# Chapter 1. Sampling Distribution of a Proportion

## Introduction

Statistical Applets
@possiblevalues:{25, 50, 75} $randomnumber: rand(0,2)$value: @possiblevalues[$randomnumber] @secondpossiblevalues:{25, 50, 75}$secondrandomnumber: rand(0,2) $secondvalue: @secondpossiblevalues[$randomnumber] @thirdpossiblevalues:{300, 500, 750, 1000} $thirdrandomnumber: rand(0,3)$thirdvalue: @thirdpossiblevalues[$randomnumber]$answertwo: 1.25*$value/25$answerfive: $secondvalue - 1.25*$secondvalue/25 $answereight: eval(round(0.95*$thirdvalue, 1)) $answernine: eval(round(0.05*$thirdvalue, 1))

Choose a population proportion of "successes" between 0 and 1 and a sample size, n, between 20 and 2000. Then click GENERATE ONE SAMPLE to generate samples one at a time or click GENERATE SAMPLES to generate 10,000 samples. Click "Show Normal Curve" to compare the sampling distribution with the Normal Distribution approximation.

Click the "Quiz Me" button to complete the activity.

If the sample size, n, is "large" and both np and n(1 - p) are large enough, the sampling distribution of the sample proportion = X/n will be approximately a Normal distribution with mean μ = p and standard deviation:

$$\sigma =\sqrt{\frac{p(1-p)}{n}}$$

This applet illustrates that important fact by allowing you to generate individual samples or thousands of samples with the specified parameters. You can then compare the distribution of sample proportion against the Normal distribution predicted from the parameters.

Yes, here, np = (formula: 1.25*$value/25) ### Question 1.3 Does this fit the rule for a sample proportion to be approximately Normal? 7zVRuryDpphPQjDfxtKbZg== 1 Incorrect. Recall, that np must be at least 10 in order for a sample proportion to be approximately Normal. When this value is small, the distribution tends to pile up close to 0. Correct. Recall, that np must be at least 10 in order for a sample proportion to be approximately Normal. When this value is small, the distribution tends to pile up close to 0. ### Question 1.4 Set p to 0.95. Set n to$secondvalue. Then “Generate samples.” What is the shape of the distribution? lW2i+QO3mFHPkZd9FAx7XxqnX3jjI1IqsCUQWnEoJ2m21EkGZ/MNQG8G4PA=

1
Incorrect. This distribution does not have that general shape. Recall that symmetric distributions are (almost) mirror images on either side of the center.
Correct. This distribution is left-skewed because the left-hand side is longer than the right. In particular 1 is an upper bound that cannot be breached because there can be no more than all “successes” in any set of trials.

### Question 1.5

What is the value of n (1 - p)? (Answer to two decimal places)
P4Z+hrXnBRud91dkBBQr4g==

2
Multiply the number of trials (n) by the probability of a success. Enter your answer with two decimal places.
Incorrect. Here, np = (formula: 1.25*$secondvalue/25) Yes, here, np = (formula: 1.25*$secondvalue/25)

### Question 1.6

Does this fit the rule for a sample proportion to be approximately Normal? 7zVRuryDpphPQjDfxtKbZg==

1
Incorrect. Recall, that n (1 - p) must be at least 10 in order for a sample proportion to be approximately Normal. When np is large (so n (1 - p) is small), the distribution tends to pile up close to 1.
Correct. Recall, that n (1 - p) must be at least 10 in order for a sample proportion to be approximately Normal. When np is large (so n (1 - p) is small), the distribution tends to pile up close to 1.

Yes, here, np = (formula: 0.95*$thirdvalue) ### Question 1.9 What is the value of n (1 - p)? (Use one decimal place in your answer.) Ausxm3tPeuklh7SNerLjmw== 2 Multiply the number of trials (n) by (1 – p), where p is the probability of a success. Enter your answer with one decimal place. Incorrect. Here, n(1 - p) = (formula: 0.05*$thirdvalue)
Yes, here, n(1 - p) = (formula: 0.05*\$thirdvalue)

### Question 1.10

Does this fit the rule for a sample proportion to be approximately Normal? esgbknumN3x7TOMm7lo9Vw==

1
Incorrect. Recall, that n (1 - p) must be at least 10 in order for a sample proportion to be approximately Normal. When np is large (so n (1 - p) is small), the distribution tends to pile up close to 1.
Correct. Recall, that n (1 - p) must be at least 10 in order for a sample proportion to be approximately Normal. When np is large (so n (1 - p) is small), the distribution tends to pile up close to 1.

### Question 1.11

Set p to 0.15 and n to 200. With these parameters, do you expect the distribution to be approximately Normal? esgbknumN3x7TOMm7lo9Vw==

1
Incorrect. Here, we have np = 30 and n (1 - p) = 170.
Correct. Here, we have np = 30 and n (1 - p) = 170.

### Question 1.12

What should be the mean of the sampling distribution? 8DZYIWMfcjSUDjuc

2
The mean of the sampling distribution is p, the population proportion.
Incorrect. The mean of the sampling distribution is p, the population proportion.
Correct. The mean of the sampling distribution is p, the population proportion.

### Question 1.13

Is the mean of the simulated distribution close? esgbknumN3x7TOMm7lo9Vw==

1
Incorrect. The mean (center) of the sampling distribution should be VERY close to 0.15.
Correct. The mean (center) of the sampling distribution should be VERY close to 0.15.

### Question 1.14

What should be the standard deviation of the sampling distribution? (Use two decimal places in your answer) BZ2bb0hmC68w2Uyb

2
Recall, the standard deviation of this distribution should be $$\sqrt{\frac{p(1-p)}{n}}$$.
Incorrect. Recall, the standard deviation of this distribution should be $$\sqrt{\frac{p(1-p)}{n}}$$.
You got it!

### Question 1.15

Increase n to 750 and generate samples. Did the center of the distribution move? 7zVRuryDpphPQjDfxtKbZg==

1
Incorrect. The mean of the sampling distribution of p-hat is always p, the population proportion.
Correct. The mean of the sampling distribution of p-hat is always p, the population proportion.

### Question 1.16

Did the standard deviation change? yLyjRnn8cuwobl0n443m6khlPqfRXlFhYAwCvMzDNZyrh/NMQLoZFjZn7doZg/m2BNXXnw==

1
Incorrect. The standard deviation of a sample proportion is $$\sqrt{\frac{p(1-p)}{n}}$$. As n gets larger, this quantity gets smaller because n is in the denominator.
Correct. The standard deviation of a sample proportion is $$\sqrt{\frac{p(1-p)}{n}}$$. As n gets larger, this quantity gets smaller because n is in the denominator.