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AP Statistics

AP Statistics Help: Interpreting P Values

Review real example questions for Interpreting P Values in AP Statistics.

Question 1

A technology company claims that the proportion of users who enable two-factor authentication is p=0.35p = 0.35p=0.35. A random sample of users is selected and a one-proportion test is performed for H0:p=0.35H_0: p = 0.35H0​:p=0.35 versus Ha:p>0.35H_a: p > 0.35Ha​:p>0.35 at α=0.05\alpha = 0.05α=0.05. The p-value is p=0.006p = 0.006p=0.006. Which interpretation of the p-value is correct?

  1. There is a 0.6% chance that ppp is greater than 0.35.
  2. If p=0.35p=0.35p=0.35, the probability of getting a sample proportion at least as large as the observed sample proportion (in the direction of HaH_aHa​) is 0.006.
  3. There is a 0.6% chance that the null hypothesis is true.
  4. A p-value of 0.006 means 0.6% of users enable two-factor authentication.
  5. 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.

Question 2

A company advertises that 60% of its customers renew their subscription. A random sample of 200 customers finds 112 renewals. A one-proportion zzz test is conducted for H0:p=0.60H_0: p = 0.60H0​:p=0.60 versus Ha:p≠0.60H_a: p \ne 0.60Ha​:p=0.60 at α=0.10\alpha = 0.10α=0.10, yielding p-value p=0.041p = 0.041p=0.041. Which interpretation of the p-value is correct?

  1. 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.
  2. There is a 4.1% chance that the true renewal rate is 60%.
  3. There is a 4.1% chance that exactly 112 out of 200 customers renew.
  4. Because p=0.041p=0.041p=0.041, 4.1% of customers will not renew.
  5. If the alternative hypothesis is true, the probability of observing a sample proportion of 0.56 is 0.041.
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.

Question 3

A manufacturer claims its light bulbs last an average of μ=1000\mu = 1000μ=1000 hours. A consumer group tests a random sample of 25 bulbs and performs a one-sample ttt test for H0:μ=1000H_0: \mu = 1000H0​:μ=1000 versus Ha:μ<1000H_a: \mu < 1000Ha​:μ<1000 at α=0.05\alpha = 0.05α=0.05. The test produces p-value p=0.002p = 0.002p=0.002. Which interpretation of the p-value is correct?

  1. There is a 0.2% chance that the manufacturer’s claim μ=1000\mu=1000μ=1000 is correct.
  2. If H0H_0H0​ is true, the probability of obtaining a sample mean as low as (or lower than) the observed sample mean is 0.002.
  3. There is a 0.2% chance that the sample mean is less than 1000 hours.
  4. A p-value of 0.002 means the probability the bulbs last less than 1000 hours is 0.002.
  5. If H0H_0H0​ is false, there is a 0.2% chance of getting the observed sample mean.
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.

Question 4

An environmental agency tests whether the mean concentration of a pollutant in a river exceeds the legal limit of 10 ppm. The hypotheses are H0:μ=10H_0: \mu=10H0​:μ=10 versus Ha:μ>10H_a: \mu>10Ha​:μ>10, and the test uses α=0.01\alpha=0.01α=0.01. The p-value is 0.009. Which interpretation of the p-value is correct?

  1. There is a 0.9% chance that the true mean concentration is at most 10 ppm.
  2. If H0H_0H0​ is true, the probability of observing a sample mean concentration at least as large as the one observed is 0.009.
  3. There is a 0.9% chance that the agency made a Type I error.
  4. If the true mean concentration exceeds 10 ppm, the probability of obtaining the observed sample mean is 0.009.
  5. A p-value of 0.009 means 0.9% of river samples exceed 10 ppm.
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.

Question 5

A researcher tests whether the mean reaction time for a task is different from 250 ms. The hypotheses are H0:μ=250H_0: \mu=250H0​:μ=250 versus Ha:μ≠250H_a: \mu\ne 250Ha​:μ=250, using α=0.05\alpha=0.05α=0.05. The p-value is 0.051. Which interpretation of the p-value is correct?

  1. There is a 5.1% chance that the mean reaction time is exactly 250 ms.
  2. If H0H_0H0​ is true, the probability of observing a result at least as extreme as the sample result (in either direction) is 0.051.
  3. Because the p-value is 0.051, the null hypothesis must be accepted as true.
  4. If the true mean is not 250 ms, the probability of getting the observed sample mean is 0.051.
  5. A p-value of 0.051 means 5.1% of individual reaction times are different from 250 ms.
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).

Question 6

A coffee shop owner believes the mean amount of coffee dispensed by a machine is μ=12\mu = 12μ=12 oz. After maintenance, a technician tests H0:μ=12H_0: \mu = 12H0​:μ=12 versus Ha:μ>12H_a: \mu > 12Ha​:μ>12 at α=0.01\alpha = 0.01α=0.01 using a random sample of 40 pours and obtains a p-value of p=0.18p = 0.18p=0.18. Which interpretation of the p-value is correct?

  1. If H0H_0H0​ is true, there is an 18% chance of getting a sample mean of 12 oz or more.
  2. There is an 18% chance that μ\muμ is greater than 12 oz.
  3. If H0H_0H0​ 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 HaH_aHa​) due to random sampling variability.
  4. There is an 18% chance that the null hypothesis is false.
  5. An 18% p-value means 18% of all pours exceed 12 oz.
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.

Question 7

A wildlife biologist tests whether the mean weight of a certain fish species in a lake differs from μ=2.5\mu = 2.5μ=2.5 kg. Using a random sample, the biologist conducts a two-sided one-sample ttt test: H0:μ=2.5H_0: \mu = 2.5H0​:μ=2.5 versus Ha:μ≠2.5H_a: \mu \ne 2.5Ha​:μ=2.5 at α=0.05\alpha = 0.05α=0.05. The p-value is p=0.08p = 0.08p=0.08. Which interpretation of the p-value is correct?

  1. There is an 8% chance the fish in the lake have mean weight exactly 2.5 kg.
  2. If H0H_0H0​ 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.
  3. There is an 8% chance that the sample mean equals the observed value.
  4. A p-value of 0.08 means 8% of fish weigh 2.5 kg.
  5. 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.

Question 8

A city planner believes the mean commute time for residents is μ=28\mu = 28μ=28 minutes. A random sample of 80 residents is used to test H0:μ=28H_0: \mu = 28H0​:μ=28 versus Ha:μ≠28H_a: \mu \ne 28Ha​:μ=28 at α=0.01\alpha = 0.01α=0.01. The p-value from the test is p=0.012p = 0.012p=0.012. Which interpretation of the p-value is correct?

  1. Because p=0.012p=0.012p=0.012, the probability the mean commute time is 28 minutes is 0.012.
  2. If μ=28\mu=28μ=28 minutes, the probability of getting a sample mean at least as extreme as the observed one (in either direction) is 0.012.
  3. There is a 1.2% chance that the alternative hypothesis is true.
  4. There is a 1.2% chance of selecting a resident with a 28-minute commute.
  5. If H0H_0H0​ is true, 1.2% of all samples will have a mean exactly equal to the observed sample mean.
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.

Question 9

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 ttt test is run for H0:μ=75H_0: \mu = 75H0​:μ=75 versus Ha:μ>75H_a: \mu > 75Ha​:μ>75 at α=0.05\alpha = 0.05α=0.05, resulting in p-value p=0.049p = 0.049p=0.049. Which interpretation of the p-value is correct?

  1. There is a 4.9% chance that the tutoring program does not increase the mean score above 75.
  2. If H0H_0H0​ is true, the probability of getting a sample mean at least as large as the observed sample mean is 0.049.
  3. There is a 4.9% chance that the sample mean is greater than 75.
  4. A p-value of 0.049 means 4.9% of students scored above 75.
  5. If the alternative hypothesis is true, the probability of observing a result at least as extreme as the sample is 0.049.
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.

Question 10

A school district claims that the mean time students spend on homework per night is μ=90\mu = 90μ=90 minutes. A random sample of 60 students reports an average of 84 minutes. A one-sample ttt test is performed for H0:μ=90H_0: \mu = 90H0​:μ=90 versus Ha:μ≠90H_a: \mu \ne 90Ha​:μ=90 at significance level α=0.05\alpha = 0.05α=0.05, and the p-value is p=0.03p = 0.03p=0.03. Which interpretation of the p-value is correct?

  1. There is a 3% chance that the null hypothesis H0H_0H0​ is true.
  2. If H0H_0H0​ 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.
  3. There is a 3% chance that the sample mean equals 84 minutes when μ=90\mu=90μ=90.
  4. Because p=0.03p=0.03p=0.03, 3% of students spend exactly 84 minutes on homework per night.
  5. If the alternative hypothesis is true, there is a 3% chance of obtaining a result like the sample mean of 84 minutes.
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.