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# Inference for Means and Proportions Free Response Practice Problems for AP Statistics (page 2)

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### Solutions

1. This is a paired study because the scores for each pair of twins are compared. Hence, it is a one-sample situation, and there are 26 pieces of data to be analyzed, which are the 26 difference scores between the twins. Hence, df = 26 – 1 = 25.
2.
• I is not correct. A confidence interval, at least in this course, cannot be used in any one-sided hypothesis test—only two-sided tests.
• II is correct. A confidence interval constructed from a random sample that does not contain the hypothesized value of the parameter can be considered statistically significant evidence against the null hypothesis.
• III is not correct. The standard error for a confidence interval is based on the sample proportions is
• .

The standard error for a significance test is based on the hypothesized population value is

.

• IV is correct.
3. The data are paired by individual students, so we need to test the difference scores for the students rather than the means for each quiz. The differences are given in the following table.
4. Let's suppose that the Stat class constructed a 95% confidence interval for the true proportion of students who plan to vote for Harvey (we are assuming that this a random sample from the population of interest, and we note that both are greater than 10). (as Harvey figured). Then a 95% confidence interval for the true proportion of votes Harvey can expect to get is 0.533 ± 1.96 . That is, we are 95% confident that between 35.5% and 71.2% of students plan to vote for Harvey. He may have a majority, but there is a lot of room between 35.5% and 50% for Harvey not to get the majority he thinks he has. (The argument is similar with a 90% CI: (0.384, 0.683); or with a 99% CI: (0.299, 0.768).)
5.
• Using a two-sample t-test, Steps I and II would not change. Step III would change to
• based on df = min{30 – 1,30 – 1} = 29. Step IV would probably arrive at a different conclusion—reject the null because the P-value is small. Large sample sizes make it easier to detect statistically significant differences.

6. H0: p = 0.98, HA: p < 0.98, .
7. .

This P-value is quite low and provides evidence against the null and in favor of the alternative that security procedures actually detect less than the claimed percentage of banned objects.

8.
1. df = min{23 – 1,27 – 1} = 22 t* = 1.717.
2. df = 22 t* = 2.074.
3. df = 23 + 27 –2 = 48 t* = 1.684 (round down to 40 degrees of freedom in the table).
4. df = 48 t* = 2.021.

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