Hypothesis Test on a Mean with Unknown Variance
A manufacturer claims that the mean mass of its cereal boxes is at least 500 g. A consumer group samples 16 boxes and obtains a sample mean of 494 g with sample standard deviation 9 g. The group suspects the true mean is lower and tests at the 5% significance level, assuming the masses are normally distributed. A student sets up H₀: μ = 500, H₁: μ < 500, and computes z = (494 − 500)/(9/√16) = −2.667, then compares with the standard normal critical value −1.645 and rejects H₀. The marking guide checks the choice of distribution, the direction of the test, and the interpretation of the p-value.
1. Critique the student's test statistic and state the correct distribution and critical value. Carry out the test correctly.
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The student used a z-statistic, which requires the population standard deviation to be known. Here σ is unknown and estimated by s = 9 from a sample of size n = 16, so a t-statistic with n − 1 = 15 degrees of freedom is required.
Correct statistic: t = (494 − 500)/(9/√16) = −6/2.25 = −2.667. The numerical value happens to match the student's z, but the reference distribution differs.
This is a one-tailed (lower-tail) test at α = 0.05. The critical value is t₁₅,₀.₀₅ ≈ −1.753. Since −2.667 < −1.753, reject H₀. The p-value is P(t₁₅ < −2.667) ≈ 0.009, which is below 0.05, consistent with rejection. The student's conclusion is correct but the justification (normal critical value) is not.
2. Explain the effect on the conclusion if the sample size had been 9 instead of 16, keeping the same sample mean and standard deviation.
Read the worked answer
With n = 9, the standard error becomes 9/√9 = 3, so t = −6/3 = −2.0 with 8 degrees of freedom. The critical value is t₈,₀.₀₅ ≈ −1.860. Since −2.0 < −1.860, H₀ is still rejected, but the margin is smaller. The p-value is P(t₈ < −2.0) ≈ 0.040, still below 0.05.
The key point is that the same mean difference and same s produce a smaller test statistic magnitude when n is smaller (because the standard error is larger), and the critical value also moves further from zero with fewer degrees of freedom. Both effects push toward non-rejection. A student should compute the new standard error and new degrees of freedom rather than assuming the conclusion is unchanged.