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 $H_0$; therefore, the population mean delivery time is exactly 2 days.
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.
Because $p=0.58$, there is a 58% chance that the null hypothesis is true.
Since $p=0.58>0.05$, we can conclude only that this sample’s mean delivery time is about 2 days.
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?
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.
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.
Fail to reject $H_0$ because $p=0.049>0.05$; there is not convincing evidence of a decrease.
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.
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.
Fail to reject $H_0$ because $p=0.006<0.05$; there is not enough evidence that the mean exceeds 8000 steps.
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.
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?
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.
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.
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.
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?
Reject $H_0$; therefore the population mean reaction time is not 250 ms.
Fail to reject $H_0$; there is not convincing evidence that the population mean reaction time differs from 250 ms.
Reject $H_0$; this shows the testing environment caused reaction times to change.
Because p-value = 0.095, there is a 9.5% chance the null hypothesis is correct.
Reject $H_0$; there is 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?
Since we did not reject $H_0$, we conclude the sample mean visit time was exactly 45 minutes.
Because p-value = 0.27, there is a 27% chance the null hypothesis is correct.
Reject $H_0$; there is convincing evidence that the population mean visit time is greater than 45 minutes.
Fail to reject $H_0$; therefore the mean visit time is 45 minutes for all members.
Fail to reject $H_0$; there is not convincing evidence that the population mean visit time is greater than 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?
Reject $H_0$; the sample proves that changing the filling machine caused the mean to differ from 500 grams.
Reject $H_0$; at the 0.01 level, there is convincing evidence that the population mean differs from 500 grams.
Fail to reject $H_0$; at the 0.01 level, there is not convincing evidence that the population mean differs from 500 grams.
Fail to reject $H_0$; therefore the population mean is exactly 500 grams.
Because the p-value is 0.043, the probability the alternative hypothesis is true is 0.043.
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 $H_0$; there is not convincing evidence that the population mean fill volume differs from 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 $H_0$; therefore the population mean is exactly 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 $p=0.048<0.05$, fail to reject $H_0$; there is not enough evidence that the mean is less than 30 minutes.
Reject $H_0$ and conclude that the average of the 45 sampled waits 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.
Because $p=0.048$, there is a 4.8% chance the staffing change reduced the mean wait time below 30 minutes.
Because $p=0.048<0.05$, reject $H_0$; there is evidence that the population mean ER wait time 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.
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?
Fail to reject $H_0$ and conclude the sample mean sodium content is exactly 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.
Because $p=0.62>0.05$, reject $H_0$; there is convincing evidence the population mean sodium content differs from 680 mg.
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.
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.