521, 522, . . .
.
520
519.5
x = more than 520
= 521, 522, 523, . . .
521
520.5
x = at most 520
= 0, 1, . . . 518, 519, 520
520
520.5
x = fewer than 520
= 0, 1, . . . 518, 519
Figure 5-26
519
519.5
x = exactly
520
x = exactly
520
520
x = exactly
520
520
519.5
520.5
Interval represents discrete number 520
ELEMENTARY
STATISTICS
Chapter 6
Estimates and Sample Sizes
MARIO F. TRIOLA
EIGHTH
EDITION
Chapter 6
Estimates and Sample Sizes
6-1
Overview
6-2
Estimating a Population Mean:
Large Samples
6-3
Estimating a Population Mean:
Small Samples
6-4
Sample Size Required to Estimate µ
6-5
Estimating a Population Proportion
6-6
Estimating a Population Variance
6-1
Overview
This chapter presents:
 methods for estimating
population means, proportions, and
variances
 methods for determining sample sizes
6-2
Estimating a Population Mean:
Large Samples
Assumptions
 n > 30
The sample must have more than 30 values.
 Simple Random Sample
All samples of the same size have an equal chance of
being selected.
Assumptions
 n > 30
The sample must have more than 30 values.
 Simple Random Sample
All samples of the same size have an equal chance of
being selected.
Data collected carelessly can be
absolutely worthless, even if the sample
is quite large.
 Estimator
Definitions
a formula or process for using sample data to estimate
a population parameter
 Estimate
a specific value or range of values used to
approximate some population parameter
 Point Estimate
a single value (or point) used to approximate a
population parameter
 Estimator
Definitions
a formula or process for using sample data to estimate
a population parameter
 Estimate
a specific value or range of values used to
approximate some population parameter
 Point Estimate
a single value (or point) used to approximate a
population parameter
The sample mean x is the best point estimate of
the population mean µ.
Definition
Confidence Interval
(or Interval Estimate)
a range (or an interval) of values used to
estimate the true value of the population
parameter
Definition
Confidence Interval
(or Interval Estimate)
a range (or an interval) of values used to
estimate the true value of the population
parameter
Lower # < population parameter < Upper #
Definition
Confidence Interval
(or Interval Estimate)
a range (or an interval) of values used to
estimate the true value of the population
parameter
Lower # < population parameter < Upper #
As an example
Lower # < m < Upper #
Definition
Degree of Confidence
(level of confidence or confidence coefficient)
the probability 1 -  (often expressed as the
equivalent percentage value) that is the relative
frequency of times the confidence interval
actually does contain the population parameter,
assuming that the estimation process is
repeated a large number of times
usually 90%, 95%, or 99%
( = 10%), ( = 5%), ( = 1%)
Interpreting a Confidence Interval
98.08 < µ < 98.32
o
o
Correct: We are 95% confident that the interval from
98.08 to 98.32 actually does contain the true value of
m.
This means that if we were to select many different
samples of size 106 and construct the confidence
intervals, 95% of them would actually contain the
value of the population mean m.
Wrong: There is a 95% chance that the true value of m
will fall between 98.08 and 98.32.
Confidence Intervals from 20 Different Samples
Figure 6-1
Definition
Critical Value
the number on the borderline separating sample
statistics that are likely to occur from those that
are unlikely to occur. The number z/2 is a critical
value that is a z score with the property that it
separates an area /2 in the right tail of the
standard normal distribution.
The Critical Value
2
2
2
-z
Figure 6-2
z
2
z=0
z
2
Found from Table A-2
(corresponds to area of
0.5 - 2 )
Finding z2 for 95% Degree of Confidence
Finding z2 for 95% Degree of Confidence
95%
 = 5%
2 = 2.5% = .025
.95
.025
-z2
.025
z2
Finding z2 for 95% Degree of Confidence
95%
 = 5%
2 = 2.5% = .025
.95
.025
.025
z2
-z2
Critical Values
Finding z2 for 95% Degree of Confidence
Finding z2 for 95% Degree of Confidence
 = 0.05
 = 0.025
.4750
.025
Use Table A-2
to find a z score of 1.96
Finding z2 for 95% Degree of Confidence
 = 0.05
 = 0.025
.4750
.025
Use Table A-2
to find a z score of 1.96
z2 =  1.96
.025
- 1.96
.025
1.96
Definition
Margin of Error
Definition
Margin of Error
is the maximum likely difference observed
between sample mean x and true population
mean µ.
denoted by E
Definition
Margin of Error
is the maximum likely difference observed
between sample mean x and true population
mean µ.
denoted by E
x -E
µ
x +E
Definition
Margin of Error
is the maximum likely difference observed
between sample mean x and true population
mean µ.
denoted by E
x -E
µ
x +E
x -E < µ < x +E
Definition
Margin of Error
is the maximum likely difference observed
between sample mean x and true population
mean µ.
denoted by E
x -E
µ
x +E
x -E < µ < x +E
lower limit
upper limit
Definition
Margin of Error
E = z/2 •
x -E
s
Formula 6-1
n
µ
x +E
Definition
Margin of Error
E = z/2 •
x -E
s
Formula 6-1
n
µ
x +E
also called the maximum error of the estimate
Calculating E When s Is Unknown
 If n > 30, we can replace s in Formula 61 by the sample standard deviation s.
 If n  30, the population must have
a normal distribution and we must know
s to use Formula 6-1.
Confidence Interval (or Interval Estimate)
for Population Mean µ
(Based on Large Samples: n >30)
x -E <µ< x +E
Confidence Interval (or Interval Estimate)
for Population Mean µ
(Based on Large Samples: n >30)
x -E <µ< x +E
µ=x +E
Confidence Interval (or Interval Estimate)
for Population Mean µ
(Based on Large Samples: n >30)
x -E <µ< x +E
µ=x +E
(x + E, x - E)
Procedure for Constructing a
Confidence Interval for µ
( Based on a Large Sample:
n
> 30 )
Procedure for Constructing a
Confidence Interval for µ
( Based on a Large Sample:
n
> 30 )
1. Find the critical value z2 that corresponds to the
desired degree of confidence.
Procedure for Constructing a
Confidence Interval for µ
( Based on a Large Sample:
n
> 30 )
1. Find the critical value z2 that corresponds to the
desired degree of confidence.
2. Evaluate the margin of error E = z2 • s / n .
If the population standard deviation s is
unknown, use the value of the sample standard
deviation s provided that n > 30.
Procedure for Constructing a
Confidence Interval for µ
( Based on a Large Sample:
n
> 30 )
1. Find the critical value z2 that corresponds to the
desired degree of confidence.
2. Evaluate the margin of error E = z2 • s / n .
If the population standard deviation s is
unknown, use the value of the sample standard
deviation s provided that n > 30.
3. Find the values of x - E and x + E. Substitute those
values in the general format of the confidence
interval: x - E < µ < x + E
Procedure for Constructing a
Confidence Interval for µ
( Based on a Large Sample:
n
> 30 )
1. Find the critical value z2 that corresponds to the
desired degree of confidence.
2. Evaluate the margin of error E = z2 • s / n .
If the population standard deviation s is
unknown, use the value of the sample standard
deviation s provided that n > 30.
3. Find the values of x - E and x + E. Substitute those
values in the general format of the confidence
interval: x - E < µ < x + E
4. Round using the confidence intervals roundoff rules.
Round-Off Rule for Confidence
Intervals Used to Estimate µ
1. When using the original set of data, round
the confidence interval limits to one more
decimal place than used in original set of data.
Round-Off Rule for Confidence
Intervals Used to Estimate µ
1. When using the original set of data, round
the confidence interval limits to one more
decimal place than used in original set of data.
2. When the original set of data is unknown
and only the summary statistics (n, x, s) are
used, round the confidence interval limits
to
the same number of decimal places used
for
the sample mean.
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval.
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval.
n = 106
x = 98.2o
s = 0.62o
 = 0.05
/2 = 0.025
z / 2 = 1.96
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval.
n = 106
x = 98.20o
s = 0.62o
 = 0.05
/2 = 0.025
z / 2 = 1.96
E = z / 2 • s
n
= 1.96 • 0.62
106
= 0.12
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval.
n = 106
x = 98.20o
s = 0.62o
 = 0.05
/2 = 0.025
z / 2 = 1.96
E = z / 2 • s
n
= 1.96 • 0.62
106
= 0.12
x -E <m< x +E
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval.
n = 106
x = 98.20o
s = 0.62o
 = 0.05
/2 = 0.025
z / 2 = 1.96
E = z / 2 • s
n
= 1.96 • 0.62
106
= 0.12
x -E <m< x +E
98.20o - 0.12
<m<
98.20o + 0.12
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval.
n = 106
x = 98.20o
s = 0.62o
 = 0.05
/2 = 0.025
z / 2 = 1.96
E = z / 2 • s
n
= 1.96 • 0.62
106
= 0.12
x -E <m< x +E
98.20o - 0.12
98.08o
<m<
<m<
98.20o + 0.12
98.32o
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval.
n = 106
x = 98.20o
s = 0.62o
 = 0.05
/2 = 0.025
z / 2 = 1.96
E = z / 2 • s
n
= 1.96 • 0.62
106
= 0.12
x -E <m< x +E
98.08o
< m <
98.32o
Based on the sample provided, the confidence interval for the
m
population mean is 98.08o <
< 98.32o. If we were to select many
different samples of the same size, 95% of the confidence intervals
would actually contain the population mean m.
Finding the Point Estimate and E
from a Confidence Interval
Point estimate of µ:
x = (upper confidence interval limit) + (lower confidence interval limit)
2
Finding the Point Estimate and E
from a Confidence Interval
Point estimate of µ:
x = (upper confidence interval limit) + (lower confidence interval limit)
2
Margin of Error:
E = (upper confidence interval limit) - (lower confidence interval limit)
2
ELEMENTARY
Section 6-3
STATISTICS
Estimating a Population Mean: Small Samples
MARIO F. TRIOLA
EIGHTH
EDITION
Small Samples
Assumptions
If 1) n  30
2) The sample is a simple random sample.
3) The sample is from a normally
distributed population.
Case 1 (s is known): Largely unrealistic; Use
methods from 6-2
Case 2 (sis unknown): Use Student t
distribution
Student t Distribution
If the distribution of a population is
essentially normal, then the distribution of
t =
x-µ
s
n
Student t Distribution
If the distribution of a population is
essentially normal, then the distribution of
t =
x-µ
s
n
 is essentially a Student t Distribution for all
samples of size n.
 is used to find critical values denoted by
t/ 2
Table A - 3
 Formulas and Tables Card
 Back cover
 Appendix
Definition
Degrees of Freedom (df )
corresponds to the number of sample values
that can vary after certain restrictions have
imposed on all data values
Definition
Degrees of Freedom (df )
corresponds to the number of sample values
that can vary after certain restrictions have
imposed on all data values
df = n - 1
in this section
Definition
Degrees of Freedom (df ) = n - 1
corresponds to the number of sample values
that can vary after certain restrictions have
imposed on all data values
Any
Any
Any
Any
Any
Any
Any
Any
Any
Specific
#
#
#
#
#
#
#
#
#
#
n = 10
df = 10 - 1 = 9
so that x = 80
Margin of Error E for Estimate of m
Based on an Unknown s and a Small Simple Random
Sample from a Normally Distributed Population
Margin of Error E for Estimate of m
Based on an Unknown s and a Small Simple Random
Sample from a Normally Distributed Population
Formula 6-2
E = t
s
2
n
where t/ 2 has n - 1 degrees of freedom
Confidence Interval for the
Estimate of E
Based on an Unknown s and a Small Simple Random
Sample from a Normally Distributed Population
Confidence Interval for the
Estimate of E
Based on an Unknown s and a Small Simple Random
Sample from a Normally Distributed Population
x-E <µ< x +E
Confidence Interval for the
Estimate of E
Based on an Unknown s and a Small Simple Random
Sample from a Normally Distributed Population
x-E <µ< x +E
where
E = t/2 s
n
Confidence Interval for the
Estimate of E
Based on an Unknown s and a Small Simple Random
Sample from a Normally Distributed Population
x-E <µ< x +E
where
E = t/2 s
n
t/2 found in Table A-3
Table A-3 t Distribution
Degrees
of
freedom
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
Large (z)
.005
(one tail)
.01
(two tails)
63.657
9.925
5.841
4.604
4.032
3.707
3.500
3.355
3.250
3.169
3.106
3.054
3.012
2.977
2.947
2.921
2.898
2.878
2.861
2.845
2.831
2.819
2.807
2.797
2.787
2.779
2.771
2.763
2.756
2.575
.01
(one tail)
.02
(two tails)
31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.718
2.681
2.650
2.625
2.602
2.584
2.567
2.552
2.540
2.528
2.518
2.508
2.500
2.492
2.485
2.479
2.473
2.467
2.462
2.327
.025
(one tail)
.05
(two tails)
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.201
2.179
2.160
2.145
2.132
2.120
2.110
2.101
2.093
2.086
2.080
2.074
2.069
2.064
2.060
2.056
2.052
2.048
2.045
1.960
.05
(one tail)
.10
(two tails)
.10
(one tail)
.20
(two tails)
.25
(one tail)
.50
(two tails)
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.796
1.782
1.771
1.761
1.753
1.746
1.740
1.734
1.729
1.725
1.721
1.717
1.714
1.711
1.708
1.706
1.703
1.701
1.699
1.645
3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.363
1.356
1.350
1.345
1.341
1.337
1.333
1.330
1.328
1.325
1.323
1.321
1.320
1.318
1.316
1.315
1.314
1.313
1.311
1.282
1.000
.816
.765
.741
.727
.718
.711
.706
.703
.700
.697
.696
.694
.692
.691
.690
.689
.688
.688
.687
.686
.686
.685
.685
.684
.684
.684
.683
.683
.675
Important Properties of the Student t Distribution
1. The Student t distribution is different for different sample sizes (see
Figure 6-5 for the cases n = 3 and n = 12).
2. The Student t distribution has the same general symmetric bell
shape as the normal distribution but it reflects the greater
variability (with wider distributions) that is expected with small
samples.
3. The Student t distribution has a mean of t = 0 (just as the standard
normal distribution has a mean of z = 0).
4. The standard deviation of the Student t distribution varies with the
sample size and is greater than 1 (unlike the standard normal
distribution, which has a s = 1).
5. As the sample size n gets larger, the Student t distribution gets
closer to the normal distribution. For values of n > 30, the
differences are so small that we can use the critical z values
instead of developing a much larger table of critical t values. (The
values in the bottom row of Table A-3 are equal to the
corresponding critical z values from the standard normal
distribution.)
Student t Distributions for
n = 3 and n = 12
Student t
Standard
normal
distribution
distribution
with n = 12
Student t
distribution
with n = 3
Figure 6-5
0
Using the Normal and t Distribution
Figure 6-6
Example:
A study of 12 Dodge Vipers involved in
collisions resulted in repairs averaging $26,227 and a
standard deviation of $15,873. Find the 95% interval
estimate of m, the mean repair cost for all Dodge Vipers
involved in collisions. (The 12 cars’ distribution appears to
be bell-shaped.)
Example:
A study of 12 Dodge Vipers involved in
collisions resulted in repairs averaging $26,227 and a
standard deviation of $15,873. Find the 95% interval
estimate of m, the mean repair cost for all Dodge Vipers
involved in collisions. (The 12 cars’ distribution appears to
be bell-shaped.)
x = 26,227
s = 15,873
 = 0.05
/2 = 0.025
Table A-3 t Distribution
Degrees
of
freedom
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
Large (z)
.005
(one tail)
.01
(two tails)
63.657
9.925
5.841
4.604
4.032
3.707
3.500
3.355
3.250
3.169
3.106
3.054
3.012
2.977
2.947
2.921
2.898
2.878
2.861
2.845
2.831
2.819
2.807
2.797
2.787
2.779
2.771
2.763
2.756
2.575
.01
(one tail)
.02
(two tails)
31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.718
2.681
2.650
2.625
2.602
2.584
2.567
2.552
2.540
2.528
2.518
2.508
2.500
2.492
2.485
2.479
2.473
2.467
2.462
2.327
.025
(one tail)
.05
(two tails)
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.201
2.179
2.160
2.145
2.132
2.120
2.110
2.101
2.093
2.086
2.080
2.074
2.069
2.064
2.060
2.056
2.052
2.048
2.045
1.960
.05
(one tail)
.10
(two tails)
.10
(one tail)
.20
(two tails)
.25
(one tail)
.50
(two tails)
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.796
1.782
1.771
1.761
1.753
1.746
1.740
1.734
1.729
1.725
1.721
1.717
1.714
1.711
1.708
1.706
1.703
1.701
1.699
1.645
3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.363
1.356
1.350
1.345
1.341
1.337
1.333
1.330
1.328
1.325
1.323
1.321
1.320
1.318
1.316
1.315
1.314
1.313
1.311
1.282
1.000
.816
.765
.741
.727
.718
.711
.706
.703
.700
.697
.696
.694
.692
.691
.690
.689
.688
.688
.687
.686
.686
.685
.685
.684
.684
.684
.683
.683
.675
Example:
A study of 12 Dodge Vipers involved in
collisions resulted in repairs averaging $26,227 and a
standard deviation of $15,873. Find the 95% interval
estimate of m, the mean repair cost for all Dodge Vipers
involved in collisions. (The 12 cars’ distribution appears to
be bell-shaped.)
x = 26,227
s = 15,873
 = 0.05
/2 = 0.025
t/2 = 2.201
E = t2 s = (2.201)(15,873) = 10,085.29
n
12
Example:
A study of 12 Dodge Vipers involved in
collisions resulted in repairs averaging $26,227 and a
standard deviation of $15,873. Find the 95% interval
estimate of m, the mean repair cost for all Dodge Vipers
involved in collisions. (The 12 cars’ distribution appears to
be bell-shaped.)
x = 26,227
s = 15,873
 = 0.05
/2 = 0.025
t/2 = 2.201
E = t2 s = (2.201)(15,873) = 10,085.3
12
n
x -E
<µ<
x +E
Example:
A study of 12 Dodge Vipers involved in
collisions resulted in repairs averaging $26,227 and a
standard deviation of $15,873. Find the 95% interval
estimate of m, the mean repair cost for all Dodge Vipers
involved in collisions. (The 12 cars’ distribution appears to
be bell-shaped.)
x = 26,227
s = 15,873
 = 0.05
/2 = 0.025
t/2 = 2.201
E = t2 s = (2.201)(15,873) = 10,085.3
n
x -E
12
<µ< x +E
26,227 - 10,085.3 < µ < 26,227 + 10,085.3
Example:
A study of 12 Dodge Vipers involved in
collisions resulted in repairs averaging $26,227 and a
standard deviation of $15,873. Find the 95% interval
estimate of m, the mean repair cost for all Dodge Vipers
involved in collisions. (The 12 cars’ distribution appears to
be bell-shaped.)
x = 26,227
s = 15,873
 = 0.05
/2 = 0.025
t/2 = 2.201
E = t2 s = (2.201)(15,873) = 10,085.3
n
x -E
12
<µ< x +E
26,227 - 10,085.3 < µ < 26,227 + 10,085.3
$16,141.7 < µ < $36,312.3
Example:
A study of 12 Dodge Vipers involved in
collisions resulted in repairs averaging $26,227 and a
standard deviation of $15,873. Find the 95% interval
estimate of m, the mean repair cost for all Dodge Vipers
involved in collisions. (The 12 cars’ distribution appears to
be bell-shaped.)
x = 26,227
s = 15,873
 = 0.05
/2 = 0.025
t/2 = 2.201
E = t2 s = (2.201)(15,873) = 10,085.3
n
12
x -E
<µ< x +E
26,227 - 10,085.3 < µ < 26,227 + 10,085.3
$16,141.7 < µ < $36,312.3
We are 95% confident that this interval contains the average cost of
repairing a Dodge Viper.
ELEMENTARY
STATISTICS
Section 6-4 Determining Sample Size Required to Estimate m
MARIO F. TRIOLA
EIGHTH
EDITION
Sample Size for Estimating Mean m
Sample Size for Estimating Mean m
s
E = z/ 2 • n
Sample Size for Estimating Mean m
s
E = z/ 2 • n
(solve for n by algebra)
Sample Size for Estimating Mean m
s
E = z/ 2 • n
(solve for n by algebra)
n=
z/ 2 s
E
2
Formula 6-3
Sample Size for Estimating Mean m
s
E = z/ 2 • n
(solve for n by algebra)
n=
z/ 2 s
2
Formula 6-3
E
z/2 = critical z score based on the desired degree of confidence
E = desired margin of error
s = population standard deviation
Round-Off Rule for Sample Size n
Round-Off Rule for Sample Size n
When finding the sample size n, if the use
of Formula 6-3 does not result in a whole
number, always increase the value of n to
the next larger whole number.
Round-Off Rule for Sample Size n
When finding the sample size n, if the use
of Formula 6-3 does not result in a whole
number, always increase the value of n to
the next larger whole number.
n = 216.09 = 217 (rounded up)
Example:
If we want to estimate the mean weight of
plastic discarded by households in one week, how many
households must be randomly selected to be 99%
confident that the sample mean is within 0.25 lb of the true
population mean? (A previous study indicates the
standard deviation is 1.065 lb.)
Example:
If we want to estimate the mean weight of
plastic discarded by households in one week, how many
households must be randomly selected to be 99%
confident that the sample mean is within 0.25 lb of the true
population mean? (A previous study indicates the
standard deviation is 1.065 lb.)
 = 0.01
z = 2.575
E = 0.25
s = 1.065
Example:
If we want to estimate the mean weight of
plastic discarded by households in one week, how many
households must be randomly selected to be 99%
confident that the sample mean is within 0.25 lb of the true
population mean? (A previous study indicates the
standard deviation is 1.065 lb.)
2
2
 = 0.01
z = 2.575
E = 0.25
s = 1.065
n = zs
E
= (2.575)(1.065)
0.25
Example:
If we want to estimate the mean weight of
plastic discarded by households in one week, how many
households must be randomly selected to be 99%
confident that the sample mean is within 0.25 lb of the true
population mean? (A previous study indicates the
standard deviation is 1.065 lb.)
2
2
 = 0.01
z = 2.575
E = 0.25
s = 1.065
n = zs
E
= (2.575)(1.065)
0.25
= 120.3 = 121 households
Example:
If we want to estimate the mean weight of
plastic discarded by households in one week, how many
households must be randomly selected to be 99%
confident that the sample mean is within 0.25 lb of the true
population mean? (A previous study indicates the
standard deviation is 1.065 lb.)
2
2
 = 0.01
z = 2.575
E = 0.25
s = 1.065
n = zs
E
= (2.575)(1.065)
0.25
= 120.3 = 121 households
If n is not a whole number, round it up
to the next higher whole number.
Example:
If we want to estimate the mean weight of
plastic discarded by households in one week, how many
households must be randomly selected to be 99%
confident that the sample mean is within 0.25 lb of the true
population mean? (A previous study indicates the
standard deviation is 1.065 lb.)
2
2
 = 0.01
z = 2.575
E = 0.25
s = 1.065
n = zs
E
= (2.575)(1.065)
0.25
= 120.3 = 121 households
We would need to randomly select 121 households and
obtain the average weight of plastic discarded in one week.
We would be 99% confident that this mean is within 1/4 lb
of the population mean.
What if sis Not Known ?
1. Use the range rule of thumb to estimate the
standard deviation as follows: s  range
4
What if sis Not Known ?
1. Use the range rule of thumb to estimate the
standard deviation as follows: s  range
4
2. Conduct a pilot study by starting the sampling
process. Based on the first collection of at least
31 randomly selected sample values, calculate
the sample standard deviation s and use it in
place of s. That value can be refined as more
sample data are obtained.
What if sis Not Known ?
1. Use the range rule of thumb to estimate the
standard deviation as follows: s  range
4
2. Conduct a pilot study by starting the sampling
process. Based on the first collection of at least
31 randomly selected sample values, calculate
the sample standard deviation s and use it in
place of s. That value can be refined as more
sample data are obtained.
3. Estimate the value of s by using the results
of some other study that was done earlier.
What happens when E is doubled ?
What happens when E is doubled ?
2
2
(z/ 2s )
z/ 2 s
E=1:
n=
1
=
1
What happens when E is doubled ?
2
2
(z/ 2s )
z/ 2 s
E=1:
n=
E=2:
(z/ 2s )
z
/ 2 s
n=
=
4
2
1
=
2
1
2
Sample size n is decreased to 1/4 of its
original value if E is doubled.
Larger errors allow smaller samples.
Smaller errors require larger samples.
ELEMENTARY
Section 6-5
Proportion
STATISTICS
Estimating a Population
MARIO F. TRIOLA
EIGHTH
EDITION
Assumptions
1. The sample is a simple random sample.
2. The conditions for the binomial
distribution
are satisfied (See Section 4-3.)
3. The normal distribution can be used
to
approximate the distribution of
sample
proportions because np  5 and nq
5
are both satisfied.
Notation for Proportions
Notation for Proportions
p=
population proportion
Notation for Proportions
p=
ˆp = xn
(pronounced
‘p-hat’)
population proportion
sample proportion
of x successes in a sample of size n
Notation for Proportions
p=
ˆp = xn
population proportion
sample proportion
of x successes in a sample of size n
(pronounced
‘p-hat’)
qˆ = 1 - pˆ = sample proportion
of x failures in a sample size of n
Definition
Point Estimate
Definition
Point Estimate
The sample proportion p
ˆ is the best
point estimate of the population
proportion p.
Margin of Error of the Estimate of p
Formula 6-4
E = z 
pˆ qˆ
n
Confidence Interval for
Population Proportion
pˆ - E < p < p̂ + E
where
E = z 
pˆ qˆ
n
Confidence Interval for
Population Proportion
pˆ - E < p < p̂ + E
p = pˆ + E
Confidence Interval for
Population Proportion
pˆ - E < p < p̂ + E
p = pˆ + E
(pˆ - E, pˆ + E)
Round-Off Rule for Confidence
Interval Estimates of p
Round the confidence
interval limits to
three significant digits.
Determining Sample Size
Determining Sample Size
E=
z 
pˆ qˆ
n
Determining Sample Size
E=
z 
pˆ qˆ
n
(solve for n by algebra)
Determining Sample Size
E=
z 
pˆ qˆ
n
(solve for n by algebra)
n=
( z pˆ qˆ
2
)
E2
Sample Size for Estimating Proportion p
ˆ
When an estimate of p is known:
n=
(
2 pq
)
z ˆ ˆ
E2
Formula 6-5
Sample Size for Estimating Proportion p
ˆ
When an estimate of p is known:
n=
(
2 pq
)
z ˆ ˆ
Formula 6-5
E2
When no estimate of p is known:
n=
(
2 0.25
)
z
E2
Formula 6-6
pˆ
qˆ
pˆ qˆ
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.09
0.16
0.21
0.24
0.25
0.24
0.21
0.16
0.09
pˆ
qˆ
pˆ qˆ
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.09
0.16
0.21
0.24
0.25
0.24
0.21
0.16
0.09
Two formulas
for proportion sample size
n=
n=
( z )2 pˆ qˆ
E2
( z (0.25)
2
)
E2
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? A 1997 study indicates 16.9% of U.S. households
used e-mail.
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? A 1997 study indicates 16.9% of U.S. households
used e-mail.
ˆ
n = [z/2 ]2 p q̂
E2
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? A 1997 study indicates 16.9% of U.S. households
used e-mail.
ˆˆ
n = [z/2 ]2 p q
E2
= [1.645]2 (0.169)(0.831)
0.042
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? A 1997 study indicates 16.9% of U.S. households
used e-mail.
ˆˆ
n = [z/2 ]2 p q
E2
= [1.645]2 (0.169)(0.831)
0.042
= 237.51965
= 238 households
To be 90% confident that our
sample percentage is within
four percentage points of the
true percentage for all
households, we should
randomly select and survey
238 households.
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? There is no prior information suggesting a
possible value for the sample percentage.
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? There is no prior information suggesting a
possible value for the sample percentage.
n = [z/2 ]2 (0.25)
E2
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? There is no prior information suggesting a
possible value for the sample percentage.
n = [z/2 ]2 (0.25)
E2
= (1.645)2 (0.25)
0.042
= 422.81641
= 423 households
Example: We want to determine, with a margin of error
of four percentage points, the current percentage of U.S.
households using e-mail. Assuming that we want 90%
confidence in our results, how many households must we
survey? There is no prior information suggesting a
possible value for the sample percentage.
n = [z/2 ]2 (0.25)
E2
= (1.645)2 (0.25)
0.042
= 422.81641
= 423 households
With no prior information,
we need a larger sample to
achieve the same results
with 90% confidence and an
error of no more than 4%.
Finding the Point Estimate and E
from a Confidence Interval
ˆ
(upper confidence interval limit) + (lower confidence interval limit)
Point estimate of p:
ˆ
p=
2
Finding the Point Estimate and E
from a Confidence Interval
ˆ
(upper confidence interval limit) + (lower confidence interval limit)
Point estimate of p:
ˆ
p=
2
Margin of Error:
E = (upper confidence interval limit) - (lower confidence interval limit)
2
ELEMENTARY
STATISTICS
Section 6-6Estimating a Population Variance
MARIO F. TRIOLA
EIGHTH
EDITION
Assumptions
1. The sample is a simple random sample.
2. The population must have
normally
distributed values (even if the
sample is
large).
Chi-Square Distribution
Chi-Square Distribution
X =
2
(n - 1) s 2
s2
Formula 6-7
where
n = sample size
s 2 = sample variance
s2 = population variance
X2 Critical Values found in Table A-4
 Formula card
 Appendix
 Degrees of freedom (df ) = n - 1
Properties of the Distribution of
the Chi-Square Statistic
1. The chi-square distribution is not symmetric, unlike
the normal and Student t distributions.
Properties of the Distribution of
the Chi-Square Statistic
1. The chi-square distribution is not symmetric, unlike
the normal and Student t distributions.
Not symmetric
0
All values are nonnegative
Figure 6-7 Chi-Square Distribution
x2
Properties of the Distribution of
the Chi-Square Statistic
1. The chi-square distribution is not symmetric, unlike
the normal and Student t distributions.
As the number of degrees of freedom increases, the
distribution becomes more symmetric. (continued)
df = 10
Not symmetric
df = 20
0
All values are nonnegative
Figure 6-7 Chi-Square Distribution
x2
0
5 10 15 20
25 30 35 40 45
Figure 6-8 Chi-Square Distribution for df = 10
and df = 20
Properties of the Distribution of
the Chi-Square Statistic
(continued)
2. The values of chi-square can be zero or positive, but
they cannot be negative.
3. The chi-square distribution is different for each
number of degrees of freedom, which is df = n - 1
in this section. As the number increases, the chisquare distribution approaches a normal
distribution.
In Table A-4, each critical value of X2 corresponds to
an area given in the top row of the table, and that area
represents the total region located to the right of the
critical value.
Degrees of
freedom
2
Table A-4 Chi-Square (x ) Distribution
Area to the Right of the Critical Value
0.995
0.99
0.975
0.95
0.90
0.10
0.05
0.025
0.01
0.005
1
2
3
4
5
_
0.010
0.072
0.207
0.412
_
0.020
0.115
0.297
0.554
0.001
0.051
0.216
0.484
0.831
0.004
0.103
0.352
0.711
1.145
0.016
0.211
0.584
1.064
1.610
2.706
4.605
6.251
7.779
9.236
3.841
5.991
7.815
9.488
11.071
5.024
7.378
9.348
11.143
12.833
6.635
9.210
11.345
13.277
15.086
7.879
10.597
12.838
14.860
16.750
6
7
8
9
10
0.676
0.989
1.344
1.735
2.156
0.872
1.239
1.646
2.088
2.558
1.237
1.690
2.180
2.700
3.247
1.635
2.167
2.733
3.325
3.940
2.204
2.833
3.490
4.168
4.865
10.645
12.017
13.362
14.684
15.987
12.592
14.067
15.507
16.919
18.307
14.449
16.013
17.535
19.023
20.483
16.812
18.475
20.090
21.666
23.209
18.548
20.278
21.955
23.589
25.188
11
12
13
14
15
2.603
3.074
3.565
4.075
4.601
3.053
3.571
4.107
4.660
5.229
3.816
4.404
5.009
5.629
6.262
4.575
5.226
5.892
6.571
7.261
5.578
6.304
7.042
7.790
8.547
17.275
18.549
19.812
21.064
22.307
19.675
21.026
22.362
23.685
24.996
21.920
23.337
24.736
26.119
27.488
24.725
26.217
27.688
29.141
30.578
26.757
28.299
29.819
31.319
32.801
16
17
18
19
20
5.142
5.697
6.265
6.844
7.434
5.812
6.408
7.015
7.633
8.260
6.908
7.564
8.231
8.907
9.591
7.962
8.672
9.390
10.117
10.851
9.312
10.085
10.865
11.651
12.443
23.542
24.769
25.989
27.204
28.412
26.296
27.587
28.869
30.144
31.410
28.845
30.191
31.526
32.852
34.170
32.000
33.409
34.805
36.191
37.566
34.267
35.718
37.156
38.582
39.997
21
22
23
24
25
8.034
8.643
9.260
9.886
10.520
8.897
9.542
10.196
10.856
11.524
10.283
10.982
11.689
12.401
13.120
11.591
12.338
13.091
13.848
14.611
13.240
14.042
14.848
15.659
16.473
29.615
30.813
32.007
33.196
34.382
32.671
33.924
35.172
36.415
37.652
35.479
36.781
38.076
39.364
40.646
38.932
40.289
41.638
42.980
44.314
41.401
42.796
44.181
45.559
46.928
26
27
28
29
30
11.160
11.808
12.461
13.121
13.787
12.198
12.879
13.565
14.257
14.954
13.844
14.573
15.308
16.047
16.791
15.379
16.151
16.928
17.708
18.493
17.292
18.114
18.939
19.768
20.599
35.563
36.741
37.916
39.087
40.256
38.885
40.113
41.337
42.557
43.773
41.923
43.194
44.461
45.722
46.979
45.642
46.963
48.278
49.588
50.892
48.290
49.645
50.993
52.336
53.672
40
50
60
70
80
90
100
20.707
27.991
35.534
43.275
51.172
59.196
67.328
22.164
29.707
37.485
45.442
53.540
61.754
70.065
24.433
32.357
40.482
48.758
57.153
65.647
74.222
26.509
34.764
43.188
51.739
60.391
69.126
77.929
29.051
37.689
46.459
55.329
64.278
73.291
82.358
51.805
63.167
74.397
85.527
96.578
107.565
118.498
55.758
67.505
79.082
90.531
101.879
113.145
124.342
59.342
71.420
83.298
95.023
106.629
118.136
129.561
63.691
76.154
88.379
100.425
112.329
124.116
135.807
66.766
79.490
91.952
104.215
116.321
128.299
140.169
Critical Values: Table A-4
Areas to the right of each tail
0.975
0.025
0.025
0.025
0
XL2 = 2.700
2
X2
(df = 9)
XR = 19.023
Estimators of s
2
The sample variance s is the best
point estimate of the population
variance s .
2
2
Confidence Interval for the
2
Population Variance s
Confidence Interval for the
2
Population Variance s
(n - 1)s 2
X
2
R
s2 
(n - 1)s 2
X
2
L
Confidence Interval for the
2
Population Variance s
(n - 1)s 2
Right-tail CV
X
2
R
s2 
(n - 1)s 2
X
2
L
Confidence Interval for the
2
Population Variance s
(n - 1)s 2
Right-tail CV
X
2
R
s2 
(n - 1)s 2
X
2
L
Left-tail CV
Confidence Interval for the
2
Population Variance s
(n - 1)s 2
X
Right-tail CV
2
R
s2 
(n - 1)s 2
X
2
L
Left-tail CV
Confidence Interval for the Population Standard Deviation
(n - 1)s 2
X
2
R
s
(n - 1)s 2
2
XL
s
Roundoff Rule for Confidence Interval
Estimates of s or s2
1. When using the original set of data to construct a
confidence interval, round the confidence interval
limits to one more decimal place than is used for
the original set of data.
2. When the original set of data is unknown and
only the summary statistics (n, s) are used, round
the confidence interval limits to the same number
of decimals places used for the sample standard
deviation or variance.
Table 6-3
Determining Sample Size
Table 6-3
Determining Sample Size
Sample Size for s2
To be 95% confident
that s2 is within
1%
5%
10%
20%
30%
40%
50%
To be 95% confident
that s2 is within
1%
5%
10%
20%
30%
40%
50%
of the value of s2 , the sample
size n should be at least
77,207
3,148
805
210
97
56
57
of the value of s2 , the sample
size n should be at least
133,448
5,457
1,401
368
171
100
67
Sample Size for s
To be 95% confident
that s is within
1%
5%
10%
20%
30%
40%
50%
To be 95% confident
that s is within
1%
5%
10%
20%
30%
40%
50%
of the value of s , the sample
size n should be at least
19,204
767
191
47
20
11
7
of the value of s , the sample
size n should be at least
33,218
1,335
335
84
37
21
13
ELEMENTARY
STATISTICS
Chapter 7
Hypothesis Testing
MARIO F. TRIOLA
EIGHTH
EDITION
Chapter 7
Hypothesis Testing
7-1 Overview
7-2 Fundamentals of Hypothesis Testing
7-3 Testing a Claim about a Mean: Large
Samples
7-4 Testing a Claim about a Mean: Small
Samples
7-5 Testing a Claim about a Proportion
7-6 Testing a Claim about a Standard
Deviation
7-1
Overview
Definition
Hypothesis
in statistics, is a claim or statement about
a property of a population
Rare Event Rule for Inferential
Statistics
If, under a given assumption, the
probability of a particular observed event
is exceptionally small, we conclude that
the assumption is probably not correct.
7-2
Fundamentals of
Hypothesis Testing
Figure 7-1
Central Limit Theorem
Figure 7-1
Central Limit Theorem
The Expected Distribution of Sample Means
Assuming that m = 98.6
Likely sample means
µx = 98.6
Figure 7-1
Central Limit Theorem
The Expected Distribution of Sample Means
Assuming that m = 98.6
Likely sample means
µx = 98.6
z = - 1.96
or
x = 98.48
z=
1.96
or
x = 98.72
Figure 7-1
Central Limit Theorem
The Expected Distribution of Sample Means
Assuming that m = 98.6
Sample data: z = - 6.64
or
x = 98.20
Likely sample means
µx = 98.6
z = - 1.96
or
x = 98.48
z=
1.96
or
x = 98.72
Components of a
Formal Hypothesis
Test
Null Hypothesis: H0
 Statement about value
of population parameter
 Must contain condition of equality
 =, , or 
 Test the Null Hypothesis directly
 Reject H0 or fail to reject H0
Alternative Hypothesis: H1
 Must be true if H0 is false
 , <, >
 ‘opposite’ of Null
Note about Forming Your Own Claims
(Hypotheses)
If you are conducting a study and want
to use a hypothesis test to support your
claim, the claim must be worded so that
it becomes the alternative hypothesis.
Note about Testing the Validity of
Someone Else’s Claim
Someone else’s claim may become the
null hypothesis (because it contains
equality), and it sometimes becomes the
alternative hypothesis (because it does
not contain equality).
Test Statistic
a value computed from the sample data that is
used in making the decision about the
rejection of the null hypothesis
Test Statistic
a value computed from the sample data that is
used in making the decision about the
rejection of the null hypothesis
For large samples, testing claims about
population means
z=
x - µx
s
n
Critical Region
Set of all values of the test statistic that
would cause a rejection of the
null hypothesis
Critical Region
Set of all values of the test statistic that
would cause a rejection of the
null hypothesis
Critical
Region
Critical Region
Set of all values of the test statistic that
would cause a rejection of the
null hypothesis
Critical
Region
Critical Region
Set of all values of the test statistic that
would cause a rejection of the
null hypothesis
Critical
Regions
Significance Level
 denoted by 
 the probability that the
test statistic will fall in the
critical region when the null
hypothesis is actually true.
 common choices are 0.05,
0.01, and 0.10
Critical Value
Value or values that separate the critical region
(where we reject the null hypothesis) from the
values of the test statistics that do not lead
to a rejection of the null hypothesis
Critical Value
Value or values that separate the critical region
(where we reject the null hypothesis) from the
values of the test statistics that do not lead
to a rejection of the null hypothesis
Critical Value
( z score )
Critical Value
Value or values that separate the critical region
(where we reject the null hypothesis) from the
values of the test statistics that do not lead
to a rejection of the null hypothesis
Reject H0
Critical Value
( z score )
Fail to reject H0
Two-tailed,Right-tailed,
Left-tailed Tests
The tails in a distribution are the
extreme regions bounded
by critical values.
Two-tailed Test
H0: µ = 100
H1: µ  100
Two-tailed Test
H0: µ = 100
H1: µ  100
 is divided equally between
the two tails of the critical
region
Two-tailed Test
H0: µ = 100
H1: µ  100
 is divided equally between
the two tails of the critical
region
Means less than or greater than
Two-tailed Test
H0: µ = 100
H1: µ  100
 is divided equally between
the two tails of the critical
region
Means less than or greater than
Reject H0
Fail to reject H0
Reject H0
100
Values that differ significantly from 100
Right-tailed Test
H0: µ  100
H1: µ > 100
Right-tailed Test
H0: µ  100
H1: µ > 100
Points Right
Right-tailed Test
H0: µ  100
H1: µ > 100
Points Right
Fail to reject H0
100
Reject H0
Values that
differ significantly
from 100
Left-tailed Test
H0: µ  100
H1: µ < 100
Left-tailed Test
H0: µ  100
H1: µ < 100
Points Left
Left-tailed Test
H0: µ  100
H1: µ < 100
Points Left
Reject H0
Values that
differ significantly
from 100
Fail to reject H0
100
Conclusions
in Hypothesis Testing
always test the null hypothesis
1. Reject the H0
2. Fail to reject the H0
need to formulate correct wording of final
conclusion
See Figure 7-4
Wording of Final Conclusion
FIGURE 7-4
Start
Does the
original claim contain
the condition of
equality
Yes
(Original claim
contains equality
and becomes H0)
No
Do
you reject
H0?.
“There is sufficient
evidence to warrant
(Reject H0) rejection of the claim
that. . . (original claim).”
Yes
No
(Fail to
reject H0)
(Original claim
does not contain
equality and
becomes H1)
Do
you reject
H0?
Yes
(Reject H0)
“There is not sufficient
evidence to warrant
rejection of the claim
that. . . (original claim).”
“The sample data
supports the claim that
. . . (original claim).”
No
(Fail to
reject H0)
(This is the
only case in
which the
original claim
is rejected).
“There is not sufficient
evidence to support
the claim
that. . . (original claim).”
(This is the
only case in
which the
original claim
is supported).
Accept versus Fail to Reject
some texts use “accept the null
hypothesis
we are not proving the null hypothesis
sample evidence is not strong enough
to warrant rejection (such as not
enough evidence to convict a suspect)
Type I Error
The mistake of rejecting the null hypothesis
when it is true.
 (alpha) is used to represent the probability
of a type I error
Example: Rejecting a claim that the mean
body temperature is 98.6 degrees when the
mean really does equal 98.6
Type II Error
the mistake of failing to reject the null
hypothesis when it is false.
ß (beta) is used to represent the probability of
a type II error
Example: Failing to reject the claim that the
mean body temperature is 98.6 degrees when
the mean is really different from 98.6
Table 7-2
Type I and Type II Errors
True State of Nature
We decide to
reject the
null hypothesis
The null
hypothesis is
true
The null
hypothesis is
false
Type I error
(rejecting a true
null hypothesis)

Correct
decision
Correct
decision
Type II error
(rejecting a false
null hypothesis)

Decision
We fail to
reject the
null hypothesis
Controlling Type I and Type II Errors
For any fixed , an increase in the sample
size n will cause a decrease in 
For any fixed sample size n , a decrease in 
will cause an increase in . Conversely, an
increase in  will cause a decrease in  .
To decrease both  and , increase the
sample size.
Definition
Power of a Hypothesis Test
is the probability (1 - ) of rejecting a
false null hypothesis, which is
computed by using a particular
significance level  and a particular
value of the mean that is an alternative
to the value assumed true in the null
hypothesis.
ELEMENTARY
Section 7-3
STATISTICS
Testing a Claim about a Mean: Large Samples
MARIO F. TRIOLA
EIGHTH
EDITION
Three Methods Discussed
1) Traditional method
2) P-value method
3) Confidence intervals
Assumptions
for testing claims about population means
1) The sample is a simple random
sample.
2) The sample is large (n > 30).
a) Central limit theorem applies
b) Can use normal distribution
3) If s is unknown, we can use
sample
standard deviation s as
estimate for s.
Traditional (or Classical) Method of
Testing Hypotheses
Goal
Identify a sample result that is significantly
different from the claimed value
The traditional (or classical) method
of hypothesis testing converts the
relevant sample statistic into a test
statistic which we compare to the
critical value.
Test Statistic for Claims about µ
when n > 30
(Step 6)
z=
x - µx
s
n
Traditional (or Classical) Method of Testing Hypotheses
Figure 7-5
1. Identify the specific claim or hypothesis to be tested, and put it in symbolic
form.
2. Give the symbolic form that must be true when the original claim is false.
3. Of the two symbolic expressions obtained so far, let null hypothesis H0 be
the one that contains the condition of equality. H1 is the other statement.
4. Select the significant level  based on the seriousness of a type I error.
Make  small if the consequences of rejecting a true H0 are severe. The
values of 0.05 and 0.01 are very common.
5. Identify the statistic that is relevant to this test and its sampling distribution.
6. Determine the test statistic, the critical values, and the critical region. Draw a
graph and include the test statistic, critical value(s), and critical region.
7. Reject H0 if the test statistic is in the critical region. Fail to reject H0 if the
test statistic is not in the critical region.
8. Restate this previous decision in simple nontechnical terms. (See Figure 7-4)
Traditional (or Classical) Method of Testing Hypotheses
Figure 7-5
1. Identify the specific claim or hypothesis to be tested, and put it in symbolic
form.
2. Give the symbolic form that must be true when the original claim is false.
3. Of the two symbolic expressions obtained so far, let null hypothesis H0 be
the one that contains the condition of equality. H1 is the other statement.
4. Select the significant level  based on the seriousness of a type I error.
Make  small if the consequences of rejecting a true H0 are severe. The
values of 0.05 and 0.01 are very common.
5. Identify the statistic that is relevant to this test and its sampling distribution.
6. Determine the test statistic, the critical values, and the critical region. Draw a
graph and include the test statistic, critical value(s), and critical region.
7. Reject H0 if the test statistic is in the critical region. Fail to reject H0 if the
test statistic is not in the critical region.
8. Restate this previous decision in simple nontechnical terms. (See Figure 7-4)
Decision Criterion (Step 7)
Reject the null hypothesis if the test
statistic is in the critical region
Fail to reject the null hypothesis if the test
statistic is not in the critical region
FIGURE 7-4
Wording of Final Conclusion
Start
Yes
Does the
Yes
(This
“There is sufficient
Do
original claim(Original
contain claim
is the
you reject
to warrant
(Rejectevidence
H0)
H0?.No
the conditioncontains
of
only
rejection of the
claim
equality
(Fail
to
No
“There
is notcase
sufficient
equality
that.
.
.
(original
claim).”
and
becomes
H
)
reject
H0) evidence to warrant
0
(Original claim
in
does not contain
rejection of the
claim
which
(This
is th
equality and
Yes
that.
. (original
claim).”
Do
the
“The . sample
data
only case
becomes H1)
you reject
(Rejectsupports
H0)
origin
the claim that
No
which the
H0?
. . . (originalalclaim).”
(Fail to
cla
“There is not original
sufficient
claim
reject H0)
support
evidence to is
support
is
the claim
reject
Example: Given a
data set of 106 healthy body temperatures,
where the mean was 98.2o and s = 0.62o , at the 0.05 significance level,
test the claim that the mean body temperature of all healthy adults is
equal to 98.6o.
Example: Given a
data set of 106 healthy body temperatures,
where the mean was 98.2o and s = 0.62o , at the 0.05 significance level,
test the claim that the mean body temperature of all healthy adults is
equal to 98.6o.
Steps:
1,2,3) Set up Claim, H0, H1
Claim: m = 98.6o
H0 : m = 98.6o
H1 : m  98.6o
Example: Given a
data set of 106 healthy body temperatures,
where the mean was 98.2o and s = 0.62o , at the 0.05 significance level,
test the claim that the mean body temperature of all healthy adults is
equal to 98.6o.
Steps:
1,2,3) Set up Claim, H0, H1
Claim: m = 98.6o
H0 : m = 98.6o
H1 : m  98.6o
4) Select if necessary  level:

= 0.05 was given
5 & 6) Identify the test statistic
z = x-µ =
s n
98.2 - 98.6
0.62 106
= - 6.64
6) Determine critical region(s) and critical
value(s)
 = 0.05
/2 = 0.025 (two tailed test)
0.4750
0.025
z = - 1.96
0.4750
0.025
1.96
Critical Values - Use Table A-2
6) Draw graph and include the test statistic, critical
value(s), and critical region
Sample data:
x = 98.2o
or
z = - 6.64
Reject
H0: µ = 98.6
z = - 1.96
Fail to
Reject
H0: µ =
98.6
µ = 98.6
Reject
H0: µ = 98.6
z = 1.96
or z = 0
z = - 6.64
Figure 7-6
7) Reject H0: if TS is in critical region
Fail to reject H0: if TS is not in critical region
Sample data:
x = 98.2o
or
z = - 6.64
Reject
H0: µ = 98.6
z = - 1.96
Fail to
Reject
H0: µ =
98.6
µ = 98.6
Reject
H0: µ = 98.6
z = 1.96
or z = 0
z = - 6.64
Figure 7-6
7) Reject H0: if TS is in critical region
Fail to reject H0: if TS is not in critical region
Sample data:
x = 98.2o
or
z = - 6.64
Reject
H0: µ = 98.6
z = - 1.96
Fail to
Reject
H0: µ =
98.6
µ = 98.6
Reject
H0: µ = 98.6
z = 1.96
or z = 0
z = - 6.64
REJECT H0
Figure 7-6
8) Restate in simple nontechnical terms - Figure 7-4
Claim: m = 98.6o
REJECT H0 : m = 98.6o
H1 : m  98.6o
8) Restate in simple nontechnical terms - Figure 7-4
Start
Yes
Does the
“There is sufficient
(This is t
Yes
Do
original claim(Original
contain claimyou reject
to warrant
(Reject Hevidence
only cas
0)
H0?
the conditioncontains
of
rejection of the which
claim th
equality
“There is not sufficient
No
No
equality?
that. . . (originaloriginal
claim).”
and
becomes
H
)
evidence to warrant
(Original claim (Fail to0
is rejecte
rejection
of
the
claim
does not containreject H0 )
(This is
Yes
Do
“The
dataclaim).”
that. sample
. . (original
equality and
only cas
(Reject support
H0)
you reject
the claim that
becomes H1 )
No
which th
H0?
. . . (original claim).”
(Fail to
original
“There is not sufficient
reject H0 ) evidence to support
Figure 7-4
is suppo
the claim
that. . . (original claim).”
8) Restate in simple nontechnical terms - Figure 7-4
Start
Yes
Does the
“There is sufficient
(This is t
Yes
Do
original claim
containclaimyou reject
to warrant
(Original
(Reject Hevidence
only cas
0)
H0?
the conditioncontains
of
rejection of the which
claim th
equality
“There is not sufficient
No
No
equality? and becomes H )
that. . . (originaloriginal
claim).”
evidence to warrant
(Original claim (Fail to0
is rejecte
rejection
of
the
claim
does not containreject H0 )
(This is
Yes
Do
“The
dataclaim).”
that. sample
. . (original
equality and
only cas
(Reject support
H0)
you reject
the claim that
becomes H1 )
No
which th
H0?
. . . (original claim).”
(Fail to
original
“There is not sufficient
reject H0 )
Figure 7-4
is suppo
evidence to support
the claim
that. . . (original claim).”
8) Restate in simple nontechnical terms - Figure 7-4
Start
Yes
Does the
“There is sufficient
(This is t
Yes
Do
original claim
containclaim
to warrant
(Original
you reject
(Reject Hevidence
only cas
0)
H0?
the conditioncontains
of
rejection of the which
claim th
equality
“There is not sufficient
No
No
equality? and becomes H )
that. . . (originaloriginal
claim).”
(Fail
to
evidence to warrant
0
(Original claim
is rejecte
reject
H
)
rejection
of
the
claim
does not contain
0
(This is
Yes
Do
“The
dataclaim).”
that. sample
. . (original
equality and
only cas
(Reject H
)
you reject
0
support the claim that
becomes H1 )
No
which th
H0?
. . . (original claim).”
(Fail to
original
“There is not sufficient
reject H0 ) evidence to support
Figure 7-4
is suppo
the claim
that. . . (original claim).”
8) Restate in simple nontechnical terms - Figure 7-4
Start
Yes
“There is sufficient
Does the
Yes
(This is t
Do
evidence to warrant
original claim
containclaim
(Original
you reject
only cas
(Reje
H0?
rejection of thewhich
claimth
the conditioncontains
of
equality
“There is not sufficient
No ct H )
No
0
that.
. . (original
claim)
equality? and becomes H )
original
evidence to warrant
0(Fail to
(Original claim
is rejecte
reject
H
)
rejection
of
the
claim
does not contain
0
(This is
Yes
Do
“The
dataclaim).”
that. sample
. . (original
equality and
only cas
(Reject H
)
you reject
0
support the claim that
becomes H1 )
No
which th
H0?
. . . (original claim).”
(Fail to
original
“There is not sufficient
reject H0 )
Figure 7-4
is suppo
evidence to support
the claim
that. . . (original claim).”
8) Restate in simple nontechnical terms - Figure 7-4
Start
There is sufficient evidence to warrant rejection of claim that
the mean body temperatures of healthy adults is equal to 98.6 o.
Yes
Does the
“There is sufficient
Yes
(This is t
Do
original claim
containclaim
(Original
you reject
only cas
(Reje evidence to warrant
H0?
the conditioncontains
of
rejection of thewhich
claimth
equality
“There is not sufficient
No ct H )
No
0
equality? and becomes H )
that. . . (original
claim)
original
(Fail
to
evidence to warrant
0
(Original claim
is rejecte
reject
H
)
rejection
of
the
claim
does not contain
0
(This is
Yes
Do
“The
dataclaim).”
that. sample
. . (original
equality and
only cas
(Reject H
)
you reject
0
support the claim that
becomes H0 )
No
which th
H0?
. . . (original claim).”
(Fail to
original
“There is not sufficient
reject H0 )
Figure 7-4
is suppo
evidence to support
the claim
that. . . (original claim).”
P-Value Method
of Testing Hypotheses
very similar to traditional method
key difference is the way in which we
decide to reject the null hypothesis
approach finds the probability (P-value) of
getting a result and rejects the null
hypothesis if that probability is very low
P-Value Method
of Testing Hypotheses
Definition
P-Value (or probability value)
the probability of getting a value of the
sample test statistic that is at least as
extreme as the one found from the
sample data, assuming that the null
hypothesis is true
P-value
Interpretation
Small P-values
(such as 0.05 or
lower)
Unusual sample results.
Significant difference from the
null hypothesis
Large P-values
(such as above
0.05)
Sample results are not unusual.
Not a significant difference from
the null hypothesis
Figure 7-8
Finding P-Values
Start
What
type of test
?
Left-tailed
Right-tailed
Two-tailed
Left
P-value = area
to the left of the
test statistic
P-value
µ
Test statistic
Is
the test statistic
to the right or left of
center
?
P-value = twice
the area to the left
of the test statistic
P-value is twice
this area
µ
Test statistic
Right
P-value = twice
the area to the right
of the test statistic
P-value = area
to the right of the
test statistic
P-value is twice
this area
P-value
µ
µ
Test statistic
Test statistic
Procedure is the same except for
steps 6 and 7
Step 6: Find the P-value (as shown in Figure 7-8)
Step 7: Report the P-value
Reject the null hypothesis if the P-value is
less than or equal to the significance level 
Fail to reject the null hypothesis if the P-value
is greater than the significance level 
Testing Claims with
Confidence Intervals
A confidence interval estimate of a
population parameter contains the likely
values of that parameter. We should
therefore reject a claim that the population
parameter has a value that is not included in
the confidence interval.
Testing Claims
with Confidence Intervals
Claim: mean body temperature = 98.6º,
where n = 106, x = 98.2º and s = 0.62º
 95% confidence interval of 106 body temperature
data (that is, 95% of samples would contain true
value µ )
 98.08º < µ < 98.32º
 98.6º is not in this interval
 Therefore it is very unlikely that µ = 98.6º
 Thus we reject claim µ = 98.6º
Underlying Rationale of
Hypotheses Testing
 If, under a given observed assumption, the
probability of getting the sample is exceptionally
small, we conclude that the assumption is
probably not correct.
 When testing a claim, we make an assumption
(null hypothesis) that contains equality. We then
compare the assumption and the sample results
and we form one of the following conclusions:
Underlying Rationale of
Hypotheses Testing
If the sample results can easily occur when the
assumption (null hypothesis) is true, we
attribute the relatively small discrepancy
between the assumption and the sample results
to chance.
If the sample results cannot easily occur when
that assumption (null hypothesis) is true, we
explain the relatively large discrepancy between
the assumption and the sample by concluding
that the assumption is not true.
ELEMENTARY
Section 7-4
STATISTICS
Testing a Claim about a Mean: Small Samples
MARIO F. TRIOLA
EIGHTH
EDITION
Assumptions
for testing claims about population means
1) The sample is a simple random sample.
2) The sample is small (n  30).
3) The value of the population
standard
deviation s is unknown.
4) The sample values come from a
population with a distribution that is
approximately normal.
Test Statistic
for a Student t-distribution
x -µx
t= s
n
Critical Values
Found in Table A-3
Degrees of freedom (df) = n -1
Critical t values to the left of the mean are
negative
Important Properties of the
Student t Distribution
1. The Student t distribution is different for different sample sizes (see
Figure 6-5 in Section 6-3).
2. The Student t distribution has the same general bell shape as the
normal distribution; its wider shape reflects the greater variability that
is expected with small samples.
3. The Student t distribution has a mean of t = 0 (just as the standard
normal distribution has a mean of z = 0).
4. The standard deviation of the Student t distribution varies with the
sample size and is greater than 1 (unlike the standard normal
distribution, which has a s = 1).
5. As the sample size n gets larger, the Student t distribution get closer to
the normal distribution. For values of n > 30, the differences are so
small that we can use the critical z values instead of developing a
much larger table of critical t values. (The values in the bottom row of
Table A-3 are equal to the corresponding critical z values from the
normal distributions.)
Figure 7-11
Choosing between the Normal and Student
t-Distributions when Testing a Claim about a Population Mean µ
Start
Use normal distribution with
Is
n > 30
?
Yes
Z
(If s is unknown use s instead.)
No
Is the
distribution of
the population essentially
normal ? (Use a
histogram.)
No
Yes
Is s
known
?
No
Use the Student t distribution
with
x - µx
t  s/
n
x - µx
s/ n
Use nonparametric methods,
which don’t require a normal
distribution.
Use normal distribution with
Z
x - µx
s/ n
(This case is rare.)
The larger Student t critical value
shows that with a small sample,
the sample evidence must be more
extreme before we consider the
difference is significant.
P-Value Method
Table A-3 includes only selected values
of 
Specific P-values usually cannot be
found
Use Table to identify limits that contain
the P-value
Some calculators and computer
programs will find exact P-values
ELEMENTARY
Section 7-5
STATISTICS
Testing a Claim about a Proportion
MARIO F. TRIOLA
EIGHTH
EDITION
Assumptions
for testing claims about population proportions
1) The sample observations are a simple
random sample.
2) The conditions for a binomial experiment
are satisfied (Section 4-3)
Assumptions
for testing claims about population proportions
1) The sample observations are a simple
random sample.
2) The conditions for a binomial experiment
are satisfied (Section 4-3)
3) The condition np  5 and nq  5 are
satisfied, so the binomial distribution of
sample proportions can be approximated by a
normal distribution with µ = np and s = npq
Notation
n = number of trials
Notation
n = number of trials

p = x/n (sample proportion)
p = population proportion (used in
the
q=1-p
null hypothesis)
Test Statistic for Testing a Claim
about a Proportion

z=
p-p
pq
n
Traditional Method
Same as described
in Sections 7-2 and 7-3
and in Figure 7-5
P-Value Method
Same as described in Section 7-3
and Figure 7-8
Reject the null hypothesis if the
P-value is less than or equal to
the significance level .

p sometimes is given directly
“10% of the observed sports cars are red”
is expressed as

p = 0.10

p sometimes is given directly
“10% of the observed sports cars are red”
is expressed as

p = 0.10

p sometimes must be calculated
“96 surveyed households have cable TV
and 54 do not” is calculated using

p
x
96
=n =
= 0.64
(96+54)
(determining the sample proportion of households with cable TV)
CAUTION
CAUTION

 When the calculation of p results in a
decimal with many places, store the
number on your calculator and use
all the decimals when evaluating the z
test statistic.
CAUTION

 When the calculation of p results in a
decimal with many places, store the
number on your calculator and use
all the decimals when evaluating the z
test statistic.

 Large errors can result from rounding p
too much.
Test Statistic for Testing a Claim
about a Proportion
z=
x-µ
z = s =

p-p
pq
n
x - np
npq
=
x
n
np
n
npq
n

=
p-p
pq
n
ELEMENTARY
Section 7-6
STATISTICS
Testing a Claim about a Standard Deviation or Variance
MARIO F. TRIOLA
EIGHTH
EDITION
Assumptions
for testing claims about a population
standard deviation or variance
1) The sample is a simple
random
sample.
2) The population has values that
are
normally distributed (a
strict requirement).
Chi-Square Distribution
Test Statistic
X2=
(n - 1) s 2
s2
Chi-Square Distribution
Test Statistic
X2=
n
(n - 1) s 2
s2
= sample size
s 2 = sample variance
s2 = population variance
(given in null hypothesis)
Critical Values for
Chi-Square Distribution
 Found in Table A-4
 Degrees of freedom = n -1
Properties of Chi-Square
Distribution
All values of X2 are nonnegative, and the
distribution is not symmetric.
There is a different distribution for each
number of degrees of freedom.
The critical values are found in Table A-4
using n-1 degrees of freedom.
Properties of Chi-Square
Distribution
Properties of the Chi-Square
Distribution
Not symmetric
x2
All values are nonnegative
Figure 7-12
Properties of Chi-Square
Distribution
Chi-Square Distribution for 10
and 20 Degrees of Freedom
Properties of the Chi-Square
Distribution
df = 10
Not symmetric
df = 20
x2
All values are nonnegative
Figure 7-12
0
5 10 15 20
25 30 35 40 45
Figure 7-13
There is a different distribution for each
number of degrees of freedom.
Example: Aircraft altimeters have measuring errors with a standard
deviation of 43.7 ft. With new production equipment, 81 altimeters measure
errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to
test the claim that the new altimeters have a standard deviation different from
the old value of 43.7 ft.
Example: Aircraft altimeters have measuring errors with a standard
deviation of 43.7 ft. With new production equipment, 81 altimeters measure
errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to
test the claim that the new altimeters have a standard deviation different from
the old value of 43.7 ft.
Claim: s 43.7
H0: s= 43.7
H1: s 43.7
Example: Aircraft altimeters have measuring errors with a standard
deviation of 43.7 ft. With new production equipment, 81 altimeters measure
errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to
test the claim that the new altimeters have a standard deviation different from
the old value of 43.7 ft.
Claim: s 43.7
H0: s= 43.7
H1: s 43.7
0.025
 = 0.05 2
0.025
= 0.025
Example: Aircraft altimeters have measuring errors with a standard
deviation of 43.7 ft. With new production equipment, 81 altimeters measure
errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to
test the claim that the new altimeters have a standard deviation different from
the old value of 43.7 ft.
Claim: s 43.7
H0: s= 43.7
H1: s 43.7
0.025
 = 0.05 2
0.025
106.629
= 0.025
n = 81
df = 80
Table A-4
Example: Aircraft altimeters have measuring errors with a standard
deviation of 43.7 ft. With new production equipment, 81 altimeters measure
errors with a standard deviation of 52.3 ft. Use the 0.05 significance level to
test the claim that the new altimeters have a standard deviation different from
the old value of 43.7 ft.
Claim: s 43.7
H0: s= 43.7
H1: s 43.7
 = 0.05 2
= 0.025
0.975
0.025
0.025
57.153
106.629
n = 81
df = 80
Table A-4
x
2
=
(n -1)s2
s
2
=
(81 -1) (52.3)2
43.72
 114.586
x
2
=
(n -1)s2
s
2
=
(81 -1) (52.3)2
43.72
 114.586
Reject H0
57.153
106.629
x2 = 114.586
x
2
=
(n -1)s2
s
2
=
(81 -1) (52.3)2
43.72
 114.586
Reject H0
57.153
106.629
x2 = 114.586
The sample evidence supports the claim that the
standard deviation is different from 43.7 ft.
x
2
=
(n -1)s2
s
2
=
(81 -1) (52.3)2
43.72
 114.586
Reject H0
57.153
106.629
x2 = 114.586
The new production method appears to be worse than the old
method. The data supports that there is more variation in the
error readings than before.
P-Value Method
Table A-4 includes only selected values
of 
Specific P-values usually cannot be
found
Use Table to identify limits that contain
the P-value
Some calculators and computer
programs will find exact P-values
Figure 7-15
Proportion,
Testing a Claim about a Mean,
Standard Deviation, or Variance
Start
Use the Chi-square
distribution with
x2 =
(n -1)s2
s2
St. Dev s
or
Variance s2
Which
parameter
does the claim
address
?
Proportion
P
Use the normal distribution
ˆ
z= P-P
pq
n
where Pˆ= x/n
Mean (µ)
Use the normal distribution with
Is
n > 30
?
Yes
z=
x - µx
sn
(If s Is unknown use s instead.)
Figure 7-15
Proportion,
Testing a Claim about a Mean,
Standard Deviation, or Variance
Use the normal distribution with
Is
n > 30
?
Yes
z=
x - µx
s n
(If s is unknown use s instead.)
No
Is the
distribution of
the population essentially
normal ? (Use a
histogram.)
No
Use nonparametric methods
which don’t require a normal
distribution. See Chapter 13.
Yes
Use the normal distribution with
Is s
known
?
Yes
z=
x - µx
sn
(This case is rare.)
No
Use the Student t
distribution with
t=
x - µx
sn
ELEMENTARY
Chapter 8
MARIO F. TRIOLA
STATISTICS
Inferences from Two Samples
EIGHTH
EDITION
Chapter 8
Inferences from Two Samples
8-1 Overview
8-2 Inferences about Two Means: Independent
and Large Samples
8-3 Inferences about Two Means: Matched Pairs
8-4 Inferences about Two Proportions
8-5 Comparing Variation in Two Samples
8-6 Inferences about Two Means:
Independent
and Small Samples
8-1
Overview
There are many important and meaningful
situations in which it becomes necessary
to compare two sets of sample data.
8-2
Inferences about Two Means:
Independent and
Large Samples
Definitions
Two Samples: Independent
The sample values selected from one
population are not related or somehow paired
with the sample values selected from the
other population.
If the values in one sample are related to the
values in the other sample, the samples are
dependent. Such samples are often referred
to as matched pairs or paired samples.
Assumptions
1. The two samples are independent.
2. The two sample sizes are large. That
is, n1 > 30 and n2 > 30.
3. Both samples are simple
random samples.
Hypothesis Tests
Test Statistic for Two Means:
Independent and Large Samples
Hypothesis Tests
Test Statistic for Two Means:
Independent and Large Samples
z
=
(x1 - x2) - (µ1 - µ2)
s
2.
1
n1
s
2
2
+ n
2
Hypothesis Tests
Test Statistic for Two Means:
Independent and Large Samples
s and s
If s and s are not known, use s1 and s2
in their places. provided that both
samples are large.
P-value:
Use the computed value of the test
statistic z, and find the P-value by following
the same procedure summarized in Figure 7-8.
Critical values:
Based on the significance level ,
find critical values by using the
procedures introduced in Section 7-2.
Coke Versus Pepsi
Data Set 1 in Appendix B includes the weights (in
pounds) of samples of regular Coke and regular
Pepsi. Sample statistics are shown. Use the 0.01
significance level to test the claim that the mean
weight of regular Coke is different from the mean
weight of regular Pepsi.
Coke Versus Pepsi
Data Set 1 in Appendix B includes the weights (in
pounds) of samples of regular Coke and regular
Pepsi. Sample statistics are shown. Use the 0.01
significance level to test the claim that the mean
weight of regular Coke is different from the mean
weight of regular Pepsi.
Regular Coke
Regular Pepsi
n
36
36
x
0.81682
0.82410
s
0.007507
0.005701
Coke Versus Pepsi
Coke Versus Pepsi
Claim: m1  m2
Ho : m 1 = m 2
H1 : m 1  m 2
 = 0.01
Reject H0
Z = - 2.575
Fail to reject H0
m1 - m = 0
or Z = 0
Reject H0
Z = 2.575
Coke Versus Pepsi
Test Statistic for Two Means:
Independent and Large Samples
z
=
(x1 - x2) - (µ1 - µ2)
s
2.
1
n1
s
2
2
+ n
2
Coke Versus Pepsi
Test Statistic for Two Means:
Independent and Large Samples
z
=
(0.81682 - 0.82410) - 0
0.0075707 2
36
= - 4.63
+
0.005701 2
36
Coke Versus Pepsi
Claim: m1  m2
Ho : m 1 = m 2
H1 : m 1  m 2
 = 0.01
Reject H0
sample data:
z = - 4.63
Z = - 2.575
Fail to reject H0
m1 - m = 0
or Z = 0
Reject H0
Z = 2.575
Coke Versus Pepsi
Claim: m1  m2
Ho : m 1 = m 2
H1 : m 1  m 2
There is significant evidence to support
the claim that there is a difference
between the mean weight of Coke and
the mean weight of Pepsi.
 = 0.01
Reject H0
Fail to reject H0
Reject H0
Reject Null
sample data:
z = - 4.63
Z = - 2.575
m1 - m = 0
or Z = 0
Z = 2.575
Confidence Intervals
Confidence Intervals
(x1 - x2) - E < (µ1 - µ2) < (x1 - x2) + E
Confidence Intervals
(x1 - x2) - E < (µ1 - µ2) < (x1 - x2) + E
where E =
z
s
2
1
n1
s
2
2
+ n
2
ELEMENTARY
Section 8-3
STATISTICS
Inferences about Two Means: Matched Pairs
MARIO F. TRIOLA
EIGHTH
EDITION
Assumptions
1. The sample data consist of matched pairs.
2. The samples are simple random samples.
3. If the number of pairs of sample data is
small (n  30), then the population of
differences in the paired values must be
approximately normally distributed.
Notation for Matched Pairs
µd
= mean value of the differences d for the
population of paired data
Notation for Matched Pairs
µd
= mean value of the differences d for the
population of paired data
d
= mean value of the differences d for the
paired sample data (equal to the mean
of the x - y values)
sd
= standard deviation of the differences d for
the paired sample data
n
= number of pairs of data.
Test Statistic for Matched Pairs of Sample Data
Test Statistic for Matched Pairs of Sample Data
t=
d - µd
sd
n
where degrees of freedom = n - 1
Critical Values
If n  30, critical values are found in
Table A-3 (t-distribution).
If n > 30, critical values are found in
Table A- 2 (normal distribution).
Confidence Intervals
Confidence Intervals
d - E < µd < d + E
Confidence Intervals
d - E < µd < d + E
where
E = t/2
sd
n
degrees of freedom = n -1
How Much Do Male Statistics Students
Exaggerate Their Heights?
Using the sample data from Table 8-1 with
the outlier excluded, construct a 95%
confidence interval estimate of md, which
is the mean of the differences between
reported heights and measured heights of
male statistics students.
Table 8-1
Reported and Measured Heights (in inches) of Male
Statistics Students
Student
A
B
C
Reported
Height
68
74
82.25
D
E
F
G
H
I
J
66.5
69
68
71
70
70
67
K
L
68
70
Measured 66.8 73.9
Height
74.3
66.1 67.2
67.9
69.4 69.9 68.6 67.9 67.6 68.8
Difference 1.2
7.95
0.4
0.1
1.6
0.1
outlier
1.8
0.1
1.4 -0.9
0.4
1.2
How Much Do Male Statistics Students
Exaggerate Their Heights?
d = 0.672727
s = 0.825943
n = 11
t/2 = 2.228 (found from Table A-3 with 10
degrees of freedom and 0.05 in two tails)
How Much Do Male Statistics Students
Exaggerate Their Heights?
E = t/2
sd
n
E = (2.228)(
0.825943
11
= 0.554841
)
How Much Do Male Statistics Students
Exaggerate Their Heights?
0.12 < µd < 1.23
In the long run, 95% o f such samples will lead to
confidence intervals that actually do contain the true
population mean of the differences. Since the interval
does not contain 0, the true value of µd is significantly
different from 0. There is sufficient evidence to
support the claim that there is a difference between
the reported heights and the measured heights of
male statistics students.
ELEMENTARY
STATISTICS
Section 8-4 Inferences about Two Proportions
MARIO F. TRIOLA
EIGHTH
EDITION
Inferences about Two Proportions
Assumptions
1. We have proportions from two
independent simple random samples.
2. For both samples, the conditions np  5
and nq  5 are satisfied.
Notation for Two Proportions
For population 1, we let:
p1 = population proportion
n1 = size of the sample
x1 = number of successes in the sample
Notation for Two Proportions
For population 1, we let:
p1 = population proportion
n1 = size of the sample
x1 = number of successes in the sample
p^1 = x1/n1 (the sample proportion)
Notation for Two Proportions
For population 1, we let:
p1 = population proportion
n1 = size of the sample
x1 = number of successes in the sample
p^1 = x1/n1 (the sample proportion)
q^1 = 1 - p^1
Notation for Two Proportions
For population 1, we let:
p1 = population proportion
n1 = size of the sample
x1 = number of successes in the sample
p^1 = x1/n1 (the sample proportion)
q^1 = 1 - p^1
The corresponding meanings are attached
^ . and q^ , which come from
to p2, n2 , x2 , p
2
2
population 2.
Pooled Estimate of p1 and p2
 The pooled estimate of p1 and
p2
is denoted by p
Pooled Estimate of p1 and p2
 The pooled estimate of p1 and
p2
is denoted by p
x1 + x2
p= n +n
1
2
Pooled Estimate of p1 and p2
 The pooled estimate of p1 and
p2
is denoted by p
x1 + x2
p= n +n
1
2

q =1- p
Test Statistic for Two Proportions
For H0: p1 = p2 , H0: p1  p2 ,
H1: p1  p2 ,
H0: p1 p2
H1: p1 < p2 , H1: p 1> p2
Test Statistic for Two Proportions
For H0: p1 = p2 , H0: p1  p2 ,
H1: p1  p2 ,
z=
H0: p1 p2
H1: p1 < p2 , H1: p 1> p2
( p^1 - p^2 ) ( p1 - p2 )
pq
pq
+
n2
n1
Test Statistic for Two Proportions
For H0: p1 = p2 , H0: p1  p2 ,
where
H0: p1 p2
H1: p1  p2 ,
H1: p1 < p2 , H1: p 1> p2
p1 - p 2 = 0
(assumed in the null hypothesis)
Test Statistic for Two Proportions
For H0: p1 = p2 , H0: p1  p2 ,
H0: p1 p2
H1: p1  p2 ,
H1: p1 < p2 , H1: p 1> p2
p1 - p 2 = 0
(assumed in the null hypothesis)
where
p^
1
p=
x1 + x2
n1 + n2
x1
= n
1
and
and
p^
2
x2
=
n2
q=1-p
Confidence Interval Estimate of p1 -
p2
(p^1 - p^2 ) - E < ( p1 - p2 ) < (p^1 - p^2 ) + E
Confidence Interval Estimate of p1 -
p2
(p^1 - p^2 ) - E < ( p1 - p2 ) < (p^1 - p^2 ) + E
where E =
z
p^1 q^1
p^2 q^2
n1 + n2
STATISTICS
ELEMENTARY
Chapter 9
Correlation and Regression
MARIO F. TRIOLA
EIGHTH
EDITION
Chapter 9
Correlation and Regression
9-1 Overview
9-2 Correlation
9-3 Regression
9-4 Variation and Prediction Intervals
9-5 Multiple Regression
9-6 Modeling
9-1
Overview
Paired Data
 is there a relationship
 if so, what is the equation
 use the equation for prediction
9-2
Correlation
Definition
Correlation
exists between two variables
when one of them is related to
the other in some way
Assumptions
1. The sample of paired data (x,y) is a
random sample.
2. The pairs of (x,y) data have a
bivariate normal distribution.
Definition
Scatterplot (or scatter diagram)
is a graph in which the paired
(x,y) sample data are plotted with
a horizontal x axis and a vertical
y axis. Each individual (x,y) pair
is plotted as a single point.
Scatter Diagram of Paired Data
Scatter Diagram of Paired Data
Positive Linear Correlation
y
y
y
(a) Positive
Figure 9-1
x
x
x
(b) Strong
positive
Scatter Plots
(c) Perfect
positive
Negative Linear Correlation
y
y
y
(d) Negative
Figure 9-1
x
x
x
(e) Strong
negative
Scatter Plots
(f) Perfect
negative
No Linear Correlation
y
y
x
(g) No Correlation
Figure 9-1
Scatter Plots
x
(h) Nonlinear Correlation
Definition
Linear Correlation Coefficient r
measures strength of the linear relationship between
paired x and y values in a sample
Definition
Linear Correlation Coefficient r
measures strength of the linear relationship between
paired x and y values in a sample
nxy - (x)(y)
r=
Formula 9-1
n(x2) - (x)2
n(y2) - (y)2
Definition
Linear Correlation Coefficient r
measures strength of the linear relationship between
paired x and y values in a sample
nxy - (x)(y)
r=
n(x2) - (x)2
n(y2) - (y)2
Formula 9-1
Calculators can compute r
(rho) is the linear correlation coefficient for all paired data in the
population.
Notation for the
Linear Correlation Coefficient
n =
number of pairs of data presented

denotes the addition of the items indicated.
x
denotes the sum of all x values.
x2
indicates that each x score should be squared and then
those squares added.
(x)2 indicates that the x scores should be added and the total
then squared.
xy
indicates that each x score should be first multiplied by its
corresponding y score. After obtaining all such products,
find their sum.
r
represents linear correlation coefficient for a sample

represents linear correlation coefficient for a population
Rounding the
Linear Correlation Coefficient r
 Round to three decimal places so
that it can be compared to critical
values in Table A-6
 Use calculator or computer if possible
Interpreting the Linear
Correlation Coefficient
If the absolute value of r exceeds the
value in Table A - 6, conclude that there
is a significant linear correlation.
Otherwise, there is not sufficient
evidence to support the conclusion of
significant linear correlation.
TABLE A-6 Critical Values of the
Pearson Correlation Coefficient r
n
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
60
70
80
90
100
 = .05
.950
.878
.811
.754
.707
.666
.632
.602
.576
.553
.532
.514
.497
.482
.468
.456
.444
.396
.361
.335
.312
.294
.279
.254
.236
.220
.207
.196
 = .01
.999
.959
.917
.875
.834
.798
.765
.735
.708
.684
.661
.641
.623
.606
.590
.575
.561
.505
.463
.430
.402
.378
.361
.330
.305
.286
.269
.256
Properties of the
Linear Correlation Coefficient r
1. -1  r  1
2. Value of r does not change if all values of
either variable are converted to a different
scale.
3. The r is not affected by the choice of x and y.
Interchange x and y and the value of r will
not
change.
4. r measures strength of a linear relationship.
Common Errors Involving Correlation
1. Causation: It is wrong to conclude that
correlation implies causality.
2. Averages: Averages suppress individual
variation and may inflate the correlation
coefficient.
3. Linearity: There may be some relationship
between x and y even when there is no
significant linear correlation.
Common Errors Involving Correlation
FIGURE 9-2
250
Distance
(feet)
200
150
100
50
0
0
1
2
3
4
5
6
7
8
Time (seconds)
Scatterplot of Distance above Ground and Time for Object Thrown Upward
Formal Hypothesis Test
 To determine whether there is a
significant linear correlation
between two variables
 Two methods
 Both methods let H0:  =
(no significant linear correlation)
H1:  
(significant linear correlation)
Method 1: Test Statistic is t
(follows format of earlier chapters)
Test statistic:
t=
r
1-r2
n-2
Method 1: Test Statistic is t
(follows format of earlier chapters)
Test statistic:
t=
r
1-r2
n-2
Critical values:
use Table A-3 with
degrees of freedom = n - 2
Method 1: Test Statistic is t
(follows format of earlier chapters)
Figure 9-4
Method 2: Test Statistic is r
(uses fewer calculations)
Test statistic: r
Critical values: Refer to Table A-6
(no degrees of freedom)
Method 2: Test Statistic is r
(uses fewer calculations)
Test statistic: r
Critical values: Refer to Table A-6
(no degrees of freedom)
Reject
 =0
-1
Figure 9-5
Fail to reject
=0
r = - 0.811
0
Reject
 =0
r = 0.811
Sample data:
r = 0.828
1
FIGURE 9-3
Start
Testing for a
Linear Correlation
Let H0:  = 0
H1:   0
Select a
significance
level 
Calculate r using
Formula 9-1
METHOD 1
METHOD 2
The test statistic is
t=
The test statistic is
r
r
Critical values of t are from
Table A-6
1-r2
n -2
Critical values of t are from Table A-3 with
n -2 degrees of freedom
If the absolute value of the
test statistic exceeds the
critical values, reject H0:  = 0
Otherwise fail to reject H0
If H0 is rejected conclude that there
is a significant linear correlation.
If you fail to reject H0, then there is
not sufficient evidence to conclude
that there is linear correlation.
Is there a significant linear correlation?
Data from the Garbage Project
x Plastic (lb)
0.27
1.41
2.19
2.83
2.19
1.81
0.85
3.05
y Household
2
3
3
6
4
2
1
5
Is there a significant linear correlation?
Data from the Garbage Project
x Plastic (lb)
0.27
1.41
2.19
2.83
2.19
1.81
0.85
3.05
y Household
2
3
3
6
4
2
1
5
n=8
 = 0.05
=0
H :  0
H0:
1
Test statistic is r = 0.842
Is there a significant linear correlation?
n=8
 = 0.05
=0
H :  0
H0:
1
Test statistic is r = 0.842
Critical values are r = - 0.707 and 0.707
(Table A-6 with n = 8 and  = 0.05)
TABLE A-6 Critical Values of the Pearson Correlation Coefficient r
n
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
60
70
80
90
100
 = .05
.950
.878
.811
.754
.707
.666
.632
.602
.576
.553
.532
.514
.497
.482
.468
.456
.444
.396
.361
.335
.312
.294
.279
.254
.236
.220
.207
.196
 = .01
.999
.959
.917
.875
.834
.798
.765
.735
.708
.684
.661
.641
.623
.606
.590
.575
.561
.505
.463
.430
.402
.378
.361
.330
.305
.286
.269
.256
Is there a significant linear correlation?
Reject
 =0
-1
r = - 0.707
Fail to reject
=0
0
Reject
 =0
r = 0.707
Sample data:
r = 0.842
1
Is there a significant linear correlation?
0.842 > 0.707, That is the test statistic does fall within the
critical region.
Reject
 =0
-1
r = - 0.707
Fail to reject
=0
0
Reject
 =0
r = 0.707
Sample data:
r = 0.842
1
Is there a significant linear correlation?
0.842 > 0.707, That is the test statistic does fall within the
critical region.
Therefore, we REJECT H0:  = 0 (no correlation) and conclude
there is a significant linear correlation between the weights of
discarded plastic and household size.
Reject
 =0
-1
r = - 0.707
Fail to reject
=0
0
Reject
 =0
r = 0.707
Sample data:
r = 0.842
1
Justification for r Formula
Justification for r Formula
Formula 9-1 is developed from
r=
 (x -x) (y -y)
(n -1) Sx Sy
Justification for r Formula
Formula 9-1 is developed from
r=
 (x -x) (y -y)
(n -1) Sx Sy
(x, y)
centroid of sample points
Justification for r Formula
Formula 9-1 is developed from
r=
 (x -x) (y -y)
(x, y)
(n -1) Sx Sy
centroid of sample points
x=3
y
x - x = 7- 3 = 4
(7, 23)
•
24
20
y - y = 23 - 11 = 12
Quadrant 1
Quadrant 2
16
•
12
8
•
Quadrant 3
••
4
y = 11
(x, y)
Quadrant 4
FIGURE 9-6
x
0
0
1
2
3
4
5
6
7
ELEMENTARY
STATISTICS
Section 9-3
Regression
MARIO F. TRIOLA
EIGHTH
EDITION
Regression
Definition
 Regression Equation
Regression
Definition
 Regression Equation
Given a collection of paired data, the regression
equation
Regression
Definition
 Regression Equation
Given a collection of paired data, the regression
equation
y^ = b0 + b1x
algebraically describes the relationship between the
two variables
Regression
Definition
 Regression Equation
Given a collection of paired data, the regression
equation
y^ = b0 + b1x
algebraically describes the relationship between the
two variables
 Regression Line
(line of best fit or least-squares line)
Regression
Definition
 Regression Equation
Given a collection of paired data, the regression
equation
y^ = b0 + b1x
algebraically describes the relationship between the
two variables
 Regression Line
(line of best fit or least-squares line)
the graph of the regression equation
Regression Line Plotted on Scatter Plot
The Regression Equation
x is the independent variable
(predictor variable)
The Regression Equation
x is the independent variable
(predictor variable)
^y is the dependent variable
(response variable)
The Regression Equation
x is the independent variable
(predictor variable)
^y is the dependent variable
(response variable)
y^ = b0 +b1x
y = mx +b
The Regression Equation
x is the independent variable
(predictor variable)
^y is the dependent variable
(response variable)
y^ = b0 +b1x
b0 = y - intercept
y = mx +b
b1 = slope
Notation for Regression
Equation
Notation for Regression
Equation
Population
Parameter
y-intercept of regression equation
Slope of regression equation
Equation of the regression line
Sample
Statistic
0
b0
1
b1
y = 0 + 1 x
^y = b + b x
0
1
Assumptions
1. We are investigating only linear relationships.
2. For each x value, y is a random variable
having a normal (bell-shaped) distribution.
All of these y distributions have the same
variance. Also, for a given value of x, the
distribution of y-values has a mean that lies
on the regression line. (Results are not
seriously affected if departures from normal
distributions and equal variances are not too
extreme.)
Formula for b0 and b1
Formula for b0 and b1
Formula 9-2
b0 =
(y) (x2) - (x) (xy)
n(x2) - (x)2
(y-intercept)
Formula for b0 and b1
Formula 9-2
Formula 9-3
b0 =
b1 =
(y) (x2) - (x) (xy)
n(x2) - (x)2
n(xy) - (x) (y)
n(x2) - (x)2
(y-intercept)
(slope)
Formula for b0 and b1
Formula 9-2
Formula 9-3
b0 =
b1 =
(y) (x2) - (x) (xy)
n(x2) - (x)2
n(xy) - (x) (y)
n(x2) - (x)2
(y-intercept)
(slope)
calculators or computers can
compute these values
If you find b1 first, then
If you find b1 first, then
b0 = y - b1x
Formula 9-4
If you find b1 first, then
b0 = y - b1x
Formula 9-4
Can be used for Formula 9-2, where y is the
mean of the y-values and x is the mean of
the x values
The regression line
fits the sample
points best.
Rounding
the y-intercept b0 and the
slope b1
 Round to three significant digits
 If you use the formulas 9-2 and 9-3,
try not to round intermediate
values.
Predictions
In predicting a value of y based on some
given value of x ...
1. If there is not a significant linear
correlation, the best predicted y-value is y.
Predictions
In predicting a value of y based on some
given value of x ...
1. If there is not a significant linear
correlation, the best predicted y-value is y.
2. If there is a significant linear correlation,
the best predicted y-value is found by
substituting the x-value into the
regression equation.
FIGURE 9-7
Predicting the Value of a Variable
Start
Calculate the value of r
and test the hypothesis
that  = 0
Is
there a
significant linear
correlation
?
Yes
No
Given any value of one
variable, the best predicted
value of the other variable
is its sample mean.
Use the regression
equation to make
predictions. Substitute
the given value in the
regression equation.
Guidelines for Using The
Regression Equation
1. If there is no significant linear correlation,
don’t use the regression equation to make
predictions.
2. When using the regression equation for
predictions, stay within the scope of the
available sample data.
3. A regression equation based on old data is
not necessarily valid now.
4. Don’t make predictions about a population
that is different from the population from
which the sample data was drawn.
What is the best predicted size of a household
that discard 0.50 lb of plastic?
Data from the Garbage Project
x Plastic (lb)
y Household
0.27 1.41
2
3
2.19
2.83
2.19
1.81
0.85
3.05
3
6
4
2
1
5
What is the best predicted size of a household
that discard 0.50 lb of plastic?
Data from the Garbage Project
x Plastic (lb)
y Household
0.27 1.41
2
3
2.19
2.83
2.19
1.81
0.85
3.05
3
6
4
2
1
5
Using a calculator:
b0 = 0.549
b1= 1.48
y = 0.549 + 1.48x
What is the best predicted size of a household
that discard 0.50 lb of plastic?
Data from the Garbage Project
x Plastic (lb)
y Household
0.27 1.41
2
3
2.19
2.83
2.19
1.81
0.85
3.05
3
6
4
2
1
5
Using a calculator:
b0 = 0.549
b1= 1.48
Because there is a significant linear
correlation, the equation can be
used for prediction.
y = 0.549 + 1.48 (0.50)
y = 1.3
What is the best predicted size of a household
that discard 0.50 lb of plastic?
Data from the Garbage Project
x Plastic (lb)
y Household
0.27 1.41
2
3
2.19
2.83
2.19
1.81
0.85
3.05
3
6
4
2
1
5
Using a calculator:
b0 = 0.549
b1= 1.48
y = 0.549 + 1.48 (0.50)
y = 1.3
A household that discards 0.50 lb of plastic has
approximately one person.
Definitions
 Marginal Change
the amount a variable changes when the
other variable changes by exactly one unit
 Outlier
a point lying far away from the other data
points
 Influential Points
points which strongly affect the graph of the
regression line
Residuals and the
Least-Squares Property
Definitions
Residuals and the
Least-Squares Property
Definitions
Residual
for a sample of paired (x,y) data, the difference (y - ^
y)
^
between an observed sample y-value and the value of y,
which is the value of y that is predicted by using the
regression equation.
Residuals and the
Least-Squares Property
Definitions
Residual
for a sample of paired (x,y) data, the difference (y - ^
y)
^
between an observed sample y-value and the value of y,
which is the value of y that is predicted by using the
regression equation.
Least-Squares Property
A straight line satisfies this property if the sum of the
squares of the residuals is the smallest sum possible.
Residuals and the
Least-Squares Property
x
y
1 2
4 24
4 5
8 32
Residuals and the
Least-Squares Property
x
y
1 2
4 24
^
y = 5 + 4x
4 5
8 32
y
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
FIGURE 9-8
• Residual = 7
• Residual = 11
•
•
Residual = -13
Residual = -5
x
1
2
3
4
5
ELEMENTARY
Section 9-4
STATISTICS
Variation and Prediction Intervals
MARIO F. TRIOLA
EIGHTH
EDITION
Definitions
Definitions
Total Deviation from the mean of the particular point (x, y)
the vertical distance y - y, which is the distance between
the point (x, y) and the horizontal line passing through
the sample mean y
Definitions
Total Deviation from the mean of the particular point (x, y)
the vertical distance y - y, which is the distance between
the point (x, y) and the horizontal line passing through
the sample mean y
Explained Deviation
the vertical distance y^ - y, which is the distance between
the predicted y value and the horizontal line passing
through the sample mean y
Definitions
Total Deviation from the mean of the particular point (x, y)
the vertical distance y - y, which is the distance between
the point (x, y) and the horizontal line passing through
the sample mean y
Explained Deviation
the vertical distance y^ - y, which is the distance between
the predicted y value and the horizontal line passing
through the sample mean y
Unexplained Deviation
the vertical distance y - ^
y, which is the vertical distance
between the point (x, y) and the regression line. (The
distance y - ^y is also called a residual, as defined in
Section 9-3.)
Figure 9-9 Unexplained, Explained, and Total Deviation
Figure 9-9 Unexplained, Explained, and Total Deviation
y
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
(5, 19)
•
Total
deviation
(y - y )
Unexplained
deviation
^)
(y - y
(5, 13)
•
Explained
deviation
(^
y - y)
•
y=9
(5, 9)
y^ = 3 + 2x
x
0
1
2
3
4
5
6
7
8
9
(total deviation) = (explained deviation) + (unexplained deviation)
^
^
(y - y) =
(y - y)
+
(y - y)
(total deviation) = (explained deviation) + (unexplained deviation)
^
^
(y - y) =
(y - y)
+
(y - y)
(total variation) = (explained variation) + (unexplained variation)
2
(y - y) =
^
(y - y)
Formula 9-5
2
+
^
(y - y)
2
Definition
Coefficient of determination
the amount of the variation in y that is
explained by the regression line
Definition
Coefficient of determination
the amount of the variation in y that is
explained by the regression line
r
2
=
explained variation
total variation
Definition
Coefficient of determination
the amount of the variation in y that is
explained by the regression line
r
2
=
explained variation.
total variation
or
simply square r
(determined by Formula 9-1, section 9-2)
Prediction Intervals
Definition
Standard error of estimate
Prediction Intervals
Definition
Standard error of estimate
a measure of the differences (or distances)
between the observed sample y values and
the predicted values y^ that are obtained
using the regression equation
Standard Error of Estimate
Standard Error of Estimate
s
(y
e
=
n-2
2
^
- y)
Standard Error of Estimate
(y
s
e
=
2
^
- y)
n-2
or
s
e
=
y
2
- b0
n-2
y - b1
xy
Formula 9-6
Prediction Interval for an Individual y
Prediction Interval for an Individual y
y^ - E < y < ^y + E
Prediction Interval for an Individual y
y^ - E < y < ^y + E
where
E=t
1
n(x0 - x)
2
1+ n +
2
2
s
2 e
n( x ) - ( x)
Prediction Interval for an Individual y
y^ - E < y < ^y + E
where
E=t
1
n(x0 - x)
2
1+ n +
2
2
s
2 e
n( x ) - ( x)
x0 represents the given value of x
t
2 has
n - 2 degrees of freedom
ELEMENTARY
Chapter 10
MARIO F. TRIOLA
STATISTICS
Multinomial Experiments and
Contingency Tables
EIGHTH
EDITION
Chapter 10
Multinomial Experiments and
Contingency Tables
10-1 Overview
10-2 Multinomial Experiments:
Goodness-of-fit
10-3 Contingency Tables:
Independence and Homogeneity
10-1
Overview
Focus on analysis of categorical (qualitative or
attribute) data that can be separated into
different categories (often called cells)
Use the X2 (chi-square) test statistic (Table A-4)
One-way frequency table (single row or column)
Two-way frequency table or contingency table
(two or more rows and columns)
10-2 Multinomial Experiments:
Goodness-of-Fit
Assumptions
when testing hypothesis that the population
proportion for each of the categories is as claimed:
1. The data have been randomly selected.
2. The sample data consist of frequency counts
for each of the different categories.
3. The expected frequency is at least 5. (There is
no requirement that the observed frequency
for each category must be at least 5.)
Definition
Multinomial Experiment
An experiment that meets the following conditions:
1. The number of trials is fixed.
2. The trials are independent.
3. All outcomes of each trial must be classified
into exactly one of several different
categories.
4. The probabilities for the different
categories
remain constant for each trial.
Definition
Goodness-of-fit test
used to test the hypothesis that an
observed frequency distribution fits
(or conforms to) some claimed
distribution
Goodness-of-Fit Test
Notation
0
represents the observed frequency of an outcome
E
represents the expected frequency of an outcome
k
represents the number of different categories or
outcomes
n
represents the total number of trials
Expected Frequencies
If all expected frequencies are equal:
E=
n
k
the sum of all observed frequencies divided
by the number of categories
Expected Frequencies
If all expected frequencies are not all equal:
E=np
each expected frequency is found by multiplying
the sum of all observed frequencies by the
probability for the category
Goodness-of-fit Test in Multinomial Experiments
Test Statistic
X =
2
(O - E)2
E
Critical Values
1. Found in Table A-4 using k-1 degrees of
freedom
where k = number of categories
2. Goodness-of-fit hypothesis tests are always
right-tailed.
A close agreement between observed
and expected values will lead to a small
value of X2 and a large P-value.
A large disagreement between observed
and expected values will lead to a large
value of X2 and a small P-value.
A significantly large value of  will cause
a rejection of the null hypothesis of no
difference between the observed and the
expected.
2
Relationships Among
Components in
Goodness-of-Fit
Hypothesis Test
Figure 10-3
Categories with Equal
Frequencies
(Probabilities)
H0: p1 = p2 = p3 = . . . = pk
H1: at least one of the probabilities is
different from the others
Categories with Unequal
Frequencies
(Probabilities)
H0: p1 , p2, p3, . . . , pk are as claimed
H1: at least one of the above proportions
is different from the claimed value
Example:
Mars, Inc. claims its M&M candies are distributed with
the color percentages of 30% brown, 20% yellow, 20% red, 10% orange,
10% green, and 10% blue. At the 0.05 significance level, test the claim
that the color distribution is as claimed by Mars, Inc.
Example:
Mars, Inc. claims its M&M candies are distributed with
the color percentages of 30% brown, 20% yellow, 20% red, 10% orange,
10% green, and 10% blue. At the 0.05 significance level, test the claim
that the color distribution is as claimed by Mars, Inc.
Claim: p1 = 0.30, p2 = 0.20, p3 = 0.20, p4 = 0.10,
p5 = 0.10, p6 = 0.10
H0 : p1 = 0.30, p2 = 0.20, p3 = 0.20, p4 = 0.10,
p5 = 0.10, p6 = 0.10
H1: At least one of the proportions is
different from the claimed value.
Example:
Mars, Inc. claims its M&M candies are distributed with
the color percentages of 30% brown, 20% yellow, 20% red, 10% orange,
10% green, and 10% blue. At the 0.05 significance level, test the claim
that the color distribution is as claimed by Mars, Inc.
Frequencies of M&Ms
Brown Yellow Red Orange Green Blue
Observed frequency
n = 100
33
26
21
8
7
5
Example:
Mars, Inc. claims its M&M candies are distributed with
the color percentages of 30% brown, 20% yellow, 20% red, 10% orange,
10% green, and 10% blue. At the 0.05 significance level, test the claim
that the color distribution is as claimed by Mars, Inc.
Frequencies of M&Ms
Brown Yellow Red Orange Green Blue
Observed frequency
n = 100
33
26
21
8
7
Brown E = np = (100)(0.30) = 30
Yellow E = np = (100)(0.20) = 20
Red E = np = (100)(0.20) = 20
Orange E = np = (100)(0.10) = 10
Green E = np = (100)(0.10) = 10
Blue E = np = (100)(0.10) = 10
5
Frequencies of M&Ms
Brown Yellow Red Orange Green Blue
Observed frequency
33
26
21
8
7
5
Expected frequency
30
20
20
10
10
10
0.3
1.8
0.05
0.4
0.9
2.5
(O -E)2/E
Frequencies of M&Ms
Brown Yellow Red Orange Green Blue
Observed frequency
33
26
21
8
7
5
Expected frequency
30
20
20
10
10
10
0.3
1.8
0.05
0.4
0.9
2.5
(O -E)2/E
Test Statistic
X =
2
(O - E)2
=
E
5.95
Frequencies of M&Ms
Brown Yellow Red Orange Green Blue
Observed frequency
33
26
21
8
7
5
Expected frequency
30
20
20
10
10
10
0.3
1.8
0.05
0.4
0.9
2.5
(O -E)2/E
Test Statistic
X2 = 
(O - E)2
=
E
2
5.95
Critical Value X =11.071
(with k-1 = 5 and  = 0.05)
Fail to Reject
Reject
 = 0.05
0
X2 = 11.071
Sample data: X2 = 5.95
Test Statistic does not fall within critical region;
Fail to reject H0: percentages are as claimed
There is not sufficient evidence to warrant rejection of the
claim that the colors are distributed with the given
percentages.
Comparison of Claimed and Observed Proportions
0.30
•
•
•
•
0.20
Proportions
Observed proportions
••
Claimed proportions
0.10
•
•
•
•
0
Orange
Yellow
Brown
Red
Green
•
•
Blue
ELEMENTARY
Section 10-3
MARIO F. TRIOLA
STATISTICS
Contingency Tables: Independence
and Homogeneity
EIGHTH
EDITIO
Definition
 Contingency Table (or two-way frequency table)
a table in which frequencies
correspond to two variables.
(One variable is used to categorize rows,
and a second variable is used to
categorize columns.)
Definition
 Contingency Table (or two-way frequency table)
a table in which frequencies
correspond to two variables.
(One variable is used to categorize rows,
and a second variable is used to
categorize columns.)
Contingency tables have at least two
rows and at least two columns.
Definition
 Test of Independence
tests the null hypothesis that
the row variable and column
variable in a contingency table are
not related.
(The null hypothesis
is the
statement that
the row and column variables are
independent.)
Assumptions
1.
The sample data are randomly selected.
2.
The null hypothesis H0 is the statement that
the row and column variables
are
independent; the alternative
hypothesis H1 is the statement that the row
and column variables are dependent.
3.
For every cell in the contingency table, the
expected frequency E is at least 5. (There is
no requirement that every observed
frequency must be at least 5.)
Test of Independence
Test Statistic
X =
2
(O - E)2
E
Critical Values
1. Found in Table A-4 using
degrees of freedom = (r - 1)(c - 1)
r is the number of rows and c is the number of columns
2. Tests of Independence are always right-tailed.
E=
(row total) (column total)
(grand total)
Total number of all observed frequencies
in the table
Tests of Independence
H0: The row variable is independent of the
column variable
H1: The row variable is dependent (related to)
the column variable
This procedure cannot be used to establish a
direct cause-and-effect link between variables in
question.
Dependence means only there is a relationship
between the two variables.
Expected Frequency for
Contingency Tables
Expected Frequency for
Contingency Tables
E=
grand total
•
row total
grand total
•
column total
grand total
Expected Frequency for
Contingency Tables
E=
grand total
n
•
•
row total
grand total
•
column total
grand total
p
(probability of a cell)
Expected Frequency for
Contingency Tables
E=
grand total
n
•
•
row total
grand total
•
column total
grand total
p
(probability of a cell)
Expected Frequency for
Contingency Tables
E=
grand total
n
•
•
row total
grand total
•
column total
grand total
p
(probability of a cell)
E=
(row total) (column total)
(grand total)
Is the type of crime independent of whether the
criminal is a stranger?
Stranger
Acquaintance
or Relative
Homicide
12
39
Robbery
Assault
379
727
106
642
Is the type of crime independent of whether the
criminal is a stranger?
Stranger
Homicide
12
Assault
Row Total
379
727
1118
Robbery
Acquaintance
or Relative
39
106
642
787
Column Total
51
485
1369
1905
Is the type of crime independent of whether the
criminal is a stranger?
Stranger
Homicide
12
Assault
Row Total
379
727
1118
Robbery
Acquaintance
or Relative
39
106
642
787
Column Total
51
485
1369
1905
E=
(row total) (column total)
(grand total)
Is the type of crime independent of whether the
criminal is a stranger?
Homicide
12
Stranger
Assault
Row Total
379
727
1118
Robbery
(29.93)
Acquaintance
or Relative
39
106
642
787
Column Total
51
485
1369
1905
E=
(row total) (column total)
(grand total)
E = (1118)(5 = 29.93
1) 1905
Is the type of crime independent of whether the
criminal is a stranger?
Stranger
Acquaintance
or Relative
Row Total
727
1118
Robbery
(29.93)
(284.64)
(803.43)
39
(21.07)
106
(200.36)
642
(565.57)
787
485
1369
1905
379
51
Column Total
E=
E
Assault
Homicide
12
= (1118)(5
1) 1905
(row total) (column total)
(grand total)
= 29.93
E = (1118)(485)= 284.64
etc.
1905
Is the type of crime independent of whether the
criminal is a stranger?
X =
2
(O - E )2
E
Homicide
Stranger
Acquaintance
12
(29.93)
Robbery
Forgery
379
(284.64)
727
(803.43)
106
(200.36)
642
(565.57
[10.741]
39
(21.07)
or Relative
(O -E )2
Upper left cell:
=
E
(12 -29.93)2
= 10.741
29.93
(E)
(O - E )2
E
Is the type of crime independent of whether the
criminal is a stranger?
X =
2
(O - E )2
E
Homicide
Stranger
Acquaintance
or Relative
12
(29.93)
[10.741]
39
(21.07)
[15.258]
(O -E )2
Upper left cell:
=
E
Robbery
Forgery
379
(284.64)
[31.281]
727
(803.43)
[7.271]
106
(200.36)
[44.439]
642
(565.57)
[10.329]
(12 -29.93)2
= 10.741
29.93
(E)
(O - E )2
E
Is the type of crime independent of whether the
criminal is a stranger?
X =
2
(O - E )2
E
Homicide
12
(29.93)
Stranger
[10.741]
Acquaintance
or Relative
39
(21.07)
[15.258]
Robbery
Forgery
379
(284.64)
[31.281]
727
(803.43)
[7.271]
106
(200.36)
[44.439]
642
(565.57)
[10.329]
2
Test Statistic X = 10.741 + 31.281 + ... + 10.329 =
119.319
(E)
(O - E )2
E
Test Statistic X2 = 119.319
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of
freedom
Critical Value X2 = 5.991 (from Table A-4)
Test Statistic X2 = 119.319
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of
freedom
Critical Value X2 = 5.991 (from Table A-4)
Fail to Reject
Independence
Reject
Independence
 = 0.05
0
X2 = 5.991
Sample data: X2 =119.319
Test Statistic X2 = 119.319
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of
freedom
Critical Value X2 = 5.991 (from Table A-4)
Fail to Reject
Independence
Reject
Independence
 = 0.05
0
X2 = 5.991
Reject independence
Sample data: X2 =119.319
Test Statistic X2 = 119.319
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of
freedom
Critical Value X2 = 5.991 (from Table A-4)
Fail to Reject
Independence
Reject
Independence
 = 0.05
0
X2 = 5.991
Reject independence
Sample data: X2 =119.319
Claim: The type of crime and knowledge of criminal are independent
Ho : The type of crime and knowledge of criminal are independent
H1 : The type of crime and knowledge of criminal are dependent
Test Statistic X2 = 119.319
with  = 0.05 and (r -1) (c -1) = (2 -1) (3 -1) = 2 degrees of
freedom
Critical Value X2 = 5.991 (from Table A-4)
Fail to Reject
Independence
Reject
Independence
 = 0.05
0
X2 = 5.991
Reject independence
Sample data: X2 =119.319
It appears that the type of crime and
knowledge of the criminal are related.
Relationships Among Components in X2 Test
of Independence
Figure 10-8
Definition
 Test of Homogeneity
test the claim that different populations
have the same proportions of some
characteristics
How to distinguish between a
test of homogeneity and a
test for independence:
Were predetermined sample sizes
used for different populations (test of
homogeneity), or was one big sample
drawn so both row and column totals
were determined randomly (test of
independence)?
ELEMENTARY
Chapter 11
STATISTICS
Analysis of Variance
MARIO F. TRIOLA
EIGHTH
EDITION
Chapter 11
Analysis of Variance
11-1 Overview
11-2 One-way ANOVA
11-3 Two-way ANOVA
11-1
Overview
An introduction of a procedure for
testing the hypothesis that three or
more population means are equal.
For example:
H0 : µ 1 = µ2 = µ3 = . . . µk
H1: At least one mean is different
11-1
Overview
Definition
Analysis of Variance (ANOVA)
a method of testing the equality
of three or more population
means by analyzing sample
variations
ANOVA methods require the
F-distribution
1. The F-distribution is not symmetric; it
is skewed to the right.
2. The values of F can be 0 or positive,
they cannot be negative.
3. There is a different F-distribution for
each pair of degrees of freedom for the
numerator and denominator.
Critical values of F are given in Table A-5
F - distribution
Not symmetric (skewed to the right)

nonnegative values only
Figure 11-1
11-2 One-Way ANOVA
Assumptions
1. The populations have normal distributions.
2. The populations have the same variance
s2 (or standard deviation s ).
3. The samples are simple random samples.
4. The samples are independent of each other.
5. The different samples are from populations
that are categorized in only one way.
Definition
Treatment (or factor)
a property or characteristic that
allows us to distinguish the different
populations from another
Use computer software or TI-83 Plus for
ANOVA calculations if possible
Procedure for testing:
H0: µ1 = µ2 = µ3 = . . .
1. Use STATDISK, Minitab, Excel, or a TI83
Calulator to obtain results.
2. Identify the P-value from the display.
3. Form a conclusion based on these criteria:
 If P-value  , reject the null hypothesis of
equal means.
 If P-value >  , fail to reject the null
hypothesis of equal means.
Relationships Among Components of ANOVA
Figure 11-2
ANOVA Fundamental Concept
Estimate the common value of s2 using
1. The variance between samples (also
called variation due to treatment) is
an
estimate of the common population
variance
s2 that is based on the variability
among the
sample means.
2. The variance within samples (also called
variation due to error) is an estimate of the
common population variance s2 based on the
sample variances.
ANOVA Fundamental Concept
Test Statistic for One-Way ANOVA
ANOVA Fundamental Concept
Test Statistic for One-Way ANOVA
F=
variance between samples
variance within samples
ANOVA Fundamental Concept
Test Statistic for One-Way ANOVA
F=
variance between samples
variance within samples
A excessively large F test statistic is
evidence against equal population means.
Calculations with Equal Sample Sizes
Variance between samples = nsx
2
Calculations with Equal Sample Sizes
Variance between samples = nsx
2
2
where sx = variance of samples means
Calculations with Equal Sample Sizes
Variance between samples = nsx
2
2
where sx = variance of samples means
Variance within samples = sp
2
Calculations with Equal Sample Sizes
Variance between samples = nsx
2
2
where sx = variance of samples means
Variance within samples = sp
2
2
where sp = pooled variance (or the mean
of the sample variances)
Critical Value of F
Right-tailed test
Degree of freedom with k samples of the
same size n
numerator df = k -1
denominator df = k(n -1)
Calculations with Unequal Sample Sizes
ni(xi - x)2
F=
variance within samples
variance between samples
=
k -1
(ni - 1)si 2
(ni - 1)
Calculations with Unequal Sample Sizes
ni(xi - x)2
F=
variance within samples
variance between samples
=
k -1
(ni - 1)si 2
(ni - 1)
where x = mean of all sample scores combined
Calculations with Unequal Sample Sizes
ni(xi - x)2
F=
variance within samples
variance between samples
=
k -1
(ni - 1)si 2
(ni - 1)
where x = mean of all sample scores combined
k = number of population means being compared
Calculations with Unequal Sample Sizes
ni(xi - x)2
F=
variance within samples
variance between samples
=
k -1
(ni - 1)si 2
(ni - 1)
where x = mean of all sample scores combined
k = number of population means being compared
ni = number of values in the ith sample
Calculations with Unequal Sample Sizes
ni(xi - x)2
F=
variance within samples
variance between samples
=
k -1
(ni - 1)si 2
(ni - 1)
where x = mean of all sample scores combined
k = number of population means being compared
ni = number of values in the ith sample
xi = mean values in the ith sample
Calculations with Unequal Sample Sizes
ni(xi - x)2
variance within samples
variance between samples
F=
=
k -1
(ni - 1)si 2
(ni - 1)
where x = mean of all sample scores combined
k = number of population means being compared
ni = number of values in the ith sample
xi = mean values in the ith sample
2
si = variance of values in the ith sample
Key Components of ANOVA Method
Key Components of ANOVA Method
SS(total), or total sum of squares, is a
measure of the total variation (around x) in
all the sample data combined.
Key Components of ANOVA Method
SS(total), or total sum of squares, is a
measure of the total variation (around x) in
all the sample data combined.
SS(total) = (x - x)
Formula 11-1
2
Key Components of ANOVA Method
SS(treatment) is a measure of the variation between the
samples. In one-way ANOVA, SS(treatment) is sometimes
referred to as SS(factor). Because it is a measure of
variability between the sample means, it is also referred to
as SS (between groups) or SS (between samples).
Key Components of ANOVA Method
SS(treatment) is a measure of the variation between the
samples. In one-way ANOVA, SS(treatment) is sometimes
referred to as SS(factor). Because it is a measure of
variability between the sample means, it is also referred to
as SS (between groups) or SS (between samples).
SS(treatment) = n1(x1 - x)2 + n2(x2 - x)2 + . . . nk(xk - x)2
Formula 11-2
= ni(xi - x)2
Key Components of ANOVA Method
SS(error) is a sum of squares representing the
variability that is assumed to be common to all the
populations being considered.
Key Components of ANOVA Method
SS(error) is a sum of squares representing the
variability that is assumed to be common to all the
populations being considered.
2
1
2
2
2
3
2
i
SS(error) = (n1 -1)s + (n2 -1)s + (n3 -1)s . . . nk(xk -1)s
= (ni - 1)s
Formula 11-3
2
i
Key Components of ANOVA Method
SS(total) = SS(treatment) + SS(error)
Formula 11-4
Mean Squares (MS)
Sum of Squares SS(treatment) and SS(error)
divided by corresponding number of degrees
of freedom.
MS (treatment) is mean square for treatment,
obtained as follows:
Mean Squares (MS)
Sum of Squares SS(treatment) and SS(error)
divided by corresponding number of degrees
of freedom.
MS (treatment) is mean square for treatment,
obtained as follows:
MS (treatment) =
Formula 11-5
SS (treatment)
k-1
Mean Squares (MS)
MS (error) is mean square for error, obtained
as follows:
Mean Squares (MS)
MS (error) is mean square for error, obtained
as follows:
MS (error) =
Formula 11-6
SS (error)
N-k
Mean Squares (MS)
MS (error) is mean square for error, obtained
as follows:
MS (error) =
SS (error)
N-k
Formula 11-6
MS (total) =
Formula 11-7
SS (total)
N-1
Test Statistic for ANOVA with
Unequal Sample Sizes
Test Statistic for ANOVA with
Unequal Sample Sizes
F=
MS (treatment)
MS (error)
Test Statistic for ANOVA with
Unequal Sample Sizes
F=
MS (treatment)
MS (error)
Formula 11-8
 Numerator
df = k -1
 Denominator df = N - k