Carrying Out Test for Population Mean
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AP Statistics › Carrying Out Test for Population Mean
An online retailer wants to check whether the mean delivery time for a certain shipping option is 2 days. A random sample of 100 deliveries is selected, and a one-sample $t$ test is performed with $H_0:\mu=2$ versus $H_a:\mu\ne 2$ at $\alpha=0.05$. The p-value is 0.58. What conclusion is appropriate?
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.
Since $p=0.58>0.05$, we can conclude only that this sample’s mean delivery time is about 2 days.
Fail to reject $H_0$; therefore, the population mean delivery time is exactly 2 days.
Because $p=0.58$, there is a 58% chance that the null hypothesis is true.
Reject $H_0$ because $p=0.58>0.05$; the population mean delivery time is not 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.
A hospital states that the mean length of stay for patients undergoing a certain procedure is $\mu=3.5$ days. A researcher suspects the mean length of stay has decreased with a new protocol. A random sample of 22 patients is collected, and a one-sample $t$ test is conducted with $H_0:\mu=3.5$ versus $H_a:\mu<3.5$ at $\alpha=0.05$. The p-value is 0.049. What conclusion is appropriate?
Because $p=0.049$, there is a 4.9% chance that $H_0$ is true.
Reject $H_0$; the new protocol caused the mean length of stay to decrease.
Because $p=0.049<0.05$, we can conclude the sample mean length of stay is less than 3.5 days.
Fail to reject $H_0$ because $p=0.049>0.05$; there is not convincing evidence of a decrease.
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.
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.
A fitness app claims that the mean number of steps per day for its users is more than $\mu=8000$. A random sample of 50 users is selected and a one-sample $t$ test is performed with $H_0:\mu=8000$ versus $H_a:\mu>8000$ at $\alpha=0.05$. The p-value is 0.006. What conclusion is appropriate?
Reject $H_0$ because $p=0.006<0.05$; there is convincing evidence that the population mean steps per day is greater than 8000.
Because $p=0.006<0.05$, we can conclude the sample mean is greater than 8000 steps per day.
Because $p=0.006$, there is a 0.6% probability that the population mean is greater than 8000.
Reject $H_0$; the app causes users to take more than 8000 steps per day on average.
Fail to reject $H_0$ because $p=0.006<0.05$; there is not enough evidence that the mean exceeds 8000 steps.
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.
A manufacturer advertises that its batteries last an average of $\mu=10$ hours. A consumer group suspects the mean lifetime is less. A random sample of 18 batteries is tested, and a one-sample $t$ test is conducted with $H_0:\mu=10$ versus $H_a:\mu<10$ at $\alpha=0.10$. The p-value is 0.12. What conclusion is appropriate?
Because $p=0.12$, there is a 12% chance that $H_0$ is false.
Since $p=0.12>0.10$, we conclude only that the sample mean is not less than 10 hours.
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.
Fail to reject $H_0$; therefore, the mean battery life is greater than or equal to 10 hours for all batteries.
Reject $H_0$ because $p=0.12>0.10$; there is evidence the mean lifetime is 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.
A researcher claims the average reaction time for a certain task is $\mu=250$ ms. A random sample of 18 participants is tested, and a one-sample $t$ test is conducted with $H_0:\mu=250$ and $H_a:\mu\neq 250$ at $\alpha=0.10$. The p-value is 0.095. What conclusion is appropriate?
Because p-value = 0.095, there is a 9.5% chance the null hypothesis is correct.
Reject $H_0$; this shows the testing environment caused reaction times to change.
Reject $H_0$; therefore the population mean reaction time is not 250 ms.
Reject $H_0$; there is convincing evidence that the population mean reaction time differs from 250 ms.
Fail to reject $H_0$; there is not convincing evidence that the population mean reaction time differs from 250 ms.
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.
A gym owner believes members spend an average of $\mu=45$ minutes per visit. A random sample of 60 visits is recorded, and a one-sample $t$ test is conducted with $H_0:\mu=45$ and $H_a:\mu>45$ at $\alpha=0.10$. The p-value is 0.27. What conclusion is appropriate?
Fail to reject $H_0$; there is not convincing evidence that the population mean visit time is greater than 45 minutes.
Because p-value = 0.27, there is a 27% chance the null hypothesis is correct.
Fail to reject $H_0$; therefore the mean visit time is 45 minutes for all members.
Reject $H_0$; there is convincing evidence that the population mean visit time is greater than 45 minutes.
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.
A cereal manufacturer advertises that boxes contain an average of $\mu=500$ grams of cereal. A quality-control analyst takes a random sample of 25 boxes and performs a one-sample $t$ test with $H_0:\mu=500$ and $H_a:\mu\neq 500$ using $\alpha=0.01$. The p-value from the test is 0.043. What conclusion is appropriate?
Fail to reject $H_0$; at the 0.01 level, there is not convincing evidence that the population mean differs from 500 grams.
Reject $H_0$; at the 0.01 level, there is convincing evidence that the population mean differs from 500 grams.
Because the p-value is 0.043, the probability the alternative hypothesis is true is 0.043.
Reject $H_0$; the sample proves that changing the filling machine caused the mean to differ from 500 grams.
Fail to reject $H_0$; therefore the population mean is exactly 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.
A bottling plant targets an average fill volume of $\mu=2.00$ liters. A random sample of 50 bottles is measured, and a one-sample $t$ test is carried out with $H_0:\mu=2.00$ and $H_a:\mu\neq 2.00$ at $\alpha=0.05$. The p-value is 0.62. What conclusion is appropriate?
Fail to reject $H_0$; therefore the population mean is exactly 2.00 liters.
Since the sample mean was close to 2.00, we conclude the sample mean fill volume equals 2.00 liters.
Fail to reject $H_0$; there is not convincing evidence that the population mean fill volume differs from 2.00 liters.
Because p-value = 0.62, there is a 62% chance that $H_0$ is true.
Reject $H_0$; there is convincing evidence the population mean fill volume differs from 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.
A hospital claims the mean emergency room (ER) wait time to see a doctor is $\mu=30$ minutes. An administrator tests whether the mean wait time is less than 30 minutes after a staffing change. A random sample of 45 ER visits is selected, and a one-sample $t$ test is conducted with $H_0:\mu=30$ versus $H_a:\mu<30$ at $\alpha=0.05$. The p-value is 0.048. What conclusion is appropriate?
Because $p=0.048<0.05$, reject $H_0$; there is evidence that the population mean ER wait time is less than 30 minutes.
Because $p=0.048<0.05$, fail to reject $H_0$; there is not enough evidence that the mean is less than 30 minutes.
Reject $H_0$ and conclude the staffing change caused the mean wait time to be less than 30 minutes for all hospitals.
Reject $H_0$ and conclude that the average of the 45 sampled waits is less than 30 minutes.
Because $p=0.048$, there is a 4.8% chance the staffing change reduced the mean wait time below 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.
A nutritionist tests whether the mean sodium content in a brand of soup is different from the stated $\mu=680$ mg per serving. A random sample of 16 cans is analyzed, and a one-sample $t$ test is performed with $H_0:\mu=680$ versus $H_a:\mu\neq 680$ at $\alpha=0.05$. The p-value is 0.62. What conclusion is appropriate?
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.
Because $p=0.62$, there is a 62% chance that the true mean sodium content equals 680 mg.
Fail to reject $H_0$ and conclude that eating this soup causes a person’s sodium intake to be 680 mg per serving.
Fail to reject $H_0$ and conclude the sample mean sodium content is exactly 680 mg.
Because $p=0.62>0.05$, reject $H_0$; there is convincing evidence the population mean sodium content differs from 680 mg.
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.