AP Statistics |
Unit Test #2 |

_____ / 50

In a poll of one thousand randomly sampled adults in the United States, 46% said that they were satisfied with the quality of K-12 education in the nation.

Use this information for all questions on this page.

1. |
Do these data provide evidence at the 5% level that a minority of adults in the United States are satisfied with the quality of education in the United States? |
(15) |

${H}_{0}:p=0.5$ vs. ${H}_{a}:p<0.5$

This requires that the sample was obtained randomly, that the sample is smaller than 10% of the population, and that both of $np$ and $n\left(1-p\right)$ are at least 10. I’m told that the sample was obtained randomly. There are obviously more than 10000 adults in the United States that could have been asked. $np=\left(1000\right)\left(0.5\right)=5000>10$ and $n\left(1-p\right)=\left(1000\right)\left(0.5\right)=500>10$. The test statistic is $z=\frac{0.46-0.5}{\sqrt{\frac{\left(0.5\right)\left(0.5\right)}{1000}}}\approx -2.5298$. $P\left(z<-2.5298\right)\approx 0.0057$. If at least 50% of adults in the United States are satisfied with the quality of K-12 education, then I can expect to find a sample of 1000 adults where 460 or less of them are satisfied in about 0.57% of trials. Since $0.0057<0.05$, this is significant at the 5% level. I reject the null hypothesis. The data provide evidence that a minority of adults in the United States are satisfied with the quality of K-12 education.

2. |
Can we determine if a 95% confidence interval for the population proportion of adults in the United States who are satisfied with K-12 education would likely contain 0.5? Explain. |
(3) |

We cannot. The connection between a test and an interval only exists if the test is conducted with a two-sided alternative…and we did not do that in this case.

3. |
A retail computer dealer is trying to decide between two methods for servicing customers’ equipment. The first method emphasizes preventive maintenance; the second emphasizes quick response to problems. The dealer randomly assigns customers to be served by one of the two methods. After six months, it finds that 171 of 200 customers served by the first method are very satisfied with the service, as compared to 153 of 200 customers served by the second method. Do these data provide evidence at the 10% level of a difference in the rate of customer satisfaction between the two methods? |
(15) |

Let ${p}_{1}$ be the population proportion of customers that will be satisfied with the service when receiving the first service method (preventative maintenance), and let ${p}_{2}$ be the population proportion of customers who will be satisfied with the service when receiving the second service method (quick response).

${H}_{0}:{p}_{1}={p}_{2}$ vs. ${H}_{a}:{p}_{1}\ne {p}_{2}$

This requires that the customers were randomly assigned to the two service methods, and that all four of ${n}_{1}\widehat{p}$, ${n}_{1}\left(1-\widehat{p}\right)$, ${n}_{2}\widehat{p}$ and ${n}_{2}\left(1-\widehat{p}\right)$ are at least five, where $\widehat{p}$ is the pooled sample proportion. I’m told that the customers were randomly assigned to the two service methods. $\widehat{p}=\frac{171+153}{200+200}=0.81$, so ${n}_{1}\widehat{p}=\left(200\right)\left(0.81\right)=162>5$, ${n}_{1}\left(1-\widehat{p}\right)=\left(200\right)\left(0.19\right)=38>5$, ${n}_{2}\widehat{p}=\left(200\right)\left(0.81\right)=162>5$ and ${n}_{2}\left(1-\widehat{p}\right)=\left(200\right)\left(0.19\right)=38>5$. The test statistic is $z=\frac{0.855-0.765}{\sqrt{\left(0.81\right)\left(0.19\right)\left(\frac{1}{200}+\frac{1}{200}\right)}}\approx 2.2942$. $P\left(\left|z\right|>2.2942\right)\approx 0.0218$. If there is no difference in the rate of customer satisfaction between the two methods, then I can expect to find two groups of 200 customers with a difference of at least 18 satisfied customers in about 2.18% of trials. Since $0.0218<0.10$, this is significant at the 10% level. I reject the null hypothesis. The data provide evidence that there is a difference in the rate of customer satisfaction between the two service methods.

4. |
The media selection manager for an advertising agency inserts the same advertisement for a client bank in two magazines, similarly placed in each. One month later, a market research study finds that 226 of 473 readers of the first magazine are aware of the banking services offered in the ad, as are 165 of 439 readers of the second magazine (readers of both magazines are excluded). Construct a 99% confidence interval for the difference in customer awareness of the client bank between the readership of the two magazines. |
(13) |

Let $\widehat{{p}_{1}}$ be the sample proportion of readers of the first magazine that are aware of the banking services, and let $\widehat{{p}_{2}}$ be the sample proportion of readers of the second magazine that are aware of the banking services. This requires that both groups of readers are random samples, that both samples are smaller than 10% of their respective population, and that all four of ${n}_{1}\widehat{{p}_{1}}$, ${n}_{1}\left(1-\widehat{{p}_{1}}\right)$, ${n}_{2}\widehat{{p}_{2}}$ and ${n}_{2}\left(1-\widehat{{p}_{2}}\right)$ are at least five. I’ll have to assume that this study has obtained random samples of readers of each magazine, and that each sample is smaller than 10% of the readership of each magazine. ${n}_{1}\widehat{{p}_{1}}=226>5$, ${n}_{1}\left(1-\widehat{{p}_{1}}\right)=247>5$, ${n}_{2}\widehat{{p}_{2}}=165>5$ and ${n}_{2}\left(1-\widehat{{p}_{2}}\right)=274>5$. With 99% confidence, ${z}^{*}\approx 2.5758$. The interval is $\left(0.4778-0.3759\right)\pm 2.5758\sqrt{\frac{\left(0.4778\right)\left(0.5222\right)}{473}+\frac{\left(0.3759\right)\left(0.6241\right)}{439}}\approx \left(0.018,0.1859\right)$. I am 99% confident that the difference in population proportion of readers that are aware of the banking services of the client bank is between 1.8% and 18.59%.

5. |
What would be the likely conclusion of a hypothesis test at the 1% level for a difference in customer awareness between the readership of the two magazines? |
(4) |

Since the interval does not contain zero, it is not a plausible value for the difference in population proportions. Since zero is not a plausible value for the difference in population proportions, a test of significance ought to reject the null hypothesis.

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