Transcript File - phsapstatistics

WARM UP
A statewide poll reveals the only 18% of Texas household
participate in giving out Halloween Treats. To examine this
you collect a Random Sample of 120 Texas households.
1. What is the Probability that less than 12% of Texas households give
out Treats?
P  pˆ  .12   ?
P  z  #  ?
P  z  1.7108  ?
z
pˆ  p
p 1  p 
n


.12  .18
P z 

.18 1  .18 

120

normalcdf ( E 99, 1.7108)  0.0436
2. Comment on the validity of your results by checking the Rule of
Thumb assumptions.
RULE OF THUMB #1: Pop. of Texas Households ≥ 10 • 120
RULE OF THUMB #2: 120 • .18 ≥ 10 AND
120 • (1 – .18) ≥ 10
3. Draw the Sampling Distribution curve.


?




N  p,

p(1  p) 

n

N(0.18, 0.035)
.075
.11
.12
.145
.18
.215
.25
.285
Chapter 18 (continued)
Sample Proportion, p̂ use Categorical Data
Data collected by COUNTING.
Sample Means,
x use Quantitative Data
Data collected by AVERAGING
SAMPLE MEANS (Chapter 18 continued)
I. THE BASICS
The unknown Population Mean is a parameter with the
symbol of:  . The Population Standard Deviation has
the symbol:

The Sample Mean is a statistic with the symbol of:
x
and the Sample Standard Deviation has the formula s 
.
The Test Statistic is:
z
x 

n

n
ASSUMPTIONS for Sample Means:
  

x
N  ,

n

#1: The sample was collected using an SRS.
#2: Approximate Normal:
1. Large n = (n ≥ 30), OR
2. Stated that its Normal, OR
3. You construct an approximately Normal
Histogram/Boxplot from the given data.
It is known that each individual Halloween Treat given out
has about 120 calories with standard deviation of 50. To
examine this you collect a Random Sample of 40 Halloween
treats.
1. What is the Probability that the sample average is less than 100
calories?
P  x  100  ?
z
x 

P  z  2.5298  ?
n


100  120 

P z 
?
50


40 

normalcdf ( E 99, z)  0.0057
2. Comment on the validity of your results by checking the
conditons/assumptions.
Stated to be a Random sample
Approximately Normal due to the Large n.
3.
Draw the Sampling Distribution curve.
II. THE CALCULATIONS
EXAMPLE: A national political poll asked a SRS of 50
adults to give an approval rating of President Obama’s
Job in office using a scale from 1(Low) to 10(High).
Suppose in fact, nationally Obama’s rating has a Mean
of 3.5 and Standard Deviation of 1.2
1. Find the Mean and Standard Deviation of the
Sampling Distribution of x .
Mean = 3.5
Std. Dev. = .1697
1.2
Std . Dev. 
50
2. Explain why you can assume normality in the
sample.
#1: The sample was collected by an SRS
#2: Approximately Normal b/c Large n
II. THE CALCULATIONS of PROBABILITY
SAME EXAMPLE: A national political poll asked a SRS of 50 adults to
give an approval rating of President Obama’s Job in office using a scale
from 1(Low) to 10(High). Suppose in fact, nationally Obama’s rating has
a Mean of 3.5 and Standard Deviation of 1.2
3. Find the probability Obama has a rating over 4.0.
x 
P  x  4.0   ?
z

Pz  #   ?


n

4.0  3.5 
P z 

P  z  2.9463  ?
1.2




50 

Probability  normalcdf (2.9463, E 99)  0.0016
SAME EXAMPLE: A national political poll asked a SRS of 50 adults to
give an approval rating of President Obama’s Job in office using a scale
from 1(Low) to 10(High). Suppose in fact, nationally Obama’s rating has
a Mean of 3.5 and Standard Deviation of 1.2
4. Find the probability Obama’s approval rating is between
3.2 and 4.0.
P  3.2  x  4.0   ?


3.2  3.5
4.0  3.5 

P
 

?

1.2
1.2


50
50 

z
z
x 

n
P  1.7678  z  2.9463  ?
Prob.=normalcdf ( 1.7678, 2.9463)  0.9598
Finding Sample Mean Observations, (x) from Probabilities
EXAMPLE: Assume that the duration of human pregnancies
is Appr. Normal with mean: 266 days and Std Dev.: 16. At
least how many days should the longest 25% of all
pregnancies last?
x 
z

1
z = invNorm(.75)= .6745
25%
218 234 250 266 282 298 314
x=?
x  266
0.6745 
16
x = 276.79
Finding Sample Mean Observations, (x) from Probabilities
EXAMPLE: Assume that the rainfall in Ithaca, NY is Appr.
Normal with mean: 35.4 inches and Std Dev.: 4.2”. Less
than how much rain falls in the driest 20% of all years
z
x 

z = invNorm(.20)= -.8416
20%
22.8 27 31.2 35.4 39.6 43.8 48
x=?
x  35.4
.8416 
4.2
x = 31.865”
1. The Test over Chapter ninteen has traditionally
had a mean of 86.2% with a std. deviation of
8.5%. Assume the data follows a Normal
distribution.
a.) If a student is selected at random, what is the
probability that he or she will score above 90%?
b.) If 16 students are selected at random, what is the
probability that their average will be above 90%?
c.) Draw the Sampling Distribution Normal Curves for both
a and b. On both graphs label all six standard deviations
(± 3s) for the N   ,  .


n


With μ = 86.2% and σ = 8.5%.


P



a.) When n = 1 the P  x  90   ?
1.Appr. Normal 
69.2
77.7


P



z

90  86.2 

?
8.5 

16 
P  z  1.7882   ?
normalcdf (1.7882, E 99)  0.0369
Stated . and 2.Data Collected Randomly  SRS
8.5 

N  86.2,

16


N  86.2, 2.125
c.).N  86.2, 8.5 


1

N  86.2,8.5
60.7
P  z  0.4471  ?
normalcdf (0.4471, E 99)  0.3274
Stated . and 2.Data Collected Randomly  SRS
b.) When n = 16 the P  x  90   ?
1.Appr. Normal 
z

90  86.2 

?
8.5 

1 
86.2 94.7 103.2 111.7