AP Statistics Difference in Proportions Question Bank

Practise calculating and interpreting the mean and standard deviation of sampling distributions for differences between two sample proportions in AP Statistics.

A large company produces an equal number of brand-name lightbulbs and generic lightbulbs. The director of

quality control sets guidelines that production will be stopped if there is evidence that the proportion of all

lightbulbs that are defective is greater than 0.10. The director also believes that the proportion of brand-name

lightbulbs that are defective is not equal to the proportion of generic lightbulbs that are defective. Therefore, the

director wants to estimate the average of the two proportions.

To estimate the proportion of brand-name lightbulbs that are defective, a simple random sample of

400 brand-name lightbulbs is taken and 44 are found to be defective. Let X represent the number of

brand-name lightbulbs that are defective in a sample of 400, and let $p_{X}$ represent the proportion of all

brand-name lightbulbs that are defective. It is reasonable to assume that X is a binomial random variable.

Consider $\hat{D}$ , the point estimate of D.

Calculate $s_{\hat{D}}$ , the standard error of $\hat{D}$ .

The standard error of $\hat{p}_{X}=0.0156$ is obtained from part (b). The standard error of $\hat{p}_{Y}$ is

$\sqrt{\frac{\hat{p}_{Y}\left(1-\hat{p}_{Y}\right)}{n}}=\sqrt{\frac{(0.26)(0.74)}{400}}=0.0219$ . So the standard error of $\hat{D}$ is

$S_{\hat{D}}=\sqrt{\frac{1}{4}\left(0.0156^{2}+0.0219^{2}\right)}=0.0134$ .

Part (e)

W=\frac{0.185-0.10}{0.0134}=6.34 .

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Question 6 (continued)

Part (f)

Suppose the true mean D is 0.10. Then the observed value of $\hat{D}=0.185$ is 6.34 standard errors from

the mean D. Using Chebyshev's inequality, the probability of observing a value of $\hat{D}$ within 6.34 standard

errors of the mean of 0.10 is at least $1-\frac{1}{6.34^{2}}=0.975$ . So the probability of observing a value as far from

0.10 as the one observed, or farther, is at most 0.025 if the true mean really is 0.10. Therefore, the p-value

for this test is at most 0.025, which is less than 0.05, so the null hypothesis can be rejected. There is

sufficient statistical evidence at the 0.05 level to conclude that the average proportion for all products that

are defective is greater than 0.10.

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3.9.A—Calculate the mean and standard deviation of the sampling distribution for the difference between two sample… question 1