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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 10n 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 set of randomly selected IQ’s of 10 people.

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(ii) Use StatCrunch to calculate the sample mean for each random sample of size 10n .

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 in row 10.

(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

10n 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 40n .

Sketch the histogram and comment on the shape of the distribution.

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For 40n most of the data is closer to the mean of 100 than it was in part (a) when 10n .

The shape of the distribution of x that came from a normally distributed population with

40n is approximately normally distributed.

Summary statistics:

Write down the results for x

= 100.019542andx

= 2.3419509

The Standard deviation has decreased to about half as much as when 10n .

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.

Column Mean Std. dev.

Row Mean 100.01954 2.3419509

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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) --> Leave Bins: 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 10n 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 40n .

Sketch the histogram and comment on the shape of the distribution.

(In order to compare the distribution shape of x for 10n versus 40n , 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 40n 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 30n .

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

wherex

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.

10

27, 6 , 15n

27 x

549.115

6

nx

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 size is 9n , 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 to be normally distributed.

Objective B : Finding Probability of x that is Normally Distributed

Standardize x to Z

Recall : Standardize x to Z : x

Z

Now : Standardize x to Z : x

x

xZ

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 36n is obtained from a population mean,

64 , and population standard deviation, 18 .

(a) Describe the sampling distribution x .

Since n ≥ 30, is normally distributed.

x

x

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(b) Find x

and x

.

64 x

336

18

nx

(c) Use StatCrunch to find ( 62.6)P x ?

Open StatCrunch → Stat → Normal → Standard → input mean = 64, Std. Dev. = 3

--> select the , input x-value= 62.6 → Compute!

( 62.6)P x = 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 )4842( xP .

Is x normally distributed?

Yes, since the population distribution is normally distributed, x is N. dist.

What is x

? What is x

?

7.43 x

212.112

2.4

nx

Open StatCrunch → Stat → Normal → Between -> Input Mean, Std. Dev. , and

the x values --> Compute!

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9194.0)4842( xP

(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 )46( xP .

Is x normally distributed?

Yes, since the population distribution is normally distributed, x is N. dist.

What is x

? What is x

?

7.43 x

084.115

2.4

nx

Open StatCrunch → Stat → Normal → Standard -> Input Mean, Std. Dev. ,

select the appropriate inequality sign, and input the x value --> Compute!

)46( xP =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.

A. Sampling distribution of sample proportion p̂ , where n

xp ˆ .

The shape of the sampling distribution of p̂ is approximately normally provided by,

(1 ) 10np p or 10npq where 1q p .

B. Finding the mean and standard deviation of p̂

pp ˆ , n

ppp

)1(ˆ

Objective B : Finding Probability of p̂ that is Normally Distributed

Standardize p̂ to Z

p

ppZ

ˆ

ˆˆ

where pp ˆ and

n

ppp

)1(ˆ

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

200n 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 , 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

10npq p̂

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Sample proportion: p̂ = 𝑥

𝑛 =

154

200 = 0.77

Looking for probability: )77.0ˆ( pP

What is p̂ ? 8.0ˆ pp

What is p̂ ? 028284.0200

)2.0)(8.0(

200

)8.01)(8.0()1(ˆ

n

ppp

Use StatCrunch to find what is the probability that 154 or fewer college 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 295.10)71.0)(29.0)(500( npq 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 )30.0ˆ25.0( pP .

Is p̂ normally distributed? Yes, based on the answer in part (a).

What is p̂ ? 29.0ˆ pp

What is p̂ ? 02029.0500

)71.0)(29.0(

500

)29.01)(29.0()1(ˆ

n

ppp

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!

)30.0ˆ25.0( pP =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?

What is p̂ ? 25.0500

125ˆ

n

xp

The question asked for )25.0ˆ( pP .

Is p̂ normally distributed? Yes, based on the answer in part (a).

What is p̂ ? 29.0ˆ pp

What is p̂ ? 02029.0500

)71.0)(29.0(

500

)29.01)(29.0()1(ˆ

n

ppp

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!

)25.0ˆ( pP =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.