Aimathic – Interpret p-values – Free Math Worksheets

Interpret p-values

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Difficulty: 1 – 5

52723512

A candidate received \(52\%\) of the vote in the previous election. In a current survey of \(200\) randomly selected eligible voters, \(98\) say they plan to vote for the candidate. Assume that survey responses are independent and have a common support probability. The question is whether support has decreased. Evaluate each statement and justify your conclusion. 1. “Because \(98\) is below the expected value of \(104\), support has definitely decreased.” 2. “If support is \(52\%\), the probability that exactly \(98\) people support the candidate is only about \(3.9\%\). Because this probability is small, the \(52\%\) claim must be false.” 3. “If support is \(52\%\), the probability of observing \(98\) or fewer supporters is about \(21.8\%\). This is a plausible result of random sampling variation.”

Hints

- An expected value is a long-run average, not a guaranteed sample result. - Distinguish a point probability from a p-value. - A left-tailed p-value includes outcomes at least as unfavorable to the null hypothesis as the observed result.

Solution

1. Statement 1 is not valid. The expected value is \(np = 200(0.52) = 104\), but individual random samples commonly fall above or below the expected value. Being below \(104\) does not by itself establish a decrease. 2. Statement 2 is not valid. For a discrete distribution with many possible outcomes, a single point probability can be small even when the model is reasonable. The relevant left-tailed p-value is the probability of an outcome at least as low as the observed one, not only \(P(X = 98)\). 3. Statement 3 is valid. Under \(H_0: p = 0.52\), \(P(X \le 98) \approx 0.21808\). This p-value is not small relative to common significance levels, so the sample does not provide convincing evidence that support has decreased.

Answer

1. Not valid. A result below the expected value can occur through ordinary sampling variation. 2. Not valid. The point probability \(P(X = 98)\) is not the relevant p-value; the left-tail probability is. 3. Valid. The p-value is \(P(X \le 98) \approx 0.21808\), so the result is not statistically significant at common levels.

52723612

A hardware manufacturer claims that at most \(5\%\) of its memory chips are defective. A large customer inspects a random sample of \(150\) chips from a shipment and finds \(12\) defective chips. Assume that chip outcomes are independent and have a common defective probability. The customer claims that the defective proportion is significantly higher than promised. Evaluate the customer’s claim at significance level \(\alpha = 0.05\). 1. State the null and alternative hypotheses. 2. Calculate the p-value. 3. State the statistical conclusion in context.

Hints

- A suspected increase requires a right-tailed test. - The p-value includes the observed count and all more extreme counts in the direction of the alternative. - Compare the p-value with \(\alpha\) before stating the conclusion.

Solution

1. The hypotheses are \(H_0: p \le 0.05\) and \(H_a: p > 0.05\). 2. Let \(X\) be the number of defective chips. At the null boundary \(p = 0.05\), the right-tailed p-value is \(P(X \ge 12) \approx 0.07400\). 3. Because \(0.07400 > 0.05\), fail to reject \(H_0\). The sample does not provide sufficient evidence at the \(5\%\) significance level that the shipment’s defective proportion exceeds \(5\%\).

Answer

1. \(H_0: p \le 0.05\); \(H_a: p > 0.05\). 2. The p-value is \(P(X \ge 12) \approx 0.07400\). 3. Fail to reject \(H_0\). There is insufficient evidence that the defective proportion exceeds \(5\%\).

52732212

An app’s marketing team claims that no more than \(10\%\) of users uninstall the app within one week. An analyst suspects that the true uninstall proportion is higher. In a random sample of \(200\) users, \(26\) uninstall the app. Show, using an exact binomial p-value, that this result is not sufficient to reject \(H_0: p = 0.10\) in favor of \(H_a: p > 0.10\) at the \(\alpha = 0.05\) significance level.

Hints

- Identify which sample outcomes are at least as supportive of \(H_a\) as the observed result. - For a right-tailed test, calculate the probability of the observed count or a larger count. - Compare the p-value with \(\alpha\). - A p-value greater than \(\alpha\) leads to failing to reject \(H_0\).

Solution

1. Let \(X\) be the number of users in the sample who uninstall the app within one week. Under \(H_0\), \(X \sim \operatorname{Bin}(200, 0.10)\). 2. Because \(H_a: p > 0.10\), the p-value is the probability of observing \(26\) or more uninstalls under \(H_0\): \(P(X \ge 26)\). 3. Using the cumulative binomial probability, \(P(X \le 25) \approx 0.8995\). Therefore, \(P(X \ge 26) = 1 - P(X \le 25) \approx 0.1005\). 4. Since \(0.1005 > 0.05\), fail to reject \(H_0\). The sample does not provide sufficient evidence that the one-week uninstall proportion exceeds \(10\%\).

Answer

The exact binomial p-value is \(P(X \ge 26) \approx 0.1005\). Because \(0.1005 > 0.05\), fail to reject \(H_0\). The result is not statistically significant at the \(0.05\) level.

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