College Statistics Quiz: Z Test For A Proportion
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Z Test For A ProportionQuestion 1 of 20

A poll of 500 likely voters was conducted to test if a candidate's support was significantly greater than 50% (H0:p=0.5,Ha:p>0.5H_0: p=0.5, H_a: p>0.5). The poll found 268 voters in support, resulting in a p-value of 0.054. While analyzing the results, one additional response from a supporter is recorded. How does this single data point change the p-value and the conclusion at the α=0.05\alpha=0.05 level?

The p-value increases slightly, and the conclusion to fail to reject H0H_0 is strengthened.
The p-value decreases to just below 0.05, changing the conclusion from fail to reject to reject H0H_0.
The p-value decreases, but remains above 0.05, so the conclusion to fail to reject H0H_0 does not change.
The change is negligible and will not impact the p-value or the conclusion in any meaningful way.
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College Statistics Quiz

College Statistics Quiz: Z Test For A Proportion

Practice Z Test For A Proportion in College Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

A poll of 500 likely voters was conducted to test if a candidate's support was significantly greater than 50% (H0:p=0.5,Ha:p>0.5H_0: p=0.5, H_a: p>0.5). The poll found 268 voters in support, resulting in a p-value of 0.054. While analyzing the results, one additional response from a supporter is recorded. How does this single data point change the p-value and the conclusion at the α=0.05\alpha=0.05 level?

  1. The p-value increases slightly, and the conclusion to fail to reject H0H_0 is strengthened.
  2. The p-value decreases to just below 0.05, changing the conclusion from fail to reject to reject H0H_0. (correct answer)
  3. The p-value decreases, but remains above 0.05, so the conclusion to fail to reject H0H_0 does not change.
  4. The change is negligible and will not impact the p-value or the conclusion in any meaningful way.
Explanation: Initially, with n=500n=500 and x=268x=268, p^=0.536\hat{p}=0.536. The p-value is 0.054, which is greater than α=0.05\alpha=0.05, so the initial conclusion is to fail to reject H0H_0. With the new data point, n=501n=501 and x=269x=269. The new sample proportion is p^new=269/501≈0.5369\hat{p}_{new} = 269/501 \approx 0.5369. The new standard error is SEnew=0.5(0.5)/501≈0.02234SE_{new} = \sqrt{0.5(0.5)/501} \approx 0.02234. The new z-statistic is znew=(0.5369−0.5)/0.02234≈1.65z_{new} = (0.5369 - 0.5) / 0.02234 \approx 1.65. The new p-value is P(Z>1.65)≈0.0495P(Z > 1.65) \approx 0.0495. Since the new p-value (0.0495) is now less than α=0.05\alpha = 0.05, the conclusion changes from failing to reject to rejecting the null hypothesis.

Question 2

A poll of 500 likely voters was conducted to test if a candidate's support was significantly greater than 50% (H0:p=0.5,Ha:p>0.5H_0: p=0.5, H_a: p>0.5). The poll found 268 voters in support, resulting in a p-value of 0.054. While analyzing the results, one additional response from a supporter is recorded. How does this single data point change the p-value and the conclusion at the α=0.05\alpha=0.05 level?

  1. The p-value increases slightly, and the conclusion to fail to reject H0H_0 is strengthened.
  2. The p-value decreases to just below 0.05, changing the conclusion from fail to reject to reject H0H_0. (correct answer)
  3. The p-value decreases, but remains above 0.05, so the conclusion to fail to reject H0H_0 does not change.
  4. The change is negligible and will not impact the p-value or the conclusion in any meaningful way.
Explanation: Initially, with n=500n=500 and x=268x=268, p^=0.536\hat{p}=0.536. The p-value is 0.054, which is greater than α=0.05\alpha=0.05, so the initial conclusion is to fail to reject H0H_0. With the new data point, n=501n=501 and x=269x=269. The new sample proportion is p^new=269/501≈0.5369\hat{p}_{new} = 269/501 \approx 0.5369. The new standard error is SEnew=0.5(0.5)/501≈0.02234SE_{new} = \sqrt{0.5(0.5)/501} \approx 0.02234. The new z-statistic is znew=(0.5369−0.5)/0.02234≈1.65z_{new} = (0.5369 - 0.5) / 0.02234 \approx 1.65. The new p-value is P(Z>1.65)≈0.0495P(Z > 1.65) \approx 0.0495. Since the new p-value (0.0495) is now less than α=0.05\alpha = 0.05, the conclusion changes from failing to reject to rejecting the null hypothesis.

Question 3

A pharmaceutical company is testing a new drug. The drug will be deemed successful and sent for regulatory approval if it is effective in more than 65% of cases. The company establishes a hypothesis test with H0:p=0.65H_0: p = 0.65 versus Ha:p>0.65H_a: p > 0.65, where pp is the true proportion of effective cases. In this context, what are the consequences of making a Type II error?

  1. The company concludes the drug is successful when it is not, potentially leading to a harmful or ineffective product on the market.
  2. The company concludes the drug is not successful when it actually is, missing a potentially valuable business and medical opportunity. (correct answer)
  3. The company correctly concludes the drug is successful, leading to a beneficial new treatment and increased revenue.
  4. The company correctly concludes the drug is not successful, saving resources that would have been spent on a failed product.
Explanation: A Type II error occurs when one fails to reject a null hypothesis that is actually false. In this scenario, the null hypothesis is H0:p=0.65H_0: p = 0.65. The alternative is Ha:p>0.65H_a: p > 0.65. A Type II error means that the test fails to provide sufficient evidence that p>0.65p > 0.65 when, in reality, it is. The consequence is that the company would fail to recognize that their drug is successful (effective in more than 65% of cases) and would therefore not pursue its approval, missing a valuable opportunity. A Type I error, described in choice (A), would be to conclude the drug is successful when it is not.

Question 4

A researcher is testing H0:p=0.5H_0: p = 0.5 against Ha:p≠0.5H_a: p \neq 0.5. From a sample of 64 individuals, 38 respond positively. The researcher calculates the standard error for the test using the formula SE=p^(1−p^)nSE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, where p^=38/64\hat{p} = 38/64. Which of the following statements correctly identifies the error in the researcher's method?

  1. The alternative hypothesis should have been one-sided, as it is more powerful.
  2. The sample size is too small to use a z-test because n<100n < 100.
  3. The standard error for a hypothesis test should be calculated using the hypothesized proportion p0p_0. (correct answer)
  4. The Large Counts Condition is not met, so the test is invalid.
Explanation: When conducting a hypothesis test for a proportion, the test statistic is calculated under the assumption that the null hypothesis is true. Therefore, the standard error of the sampling distribution of p^\hat{p} should be calculated using the hypothesized value p0p_0, not the sample value p^\hat{p}. The correct formula is SE=p0(1−p0)nSE = \sqrt{\frac{p_0(1-p_0)}{n}}. The formula using p^\hat{p} is appropriate for constructing a confidence interval. (A) is a judgment call, not a procedural error. (B) is incorrect, there is no rule that n must be at least 100. (D) is incorrect; the condition is np0=64(0.5)=32≥10np_0 = 64(0.5) = 32 \geq 10 and n(1−p0)=64(0.5)=32≥10n(1-p_0) = 64(0.5) = 32 \geq 10, which is met.

Question 5

A political analyst constructs a 99% confidence interval for the proportion of voters favoring a certain policy, which is found to be (0.52, 0.68). Using the same sample data, the analyst also wants to perform a hypothesis test with H0:p=0.50H_0: p = 0.50 versus Ha:p≠0.50H_a: p \neq 0.50. What is the appropriate decision for this test at the α=0.01\alpha = 0.01 significance level?

  1. Reject H0H_0, because the hypothesized value of 0.50 falls outside the 99% confidence interval. (correct answer)
  2. Fail to reject H0H_0, because the sample proportion is not exactly 0.50.
  3. Reject H0H_0, because the entire confidence interval is above the hypothesized value of 0.50.
  4. The decision cannot be made without knowing the sample size and the exact sample proportion.
Explanation: For a two-sided hypothesis test, a confidence interval can be used to make a decision. A CC% confidence interval corresponds to a significance level of α=1−C/100\alpha = 1 - C/100. Here, a 99% confidence interval corresponds to an α=0.01\alpha = 0.01 significance level. The rule is: if the hypothesized value p0p_0 falls outside the confidence interval, we reject H0H_0. Since p0=0.50p_0 = 0.50 is not within the interval (0.52, 0.68), the null hypothesis should be rejected at the 0.01 level. Choice (C) gives the correct reason for the rejection specified in (A). (D) is incorrect because the confidence interval provides sufficient information.

Question 6

A marketing agency is testing a new advertisement. They will launch the ad campaign if they have evidence that it will be effective for more than 30% of the target audience. They conduct a hypothesis test, H0:p=0.30H_0: p = 0.30 vs. Ha:p>0.30H_a: p > 0.30. Which of the following scenarios would provide the greatest power to detect a true effectiveness of p=0.35p = 0.35?

  1. Testing at a significance level of α=0.01\alpha = 0.01 with a sample size of n=500n=500.
  2. Testing at a significance level of α=0.05\alpha = 0.05 with a sample size of n=500n=500.
  3. Testing at a significance level of α=0.01\alpha = 0.01 with a sample size of n=1000n=1000.
  4. Testing at a significance level of α=0.05\alpha = 0.05 with a sample size of n=1000n=1000. (correct answer)
Explanation: The power of a hypothesis test is the probability of correctly rejecting a false null hypothesis. Power is increased by (1) increasing the sample size (n), (2) increasing the significance level (α\alpha), and (3) increasing the effect size (the distance between the true parameter and the hypothesized parameter). Comparing the four options, we want the largest nn and the largest α\alpha. Both (C) and (D) have the larger sample size (n=1000n=1000). Between these two, (D) has the larger significance level (α=0.05\alpha=0.05 vs. α=0.01\alpha=0.01). Therefore, the combination of the largest sample size and the largest significance level will yield the greatest power.

Question 7

A researcher performs a one-proportion z-test with H0:p=0.65H_0: p = 0.65 using data from a sample of size n=100n=100. The resulting test statistic is z1z_1. A colleague suggests that the null hypothesis should have been H0:p=0.70H_0: p = 0.70. Using the same sample data, the researcher calculates a new test statistic, z2z_2. If the sample proportion p^\hat{p} was 0.72, how would z2z_2 compare to z1z_1?

  1. z2z_2 would be smaller than z1z_1 because the numerator of the test statistic decreases while the standard error increases.
  2. z2z_2 would be smaller than z1z_1 because the numerator of the test statistic decreases while the standard error decreases. (correct answer)
  3. z2z_2 would be larger than z1z_1 because the numerator of the test statistic increases while the standard error decreases.
  4. z2z_2 would be larger than z1z_1 because the numerator of the test statistic decreases while the standard error also decreases.
Explanation: Let's analyze the components of the z-statistic, z=(p^−p0)/SEz = (\hat{p} - p_0) / SE. The sample proportion is p^=0.72\hat{p}=0.72. For z1z_1, p0=0.65p_0=0.65. Numerator is 0.72−0.65=0.070.72 - 0.65 = 0.07. For z2z_2, p0=0.70p_0=0.70. Numerator is 0.72−0.70=0.020.72 - 0.70 = 0.02. The numerator decreases. Now, the standard error: SE=p0(1−p0)/nSE = \sqrt{p_0(1-p_0)/n}. For z1z_1, SE1=0.65(0.35)/100≈0.0477SE_1 = \sqrt{0.65(0.35)/100} \approx 0.0477. For z2z_2, SE2=0.70(0.30)/100≈0.0458SE_2 = \sqrt{0.70(0.30)/100} \approx 0.0458. The standard error also decreases (since 0.7 is further from 0.5 than 0.65 is). Since the numerator experiences a large relative decrease (from 0.07 to 0.02) and the denominator experiences a small relative decrease, the overall fraction z2z_2 will be smaller than z1z_1. Let's calculate: z1=0.07/0.0477≈1.47z_1 = 0.07/0.0477 \approx 1.47 and z2=0.02/0.0458≈0.44z_2 = 0.02/0.0458 \approx 0.44. Thus, z2z_2 is smaller.

Question 8

A large-scale study on consumer behavior finds that with a sample of 25,000 shoppers, a change in product placement led to a statistically significant increase in sales proportion (p-value < 0.001) from the historical baseline of 2.0%. The new sample proportion was 2.2%. Which of the following statements provides the most important critique of a manager's conclusion that the change was a major success?

  1. The result is not reliable because the p-value is extremely small, suggesting a calculation error.
  2. A Type I error may have occurred, meaning there was no actual increase in the sales proportion.
  3. The sample size is so large that a very small, perhaps practically unimportant, increase was detected as statistically significant. (correct answer)
  4. The study should have used a confidence interval instead of a hypothesis test to determine success.
Explanation: This question highlights the distinction between statistical significance and practical significance. With a very large sample size (n=25,000), even a tiny effect can be statistically significant (i.e., have a small p-value). The observed increase from 2.0% to 2.2% is a difference of only 0.2 percentage points. This might not be a large enough increase to be considered a 'major success' from a business perspective, especially if the change in product placement was costly. The most salient critique is that the statistical significance doesn't imply practical importance. (A) is incorrect. (B) is always a possibility but less specific to the problem presented. (D) is a matter of preference; both methods provide useful information.

Question 9

A z-test for the proportion of students who own a car was conducted under the hypotheses H0:p=0.35H_0: p = 0.35 and Ha:p≠0.35H_a: p \neq 0.35. The test, based on a sample of 250 students, yielded a test statistic of z=2.15z=2.15. Which of the following changes to the study would have resulted in a larger z-statistic?

  1. Changing the alternative hypothesis to Ha:p>0.35H_a: p > 0.35.
  2. Observing the same number of car owners but from a sample of 300 students.
  3. Changing the null hypothesis to H0:p=0.40H_0: p = 0.40, with the same sample results.
  4. Observing the same sample proportion but from a sample of 300 students. (correct answer)
Explanation: When analyzing what affects the magnitude of a z-statistic, you need to understand its formula: z=p^−p0p0(1−p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}, where p^\hat{p} is the sample proportion, p0p_0 is the hypothesized proportion, and nn is the sample size. A larger z-statistic results from either a larger numerator (bigger difference between sample and hypothesized proportions) or a smaller denominator (smaller standard error). Since the standard error decreases as sample size increases, option D is correct: keeping the same sample proportion while increasing the sample size from 250 to 300 would decrease the standard error, making the z-statistic larger. Option A is wrong because changing from a two-tailed to one-tailed test doesn't affect the z-statistic calculation—it only changes the critical value for decision-making. Option B is incorrect because if you observe the same number of car owners (say, 87) but increase the sample to 300, your sample proportion p^\hat{p} would decrease (from 87/250 = 0.348 to 87/300 = 0.29), reducing the numerator more than the denominator shrinks. Option C is incorrect because moving the null hypothesis value further from the likely sample proportion (if p^>0.35\hat{p} > 0.35) would actually decrease the numerator. Remember: for proportion z-tests, larger sample sizes with the same sample proportion always yield larger test statistics due to reduced sampling variability. Focus on how changes affect both the numerator and denominator of the z-formula.

Question 10

A one-proportion z-test is conducted for H0:p=p0H_0: p=p_0 versus Ha:p>p0H_a: p > p_0. The sample proportion p^\hat{p} was found to be less than p0p_0, resulting in a z-statistic of z=−1.5z = -1.5 and a p-value of pval≈0.933p_{val} \approx 0.933. If the test had been two-tailed (Ha:p≠p0H_a: p \neq p_0) using the same sample data, what would be the correct p-value?

  1. 2×pval≈1.8662 \times p_{val} \approx 1.866
  2. pval/2≈0.467p_{val} / 2 \approx 0.467
  3. 1−pval≈0.0671 - p_{val} \approx 0.067
  4. 2×(1−pval)≈0.1342 \times (1 - p_{val}) \approx 0.134 (correct answer)
Explanation: The initial test was right-tailed, so pval=P(Z>−1.5)≈0.933p_{val} = P(Z > -1.5) \approx 0.933. A two-tailed p-value is the probability of getting a result as extreme or more extreme than the observed result in either direction. This is calculated as 2×P(Z>∣zobs∣)2 \times P(Z > |z_{obs}|). Here, zobs=−1.5z_{obs} = -1.5, so ∣zobs∣=1.5|z_{obs}| = 1.5. The two-tailed p-value is 2×P(Z>1.5)2 \times P(Z > 1.5). Since P(Z>1.5)=1−P(Z<1.5)=1−0.933=0.067P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.933 = 0.067, the two-tailed p-value is 2×0.067=0.1342 \times 0.067 = 0.134. This can also be expressed as 2×(1−pval)2 \times (1 - p_{val}), since pvalp_{val} was the area to the right of -1.5.

Question 11

A city planner investigates if the proportion of residents who commute via public transit has changed from the 0.08 value recorded in a census ten years ago. A new survey of 100 randomly selected residents is planned. The planner will conduct a hypothesis test with H0:p=0.08H_0: p = 0.08 and Ha:p≠0.08H_a: p \neq 0.08. Which of the following is the most critical consequence if the assumptions for a one-proportion z-test are not met in this scenario?

  1. The test will likely result in a Type I error, leading the planner to a false conclusion.
  2. The sampling distribution of the sample proportion cannot be reliably approximated by a normal distribution. (correct answer)
  3. The sample size is too small to estimate the population proportion, regardless of the hypothesis test.
  4. The standard error of the sample proportion will be overestimated, making the test too conservative.
Explanation: The primary assumption for a one-proportion z-test is the Large Counts Condition, which states that np0np_0 and n(1−p0)n(1-p_0) must both be at least 10. In this case, np0=100(0.08)=8np_0 = 100(0.08) = 8. Since this is less than 10, the condition is not met. The consequence is that the sampling distribution of p^\hat{p} is not sufficiently close to a normal distribution, which invalidates the use of the z-statistic and the calculation of the p-value based on the normal model. (A) is not guaranteed; the test is simply invalid. (C) is too strong; estimation is possible, but this specific test is inappropriate. (D) is not necessarily true and is a secondary issue to the invalidity of the normal approximation.

Question 12

In a hypothesis test for a proportion, the z-test statistic was calculated to be z=−2.05z = -2.05 for the alternative hypothesis Ha:p<p0H_a: p < p_0. The sample size was n=400n=400 and the null hypothesis was H0:p=0.25H_0: p = 0.25. What was the sample proportion, p^\hat{p}, observed in the study?

  1. 0.199
  2. 0.207 (correct answer)
  3. 0.293
  4. 0.301
Explanation: The formula for the z-test statistic is z=p^−p0p0(1−p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}. We are given z=−2.05z = -2.05, p0=0.25p_0 = 0.25, and n=400n=400. First, calculate the standard error: SE=0.25(0.75)400=0.1875400=0.00046875≈0.02165SE = \sqrt{\frac{0.25(0.75)}{400}} = \sqrt{\frac{0.1875}{400}} = \sqrt{0.00046875} \approx 0.02165. Now, solve for p^\hat{p}: −2.05=p^−0.250.02165-2.05 = \frac{\hat{p} - 0.25}{0.02165}. This gives p^−0.25=−2.05×0.02165≈−0.04438\hat{p} - 0.25 = -2.05 \times 0.02165 \approx -0.04438. Finally, p^≈0.25−0.04438=0.20562\hat{p} \approx 0.25 - 0.04438 = 0.20562. The closest answer is 0.207. Distractor (A) is a miscalculation. Distractors (C) and (D) would result from an incorrectly positive z-statistic.

Question 13

A news organization reported that a study on a new law found that more than half of the population supported it, with a p-value of 0.02 for the corresponding hypothesis test. Which of the following is the correct interpretation of this p-value?

  1. There is a 2% chance that the null hypothesis of 50% support is true.
  2. There is a 98% probability that the proportion of the population supporting the law is greater than 50%.
  3. If the true proportion of support is 50%, the probability of observing a sample proportion as high or higher than the one obtained is 2%. (correct answer)
  4. The probability that the new law is actually effective and supported by the population is 98%.
Explanation: The p-value is the probability of observing a test statistic as extreme or more extreme than the one actually observed, assuming the null hypothesis is true. The null hypothesis here would be H0:p=0.50H_0: p = 0.50. The alternative is Ha:p>0.50H_a: p > 0.50. Therefore, the p-value of 0.02 means that if the true support level were exactly 50%, there would be a 2% chance of getting the observed sample result (or one even more favorable to the alternative hypothesis) just by random sampling variation. (A) and (B) are common misinterpretations that incorrectly assign a probability to the hypothesis itself. (D) is a vague and incorrect interpretation.

Question 14

Historically, 15% of customers at a coffee shop order a specialty latte. After a promotion, the manager takes a random sample of 150 customers and finds that 30 of them ordered a specialty latte. The manager performs a hypothesis test for H0:p=0.15H_0: p = 0.15 versus Ha:p>0.15H_a: p > 0.15 at the α=0.05\alpha = 0.05 significance level. Which of the following is the most appropriate conclusion?

  1. Reject H0H_0; there is significant evidence that the proportion of customers ordering lattes has increased. (correct answer)
  2. Fail to reject H0H_0; there is not significant evidence that the proportion of customers ordering lattes has increased.
  3. Reject H0H_0; the sample proportion of 20% is greater than 15%, so the promotion was effective.
  4. Fail to reject H0H_0; the test proves that the proportion of customers ordering lattes is still 15%.
Explanation: First, calculate the sample proportion: p^=30/150=0.20\hat{p} = 30/150 = 0.20. Next, calculate the z-test statistic: z=p^−p0p0(1−p0)n=0.20−0.150.15(0.85)150=0.050.00085≈0.050.02915≈1.715z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{0.20 - 0.15}{\sqrt{\frac{0.15(0.85)}{150}}} = \frac{0.05}{\sqrt{0.00085}} \approx \frac{0.05}{0.02915} \approx 1.715. The p-value for this right-tailed test is P(Z>1.715)P(Z > 1.715). Using a standard normal table, this value is approximately 0.043. Since the p-value (0.043) is less than the significance level (α=0.05\alpha = 0.05), we reject the null hypothesis. There is sufficient evidence to conclude the proportion has increased. (B) reaches the wrong conclusion. (C) makes the right decision but for the wrong reason; simply observing p^>p0\hat{p} > p_0 is not sufficient. (D) incorrectly 'proves' the null hypothesis.

Question 15

A state legislature is considering a bill. A political group wants to survey residents to test if the proportion who support the bill, pp, is less than the 40% claimed by the bill's sponsors. They plan to use a one-proportion z-test. What is the minimum sample size needed to satisfy the conditions for this test, assuming the sponsors' claim is true for the purposes of planning?

  1. 17
  2. 25 (correct answer)
  3. 30
  4. 40
Explanation: The conditions for a one-proportion z-test require that np0≥10np_0 \geq 10 and n(1−p0)≥10n(1-p_0) \geq 10. The hypotheses are H0:p=0.40H_0: p = 0.40 versus Ha:p<0.40H_a: p < 0.40. We use p0=0.40p_0 = 0.40 to check the conditions. We must satisfy both: 1) n(0.40)≥10⇒n≥10/0.40⇒n≥25n(0.40) \geq 10 \Rightarrow n \geq 10/0.40 \Rightarrow n \geq 25. 2) n(1−0.40)≥10⇒n(0.60)≥10⇒n≥10/0.60⇒n≥16.67n(1 - 0.40) \geq 10 \Rightarrow n(0.60) \geq 10 \Rightarrow n \geq 10/0.60 \Rightarrow n \geq 16.67. To satisfy both inequalities, the sample size nn must be at least 25. Therefore, the minimum required sample size is 25.

Question 16

A one-proportion z-test is conducted for H0:p=p0H_0: p=p_0 versus Ha:p>p0H_a: p > p_0. The sample proportion p^\hat{p} was found to be less than p0p_0, resulting in a z-statistic of z=−1.5z = -1.5 and a p-value of pval≈0.933p_{val} \approx 0.933. If the test had been two-tailed (Ha:p≠p0H_a: p \neq p_0) using the same sample data, what would be the correct p-value?

  1. 2×pval≈1.8662 \times p_{val} \approx 1.866
  2. pval/2≈0.467p_{val} / 2 \approx 0.467
  3. 1−pval≈0.0671 - p_{val} \approx 0.067
  4. 2×(1−pval)≈0.1342 \times (1 - p_{val}) \approx 0.134 (correct answer)
Explanation: The initial test was right-tailed, so pval=P(Z>−1.5)≈0.933p_{val} = P(Z > -1.5) \approx 0.933. A two-tailed p-value is the probability of getting a result as extreme or more extreme than the observed result in either direction. This is calculated as 2×P(Z>∣zobs∣)2 \times P(Z > |z_{obs}|). Here, zobs=−1.5z_{obs} = -1.5, so ∣zobs∣=1.5|z_{obs}| = 1.5. The two-tailed p-value is 2×P(Z>1.5)2 \times P(Z > 1.5). Since P(Z>1.5)=1−P(Z<1.5)=1−0.933=0.067P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.933 = 0.067, the two-tailed p-value is 2×0.067=0.1342 \times 0.067 = 0.134. This can also be expressed as 2×(1−pval)2 \times (1 - p_{val}), since pvalp_{val} was the area to the right of -1.5.

Question 17

A university claims that at least 75% of its students graduate in four years. A student journalist believes this rate is an exaggeration and plans a hypothesis test. Which of the following represents the correct null and alternative hypotheses for the journalist's investigation?

  1. H0:p=0.75H_0: p = 0.75 vs. Ha:p<0.75H_a: p < 0.75 (correct answer)
  2. H0:p=0.75H_0: p = 0.75 vs. Ha:p>0.75H_a: p > 0.75
  3. H0:p=0.75H_0: p = 0.75 vs. Ha:p≠0.75H_a: p \neq 0.75
  4. H0:p^=0.75H_0: \hat{p} = 0.75 vs. Ha:p^<0.75H_a: \hat{p} < 0.75
Explanation: The journalist is investigating if the graduation rate is an 'exaggeration,' meaning they suspect the true proportion pp is lower than the claimed 75%. This suspicion dictates the alternative hypothesis. Therefore, the alternative hypothesis is Ha:p<0.75H_a: p < 0.75. The null hypothesis is a statement of no effect or the status quo, which is the university's claim, so H0:p=0.75H_0: p = 0.75. The claim 'at least 75%' (p≥0.75p \ge 0.75) is technically the null, but in practice, the test is conducted using the boundary value p=0.75p=0.75. (B) would test if the rate is higher. (C) would test for any difference. (D) is incorrect because hypotheses are always stated in terms of the population parameter pp, not the sample statistic p^\hat{p}.

Question 18

A pollster is investigating voter preference in a large country. A z-test for the proportion of voters supporting candidate A is to be conducted. The pollster checks the Large Counts condition (np0≥10np_0 \ge 10 and n(1−p0)≥10n(1-p_0) \ge 10) and the 10% condition (sample size is no more than 10% of the population). However, the poll is conducted by randomly sampling names from a published phone book. Which of the following is the most significant threat to the validity of the poll's results?

  1. The 10% condition may be violated if the sample size is too large.
  2. The normal approximation may be poor if the Large Counts condition is barely met.
  3. The sample may not be representative of the population of interest due to undercoverage. (correct answer)
  4. The test may have low power if the true proportion is very close to the hypothesized proportion.
Explanation: The most fundamental assumption for inference is that the sample is representative of the population. Sampling from a phone book introduces severe undercoverage bias, as it excludes individuals with unlisted numbers, those who only use cell phones, and those without phones. This means the sample is not truly random and likely not representative of all voters. While (A), (B), and (D) describe potential statistical issues, the flawed sampling method (C) is the most critical threat to the validity of the entire study, as it means the results may not be generalizable to the target population regardless of how the numbers are calculated.

Question 19

A z-test for the proportion of students who own a car was conducted under the hypotheses H0:p=0.35H_0: p = 0.35 and Ha:p≠0.35H_a: p \neq 0.35. The test, based on a sample of 250 students, yielded a test statistic of z=2.15z=2.15. Which of the following changes to the study would have resulted in a larger z-statistic?

  1. Changing the alternative hypothesis to Ha:p>0.35H_a: p > 0.35.
  2. Observing the same number of car owners but from a sample of 300 students.
  3. Changing the null hypothesis to H0:p=0.40H_0: p = 0.40, with the same sample results.
  4. Observing the same sample proportion but from a sample of 300 students. (correct answer)
Explanation: When analyzing what affects the magnitude of a z-statistic, you need to understand its formula: z=p^−p0p0(1−p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}, where p^\hat{p} is the sample proportion, p0p_0 is the hypothesized proportion, and nn is the sample size. A larger z-statistic results from either a larger numerator (bigger difference between sample and hypothesized proportions) or a smaller denominator (smaller standard error). Since the standard error decreases as sample size increases, option D is correct: keeping the same sample proportion while increasing the sample size from 250 to 300 would decrease the standard error, making the z-statistic larger. Option A is wrong because changing from a two-tailed to one-tailed test doesn't affect the z-statistic calculation—it only changes the critical value for decision-making. Option B is incorrect because if you observe the same number of car owners (say, 87) but increase the sample to 300, your sample proportion p^\hat{p} would decrease (from 87/250 = 0.348 to 87/300 = 0.29), reducing the numerator more than the denominator shrinks. Option C is incorrect because moving the null hypothesis value further from the likely sample proportion (if p^>0.35\hat{p} > 0.35) would actually decrease the numerator. Remember: for proportion z-tests, larger sample sizes with the same sample proportion always yield larger test statistics due to reduced sampling variability. Focus on how changes affect both the numerator and denominator of the z-formula.

Question 20

The p-value for a one-proportion z-test is calculated as P(Z≥zobs)P(Z \ge z_{obs}), where zobs=p^−p0SEz_{obs} = \frac{\hat{p} - p_0}{SE}. If the sample size nn is quadrupled while p^\hat{p} and p0p_0 remain the same, what is the effect on the z-test statistic?

  1. The z-test statistic is quadrupled.
  2. The z-test statistic remains the same.
  3. The z-test statistic is halved.
  4. The z-test statistic is doubled. (correct answer)
Explanation: When analyzing how changes in sample size affect test statistics, you need to understand the relationship between sample size and the standard error component of the z-statistic. The z-test statistic formula is zobs=p^−p0SEz_{obs} = \frac{\hat{p} - p_0}{SE}, where the standard error is SE=p0(1−p0)nSE = \sqrt{\frac{p_0(1-p_0)}{n}}. Notice that the standard error has sample size nn in the denominator under the square root. When sample size is quadrupled (multiplied by 4), the new standard error becomes SEnew=p0(1−p0)4n=12p0(1−p0)n=SEoriginal2SE_{new} = \sqrt{\frac{p_0(1-p_0)}{4n}} = \frac{1}{2}\sqrt{\frac{p_0(1-p_0)}{n}} = \frac{SE_{original}}{2}. Since p^\hat{p} and p0p_0 remain unchanged, the new z-statistic is znew=p^−p0SEoriginal/2=2×p^−p0SEoriginal=2zoriginalz_{new} = \frac{\hat{p} - p_0}{SE_{original}/2} = 2 \times \frac{\hat{p} - p_0}{SE_{original}} = 2z_{original}. The z-test statistic doubles. Choice A incorrectly assumes the z-statistic increases by the same factor as sample size (4x). Choice B wrongly suggests no relationship exists between sample size and the test statistic. Choice C represents the opposite effect – this would happen if you confused how standard error changes with sample size. Study tip: Remember that standard error decreases as sample size increases, specifically by the square root relationship. When sample size increases by a factor of k2k^2, the standard error decreases by a factor of kk, making the test statistic kk times larger. This makes larger samples more sensitive to detecting true differences.