AP Statistics Quiz: Carrying Out Test For Population Mean

What this quiz covers

This quiz focuses on Carrying Out Test For Population Mean, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Reject $H_0$ because $p=0.58>0.05$; the population mean delivery time is not 2 days.
2. Fail to reject $H_0$ because $p=0.58>0.05$; there is not convincing evidence that the population mean delivery time differs from 2 days. (correct answer)
3. Fail to reject $H_0$; therefore, the population mean delivery time is exactly 2 days.
4. Because $p=0.58$, there is a 58% chance that the null hypothesis is true.
5. Since $p=0.58>0.05$, we can conclude only that this sample's mean delivery time is about 2 days.

Explanation: This question evaluates interpreting a one-sample t-test for a population mean. With p=0.58 > α=0.05, we fail to reject H₀: μ=2, indicating no convincing evidence that the population mean delivery time differs from 2 days. Choice D is a distractor that wrongly takes p as the chance H₀ is true, whereas p assumes H₀ for its calculation. Choice C overclaims that the mean is exactly 2, but failing to reject does not prove this. In a mini-lesson: for mean tests, p > α means insufficient evidence against H₀, not confirmation of it; always conclude about the population, maintain probabilistic phrasing, and differentiate from sample-specific statements.

Question 2

1. Fail to reject $H_0$ because $p=0.049>0.05$; there is not convincing evidence of a decrease.
2. Reject $H_0$; the new protocol caused the mean length of stay to decrease.
3. Because $p=0.049$, there is a 4.9% chance that $H_0$ is true.
4. Reject $H_0$ because $p=0.049<0.05$; there is convincing evidence that the population mean length of stay is less than 3.5 days. (correct answer)
5. Because $p=0.049<0.05$, we can conclude the sample mean length of stay is less than 3.5 days.

Explanation: This problem tests skills in performing a one-sample t-test for a population mean and concluding appropriately. Since p=0.049 < α=0.05, we reject H₀: μ=3.5, concluding convincing evidence that the population mean length of stay is less than 3.5 days. Distractor choice C misinterprets p as the chance H₀ is true, but p gauges data probability assuming H₀. Choice B attributes causation to the protocol, which is not supported by the test alone. Mini-lesson on conclusions: rejecting H₀ supports H_a with evidence, but avoid causal claims unless the study design allows; conclusions target the population mean, use precise language about evidence, and never confuse p with the probability of hypotheses.

Question 3

1. Fail to reject $H_0$ because $p=0.006<0.05$; there is not enough evidence that the mean exceeds 8000 steps.
2. Reject $H_0$; the app causes users to take more than 8000 steps per day on average.
3. Reject $H_0$ because $p=0.006<0.05$; there is convincing evidence that the population mean steps per day is greater than 8000. (correct answer)
4. Because $p=0.006$, there is a 0.6% probability that the population mean is greater than 8000.
5. Because $p=0.006<0.05$, we can conclude the sample mean is greater than 8000 steps per day.

Explanation: This problem tests the skill of conducting a one-sample t-test for a population mean and drawing appropriate conclusions. Since the p-value of 0.006 is less than α=0.05, we reject H₀: μ=8000 in favor of H_a: μ>8000, providing convincing evidence that the population mean steps per day exceeds 8000. A typical distractor is choice D, which misstates the p-value as the probability that the mean is greater than 8000, but it actually assumes H₀ and assesses data extremity. Choice E incorrectly shifts the conclusion to the sample mean, missing that inference is about the population. Mini-lesson on mean test conclusions: rejecting H₀ supports H_a with evidence at the given α level, but does not imply causation; always frame conclusions in terms of the population and evidence strength, avoiding misinterpretation of p as the probability of H_a.

Question 4

1. Reject $H_0$ because $p=0.12>0.10$; there is evidence the mean lifetime is less than 10 hours.
2. Fail to reject $H_0$ because $p=0.12>0.10$; there is not convincing evidence that the population mean lifetime is less than 10 hours. (correct answer)
3. Fail to reject $H_0$; therefore, the mean battery life is greater than or equal to 10 hours for all batteries.
4. Because $p=0.12$, there is a 12% chance that $H_0$ is false.
5. Since $p=0.12>0.10$, we conclude only that the sample mean is not less than 10 hours.

Explanation: This question focuses on interpreting a one-sample t-test for a population mean. The p-value of 0.12 exceeds α=0.10, so we fail to reject H₀: μ=10, meaning there is not convincing evidence that the population mean lifetime is less than 10 hours. Choice D is a common distractor, incorrectly presenting the p-value as the chance H₀ is false, when it is conditional on H₀ being true. Choice C overreaches by claiming the mean is >=10 for all batteries, but failing to reject only indicates insufficient evidence for H_a. In a mini-lesson for t-test conclusions: compare p to α carefully; if p > α, do not support H_a, but refrain from affirming H₀ as definitively true—conclusions are probabilistic and apply to the population, not guaranteeing outcomes for every individual case.

Question 5

1. Reject $H_0$; there is convincing evidence that the population mean reaction time differs from 250 ms. (correct answer)
2. Fail to reject $H_0$; there is not convincing evidence that the population mean reaction time differs from 250 ms.
3. Reject $H_0$; therefore the population mean reaction time is not 250 ms.
4. Because p-value = 0.095, there is a 9.5% chance the null hypothesis is correct.
5. Reject $H_0$; this shows the testing environment caused reaction times to change.

Explanation: This question requires careful comparison of p-value to significance level in a two-tailed test. With p-value = 0.095 and α = 0.10, we reject H₀ because 0.095 < 0.10, providing convincing evidence that the population mean reaction time differs from 250 ms. Choice B would be correct if α were 0.05, but with α = 0.10, we do reject H₀. Choice D misinterprets the p-value as the probability H₀ is correct. Choice E makes an unsupported causal claim. Always compare the p-value to the stated significance level—here, 0.095 < 0.10 leads to rejection of H₀.

Question 6

1. Reject $H_0$; there is convincing evidence that the population mean visit time is greater than 45 minutes.
2. Fail to reject $H_0$; there is not convincing evidence that the population mean visit time is greater than 45 minutes. (correct answer)
3. Because p-value = 0.27, there is a 27% chance the null hypothesis is correct.
4. Fail to reject $H_0$; therefore the mean visit time is 45 minutes for all members.
5. Since we did not reject $H_0$, we conclude the sample mean visit time was exactly 45 minutes.

Explanation: This problem tests understanding of a one-tailed test with a large p-value. With p-value = 0.27 and α = 0.10, we fail to reject H₀ because 0.27 > 0.10, indicating insufficient evidence that the population mean visit time is greater than 45 minutes. Choice C incorrectly interprets the p-value as the probability H₀ is correct. Choice D wrongly concludes that failing to reject H₀ proves the mean equals 45 minutes. Choice E confuses the sample mean with our conclusion about H₀. Remember that a large p-value means our sample result is consistent with H₀, but doesn't prove H₀ is true—we simply lack evidence to reject it.

Question 7

1. Fail to reject $H_0$; at the 0.01 level, there is not convincing evidence that the population mean differs from 500 grams. (correct answer)
2. Reject $H_0$; at the 0.01 level, there is convincing evidence that the population mean differs from 500 grams.
3. Because the p-value is 0.043, the probability the alternative hypothesis is true is 0.043.
4. Fail to reject $H_0$; therefore the population mean is exactly 500 grams.
5. Reject $H_0$; the sample proves that changing the filling machine caused the mean to differ from 500 grams.

Explanation: This question requires comparing a p-value to the significance level in a two-tailed test. With p-value = 0.043 and α = 0.01, we fail to reject H₀ because 0.043 > 0.01, meaning there is not convincing evidence at the 0.01 level that the population mean differs from 500 grams. Choice C misinterprets the p-value as the probability of H₁ being true. Choice D incorrectly concludes that failing to reject H₀ means the population mean equals exactly 500 grams. Choice E makes a causal claim that cannot be supported by this observational study. When the p-value exceeds α, we fail to reject H₀ but cannot conclude H₀ is true—we simply lack sufficient evidence against it.

Question 8

1. Reject $H_0$; there is convincing evidence the population mean fill volume differs from 2.00 liters.
2. Fail to reject $H_0$; there is not convincing evidence that the population mean fill volume differs from 2.00 liters. (correct answer)
3. Fail to reject $H_0$; therefore the population mean is exactly 2.00 liters.
4. Because p-value = 0.62, there is a 62% chance that $H_0$ is true.
5. Since the sample mean was close to 2.00, we conclude the sample mean fill volume equals 2.00 liters.

Explanation: This problem tests understanding of a two-tailed test with a large p-value. With p-value = 0.62 and α = 0.05, we fail to reject H₀ because 0.62 > 0.05, indicating no convincing evidence that the population mean fill volume differs from 2.00 liters. Choice C incorrectly concludes that failing to reject H₀ proves the mean equals exactly 2.00 liters. Choice D misinterprets the p-value as the probability H₀ is true. Choice E discusses only the sample mean rather than making an inference about the population. A large p-value suggests our sample data is consistent with H₀, but we cannot conclude H₀ is definitely true.

Question 9

1. Because $p=0.048<0.05$, reject $H_0$; there is evidence that the population mean ER wait time is less than 30 minutes. (correct answer)
2. Because $p=0.048<0.05$, fail to reject $H_0$; there is not enough evidence that the mean is less than 30 minutes.
3. Because $p=0.048$, there is a 4.8% chance the staffing change reduced the mean wait time below 30 minutes.
4. Reject $H_0$ and conclude the staffing change caused the mean wait time to be less than 30 minutes for all hospitals.
5. Reject $H_0$ and conclude that the average of the 45 sampled waits is less than 30 minutes.

Explanation: This problem tests whether the mean ER wait time is less than 30 minutes after a staffing change. With p-value (0.048) < α (0.05), we reject the null hypothesis and conclude there is evidence that the population mean ER wait time is less than 30 minutes. Choice B incorrectly fails to reject when p < α. Choice C misinterprets the p-value as a probability about the staffing change's effect. Choice D overgeneralizes to all hospitals and incorrectly implies causation. While the test provides evidence of a difference, establishing causation would require a controlled experiment. The p-value close to α indicates borderline evidence against H₀.

Question 10

1. Because $p=0.62>0.05$, fail to reject $H_0$; there is not convincing evidence that the population mean sodium content differs from 680 mg. (correct answer)
2. Because $p=0.62>0.05$, reject $H_0$; there is convincing evidence the population mean sodium content differs from 680 mg.
3. Because $p=0.62$, there is a 62% chance that the true mean sodium content equals 680 mg.
4. Fail to reject $H_0$ and conclude the sample mean sodium content is exactly 680 mg.
5. Fail to reject $H_0$ and conclude that eating this soup causes a person's sodium intake to be 680 mg per serving.

Explanation: In this two-tailed test for mean sodium content, the p-value (0.62) is much larger than α (0.05), so we fail to reject the null hypothesis. This means there is not convincing evidence that the population mean sodium content differs from 680 mg. Choice B incorrectly rejects H₀ when p > α. Choice C grossly misinterprets the p-value as P(μ = 680). Choice D confuses failing to reject H₀ with proving the sample mean equals 680 mg. A large p-value like 0.62 indicates the observed sample result is quite consistent with the null hypothesis, providing no evidence against the claimed population mean.

Question 11

1. Because the p-value is 0.018, there is a 1.8% chance that the null hypothesis is true.
2. Reject $H_0$; there is convincing evidence that the population mean battery life is less than 10 hours. (correct answer)
3. Fail to reject $H_0$; there is not convincing evidence that the population mean battery life is less than 10 hours.
4. Reject $H_0$; we have proven that the mean battery life is less than 10 hours.
5. Since the sample mean was below 10, we can conclude the sample mean battery life is less than 10 hours.

Explanation: This question tests your ability to interpret a one-sample t-test for a population mean. With a p-value of 0.018 and significance level α = 0.05, we reject H₀ because 0.018 < 0.05, providing convincing evidence that the population mean battery life is less than 10 hours. Choice A incorrectly interprets the p-value as the probability that H₀ is true, when it actually represents the probability of obtaining our sample result if H₀ were true. Choice D overstates the conclusion by claiming we've "proven" something, which is inappropriate in hypothesis testing. Choice E only discusses the sample mean, not the population parameter. Remember that hypothesis test conclusions always refer to the population parameter, not the sample statistic.

Question 12

1. Fail to reject $H_0$ because the p-value is less than 0.05, which means the null is likely true.
2. Reject $H_0$; there is convincing evidence that the population mean quiz score for her class is greater than 72. (correct answer)
3. Reject $H_0$; this proves her teaching method caused higher scores than the district average.
4. Because p-value = 0.004, there is a 0.4% chance that the alternative hypothesis is false.
5. Reject $H_0$; therefore every student in her class scored above 72 on the quiz.

Explanation: This question involves a one-tailed test with a very small p-value. With p-value = 0.004 and α = 0.05, we reject H₀ because 0.004 < 0.05, providing convincing evidence that the population mean quiz score for her class is greater than 72. Choice A incorrectly states we should fail to reject when p < 0.05. Choice C makes an unsupported causal claim about the teaching method. Choice D misinterprets what the p-value represents. Choice E overgeneralizes to individual students rather than the population mean. A small p-value provides strong evidence against H₀, leading us to conclude there is convincing evidence for H₁.

Question 13

1. Fail to reject $H_0$ because the p-value is close to 0.05, so there is not convincing evidence of a difference.
2. Reject $H_0$; there is convincing evidence that the population mean speed is different from 35 mph. (correct answer)
3. Reject $H_0$; this shows the new signage caused drivers to change their mean speed.
4. Because p-value = 0.049, there is a 4.9% chance that $\mu$ is not 35 mph.
5. Reject $H_0$; therefore the sample mean speed is definitely not 35 mph.

Explanation: This question involves a two-tailed test where the p-value barely falls below α. With p-value = 0.049 and α = 0.05, we reject H₀ because 0.049 < 0.05, providing convincing evidence that the population mean speed differs from 35 mph. Choice A incorrectly suggests we should fail to reject because the p-value is "close" to 0.05, but any p-value less than α leads to rejection. Choice C makes an unsupported causal claim about the signage. Choice D misinterprets the p-value's meaning. When the p-value is less than α (even by a small amount), we reject H₀ and conclude there is convincing evidence for H₁.

Question 14

1. Reject $H_0$; there is convincing evidence that the population mean wait time is greater than 30 minutes.
2. Fail to reject $H_0$; there is not convincing evidence that the population mean wait time is greater than 30 minutes. (correct answer)
3. Because p-value = 0.012, there is a 1.2% chance that $\mu$ is greater than 30 minutes.
4. Fail to reject $H_0$; therefore the mean wait time is 30 minutes for every patient.
5. Reject $H_0$; this proves the hospital's staffing changes caused longer wait times.

Explanation: This problem tests interpretation when p-value slightly exceeds α in a one-tailed test. With p-value = 0.012 and α = 0.01, we fail to reject H₀ because 0.012 > 0.01, meaning there is not convincing evidence at the 0.01 level that the population mean wait time is greater than 30 minutes. Choice A would be correct if α were 0.05, but the stricter 0.01 level requires stronger evidence. Choice C misinterprets what the p-value represents. Choice E makes an unsupported causal claim about staffing changes. When using α = 0.01, we require very strong evidence (p < 0.01) to reject H₀.

Question 15

1. Reject $H_0$; there is convincing evidence that the population mean sodium content is less than 160 mg.
2. Fail to reject $H_0$; there is not convincing evidence that the population mean sodium content is less than 160 mg. (correct answer)
3. Fail to reject $H_0$; therefore the population mean sodium content is 160 mg.
4. Because the p-value is 0.081, the probability that $H_a$ is true is 0.081.
5. Since the sample was small, we can only conclude the sample mean sodium content is less than 160 mg.

Explanation: This problem tests interpretation of a one-tailed test where p-value exceeds α. With p-value = 0.081 and α = 0.05, we fail to reject H₀ because 0.081 > 0.05, meaning there is not convincing evidence that the population mean sodium content is less than 160 mg. Choice C incorrectly concludes that failing to reject H₀ proves the mean equals 160 mg. Choice D misinterprets the p-value as the probability of H₁ being true. Choice E inappropriately limits the conclusion to the sample rather than the population. Remember that hypothesis tests always make inferences about population parameters, and failing to reject H₀ doesn't prove H₀ is true.

Question 16

1. Because $p=0.14>0.05$, fail to reject $H_0$; there is not convincing evidence that the population mean sleep is less than 7 hours. (correct answer)
2. Because $p=0.14>0.05$, reject $H_0$; there is convincing evidence the population mean sleep is less than 7 hours.
3. Because $p=0.14$, there is a 14% probability that the null hypothesis is correct.
4. Fail to reject $H_0$ and conclude the sample mean sleep time is 7 hours.
5. Fail to reject $H_0$ and conclude that studying more causes students to sleep less than 7 hours.

Explanation: This one-tailed test examines whether college students' mean sleep is less than 7 hours. Since p-value (0.14) > α (0.05), we fail to reject the null hypothesis. The correct conclusion is that there is not convincing evidence that the population mean sleep is less than 7 hours. Choice B incorrectly rejects H₀ when p > α. Choice C misinterprets the p-value as the probability that H₀ is correct. Choice D confuses the sample mean with conclusions about the population mean. When we fail to reject H₀, we're not proving the null hypothesis is true; we're simply stating that the sample doesn't provide sufficient evidence against it.

Question 17

1. Because $p=0.043>0.01$, fail to reject $H_0$; there is not sufficient evidence that the population mean battery life differs from 20 hours. (correct answer)
2. Because $p=0.043>0.01$, reject $H_0$; there is convincing evidence the population mean battery life differs from 20 hours.
3. Because $p=0.043$, there is a 4.3% chance that the true mean battery life is 20 hours.
4. Since the sample did not match 20 hours exactly, the manufacturer's claim is false for all batteries.
5. Fail to reject $H_0$ and conclude the manufacturer's testing procedure causes batteries to last 20 hours.

Explanation: This problem involves a two-tailed test for a population mean at significance level α = 0.01. Since the p-value (0.043) is greater than α (0.01), we fail to reject the null hypothesis. This means there is not sufficient evidence that the population mean battery life differs from 20 hours. Choice B incorrectly rejects H₀ when the p-value exceeds α. Choice C misinterprets the p-value as the probability that μ = 20, rather than the probability of the observed data given that μ = 20. In hypothesis testing, we compare the p-value to α: reject H₀ when p < α, and fail to reject when p ≥ α. Failing to reject H₀ does not prove it's true; it simply means we lack sufficient evidence against it.

Question 18

1. Because $p=0.26>0.05$, fail to reject $H_0$; there is not convincing evidence that the population mean weight is less than 18 ounces. (correct answer)
2. Because $p=0.26>0.05$, reject $H_0$; there is convincing evidence the population mean weight is less than 18 ounces.
3. Because $p=0.26$, there is a 26% chance that $H_0$ is true.
4. Fail to reject $H_0$ and conclude the mean weight of the sampled 12 boxes is 18 ounces.
5. Fail to reject $H_0$ and conclude that changing the factory machine will not affect box weights.

Explanation: In this one-tailed test for a population mean, the p-value (0.26) is much greater than the significance level (0.05), so we fail to reject the null hypothesis. This means there is not convincing evidence that the population mean weight is less than 18 ounces. Choice B incorrectly rejects H₀ when p > α. Choice C misinterprets the p-value as P(H₀ is true) rather than P(data | H₀). Choice D confuses the sample mean with the population mean claim. When we fail to reject H₀, we're saying the data doesn't provide strong enough evidence against the null hypothesis claim about the population parameter. A large p-value suggests the observed sample result is reasonably likely under the null hypothesis.

Question 19

1. Because $p=0.009<0.01$, reject $H_0$; there is convincing evidence that the population mean resting heart rate after 8 weeks is less than 70 bpm. (correct answer)
2. Because $p=0.009<0.01$, fail to reject $H_0$; there is not convincing evidence the mean is less than 70 bpm.
3. Because $p=0.009$, there is a 0.9% chance the true mean is less than 70 bpm.
4. Reject $H_0$ and conclude that using the app causes every user's resting heart rate to drop below 70 bpm.
5. Reject $H_0$ and conclude the sample mean resting heart rate is less than 70 bpm.

Explanation: This problem involves testing whether the population mean resting heart rate is less than 70 bpm. Since p-value (0.009) < α (0.01), we reject the null hypothesis and conclude there is convincing evidence that the population mean resting heart rate after 8 weeks is less than 70 bpm. Choice B incorrectly fails to reject when p < α. Choice C misinterprets the p-value as the probability that μ < 70. Choice D overgeneralizes to every individual user rather than making an inference about the population mean. When conducting hypothesis tests, we make conclusions about population parameters based on sample evidence, not about individual values or certainties.

Question 20

1. Because $p=0.051>0.05$, fail to reject $H_0$; there is not quite sufficient evidence that the population mean wait time is greater than 8 minutes. (correct answer)
2. Because $p=0.051>0.05$, reject $H_0$; there is convincing evidence the population mean wait time is greater than 8 minutes.
3. Because $p=0.051$, there is a 5.1% chance that the mean wait time exceeds 8 minutes.
4. Fail to reject $H_0$ and conclude the mean wait time for the 50 observed waits is 8 minutes.
5. Fail to reject $H_0$ and conclude that longer traffic lights cause longer bus waits.

Explanation: In this one-tailed test for mean wait time, the p-value (0.051) is just slightly greater than α (0.05), so we fail to reject the null hypothesis. The conclusion is that there is not quite sufficient evidence that the population mean wait time is greater than 8 minutes. Choice B incorrectly rejects H₀ when p > α. Choice C misinterprets the p-value as P(μ > 8) rather than P(data | μ = 8). Choice D confuses inference about the sample with inference about the population. This case illustrates the importance of the significance level as a decision threshold - even a p-value very close to α still leads to failing to reject H₀ when p > α.