AP Statistics Unit 6: Inference for Categorical Data: Proportions Practice Exam

Assessment for Unit 6: Inference for Categorical Data: Proportions

Estimated Time: 60 minutes

Exam Details

Questions: 54 total (54 multiple choice)
Time: 60 minutes (recommended)
Topics Covered: Unit 6: Inference for Categorical Data: Proportions
Format: Multiple Choice questions matching real exam difficulty
Scoring: Instant results with detailed explanations

A. Multiple-Choice Questions (54 Questions)

Select the one best answer for each question.

1. A market researcher wants to construct a 95 percent confidence interval for the proportion of all adults in a city who support a proposed tax increase. A random sample of 200 adults is selected, and 120 of them indicate support for the tax. Which of the following conditions must be satisfied to justify the use of the Normal distribution for the sampling distribution of the sample proportion?

(A)The sample size must be less than 10 percent of the population size.(B)The sample must be selected randomly.(C)The number of successes (120) and the number of failures (80) must both be at least 10.(D)The population distribution of support for the tax must be approximately Normal.

2. A university claims that 80 percent of its graduates find employment within six months of graduation. A student group believes the true proportion is lower. They survey a random sample of 150 recent graduates and find that 114 of them found employment within six months. Which of the following represents the correct test statistic for testing the null hypothesis H_0: p = 0.80 against the alternative hypothesis H_a: p < 0.80?

(A)z = $\frac{0.76 - 0.80}{\sqrt{\frac{0.76(0.24)}{150}}}$(B)z = $\frac{0.76 - 0.80}{\sqrt{\frac{0.80(0.20)}{150}}}$(C)z = $\frac{0.80 - 0.76}{\sqrt{\frac{0.76(0.24)}{150}}}$(D)z = $\frac{0.80 - 0.76}{\sqrt{\frac{0.80(0.20)}{150}}}$

3. A study was conducted to compare the proportion of smartphone users who use Brand A versus Brand B. A 95 percent confidence interval for the difference in population proportions (p_A - p_B) was calculated to be (-0.04, 0.12). Based on this interval, which of the following is the most appropriate conclusion?

(A)There is convincing evidence that the proportion of users for Brand A is different from Brand B because the interval contains 0.(B)There is convincing evidence that the proportion of users for Brand A is greater than Brand B because the center of the interval is positive.(C)There is not convincing evidence of a difference in the proportions because the interval includes 0.(D)There is not convincing evidence because the interval width is too large to be precise.

4. Researchers are conducting a two-sample z-test to determine if there is a significant difference in the success rates of two medical treatments. Sample 1 has 40 successes out of 100 trials, and Sample 2 has 75 successes out of 250 trials. When calculating the test statistic, what is the value of the pooled proportion, $\hat{p}_c$, that should be used in the standard error formula?

(A)0.300(B)0.329(C)0.350(D)0.400

5. A store manager performs a hypothesis test to determine if a new advertising campaign increased the proportion of customers who buy a warranty. The null hypothesis is H_0: p = 0.20 and the alternative is H_a: p > 0.20. The calculated p-value is 0.034. Which of the following is the correct interpretation of this p-value?

(A)There is a 0.034 probability that the null hypothesis is true.(B)Assuming the true proportion of customers who buy a warranty is 0.20, there is a 0.034 probability of obtaining a sample proportion as high as or higher than the one observed.(C)There is a 0.034 probability that the new advertising campaign is effective.(D)Assuming the new advertising campaign is effective, there is a 0.034 probability of observing the sample result.

6. A marketing consultant wants to determine if there is a statistically significant difference in the proportion of customers who prefer the new packaging design compared to the old packaging design. The consultant takes a random sample of 150 customers and shows them the new design, finding that 90 prefer it. Independent of the first sample, another random sample of 200 customers is shown the old design, and 110 prefer it. Which of the following represents the correct null and alternative hypotheses for a two-proportion z-test?

(A)$H_0: \hat{p}_{new} = \hat{p}_{old}$; $H_a: \hat{p}_{new} \neq \hat{p}_{old}$(B)$H_0: p_{new} = p_{old}$; $H_a: p_{new} > p_{old}$(C)$H_0: p_{new} = p_{old}$; $H_a: p_{new} \neq p_{old}$(D)$H_0: p_{new} \neq p_{old}$; $H_a: p_{new} = p_{old}$

7. A school district wants to compare the proportion of high school students who drive to school in District North ($p_N$) versus District South ($p_S$). They plan to test the hypothesis $H_0: p_N - p_S = 0$ against $H_a: p_N - p_S \neq 0$. They select a random sample of 100 students from District North and 100 students from District South. Which of the following best describes the parameter of interest in this study?

(A)The difference in the number of students who drive to school in the two samples.(B)The difference in the proportion of students who drive to school between the two specific samples collected.(C)The true difference in the proportion of all students in District North and all students in District South who drive to school.(D)The probability that a randomly selected student from District North drives to school minus the probability that a randomly selected student from District South drives to school, assuming the null hypothesis is true.

8. Researchers are setting up a test to see if the proportion of defective parts differs between Machine 1 and Machine 2. They collect a random sample of 200 parts from Machine 1 and a separate random sample of 200 parts from Machine 2. The parts are produced in a factory that manufactures 5,000 parts per day per machine. Why is it important to verify the 10% condition ($n < 0.10N$) in this context?

(A)To ensure that the distribution of the difference in sample proportions is approximately Normal.(B)To ensure that the samples are independent of each other.(C)To ensure that the individual observations within each sample can be treated as independent, allowing for the calculation of the standard deviation.(D)To ensure that the sample sizes are large enough to detect a significant difference.

9. A city official wants to test the claim that the proportion of residents who support a new park initiative is higher in the West Side neighborhood than in the East Side neighborhood. Let $p_W$ represent the true support proportion in the West Side and $p_E$ represent the true support proportion in the East Side. A random sample of 150 residents from the West Side shows 85 supporters. A random sample of 120 residents from the East Side shows 60 supporters. Which of the following is the correct setup for the test statistic denominator (standard error) assuming the null hypothesis $H_0: p_W = p_E$ is true?

(A)$\sqrt{\frac{0.567(1-0.567)}{150} + \frac{0.500(1-0.500)}{120}}$(B)$\sqrt{\frac{0.537(1-0.537)}{150} + \frac{0.537(1-0.537)}{120}}$(C)$\sqrt{\frac{85}{150} + \frac{60}{120}}$(D)$\sqrt{\frac{0.567}{150} + \frac{0.500}{120}}$

10. A marketing firm wants to test if a new advertisement design results in a higher click-through rate than the current design. They display the current design to a random sample of 100 users, resulting in 30 clicks. They display the new design to an independent random sample of 100 users, resulting in 40 clicks. Which of the following represents the correct calculation for the z-test statistic to test the null hypothesis $H_0: p_{new} = p_{current}$?

(A)$\frac{0.40 - 0.30}{\sqrt{\frac{0.40(0.60)}{100} + \frac{0.30(0.70)}{100}}}$(B)$\frac{0.40 - 0.30}{\sqrt{0.35(0.65)(\frac{1}{100} + \frac{1}{100})}}$(C)$\frac{0.40 - 0.30}{\frac{0.35(0.65)}{\sqrt{200}}}$(D)$\frac{0.40 - 0.30}{\sqrt{0.35(0.65)(\frac{1}{200})}}$

11. A school district is comparing the proportion of high school students who play a musical instrument in School A versus School B. A significance test is performed with the null hypothesis $H_0: p_A = p_B$ and the alternative $H_a: p_A \neq p_B$. The resulting p-value is 0.034. Which of the following is the correct interpretation of this p-value?

(A)There is a 0.034 probability that the proportion of students who play an instrument is the same in both schools.(B)There is a 0.034 probability that the alternative hypothesis is true given the observed sample data.(C)Assuming the true proportion of students playing an instrument is the same in both schools, there is a 0.034 probability of observing a difference in sample proportions as large as or larger than the one observed.(D)Assuming the true proportion of students playing an instrument is the same in both schools, there is a 0.966 probability that the observed difference is due to random chance.

12. Researchers conducted a two-sample z-test to determine if there is a significant difference in the recovery rates of two different physical therapy treatments. The test yielded a z-statistic of $z = 2.05$. If the researchers are testing the alternative hypothesis $H_a: p_1 \neq p_2$ at a significance level of $\alpha = 0.05$, which of the following describes the correct calculation of the p-value and the appropriate conclusion?

(A)p-value = $P(Z > 2.05) \approx 0.0202$; since $0.0202 < 0.05$, reject $H_0$.(B)p-value = $2 \cdot P(Z > 2.05) \approx 0.0404$; since $0.0404 < 0.05$, reject $H_0$.(C)p-value = $2 \cdot P(Z > 2.05) \approx 0.0404$; since $0.0404 < 0.05$, fail to reject $H_0$.(D)p-value = $P(Z < 2.05) \approx 0.9798$; since $0.9798 > 0.05$, fail to reject $H_0$.

13. In a test of $H_0: p_1 = p_2$ versus $H_a: p_1 > p_2$, why is the combined (pooled) sample proportion $\hat{p}_c$ used to calculate the standard error of the test statistic, rather than the individual sample proportions $\hat{p}_1$ and $\hat{p}_2$?

(A)Because the sample sizes $n_1$ and $n_2$ are assumed to be equal in the calculation.(B)Because the null hypothesis assumes that the two population proportions are equal, so the best estimate for the common population proportion comes from combining the data.(C)Because using the pooled proportion guarantees that the sampling distribution will be exactly Normal regardless of sample size.(D)Because the pooled proportion provides a larger standard error, which reduces the Type I error rate.

14. A botanist claims that a new fertilizer increases the proportion of seeds that germinate compared to a standard fertilizer. A study yields a p-value of 0.003. Based on this p-value and a significance level of $\alpha = 0.01$, which of the following is the best justification for the conclusion?

(A)Since $0.003 < 0.01$, there is convincing evidence that the new fertilizer increases the germination proportion.(B)Since $0.003 < 0.01$, the null hypothesis is proven to be false, confirming the botanist's claim.(C)Since $0.003 < 0.01$, there is not convincing evidence that the new fertilizer increases the germination proportion.(D)Since $0.003 > 0$, we accept the null hypothesis and conclude the fertilizers are equally effective.

15. A marketing firm wants to estimate the proportion of customers who are satisfied with a new product. They select a simple random sample of 40 customers from a database of 2,500 customers. In the sample, 3 customers indicate they are satisfied. Which of the following conditions for constructing a one-sample z-interval for a population proportion has NOT been met?

(A)The Random condition(B)The 10% condition(C)The Large Counts condition(D)All conditions for inference have been met.

16. A political pollster wants to estimate the proportion of voters in a large city who support a specific ballot initiative. The pollster wants to construct a 95% confidence interval with a margin of error of at most 0.04. Assuming a conservative estimate where $p = 0.5$, which of the following is the minimum sample size required?

(A)423(B)601(C)1041(D)2401

17. In a random sample of 200 high school students, 120 stated that they have a part-time job. Which of the following represents the 99% confidence interval for the proportion of all high school students who have a part-time job?

(A)0.60 $\pm $2.576 $\sqrt{\frac{0.60(0.40)}{200}}$(B)0.60 $\pm $1.96 $\sqrt{\frac{0.60(0.40)}{200}}$(C)0.60 $\pm $2.576 $\frac{0.60(0.40)}{\sqrt{200}}$(D)120 $\pm $2.576 $\sqrt{\frac{120(80)}{200}}$

18. A 95% confidence interval for the proportion of adults in a certain region who use a specific social media platform is calculated to be (0.64, 0.72). Which of the following is the correct interpretation of the confidence level?

(A)There is a 95% probability that the sample proportion of adults who use the platform is between 0.64 and 0.72.(B)There is a 95% probability that the true population proportion of adults who use the platform is between 0.64 and 0.72.(C)If we were to take many random samples of the same size and calculate a 95% confidence interval for each, about 95% of those intervals would capture the true population proportion.(D)We are 95% confident that the true population proportion lies exactly in the middle of the interval (0.64, 0.72).

19. A researcher constructs a 90% confidence interval for a population proportion using a sample size of $n = 100$. If the researcher wishes to reduce the margin of error by half while maintaining the same 90% confidence level and assuming the sample proportion remains the same, the new sample size should be closest to which of the following?

(A)50(B)141(C)200(D)400

20. A city council member claims that more than 60% of the city's residents support a new zoning ordinance. To test this claim, a random sample of 200 residents is selected, and a 95% confidence interval for the proportion of residents who support the ordinance is calculated to be $(0.58, 0.72)$. Based on this interval, which of the following is the correct conclusion regarding the council member's claim?

(A)The claim is supported because the interval contains values greater than 0.60.(B)The claim is supported because the center of the interval, 0.65, is greater than 0.60.(C)The claim is not supported because the interval contains the value 0.60 and values less than 0.60.(D)The claim is not supported because the interval does not include 100% of the possible population proportions.

21. A marketing firm wants to estimate the proportion of consumers who recognize a new brand logo. They conduct a pilot study and calculate a 90% confidence interval with a margin of error of 0.08. The firm wants to reduce the margin of error to 0.04 while maintaining the same 90% level of confidence. Which of the following describes the necessary change to the sample size, assuming the sample proportion remains roughly the same?

(A)The sample size should be multiplied by 2.(B)The sample size should be divided by 2.(C)The sample size should be multiplied by 4.(D)The sample size should be divided by 4.

22. A statistics student calculated a 95% confidence interval for the proportion of students at a large university who commute by bicycle to be $(0.12, 0.18)$. Which of the following is the correct interpretation of the confidence level?

(A)There is a 0.95 probability that the true proportion of students who commute by bicycle is between 0.12 and 0.18.(B)If this sampling procedure were repeated many times, approximately 95% of the resulting confidence intervals would contain the true proportion of students who commute by bicycle.(C)95% of all random samples of students from this university will yield a sample proportion between 0.12 and 0.18.(D)We are 95% confident that the sample proportion of students who commute by bicycle falls between 0.12 and 0.18.

23. A quality control engineer inspects a random sample of component parts to estimate the proportion of parts that are defective. Using the same sample data, the engineer calculates both a 90% confidence interval and a 99% confidence interval for the true proportion of defective parts. Which of the following compares the two intervals correctly?

(A)The 99% confidence interval will be narrower than the 90% confidence interval because the higher confidence reduces the variability.(B)The 99% confidence interval will be wider than the 90% confidence interval because a wider range of values is needed to have higher confidence that the parameter is captured.(C)The widths of the two intervals will be the same because the sample size and sample proportion are unchanged.(D)The 99% confidence interval will have a smaller margin of error than the 90% confidence interval.

24. A newspaper article reports that 'only 30% of local voters favor the proposed tax increase.' A community group believes the support is actually higher. They survey a random sample of voters and calculate a 95% confidence interval for the proportion of voters who favor the tax increase to be $(0.32, 0.41)$. Does the confidence interval provide convincing evidence against the newspaper's report?

(A)No, because the interval is fairly wide, indicating high variability in the sample.(B)No, because 0.30 is close to the lower bound of 0.32.(C)Yes, because the center of the interval, 0.365, is higher than 0.30.(D)Yes, because all values in the confidence interval are greater than 0.30.

25. [Skill: 1.F | Topic: 6.4] A university administrator claims that more than 75% of the student body approves of the new dining hall meal plan. To investigate this claim, a random sample of 200 students is selected, and 160 of them express approval. Which of the following pairs of hypotheses is appropriate for this test?

(A)H₀: p = 0.75 and Hₐ: p > 0.75(B)H₀: p = 0.75 and Hₐ: p ≠ 0.75(C)H₀: p = 0.80 and Hₐ: p > 0.80(D)H₀: p̂ = 0.80 and Hₐ: p̂ > 0.80

26. [Skill: 4.C | Topic: 6.4] A manufacturer claims that 90% of its LED light bulbs last longer than 10,000 hours. A quality control manager wants to test this claim to see if the true proportion is actually lower. A random sample of 50 bulbs is selected, and 42 of them last longer than 10,000 hours. The manager intends to conduct a one-sample z-test for a population proportion. Which of the following describes the status of the Large Counts condition for this test?

(A)The condition is satisfied because n = 50, which is greater than 30.(B)The condition is satisfied because the number of successes is 42 and the number of failures is 8, and both are close enough to the expected values.(C)The condition is not satisfied because np̂ = 42 and n(1 - p̂) = 8; since 8 is less than 10, the sampling distribution may not be approximately normal.(D)The condition is not satisfied because np₀ = 45 and n(1 - p₀) = 5; since 5 is less than 10, the sampling distribution of the sample proportion cannot be assumed to be approximately normal.

27. [Skill: 1.E | Topic: 6.4] A sociologist wants to determine if the proportion of adults in a city who commute by bicycle differs from the national average of 0.04. She collects a random sample of 500 adults from the city and finds that 28 of them commute by bicycle. Which of the following is the most appropriate inference procedure for this study?

(A)One-sample z-test for a population proportion(B)Two-sample z-test for the difference in population proportions(C)One-sample t-test for a population mean(D)Chi-square goodness-of-fit test

28. [Skill: 4.C | Topic: 6.4] A small private high school has a total student population of 300 students. The principal wants to test the hypothesis that more than 50% of the students plan to attend the upcoming prom. She selects a random sample of 40 students without replacement and asks them if they plan to attend. Which condition for performing a one-sample z-test for a proportion has not been met?

(A)The Random condition(B)The 10% condition (Independence)(C)The Large Counts condition(D)All conditions have been met.

29. [Skill: 1.F | Topic: 6.4] A researcher is studying a specific genetic trait found in 15% of the general population. The researcher believes that the prevalence of this trait is different in a specific remote community. Let p represent the true proportion of residents in the remote community with the trait. Which of the following represents the correct null and alternative hypotheses?

(A)H₀: p = 0.15 and Hₐ: p < 0.15(B)H₀: p = 0.15 and Hₐ: p > 0.15(C)H₀: p = 0.15 and Hₐ: p ≠ 0.15(D)H₀: p ≠ 0.15 and Hₐ: p = 0.15

30. A school administrator claims that 80% of students support a new schedule change. A student council member believes the true support is less than 80%. They survey a random sample of 100 students and find that 74% support the change. A significance test is performed, resulting in a p-value of 0.067. Which of the following is the correct interpretation of this p-value?

(A)There is a 0.067 probability that the null hypothesis is true.(B)There is a 0.067 probability that the alternative hypothesis is true.(C)Assuming the true proportion of students who support the change is 0.80, there is a 0.067 probability of observing a sample proportion of 0.74 or less.(D)Assuming the true proportion of students who support the change is 0.74, there is a 0.067 probability of observing a sample proportion of 0.80 or more.

31. A manufacturing company claims that 95% of its components are defect-free. A quality control manager believes the true percentage is different. A random sample of 200 components is inspected, and 184 are found to be defect-free. Which of the following is the correct test statistic for a one-sample z-test for a proportion?

(A)$z = \frac{0.92 - 0.95}{\sqrt{\frac{0.92(0.08)}{200}}}$(B)$z = \frac{0.92 - 0.95}{\sqrt{\frac{0.95(0.05)}{200}}}$(C)$z = \frac{0.95 - 0.92}{\sqrt{\frac{0.95(0.05)}{200}}}$(D)$z = \frac{0.92 - 0.95}{\frac{0.95(0.05)}{\sqrt{200}}}$

32. Researchers are conducting a two-tailed hypothesis test for a population proportion with $H_0: p = 0.45$ and $H_a: p \neq 0.45$. The calculated test statistic is $z = 2.15$. Which of the following expressions correctly represents the p-value for this test?

(A)$P(Z > 2.15)$(B)$P(Z < 2.15)$(C)$1 - P(Z < 2.15)$(D)$2 \times P(Z > 2.15)$

33. A nutritionist claims that more than 30% of adults consume the recommended daily amount of vegetables. A hypothesis test is conducted with $H_0: p = 0.30$ and $H_a: p > 0.30$ at a significance level of $\alpha = 0.05$. The resulting p-value is 0.035. Which of the following is the correct conclusion?

(A)Fail to reject $H_0$. There is not convincing evidence that more than 30% of adults consume the recommended amount.(B)Fail to reject $H_0$. There is convincing evidence that more than 30% of adults consume the recommended amount.(C)Reject $H_0$. There is convincing evidence that more than 30% of adults consume the recommended amount.(D)Reject $H_0$. There is not convincing evidence that more than 30% of adults consume the recommended amount.

34. A statistics student wants to test if a coin is biased towards heads ($H_0: p = 0.5$ vs $H_a: p > 0.5$). She flips the coin 50 times and gets 30 heads ($\hat{p} = 0.60$). To calculate the p-value, she performs a simulation by flipping a fair coin 50 times, recording the proportion of heads, and repeating this process 100 times. The dotplot of her simulation results shows 4 dots at 0.60 or higher. Based on this simulation, what is the estimated p-value and the appropriate conclusion at $\alpha = 0.05$?

(A)p-value $\approx 0.04$; Reject $H_0$.(B)p-value $\approx 0.04$; Fail to reject $H_0$.(C)p-value $\approx 0.30$; Fail to reject $H_0$.(D)p-value $\approx 0.60$; Reject $H_0$.

35. A manufacturing company claims that at least 95% of the components it produces are defect-free. A quality control manager suspects that the actual proportion of defect-free components is less than 95%. The manager tests the hypotheses $H_0: p = 0.95$ versus $H_a: p < 0.95$ using a random sample of 200 components. The test yields a $P$-value of 0.032. Assuming a significance level of $\alpha = 0.05$, which of the following is the correct conclusion?

(A)Because $0.032 < 0.05$, the manager should fail to reject $H_0$. There is insufficient evidence to conclude that the proportion of defect-free components is less than 0.95.(B)Because $0.032 < 0.05$, the manager should reject $H_0$. There is convincing evidence that the proportion of defect-free components is less than 0.95.(C)Because $0.032 < 0.05$, the manager should reject $H_0$. There is convincing evidence that the proportion of defect-free components is equal to 0.95.(D)Because $0.032 < 0.05$, the manager should fail to reject $H_0$. There is convincing evidence that the proportion of defect-free components is 0.95 or greater.

36. A pollster claims that 40% of voters in a city support a specific ballot initiative. A local news organization believes the support level is different from 40%. They conduct a hypothesis test with $H_0: p = 0.40$ and $H_a: p \neq 0.40$. The results of the test yield a $P$-value of 0.18. Based on this result, which of the following statements is the most appropriate conclusion at the $\alpha = 0.05$ level?

(A)There is a 18% chance that the true proportion of voters who support the initiative is 40%.(B)The null hypothesis is proven true; exactly 40% of voters support the initiative.(C)There is insufficient evidence to conclude that the proportion of voters who support the initiative is different from 40%.(D)There is convincing evidence that the proportion of voters who support the initiative is different from 40%.

37. A researcher is conducting a one-proportion $z$-test to determine if a new drug causes a side effect in more than 10% of patients. The researcher sets the significance level at $\alpha = 0.01$. Which of the following best describes the meaning of $\alpha = 0.01$ in this context?

(A)There is a 0.01 probability that the null hypothesis is true.(B)There is a 0.01 probability of rejecting the null hypothesis when the true proportion of patients with side effects is actually 0.10.(C)There is a 0.01 probability of failing to reject the null hypothesis when the true proportion of patients with side effects is actually greater than 0.10.(D)There is a 0.99 probability that the alternative hypothesis is true if the null hypothesis is rejected.

38. A marketing firm tests the claim that more than 50% of consumers recognize a new brand logo. They test $H_0: p = 0.50$ versus $H_a: p > 0.50$ and calculate a $P$-value of 0.042. Which of the following justifies the correct conclusion based on the significance level $\alpha$?

(A)If $\alpha = 0.05$, the firm should reject $H_0$ and conclude there is convincing evidence that more than 50% of consumers recognize the logo.(B)If $\alpha = 0.05$, the firm should fail to reject $H_0$ and conclude there is insufficient evidence that more than 50% of consumers recognize the logo.(C)If $\alpha = 0.01$, the firm should reject $H_0$ and conclude there is convincing evidence that more than 50% of consumers recognize the logo.(D)If $\alpha = 0.10$, the firm should fail to reject $H_0$ and conclude there is insufficient evidence that more than 50% of consumers recognize the logo.

39. A recent study on internet usage claimed that 85% of teenagers use social media daily. A sociologist believes the actual proportion is lower. After collecting data from a random sample of teenagers, the sociologist calculates a standardized test statistic of $z = -1.15$ for the hypotheses $H_0: p = 0.85$ versus $H_a: p < 0.85$. Using the standard normal distribution, which of the following is the correct justification for the conclusion at the $\alpha = 0.05$ level?

(A)The $P$-value is approximately 0.125. Since $0.125 > 0.05$, there is convincing evidence that the proportion is less than 0.85.(B)The $P$-value is approximately 0.125. Since $0.125 > 0.05$, there is insufficient evidence to conclude that the proportion is less than 0.85.(C)The $P$-value is approximately 0.875. Since $0.875 > 0.05$, there is convincing evidence that the proportion is 0.85.(D)The $P$-value is approximately 0.05. Since $0.05 \le 0.05$, there is convincing evidence that the proportion is less than 0.85.

40. [Skill: 4.B | Topic: 6.7] A local school board claims that 60% of voters support a new budget proposal. A parent group believes the true support is less than 60%. They plan to conduct a hypothesis test at the significance level of $\alpha = 0.05$ using the hypotheses $H_0: p = 0.60$ and $H_a: p < 0.60$, where $p$ is the true proportion of voters who support the proposal. Which of the following describes a Type I error in this context?

(A)The parent group concludes that the support is less than 60% when, in reality, 60% of voters support the proposal.(B)The parent group concludes that the support is 60% when, in reality, less than 60% of voters support the proposal.(C)The parent group concludes that the support is less than 60% when, in reality, less than 60% of voters support the proposal.(D)The parent group concludes that the support is 60% when, in reality, 60% of voters support the proposal.

41. [Skill: 3.A | Topic: 6.7] A pharmaceutical company is testing a new drug to see if it causes side effects in more than 5% of patients. They test the hypotheses $H_0: p = 0.05$ versus $H_a: p > 0.05$ at a significance level of $\alpha = 0.01$. If the true proportion of patients who experience side effects is actually $p = 0.08$, the power of the test is calculated to be 0.64. What is the probability of making a Type II error?

(A)0.01(B)0.36(C)0.64(D)0.99

42. [Skill: 4.E | Topic: 6.7] A quality control manager at a factory performs a significance test to detect if the proportion of defective items produced by a machine is greater than the acceptable limit of 0.02. The manager wants to increase the power of the test to detect a shift to a defective rate of 0.04. Which of the following changes to the test design would achieve this goal?

(A)Decrease the significance level ($\alpha$) from 0.05 to 0.01.(B)Decrease the sample size used in the test.(C)Increase the sample size used in the test.(D)Test against an alternative hypothesis value closer to the null hypothesis (e.g., shift to 0.025 instead of 0.04).

43. [Skill: 4.B | Topic: 6.7] A city health inspector is testing water samples from a public swimming pool. The null hypothesis states that the water is safe ($p \le 0.01$, where $p$ is the proportion of bacteria), and the alternative hypothesis states that the water is unsafe ($p > 0.01$). If the null hypothesis is rejected, the pool will be closed for cleaning. Which of the following describes the consequence of a Type II error in this setting?

(A)The pool is closed for cleaning when the water is actually safe, causing unnecessary financial loss.(B)The pool remains open when the water is actually safe, causing no issues.(C)The pool remains open when the water is actually unsafe, potentially causing swimmers to get sick.(D)The pool is closed for cleaning when the water is actually unsafe, protecting the public health.

44. [Skill: 4.E | Topic: 6.7] A researcher is conducting a one-sample z-test for a proportion. If the researcher decides to increase the significance level ($\alpha$) of the test from 0.05 to 0.10, which of the following correctly describes the impact on the probabilities of Type I and Type II errors?

(A)The probability of a Type I error decreases, and the probability of a Type II error increases.(B)The probability of a Type I error increases, and the probability of a Type II error decreases.(C)Both the probability of a Type I error and the probability of a Type II error increase.(D)Both the probability of a Type I error and the probability of a Type II error decrease.

45. A university administrator wants to compare the proportion of students who support a new campus housing policy at the main campus versus the satellite campus. The administrator takes a random sample of 200 students from the main campus and a separate random sample of 150 students from the satellite campus. Which of the following is the most appropriate procedure for analyzing the difference in support for the policy between the two campuses?

(A)A one-sample z-interval for a proportion(B)A two-sample z-interval for a difference between proportions(C)A two-sample t-interval for a difference between means(D)A matched pairs t-interval for a mean difference

46. A biologist is studying the effect of a fungus on two different species of maple trees. In a random sample of 80 red maple trees, 20 exhibit signs of the fungus. In a random sample of 100 sugar maple trees, 15 exhibit signs of the fungus. The biologist intends to construct a 95% confidence interval for the difference in the proportion of trees infected ($p_R - p_S$). Which of the following represents the correct verification of the condition necessary to assume the sampling distribution of the difference in sample proportions is approximately normal?

(A)Both sample sizes are greater than 30 ($80 > 30$ and $100 > 30$).(B)$80(0.25) \geq 10$, $80(0.75) \geq 10$, $100(0.15) \geq 10$, and $100(0.85) \geq 10$.(C)The samples are chosen randomly and independently from the populations.(D)The combined sample proportion $\hat{p}_c = \frac{20+15}{80+100}$ satisfies $180(\hat{p}_c) \geq 10$ and $180(1-\hat{p}_c) \geq 10$.

47. A poll was conducted to estimate the difference in the proportion of voters who support a tax increase in City A versus City B. In City A, 450 out of 1,000 randomly selected voters supported the increase. In City B, 300 out of 800 randomly selected voters supported the increase. Which of the following expressions represents the 99% confidence interval for the difference in the proportion of voters supporting the tax increase ($p_A - p_B$)?

(A)$(0.45 - 0.375) \pm 2.576 \sqrt{\frac{0.45(0.55)}{1000} + \frac{0.375(0.625)}{800}}$(B)$(0.45 - 0.375) \pm 2.576 \sqrt{\frac{0.45(0.55)}{1000} - \frac{0.375(0.625)}{800}}$(C)$(0.45 - 0.375) \pm 2.326 \sqrt{\frac{0.417(0.583)}{1000} + \frac{0.417(0.583)}{800}}$(D)$(0.45 - 0.375) \pm 2.576 \left( \frac{0.45}{\sqrt{1000}} + \frac{0.375}{\sqrt{800}} \right)$

48. A 95% confidence interval for the difference in the proportion of customers who prefer Brand X over Brand Y ($p_X - p_Y$) is calculated to be $(-0.04, 0.12)$. Which of the following is the correct interpretation of this interval?

(A)We are 95% confident that the true difference in the proportion of customers who prefer Brand X over Brand Y is between -0.04 and 0.12.(B)There is a 95% probability that the sample difference in proportions falls between -0.04 and 0.12.(C)We are 95% confident that the proportion of customers who prefer Brand X is 0.08 higher than the proportion who prefer Brand Y.(D)Since the interval contains 0, there is a 95% chance that the proportions are equal.

49. Researchers constructed a 90% confidence interval for the difference between the proportion of teenagers ($p_T$) and the proportion of adults ($p_A$) who use a specific social media platform daily. The resulting interval was $(0.15, 0.25)$. Based on this interval, which of the following conclusions is supported?

(A)Because the interval contains only positive values, there is convincing evidence that the proportion of teenagers who use the platform daily is greater than the proportion of adults.(B)Because the interval contains only positive values, there is convincing evidence that the proportion of adults who use the platform daily is greater than the proportion of teenagers.(C)Because the center of the interval is 0.20, we can conclude that exactly 20% more teenagers use the platform than adults.(D)There is no convincing evidence of a difference because the interval width is only 0.10.

50. A statistics student constructs a 95% confidence interval for the difference in the proportion of students who pack their lunch at High School A versus High School B ($p_A - p_B$). The resulting interval is $(0.04, 0.12)$. Which of the following is the correct interpretation of the confidence level?

(A)There is a 95% probability that the true difference in the proportion of all students who pack their lunch at High School A and High School B is between 0.04 and 0.12.(B)In repeated random sampling from the same populations, approximately 95% of the confidence intervals constructed will capture the true difference in population proportions of students who pack their lunch.(C)We are 95% confident that the difference in the sample proportions of students who pack their lunch at High School A and High School B is between 0.04 and 0.12.(D)In repeated random sampling from the same populations, approximately 95% of the sample differences will fall between 0.04 and 0.12.

51. A marketing firm wants to estimate the difference in the proportion of customers who prefer Brand X over Brand Y in two different regions, North and South. Let $p_N$ represent the proportion of all customers in the North who prefer Brand X, and $p_S$ represent the proportion of all customers in the South who prefer Brand X. A 90% confidence interval for the difference $p_N - p_S$ is calculated to be $(-0.15, -0.05)$. Which of the following is the correct interpretation of this interval?

(A)We are 90% confident that the proportion of customers in the North who prefer Brand X is between 0.05 and 0.15 lower than the proportion of customers in the South who prefer Brand X.(B)We are 90% confident that the proportion of customers in the North who prefer Brand X is between 0.05 and 0.15 higher than the proportion of customers in the South who prefer Brand X.(C)We are 90% confident that the difference in sample proportions of customers who prefer Brand X between the North and South regions is between -0.15 and -0.05.(D)There is a 90% chance that the difference in population proportions is equal to -0.10.

52. A botanist is investigating the germination rates of two different types of seeds, Type A and Type B. Let $p_A$ and $p_B$ represent the true germination proportions for Type A and Type B seeds, respectively. A 95% confidence interval for the difference $p_A - p_B$ is found to be $(-0.03, 0.09)$. Based on this interval, which of the following conclusions is supported?

(A)There is convincing evidence that the germination rate of Type A seeds is greater than that of Type B seeds because the center of the interval is positive.(B)There is convincing evidence that the germination rate of Type A seeds is different from that of Type B seeds because the interval includes positive values.(C)There is not convincing evidence of a difference in the germination rates because the interval contains 0.(D)There is not convincing evidence of a difference because the interval width is small.

53. A city official claims that the proportion of residents who support a new recycling program is higher in District 1 than in District 2. A random sample is taken from each district, and a 99% confidence interval for the difference in population proportions ($p_1 - p_2$) is calculated to be $(0.02, 0.14)$. Does the confidence interval support the official's claim?

(A)Yes, because the interval contains 0.02, which represents a 2% difference.(B)Yes, because the entire interval consists of positive values, suggesting that $p_1$ is greater than $p_2$.(C)No, because the interval does not contain 0.(D)No, because the interval suggests the difference could be as small as 0.02, which is practically insignificant.

54. Researchers are comparing the effectiveness of two pain relief medications, Drug A and Drug B. They construct a 95% confidence interval for the difference in the proportion of patients reporting relief ($p_A - p_B$) and obtain the interval $(-0.12, -0.04)$. Which of the following statements is justified based on this interval?

(A)There is convincing evidence that Drug A is less effective than Drug B because the entire interval is negative.(B)There is convincing evidence that Drug A is more effective than Drug B because the interval does not contain 0.(C)The researchers can be 95% confident that the proportion of patients finding relief with Drug A is equal to the proportion finding relief with Drug B.(D)The results are inconclusive because the interval values are close to 0.

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