Transcript Example - udcompsci

Chapter 14: Statistics
Introductory Question: On the most
recent Chemistry Test, Mrs. Jones’
class had the following scores: 81, 45,
67, 88, 72, 97, 59, 82, 67, 86.
How many students scored above the
class average for this Test?
What are the maximum and minimum
scores?
What are the mode and median?
Statistics
Statistics: The mathematics of the
collection, organization, and
interpretation of numerical data,
especially the analysis of population
characteristics by inference from
sampling.
Characteristics of the Mean
The arithmetic mean is the most widely used
measure of location.
It is calculated by summing the values and dividing
by the number of values (the average).
Sample Mean
The sample mean is the sum of all the sample
values divided by the number of sample values:
X
X 
n
Where n is the total number of values in the
sample.
EXAMPLE
A sample of five executives received the
following bonus last year ($000):
14.0, 15.0, 17.0, 16.0, 15.0
What is the mean for this data?
X 
X 14 .0  ...  15 .0 77


 15 .4
n
5
5
The Median
The Median is the midpoint of the values
after they have been ordered from the
smallest to the largest.
There are as many values above the
median as below it in the data array.
 For an even set of values, the median will
be the arithmetic average of the two middle
numbers.

EXAMPLE
The ages for a sample of five college students
are:
21, 25, 19, 20, 22
Arranging the data in ascending order
gives: 19, 20, 21, 22, 25. Thus the median
is 21.
Example
The heights of four basketball players, in
inches, are:
76, 73, 80, 75
Arranging the data in ascending order
gives: 73, 75, 76, 80. Thus the median is
75.5, found by (75+76)/2.
The Mode
The mode is the value of the observation that
appears most frequently.
EXAMPLE 6: The exam scores for ten students are:
81, 93, 84, 75, 68, 87, 81, 75, 81, 87. Because the
score of 81 occurs the most often, it is the mode.
Stem-and-leaf Displays
Stem-and-leaf display: A statistical
technique for displaying a set of data.
Each numerical value is divided into two
parts: the leading digits become the
stem and the trailing digits the leaf.
EXAMPLE
Colin achieved the following
scores on his twelve Accounting
quizzes this semester:
86, 79, 92, 84, 69, 88,
91, 83, 96, 78, 82, 85.
Construct a stem-and-leaf chart.
Example continued
stem
leaf
6
9
7
89
8
234568
9
126
EXAMPLE
The top ten rushers in the NFL
this past season had the
following number of total rushes
for the season:
360, 335, 330, 290, 323, 282,
307, 300, 305, 372
Construct a stem-and-leaf chart.
Percentiles and Quartiles
The Percentile gives us the location, or ranking, of a
data point in relation to the data set.
Example: the 9th percentile is the value that is above
exactly 9% of all the data points.
A special percentile is the Quartile.
The first quartile, Q1, is the value that is above one
quarter, or 25% of the data values.
The third quartile, Q3, is the value that is above three
quarters, or 75% of the data values.
Location of a Percentile
To find the location of the percentile, p, in a data set
containing n data points, first order the data from
smallest to largest. Then, to find the location in
the ordered set, use the following formulas.
P
L p  (n  1)
100
If the location falls between two data points, you will
find a value between those data points.
EXAMPLE
Find the 18th percentile for the following data set:
30, 32, 37, 39, 41, 43, 44, 46, 48, 48, 53
In this problem, n = 11. Therefore the location of the
18th percentile is
18
L p  (11  1)
 2.16
100
and is between the 2nd and 3rd data points. With a
difference of 5, the 18th percentile is 32 + .16*5 or
p18 = 32.80
EXAMPLE (cont)
Find the first quartile for the following data set:
30, 32, 37, 39, 41, 43, 44, 46, 48, 48, 53
To find the first quartile, we need to find the 25th
percentile. It’s location is
25
L p  (11  1)
3
100
Which is the 3rd data point, or Q1 = 37
Quartiles
The first quartile, Q1, is essentially the median for
the first half of the data.
The third quartile, Q3, is essentially the median for
the second half of the data.
Range
The range is the difference between
the largest and the smallest value.


Only two values are used in its calculation.
To calculate, range = maximum-minimum
Interquartile Range
The Interquartile range is the distance
between the third quartile Q3 and the first
quartile Q1.
This distance will include the middle 50
percent of the observations.
Interquartile range = Q3 - Q1
Example
Given the following set of data:
52, 26, 33, 40, 35, 29, 26, 37, 28
What is the median, Q1, and Q3?
Arranging the data in ascending order
gives: 26, 26, 28, 29, 33, 35, 37, 40, 52.
Thus the median is 33, Q1 is 27, and Q3
is 38.5
What is the inter-quartile range?
Q3 - Q1 = 38.5 – 27 = 11.5
EXAMPLE
For a set of observations the third quartile
is 24 and the first quartile is 10. What is
the interquartile range?
The interquartile range is 24 - 10 = 14.
Fifty percent of the observations will occur
between 10 and 24.
Box Plots
A box plot is a graphical display, based
on quartiles, that helps to picture a set
of data.
Five pieces of data are needed to construct
a box plot:
the Minimum Value, the First Quartile, the
Median, the Third Quartile, and the
Maximum Value.
EXAMPLE
min
12
Q1
14
median
16
18
Q3
20
22
max
24
26
28
30
32
Box Plots
A box plot sometimes includes an
outlier.
An outlier is an extreme value that are more
than 1.5 times the interquartile range
beyond the upper or lower quartiles.
If an outlier exists, it is marked by a single
point, and each whisker is extended to the
last value of the data that is not an outlier.
Mean Deviation
The Mean Deviation is the arithmetic mean of
the absolute values of the deviations from the
arithmetic mean.
The formula is:
MD 
 X X
n
EXAMPLE
The weights of a sample of crates containing books
for the bookstore (in pounds ) are:
103, 97, 101, 106, 103
Find the mean deviation.
Example (cont)
To find the mean deviation, first find the
mean weight.
X 510
X

 102
n
5
Example (cont)
The mean deviation is:
MD 
X X

103  102  ...  103  102
n
1 5 1 4 1

 2.4
5
5
Variance
The variance is the arithmetic mean of
the squared deviations from the mean.
The formula for the variance is:
( X  X )
 
N
2
2
EXAMPLE
The ages of the Dunn family are:
2, 18, 34, 42
What is the variance?
X 96
X
  24
n
4
( X  X ) 2  24  ...  42  24
 

N
4
944

 236
4
2
2
2
2
The Standard Deviation
The standard deviation σ is the square root
of the variance.
For the previous example, the standard
deviation is 15.36, found by
    236  15 .36
2
EXAMPLE
Consider the test scores: 100, 98, 95, 88, 84,
77, 75, 72, 70, 66. Find the standard
deviation.
EXAMPLE
Consider the test scores: 100, 98, 95, 88, 84,
77, 75, 72, 70, 66. Find the standard
deviation. Create a Chart (see below)
X
X X
(X  X )
2
EXAMPLE
Consider the test scores: 100, 98, 95, 88, 84, 77,
75, 72, 70, 66. How many scores were within 1
standard deviation from the mean? How many
were within 2 standard deviations?
EXAMPLE
The hourly wages earned by a sample
of five students are: $7, $5, $11, $8, $6.
Find the variance.
X 37
X 

 7.40
n
5

 X  X 
7  7.4  ...  6  7.4
 

n 1
5 1
21.2

 5.30
5 1
2
2
2
2
Frequency Distribution
A Frequency distribution is a grouping
of data into mutually exclusive
categories showing the number of
observations in each class.
Frequency Distribution
Class frequency: The number of
observations in each class.
Class interval: The class interval is obtained by
subtracting the lower limit of a class from the
lower limit of the next class.
Class Mark: The midpoint of a class interval.
Number of Classes: Should use at least k classes,
where 2k > n ( the number of data points).
(This is the 2k rule)
Suggestions on Constructing a
Frequency Distribution
The class intervals used in the frequency distribution
should be equal.
Determine
a suggested class interval
by using the formula:
(Highest va lue - Lowest val ue)
i
Number of classes
Note: this is a suggested class interval; if the computed class interval
is ’97’, it may be better to use ‘100’.
Example: Body Temperatures of 44
Healthy Adults
98.6
98.4
97.6
97.8
98.2
98.6
98.6
97.7
97.4
99.6
98.0
98.6
98.8
98.9
98.7
98.0
98.8
98.0
98.6
99.4
99.0
98.6
98.0
99.5
98.2
98.4
97.0
98.3
97.5
98.0
98.4 98.4
97.0 98.8
98.5 97.3
97.3 97.6
98.6 98.6
97.2 98.4 98.6 98.2
Construct a frequency table with 6 classes.
EXAMPLE 1
Dr. Tillman is Dean of the School of Business at
Hampton University. He wishes to prepare a
report showing the number of hours per week
students spend studying. He selects a random
sample of 30 students and determines the
number of hours each student studied last week.
15.0, 23.7, 19.7, 15.4, 18.3, 23.0, 14.2, 20.8, 13.5,
20.7, 17.4, 18.6, 12.9, 20.3, 13.7, 21.4, 18.3, 29.8,
17.1, 18.9, 10.3, 26.1, 15.7, 14.0, 17.8, 33.8, 23.2,
12.9, 27.1, 16.6.
Organize the data into a frequency distribution.
Example 1 continued
There are 30 observations
Two raised to the fifth power is 32.
Therefore, we should have at least 5 classes. It
turns out we will use 6 classes.
The range is 23.5 hours, (found by 33.8 hours – 10.3 hours).
We choose an interval of 5 hours.
The lower limit of the first class is 7.5 hours.
EXAMPLE 1 continued
Hours studying
7.5 up to 12.5
12.5 up to 17.5
17.5 up to 22.5
22.5 up to 27.5
27.5 up to 32.5
Frequency, f
1
12
10
5
1
32.5 up to 37.5
1
Relative Frequency Distributions
A relative frequency distribution shows
the percent of observations in each
class.
Example 1
Hours
f
Relative
Frequency
7.5 up to 12.5
1
1/30=.0333
12.5 up to 17.5
12
12/30=.400
17.5 up to 22.5
22.5 up to 27.5
10
5
10/30=.333
5/30=.1667
27.5 up to 32.5
32.5 up to 37.5
1
1
1/30=.0333
1/30=.0333
TOTAL
30
30/30=1
Graphic Presentation of
a Frequency Distribution
A Histogram is a graph in which the
classes are marked on the horizontal axis
and the class frequencies on the vertical
axis. The class frequencies are represented
by the heights of the bars and the bars are
drawn adjacent to each other.
Frequency
Histogram for Hours Spent Studying
14
12
10
8
6
4
2
0
7.5-12.5
12.5-17.5
17.5-22.5
22.5-27.5
Hours spent studying
27.5-32.5
32.5-37.5
Normal Distribution
Normal Distributions are really a
family of frequency distributions
that have the same general
“Bell”shape when shown
graphically. They are symmetric
with scores more concentrated in
the middle than in the tails.
A Normal Distribution often occurs
when there is a large data set.
Normal Distribution
Normal Distributions have the following properties:
1. The maximum point of the curve is the MEAN.
2. About 68.3% of the data are within 1 standard
deviation from the mean
3. About 95.5% of the data are within 2 standard
deviations from the mean.
4. About 99.7% of the data are within 3 standard
deviations from the mean
Lesson Overview 14-4B
Normal Distribution
Example: A data set of 250 values has a normal
distribution. The mean of the data is 45 and
the standard deviation is 3.
a) What percent of the data is in the range 39 to
51?
b) What is the range of data that includes 68.3%
of the data?
Normal Distribution
Example: A data set of 250 values has a normal
distribution. The mean of the data is 45 and
the standard deviation is 3.
c) Find the probability that a value selected at
random will be within the limits 36 to 54.
Normal Distribution
Example: A data set of 250 values has a normal
distribution. The mean of the data is 45 and
the standard deviation is 3.
d) Find the probability that a value selected at
random will be less than 48.
e) Find the probability that the value selected will
be greater than 48.
Normal Distribution
Example: A data set of 250 values has a normal
distribution. The mean of the data is 45 and
the standard deviation is 3.
f) Find the probability that the score is between 33
and 48.
Normal Distribution
Example: A sample of 600 young people are
weighed at a clinic. If 100 pounds is the
average weight, and the weights are normally
distributed, determine how many young
people are within 1 standard deviation from
the mean.
How many are within 2 standard deviations?
Normal Distribution
Example: A company manufactures light bulbs
that have a life expectancy that is normally
distributed with a mean of 750 hours and a
standard deviation of 40 hours. Find the
probability that a bulb burns between 728 and
784 hours.
Normal Distribution
Example: On a SAT exam, the mean math score
was 475 with a standard deviation of 130. If a
scholarship is available to students with
scores above the 85th percentile, what is the
score needed to be eligible for the
scholarship?
Normal Distribution
Example: The heights of a group of students are
taken, and the mean is 52 inches with a
standard deviation of 2.5 inches. Assuming
the heights are normally distributed, what is
the probability that a student selected at
random will have a height less than 50
inches?
5-Minute Check Lesson 14-5A
5-Minute Check Lesson 14-5B
Scatter Plots
Comparing two variables (like time vs distance)
involves bivariate statistics.
A “picture” or graph of the data can be shown by a
scatterplot.
Label the axes and plot the points, just like the
rectangular coordinate system (but do NOT
connect the dots-that is why it is called a
‘scatter’ plot; it gives you an indication of the
relationship that exists between the two sets of
variables)
Linear Regression
Some data is related linearly; i.e. the scatterplot of
the data most closely resembles a line. Not all
data is linear in nature, but we can run a linear
regression on the data to see if a linear
equation could be used for a given situation.
If data is linear, then the equation should be of the
form:
y = mx + b
(where m is the slope and b is the y-intercept)
Linear Regression
We will use the graphing calculator to run the
regression. First, we must type in the data for
each variable set, storing them in L1 and L2.
Next, we use the ‘Stats’ button and choose
‘linreg’.
The closer the ‘r’ value (known as the correlation
coefficient) is to 1, the more appropriate a
linear equation would be to relate the two sets
of data. Notice that the calculator actually tells
what the best linear equation would be to use
for the data.
Example
Example: Scientists have monitored the number of
chirps per minute made by crickets and the
corresponding temperature.
# of chirps/min136 165 98 110 150 210 84 158 221 178
Temp in F
72 84 68 75 80 94 60 75 92 89
Make a scatter plot of the data using appropriate scales
for the x and y axes.
Example (continued)
-Find the "line of best fit" for the data and draw that
line.
-Pick two points of your line (not necessarily of the
data points) and write the equation of the line.
-What does the slope indicate? What does the yintercept represent?
-Predict - if a cricket chirps 90 times/min, what is
the temperature?
-If the temperature is 78, how many times will the
cricket chirp?
Example (continued)
Now, we will run the Linear Regression on the
calculator and record the values of a, b, and r,
where y = ax + b, and r represents the
correlation coefficient
a:
b:
r:
How close does your equation match the one that
the calculator came up with?
Other Regressions
If your scatterplot does not suggest a linear
relationship, there are other types of
regressions you can run.
Expreg (if the relationship is exponential)
Powreg (if the relationship is a polynomial
function)
Lnreg (if the relationship is logarithmic)
Example
Example: Year vs. Cost of Postage Stamps
Year
1919 1932 1958 1963 1968 1971 1974 1978 1981 1983 1988 1991
Cost of 2
3
4
5
6
8
10 15 18 22 25 29
Stamps
Make a scatter plot of the data using appropriate scales for the x
and y axes. Then run the 4 regressions we mentioned to
determine which type of equation would correlate most to the
given data.
Example
Example: Year vs. Cost of Postage Stamps
Year
1919 1932 1958 1963 1968 1971 1974 1978 1981 1983 1988 1991
Cost of 2
3
4
5
6
8
10 15 18 22 25 29
Stamps
Based on the Regression equation you came up with, calculate the
price of a postage stamp in the current year. Does that match
up with what a postage stamp actually costs?
Bell - Shaped Curve showing the relationship between  and m.
m3
m2 m1
m
m1 m2 m3
Mean of the Data in a Frequency
Distribution
First, find the class marks for each class in the
distribution.
Next, for each class mark, multiply it by its
corresponding frequency.
Then, take each of these products and add them
together.
Finally, take that sum and divide by the total number of
frequencies the distribution has.
Standard Deviation of the Data in a
Frequency Distribution
First, find the class marks for each class in the distribution.
Next, find the mean for the distribution (see previous information).
Next, take each class mark and subtract the mean from it.
Next, take those results and square them.
Next, take those numbers and multiply them by their corresponding
frequencies.
Next, take those values and add them together.
Finally, take that sum and divide it by the total number of frequencies
you have, then take the square root.