Interpreting p-Values
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AP Statistics › Interpreting p-Values
A technology company claims that the proportion of users who enable two-factor authentication is $p = 0.35$. A random sample of users is selected and a one-proportion test is performed for $H_0: p = 0.35$ versus $H_a: p > 0.35$ at $\alpha = 0.05$. The p-value is $p = 0.006$. Which interpretation of the p-value is correct?
A p-value of 0.006 means 0.6% of users enable two-factor authentication.
There is a 0.6% chance that $p$ is greater than 0.35.
If $p=0.35$, the probability of getting a sample proportion at least as large as the observed sample proportion (in the direction of $H_a$) is 0.006.
There is a 0.6% chance that the null hypothesis is true.
If the alternative hypothesis is true, there is a 0.6% chance of observing the sample result.
Explanation
This question evaluates interpreting p-values in a one-sided right-tailed test for a proportion in AP Statistics. The p-value of 0.006 is the conditional probability of getting a sample proportion at least as large as observed, given H0: p = 0.35 is true, in the direction of Ha: p > 0.35. A frequent distractor is choice C, which mistakes the p-value for the probability that H0 is true. As a mini-lesson, p-values assess evidence by computing data extremity under H0: a small 0.006 provides strong grounds to reject H0 at α = 0.05, suggesting more than 35% enable authentication. Choice B accurately specifies 'at least as large... (in the direction of Ha)'. Misinterpretations in A, D, and E include reversing conditioning or applying to population subsets.
A company advertises that 60% of its customers renew their subscription. A random sample of 200 customers finds 112 renewals. A one-proportion $z$ test is conducted for $H_0: p = 0.60$ versus $H_a: p \ne 0.60$ at $\alpha = 0.10$, yielding p-value $p = 0.041$. Which interpretation of the p-value is correct?
If the alternative hypothesis is true, the probability of observing a sample proportion of 0.56 is 0.041.
If the true renewal rate is 60%, the probability of getting a sample proportion at least as far from 0.60 as 0.56 (in either direction) is 0.041.
Because $p=0.041$, 4.1% of customers will not renew.
There is a 4.1% chance that the true renewal rate is 60%.
There is a 4.1% chance that exactly 112 out of 200 customers renew.
Explanation
This question tests the skill of interpreting p-values in a two-sided one-proportion z-test in AP Statistics. The p-value of 0.041 is the conditional probability of getting a sample proportion at least as far from 0.60 as 0.56 (in either direction) if H0: p = 0.60 is true. A common distractor is choice B, which wrongly treats the p-value as the probability that the null hypothesis is true, a classic error mixing conditional and posterior probabilities. In a mini-lesson, p-values indicate the likelihood of the observed data or more extreme under H0: here, 0.041 is small enough to reject H0 at α = 0.10, suggesting evidence against the claimed 60% renewal rate. Choice A correctly includes the two-sided extremeness and the assumption of H0. Avoid errors like those in C, D, and E, which misapply the p-value to exact outcomes or alternative hypotheses.
A manufacturer claims its light bulbs last an average of $\mu = 1000$ hours. A consumer group tests a random sample of 25 bulbs and performs a one-sample $t$ test for $H_0: \mu = 1000$ versus $H_a: \mu < 1000$ at $\alpha = 0.05$. The test produces p-value $p = 0.002$. Which interpretation of the p-value is correct?
If $H_0$ is true, the probability of obtaining a sample mean as low as (or lower than) the observed sample mean is 0.002.
There is a 0.2% chance that the sample mean is less than 1000 hours.
There is a 0.2% chance that the manufacturer’s claim $\mu=1000$ is correct.
If $H_0$ is false, there is a 0.2% chance of getting the observed sample mean.
A p-value of 0.002 means the probability the bulbs last less than 1000 hours is 0.002.
Explanation
This question examines the skill of interpreting p-values in a one-sided left-tailed t-test for a mean in AP Statistics. The p-value of 0.002 is the conditional probability of obtaining a sample mean as low as or lower than observed, assuming H0: μ = 1000 is true. A typical distractor is choice A, which incorrectly states the p-value as the probability that H0 is correct, reversing the conditioning. For a mini-lesson, p-values assess evidence against H0 by calculating the probability of data extremes under it: a very small p-value like 0.002 provides strong evidence to reject H0 at α = 0.05, indicating the bulbs likely last less than claimed. Choice B properly specifies the one-tailed direction 'as low as (or lower than)'. Choices C, D, and E exemplify errors like confusing p-values with unconditional probabilities or switching to the alternative hypothesis.
An environmental agency tests whether the mean concentration of a pollutant in a river exceeds the legal limit of 10 ppm. The hypotheses are $H_0: \mu=10$ versus $H_a: \mu>10$, and the test uses $\alpha=0.01$. The p-value is 0.009. Which interpretation of the p-value is correct?
If $H_0$ is true, the probability of observing a sample mean concentration at least as large as the one observed is 0.009.
A p-value of 0.009 means 0.9% of river samples exceed 10 ppm.
If the true mean concentration exceeds 10 ppm, the probability of obtaining the observed sample mean is 0.009.
There is a 0.9% chance that the true mean concentration is at most 10 ppm.
There is a 0.9% chance that the agency made a Type I error.
Explanation
This question involves interpreting a p-value in a right-tailed test about pollutant concentration. The p-value represents the conditional probability of observing a sample mean at least as extreme as what was observed, assuming the null hypothesis is true. For this right-tailed test (H_a: μ > 10), the p-value of 0.009 means that if the true mean concentration is exactly 10 ppm, there's a 0.9% chance of observing a sample mean as large as or larger than what was observed. Choice B correctly captures this interpretation. Since 0.009 < 0.01 (the significance level), we would reject H_0, providing strong evidence that the mean exceeds the legal limit. Common mistakes include thinking the p-value represents probabilities about hypotheses being true (Choice A) or about Type I error rates for specific tests (Choice C). Remember: p-values are calculated assuming H_0 is true.
A researcher tests whether the mean reaction time for a task is different from 250 ms. The hypotheses are $H_0: \mu=250$ versus $H_a: \mu\ne 250$, using $\alpha=0.05$. The p-value is 0.051. Which interpretation of the p-value is correct?
Because the p-value is 0.051, the null hypothesis must be accepted as true.
There is a 5.1% chance that the mean reaction time is exactly 250 ms.
If the true mean is not 250 ms, the probability of getting the observed sample mean is 0.051.
A p-value of 0.051 means 5.1% of individual reaction times are different from 250 ms.
If $H_0$ is true, the probability of observing a result at least as extreme as the sample result (in either direction) is 0.051.
Explanation
This question tests understanding of p-value interpretation in a two-tailed test about reaction times. The p-value represents the conditional probability of obtaining a test statistic at least as extreme as observed in either direction, given that the null hypothesis is true. With H_0: μ = 250 and a two-tailed alternative, the p-value of 0.051 means that if the true mean reaction time is 250 ms, there's a 5.1% chance of observing a sample mean at least as far from 250 ms as what was observed. Choice B correctly states this interpretation. With α = 0.05, we would fail to reject H_0 since 0.051 > 0.05, but this doesn't mean we "accept" H_0 as true (Choice C is incorrect). The p-value doesn't tell us about individual measurements (Choice E) or probabilities under the alternative hypothesis (Choice D).
A coffee shop owner believes the mean amount of coffee dispensed by a machine is $\mu = 12$ oz. After maintenance, a technician tests $H_0: \mu = 12$ versus $H_a: \mu > 12$ at $\alpha = 0.01$ using a random sample of 40 pours and obtains a p-value of $p = 0.18$. Which interpretation of the p-value is correct?
An 18% p-value means 18% of all pours exceed 12 oz.
There is an 18% chance that the null hypothesis is false.
There is an 18% chance that $\mu$ is greater than 12 oz.
If $H_0$ is true, there is an 18% chance of getting a sample mean at least as large as the observed sample mean (in the direction of $H_a$) due to random sampling variability.
If $H_0$ is true, there is an 18% chance of getting a sample mean of 12 oz or more.
Explanation
This question evaluates the skill of interpreting p-values in a one-sided hypothesis test for a population mean in AP Statistics. The p-value of 0.18 is the conditional probability of observing a sample mean at least as large as the one obtained, given that H0: μ = 12 is true, accounting for random sampling variability in the direction of Ha: μ > 12. A frequent distractor is choice A, which omits the directionality of the alternative hypothesis, making it seem like a two-sided interpretation instead of one-sided. As a mini-lesson, p-values quantify how compatible the data is with the null hypothesis: a larger p-value like 0.18 suggests the data is not surprising under H0, so we fail to reject it at α = 0.01. Choice C accurately reflects the one-tailed nature by specifying 'at least as large as the observed sample mean (in the direction of Ha)'. Misinterpretations like those in B, D, and E confuse p-values with probabilities of hypotheses or population parameters.
A wildlife biologist tests whether the mean weight of a certain fish species in a lake differs from $\mu = 2.5$ kg. Using a random sample, the biologist conducts a two-sided one-sample $t$ test: $H_0: \mu = 2.5$ versus $H_a: \mu \ne 2.5$ at $\alpha = 0.05$. The p-value is $p = 0.08$. Which interpretation of the p-value is correct?
There is an 8% chance the fish in the lake have mean weight exactly 2.5 kg.
There is an 8% chance that the sample mean equals the observed value.
A p-value of 0.08 means 8% of fish weigh 2.5 kg.
If $H_0$ is true, the probability of getting a sample mean at least as far from 2.5 kg as the observed sample mean (in either direction) is 0.08.
There is an 8% chance that the null hypothesis is false.
Explanation
This question assesses the skill of interpreting p-values in a two-sided t-test for a mean in AP Statistics. The p-value of 0.08 is the conditional probability of a sample mean at least as far from 2.5 kg as observed (in either direction) given H0: μ = 2.5 is true. A common distractor is choice E, which wrongly interprets the p-value as the probability that H0 is false, a frequent misunderstanding. In a mini-lesson, p-values evaluate surprise under H0: 0.08 is above α = 0.05, so we fail to reject H0, indicating insufficient evidence of a difference in mean weight. Choice B correctly includes the two-sided aspect with 'at least as far... (in either direction)'. Avoid errors in A, C, and D, such as equating p-values to chances of exact values or population proportions.
A city planner believes the mean commute time for residents is $\mu = 28$ minutes. A random sample of 80 residents is used to test $H_0: \mu = 28$ versus $H_a: \mu \ne 28$ at $\alpha = 0.01$. The p-value from the test is $p = 0.012$. Which interpretation of the p-value is correct?
If $\mu=28$ minutes, the probability of getting a sample mean at least as extreme as the observed one (in either direction) is 0.012.
Because $p=0.012$, the probability the mean commute time is 28 minutes is 0.012.
If $H_0$ is true, 1.2% of all samples will have a mean exactly equal to the observed sample mean.
There is a 1.2% chance that the alternative hypothesis is true.
There is a 1.2% chance of selecting a resident with a 28-minute commute.
Explanation
This question assesses interpreting p-values in a two-sided test for a population mean in AP Statistics. The p-value of 0.012 is the conditional probability of a sample mean at least as extreme as observed (in either direction) given H0: μ = 28 is true. A common distractor is choice C, which misinterprets the p-value as the probability that the alternative hypothesis is true, a misunderstanding of hypothesis testing logic. In a mini-lesson, p-values help decide if data is surprising under H0: here, 0.012 exceeds α = 0.01 slightly, so we fail to reject H0, but it would be significant at higher α. Choice B correctly notes the two-sided nature with 'at least as extreme... (in either direction)'. Avoid pitfalls in A, D, and E, such as equating p-values to probabilities of specific values or sample equalities.
A researcher tests whether a new tutoring program increases the mean math score above 75. For a random sample of students in the program, a one-sample $t$ test is run for $H_0: \mu = 75$ versus $H_a: \mu > 75$ at $\alpha = 0.05$, resulting in p-value $p = 0.049$. Which interpretation of the p-value is correct?
If the alternative hypothesis is true, the probability of observing a result at least as extreme as the sample is 0.049.
If $H_0$ is true, the probability of getting a sample mean at least as large as the observed sample mean is 0.049.
A p-value of 0.049 means 4.9% of students scored above 75.
There is a 4.9% chance that the tutoring program does not increase the mean score above 75.
There is a 4.9% chance that the sample mean is greater than 75.
Explanation
This question tests interpreting p-values in a one-sided right-tailed t-test in AP Statistics. The p-value of 0.049 is the conditional probability of getting a sample mean at least as large as observed, assuming H0: μ = 75 is true. A common distractor is choice E, which incorrectly conditions on the alternative hypothesis instead of the null. In a mini-lesson, p-values provide evidence against H0 by quantifying extremeness under it: 0.049 is just below α = 0.05, suggesting borderline evidence to reject H0 and conclude the program increases scores. Choice B correctly specifies the one-tailed direction 'at least as large'. Avoid mistakes like those in A, C, and D, which confuse p-values with probabilities of hypotheses or direct population percentages.
A school district claims that the mean time students spend on homework per night is $\mu = 90$ minutes. A random sample of 60 students reports an average of 84 minutes. A one-sample $t$ test is performed for $H_0: \mu = 90$ versus $H_a: \mu \ne 90$ at significance level $\alpha = 0.05$, and the p-value is $p = 0.03$. Which interpretation of the p-value is correct?
If the alternative hypothesis is true, there is a 3% chance of obtaining a result like the sample mean of 84 minutes.
There is a 3% chance that the sample mean equals 84 minutes when $\mu=90$.
Because $p=0.03$, 3% of students spend exactly 84 minutes on homework per night.
There is a 3% chance that the null hypothesis $H_0$ is true.
If $H_0$ is true, there is a 3% chance of obtaining a sample mean at least as far from 90 minutes as the one observed (in either direction) just by random sampling variability.
Explanation
This question assesses the skill of interpreting p-values in the context of a two-sided hypothesis test for a population mean in AP Statistics. The p-value of 0.03 represents the conditional probability of obtaining a sample mean at least as extreme as 84 minutes (in either direction from 90) given that the null hypothesis H0: μ = 90 is true, due to random sampling variability. A common distractor is choice A, which incorrectly interprets the p-value as the probability that H0 is true, confusing it with posterior probability rather than the conditional probability under H0. In a mini-lesson on p-values, remember that they measure the strength of evidence against the null hypothesis: a small p-value like 0.03 indicates the observed data would be rare if H0 were true, potentially leading to rejection at α = 0.05. Choice B correctly captures this by emphasizing the assumption of H0 and the extremeness in both tails for a two-sided test. Avoid mistaking p-values for probabilities of hypotheses or specific sample outcomes, as seen in choices C, D, and E.