One-sided underfill test and its two-sided interpretation
A Galician bottled-water company labels its bottles as containing 1,500 ml. A consumer association takes a random sample of 36 bottles and finds a sample mean of 1,497.2 ml with a sample standard deviation of 8.5 ml. The association suspects systematic underfilling. The company replies that the label refers to an average of at least 1,500 ml, so a small sample mean is normal variation. The association plans a one-sided test at the 5% significance level. Its lawyer asks whether the same data would also support a two-sided claim that the mean is not exactly 1,500 ml, and what a 95% confidence interval shows. The association must decide whether to publish a warning: a false warning would damage its credibility, while missing underfilling would leave consumers unprotected. Assume the fill volumes are approximately normal.
1. State the hypotheses, compute the test statistic, and carry out the one-sided test at α=0.05.
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Let μ be the population mean fill. H0: μ=1500 ml (or μ ≥ 1500) against H1: μ < 1500 ml. With n=36, xbar=1497.2, s=8.5, the standard error is s/sqrt(n)=8.5/6≈1.4167. t=(1497.2-1500)/1.4167≈-1.976. With df=35, the one-sided 5% critical value is -1.690. Since -1.976 < -1.690, reject H0. The p-value is about 0.028, below 0.05. There is significant evidence at the 5% level that the mean fill is below 1500 ml.
2. Find a 95% confidence interval for μ and reconcile it with the one-sided conclusion.
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For df=35, t(0.025)=2.030. The interval is 1497.2±2.030*1.4167=1497.2±2.876, i.e. (1494.32, 1500.08) ml. This interval contains 1500, so a two-sided test of H0: μ=1500 at 5% would not reject. There is no contradiction: the one-sided test puts all 5% in the lower tail and is more sensitive to underfilling, while the 95% two-sided interval corresponds to a two-sided test. The data support a one-sided underfill claim but not a two-sided claim that the mean differs from 1500.
3. If the true mean were 1498 ml, estimate the probability that the test fails to detect underfilling. What does this imply about the sample size?
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Under α=0.05 one-sided, reject if xbar < 1500 - 1.690*1.4167 ≈ 1497.61 ml. If μ=1498 and we approximate the distribution of xbar by N(1498, 1.4167^2), then z=(1497.61-1498)/1.4167≈-0.275. The probability of not rejecting is P(Z > -0.275)≈0.608, so power is only about 0.392. There is roughly a 61% chance of missing a 2 ml deficit. The sample of 36 is too small to reliably detect such a small underfill; a larger sample would be needed if the association wants high power against 2 ml.