Testing a Claim of Underfilled Milk Bottles
A consumer group in Yaoundé suspects that a dairy company is underfilling its 500 ml milk bottles. The company claims the mean fill is 500 ml. The group takes a random sample of 12 bottles and measures the contents, obtaining a sample mean of 496 ml and a sample standard deviation of 6.5 ml. The fill volumes are assumed to be approximately normally distributed. The group plans a one-tailed test at the 5% significance level because only underfilling is harmful to consumers; overfilling is not the complaint. The company replies that its own 95% two-sided confidence interval for the mean, calculated from the same sample, includes 500 ml, so it argues there is no evidence of underfilling. The consumer group must decide whether the sample supports its suspicion, how to interpret the p-value, and how to answer the company's confidence-interval argument without ignoring the difference between one-sided and two-sided procedures.
1. State the null and alternative hypotheses, calculate the test statistic, and carry out the test at the 5% level. Use the t-distribution with 11 degrees of freedom; the relevant one-tailed critical value is 1.796.
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Let μ be the population mean fill. The hypotheses are H0: μ = 500 ml, H1: μ < 500 ml.
The sample size is n = 12, so df = 11. The standard error is SE = s/√n = 6.5/√12 ≈ 1.876 ml.
The test statistic is t = (x̄ − μ0)/SE = (496 − 500)/1.876 ≈ −2.132.
For a one-tailed test at the 5% level with df = 11, the critical value is −1.796. Since −2.132 < −1.796, the test statistic lies in the rejection region. Reject H0. There is sufficient evidence at the 5% level to support the claim that the mean fill is less than 500 ml.
2. Explain why a one-tailed test is appropriate here and interpret the p-value in context.
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A one-tailed test is appropriate because the consumer group's complaint is specifically about underfilling. A mean above 500 ml would not support the complaint, so the rejection region should be placed entirely in the lower tail. This gives the test more power to detect the effect of interest than a two-tailed test, which would divide the 5% significance level between both tails.
The p-value is approximately 0.028. This means that, if the true mean fill were 500 ml, there would be about a 2.8% chance of obtaining a sample mean as low as 496 ml, or lower, with n = 12 and s = 6.5 ml. Since 0.028 < 0.05, the sample result is statistically significant at the 5% level in the one-sided sense. In context, the data provide moderate evidence that the company is underfilling the bottles.
3. The company says its 95% two-sided confidence interval includes 500 ml. Reconcile this with your conclusion.
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The 95% two-sided confidence interval is x̄ ± t(0.025,11) × SE = 496 ± 2.201 × 1.876 ≈ 496 ± 4.13 = (491.87, 500.13) ml.
This interval does include 500 ml. That is consistent with a two-sided test at the 5% level: the two-sided critical value is 2.201, and the observed |t| = 2.132 is just below it, so a two-sided test would not reject μ = 500.
However, the consumer group pre-specified a one-sided alternative, μ < 500. For that directional question, the critical value is 1.796, and t = −2.132 does fall in the lower rejection region. The one-sided 95% upper confidence bound would be 496 + 1.796(1.876) ≈ 499.4 ml, which is below 500 ml. The two procedures answer different questions, so the company's two-sided interval does not overturn the one-sided conclusion; it does show, however, that the evidence is moderate rather than overwhelming.