Wait-time claim, p-value interpretation, and sample size
A coffee chain advertises that the mean wait at a city branch is no more than 4.5 minutes. A consumer group records waiting times for a random sample of 36 customers at a busy period. The sample mean is 4.90 minutes and the sample standard deviation is 1.20 minutes. The group plans a one-tailed test at the 5% significance level to see whether the mean wait is greater than 4.5 minutes. It will assume waiting times are approximately normally distributed; the population standard deviation is unknown. After the test, the branch manager says: 'If you reject my claim, there is only a 5% chance that the claim is true.' The group also wants to plan a future study that estimates the mean wait to within 0.20 minutes with 95% confidence, using the same estimated standard deviation. Discuss whether a sample of 36 is adequate and state any assumptions.
1. Carry out the one-tailed test at the 5% significance level and state the conclusion.
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Let μ be the population mean wait. The hypotheses are H0: μ = 4.5 min (or μ ≤ 4.5) against H1: μ > 4.5 min. Because the population standard deviation is unknown and n = 36, use a one-sample t-test with df = 35. The standard error is 1.20/√36 = 0.200. The test statistic is t = (4.90 − 4.5)/0.200 = 2.00. At the 5% one-tailed level, the critical value is t(0.95, 35) ≈ 1.690. Since 2.00 > 1.690, reject H0. Equivalently, the p-value is P(T35 ≥ 2.00) ≈ 0.027, which is less than 0.05. There is sufficient evidence at the 5% level that the population mean wait exceeds 4.5 minutes, subject to random sampling and approximate normality.
2. Critique the manager's statement about the 5% chance and explain what the p-value does and does not measure.
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The manager's statement is a misinterpretation. The 5% is the significance level: if H0 were true, a test would reject it in about 5% of repeated samples, which is the Type I error rate. It is not the probability that H0 is true after rejection. The p-value ≈ 0.027 is the probability, assuming μ = 4.5, of getting a sample mean at least as large as 4.90; it is not P(μ = 4.5 | data). A posterior probability would require a prior and is not given. Also, failing to reject H0 would not prove that μ ≤ 4.5; it would only mean the sample did not provide sufficient evidence against it.
3. Determine the sample size needed for a 95% confidence interval with margin of error 0.20 minutes, and comment on whether n = 36 is adequate.
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For margin of error E = 0.20, a 95% confidence level, and estimated σ ≈ 1.20, use n = (zσ/E)² = (1.96 × 1.20/0.20)² = (11.76)² ≈ 138.3, so at least 139 customers. If using a t critical value near 1.98, the required n is about 141. Either way, n = 36 is far too small. The current margin of error is roughly 1.96(1.20/6) ≈ 0.392 min, about double the target. The planning assumes that the estimated standard deviation 1.20 is a reliable stand-in for the population standard deviation and that future samples remain random; if waiting times are strongly skewed, a larger sample or a bootstrap interval may be preferable.