A Hypothesis Test With a One-Sided Story
A manufacturer claims batteries last at least 500 minutes on average. A sample of 40 batteries gives mean 492 minutes with sample standard deviation 18 minutes. A student runs a two-tailed z-test at α = 0.05 and obtains z = (492 − 500)/(18/√40) ≈ −2.81, p ≈ 0.005, and rejects H₀. A second student argues the test should be one-tailed because only low lifetimes are bad, and that the p-value should be halved. A third notes the sample size is large but the population is skewed, and that the sample standard deviation was used.
1. State the appropriate hypotheses, compute the test statistic and p-value, and justify one- versus two-tailed.
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The claim to challenge is μ ≥ 500, so H₀: μ = 500 and H₁: μ < 500 (one-tailed), since only unusually low means would contradict the manufacturer.
z = (492 − 500)/(18/√40) = −8/2.846 ≈ −2.81.
One-tailed p = P(Z < −2.81) ≈ 0.0025. Two-tailed p ≈ 0.005.
At α = 0.05 both reject, but the correct p-value is the one-tailed 0.0025. The judgement call is that the alternative direction is fixed by the practical concern before seeing data; halving after the fact is only valid if the direction was pre-specified.
2. Evaluate the third student's concern about using s and a skewed population with n = 40.
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Because σ is unknown and replaced by s, a t-test with 39 degrees of freedom is technically correct; with n = 40 the t and z critical values are close (t ≈ 2.02 vs z ≈ 1.96 for two-tailed 0.05), so the conclusion is unchanged here.
The skew matters for the sampling distribution of the mean, but the Central Limit Theorem makes the distribution of x̄ approximately normal for n = 40 unless skew is severe. The safe statement is that the test is approximate, and the very small p-value makes the conclusion robust to these approximations. The judgement is to report the t-test and note the assumption rather than to abandon inference.