Chapter 8 Sampling Distribution
Ch 8.1 Distribution of Sample Mean
Objective A: Shape, Center, and Spread of the Distributions of x
A1. Sampling Distributions of Mean
A1.1 Sampling Distribution of the Sample Mean: Normal Population
Example 1: IQ is a measurement of intelligence derived from the Stanford Binet IQ test. Scores on this test
are normally distributed with a mean score of 100 and a standard deviation of 15.
(a) (i)
Use StatCrunch to simulate 1000 random samples of size n  10 from this population.
StatCrunch -->StatCrunch Website -->Open StatCrunch --> Data --> Simulate --> Normal
Click Options and Save.
Explain the representation of entries in row 1 of the StatCrunch spreadsheet.
The entries in row one represent the 1st set of randomly selected IQ’s of 10 people.
(ii) Use StatCrunch to calculate the sample mean for each random sample of size n  10 .
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Stat --> Summary Stats --> Rows ---> Select all Normal1 to Normal10 for Select column(s)
--> Select Mean for Statistics --> Check Store in data table --> Compute!
Explain the representation of entries in column 11 of the StatCrunch spreadsheet.
Column 11 entries represent the mean for the 10 IQ scores for each row.
(iii) Use StatCrunch to draw a histogram for the 1000 sample means.
What is the sampling distribution of the sample means of sample size n = 10?
Let's use a lower class limit of the first class of 65 and a class width of 5.
Graph --> Histogram --> Select Row Mean for Select column(s) --> Input Start at: 65
and Width: 5 for Bins ---> Compute ---> Options ---> Save.
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Sketch the histogram and comment on the shape of the distribution.
The shape of the distribution of x that came from a normally distributed population with
n  10 is approximately normally distributed.
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(b) Use StatCrunch to find the mean and standard deviation of the sampling distribution of
the 1000 sample means?
Stat ---> Summary Stats ----> Columns ---> Select Row Mean ---> Select Mean and Std. dev.-->
for Statistics ---> Compute!
Summary statistics:
Column
Mean
Std. dev.
Row Mean 99.876239 4.6205712
Write down the results.
 x =99.8762394  x = 4.6205712
(c) Repeat part (a) and (b) with size n  40 .
Sketch the histogram and comment on the shape of the distribution.
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For n  40 most of the data is closer to the mean of 100 than it was in part (a) when n  10 .
The shape of the distribution of x that came from a normally distributed population with
n  40 is approximately normally distributed.
Summary statistics:
Column
Mean
Std. dev.
Row Mean 100.01954 2.3419509
Write down the results for  x = 100.019542and  x = 2.3419509
The Standard deviation has decreased to about half as much as when n  10 .
A1. Sampling Distributions of Mean
A1.2 Sampling Distribution of the Sample Mean: Non-normal Population
Example 1: The waiting time in line can be modeled by an exponential distribution which is similar to skewed to
the right with a mean of 5 minutes and a standard deviation of 5 minutes.
(a) Repeat Example 1 of A1.1 or part (a) but using exponential distribution instead of normal distribution.
StatCrunch -->StatCrunch Website -->Open StatCrunch --> Data --> Simulate -->Exponential,
then follow the steps given in (i) to (iii) to construct a histogram. If you are not sure about the start
at value and the class width, just leave them blank and StatCrunch will figure it out based on
the 1000 sample means.
Sketch the histogram and comment on the shape of the distribution.
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StatCrunch -->StatCrunch Website -->Open StatCrunch --> Data --> Simulate --> Exponential
Stat --> Summary Stats --> Rows ---> Select all Exponential1 to Exponential10 for Select column(s)
--> Select Mean for Statistics --> Check Store in data table --> Compute!.
Graph --> Histogram --> Select Row Mean for Select column(s) --> LeaveBins: as optional -->
--> Compute!
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The shape of the distribution of x that came from a non-normally distributed population
(exponential distribution is skewed to the right) with n  10 is approximately right skewed.
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(b) Repeat Example 1 of A1.1 of part (b) using the sample means obtained from A1.2 of part (a).
Write down the results.
Stat ---> Summary Stats ----> Columns ---> Select Row Mean ---> Select Mean and Std. dev.--> for Statistics ---> Compute!
 x = 5.0944972  x = 1.6251305
(c) Repeat part (a) and (b) with size n  40 .
Sketch the histogram and comment on the shape of the distribution.
(In order to compare the distribution shape of x for n  10 versus n  40 , we will use the same starting
point of 0 and bin width of 2)
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The shape of the distribution of x that came from a non-normally distributed population
(exponential distribution is skewed to the right) with n  40 is approximately normally distributed.
Write down the results for  x = 4.9381378_ and  x = 0.77093207_
A2. Central Limit Theorem
A.
If the population distribution of x is normally distributed, the sampling distribution of x is normally
distributed regardless of the sample size n .
If the population distribution x is not normally distributed, the sampling distribution of x is
guaranteed to be normally distributed if n  30 .
Use the distribution shapes obtained from the two simulation Example 1results of objective A1
to verify the statement A of the Central Limit Theorem.
B.
Mean/standard deviation of a sampling distribution of x versus mean/standard deviation of a
population distribution of x .
The mean and standard deviation of population distribution are  and  respectively.
The mean of the sampling distribution of x is  x where  x   .
The standard deviation of the sampling distribution of x is  x where  x 

n.
Use the  x and  x obtained from the two simulation Example 1 of objective A1 to verify
the statement B of the Central Limit Theorem.
Example 1 : Determine  x and  x from the given parameters of the population and
the sample size.
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  27,   6 , n  15
 x    27
x 

n

6
15
 1.549
Example 2 : A simple random sample is obtained from a population with   64 and   18 .
(a) If the population distribution is skewed to the right, what condition must be
applied in order to guarantee the sampling distribution of x is normally distributed?
Since the population distribution is not normally distributed, the selected sample size
must be greater than or equal to 30 (i.e.n ≥ 30).
(b) If the sample sizeis n  9 , what must be true regarding the distribution of the
population in order to guarantee the sampling distribution of x to be normally
distributed?
For a small sample size, the population distribution must be normally distributed in order
to guarantee the sampling distribution of x to be normally distributed.
Objective B : Finding Probability of x that is Normally Distributed
Standardize x to Z
Recall : Standardize x to Z : Z 
Now :Standardize x to Z :
Z
x

x  x
x
StatCrunch will perform the standardizing of the statistics and calculate the appropriate area under
the standard normal curve. We must verify a given sampling distribution is normally distributed and
provide the appropriate mean and standard deviation for the sampling distribution.
Example 1 :
A simple random sample of size n  36 is obtained from a population mean,
  64 , and population standard deviation,   18 .
(a) Describe the sampling distribution x .
Since n ≥ 30, x is normally distributed.
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(b) Find  x and  x .
 x    64
x 

n

18
36
3
(c) Use StatCrunch to find P( x  62.6) ?
Open StatCrunch → Stat → Normal → Standard → input mean = 64, Std. Dev.= 3
-->select the  , input x-value= 62.6 → Compute!
P( x  62.6) = 0.3204
Example 2 : The upper leg of 20 to 29 year old males is normally distributed with a mean
length of 43.7cm and a standard deviation of 4.2cm.
(a) Use StatCrunch to computethe probability that a random sample of 12 males
who are 20 to 29 years old results in a mean upper leg length that is between
42cm and 48cm?
The question asked for P(42  x  48) .
Is x normally distributed?
Yes, since the population distribution is normally distributed, x is N. dist.
What is  x ? What is  x ?
 x    43.7
x 

n

4.2
 1.212
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Open StatCrunch → Stat → Normal → Between -> Input Mean, Std. Dev. , and
the x values --> Compute!
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P(42  x  48)  0.9194
(b) A random sample of 15 males who are 20 to 29 years old results in a mean upper
leg length greater than 46 cm. Do you find the result unusual? Why?
The question asked for P ( x  46) .
Is x normally distributed?
Yes, since the population distribution is normally distributed, x is N. dist.
What is  x ? What is  x ?
 x    43.7
x 


4.2
 1.084
n
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Open StatCrunch → Stat → Normal → Standard -> Input Mean, Std. Dev. ,
select the appropriate inequality sign, and input the x value --> Compute!
P ( x  46) =0.0169. Since 0.0169 is smaller than 0.05, finding a random sample
of 15 males who are 20 to 29 years old results in a mean upper leg length greater
than 46 cm is an unusual event.
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Ch 8.2 Distribution of the Sample Proportion
Objective A: Shape, Center and Spread of the Distribution of p̂ .
Distribution of the Sample Proportions - Explain what is a p̂ distribution.
x
.
n
The shape of the sampling distribution of p̂ is approximately normally provided by,
A. Sampling distribution of sample proportion p̂ , where pˆ 
np(1  p)  10 or npq  10 where q  1  p .
B. Finding the mean and standard deviation of p̂
 pˆ  p
,
 pˆ 
p(1  p)
n
Objective B : Finding Probability of p̂ that is Normally Distributed
Standardize p̂ to Z
Z
pˆ   pˆ
 pˆ
where  pˆ  p and  pˆ 
p(1  p)
n
provided p̂ is approximately normally distributed.
Example 1: A nationwide study indicated that 80% of college students, who use a cell phone,
send and receive text messages on their phone. A simple random sample of
n  200 college students using a cell phone is obtained.
(a) Describe sampling distribution of p̂ .
p = 0.8, n = 200
Check to see if np (1- p) ≥ 10
(200 )(0.8) (1 - 0.8) ≥ 10
(200)(0.8) (0.2) ≥ 10
32 ≥ 10
Since npq  10 , p̂ is normally distributed.
(b) If 154 college students in the sample send and receive text messages on the cell phone,
find p̂ ,  p̂ and  p̂
Sample proportion: p̂
𝑥
154
𝑛
200
= =
= 0.77
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Looking for probability: P( pˆ  0.77)
What is  p̂ ?  pˆ  p  0.8
What is  p̂ ?  pˆ 
p(1  p)
(0.8)(1  0.8)
(0.8)(0.2)


 0.028284
n
200
200
Use StatCrunch to find what is the probability that 154 or fewercollege students in
the sample send and receive text messages on the cell phone? Is this unusual?
Open StatCrunch → Stat → Normal → Standard -> Input Mean (  p̂ ), Std. Dev.(  p̂ ),
select the appropriate inequality sign, and input the x value (in this case it is p̂ )--> Compute!
Since 0.1444 is bigger than 0.05, this result is not usual.
Example 2: According to creditcard.com, 29% of adults do not own a credit card.
(a) Suppose a random sample of 500 adults is asked, "Do you own a credit card?"
Describe the sampling distribution of p̂ , the proportion of adults who own a credit card.
n = 500
p = 0.29
q = (1-0.29) = 0.71
500 (0.29) (0.71) = 10.295
Since npq  (500)(0.29)(0.71)  10.295 which is  10, p̂ is normally distributed.
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(b) Use StatCrunch to find the probability that in a random sample of 500 adults between
25% and 30% do not own a credit card?
The question asked for P(0.25  pˆ  0.30) .
Is p̂ normally distributed? Yes, based on the answer in part (a).
What is  p̂ ?
What is  p̂ ?
 pˆ  p  0.29
 pˆ 
p(1  p)
(0.29)(1  0.29)
(0.29)(0.71)


 0.02029
n
500
500
Open StatCrunch → Stat → Normal → Between -> Input Mean (  p̂ ), Std. Dev.(  p̂ ),
and input the x values (in this case they are the two given p̂ s)--> Compute!
P(0.25  pˆ  0.30) =0.7682
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(c) Would it be unusual for a random sample of 500 adults to result in 125 or fewer who
do not own a credit card? Why?
x 125

 0.25
n 500
The question asked for P( pˆ  0.25) .
What is p̂ ?
pˆ 
Is p̂ normally distributed? Yes, based on the answer in part (a).
What is  p̂ ?
What is  p̂ ?
 pˆ  p  0.29
 pˆ 
p(1  p)
(0.29)(1  0.29)
(0.29)(0.71)


 0.02029
n
500
500
Open StatCrunch → Stat → Normal → Standard -> Input Mean (  p̂ ), Std. Dev.(  p̂ ),
and input the x value (in this case it is the given p̂ )--> Compute!
P( pˆ  0.25) =0.0243
Since 0.0243 is smaller than 0.05, it would be unusual for a random sample of 500 adults
to result in 125 or fewer who do not own a credit card.
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