Bottle fill: one-tailed test and confidence intervals
A bottling plant claims that its large bottles contain a mean of 500 ml. A quality inspector takes a random sample of 20 bottles from a production run and measures the contents. The sample mean is 496.5 ml and the sample standard deviation is 5.8 ml. The plant manager says that a shortfall of 3.5 ml is small and could easily be due to random variation. The inspector is worried that the plant is systematically underfilling. Previous checks suggest that the fill volume is approximately normally distributed, but the population standard deviation is not known. The sample was taken during one shift, and the bottles were selected from the end of the production line. The inspector must decide whether there is evidence at the 1% significance level that the mean fill is below 500 ml. The manager also wants a 95% confidence interval for the mean fill. A colleague points out that a 99% confidence interval would include 500 ml, so the inspector should not reject the claim.
1. Carry out a one-tailed t-test at the 1% level for H0: μ = 500 against H1: μ < 500. State the test statistic, critical value, and conclusion.
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Standard error = s/√n = 5.8/√20 = 1.297.
t = (496.5 − 500)/1.297 = −2.70 with df = 19.
For a one-tailed test at the 1% level, the critical value is −2.539.
Since −2.70 < −2.539, reject H0. There is evidence at the 1% level that the mean fill is below 500 ml. The p-value is about 0.007.
2. Construct a 95% confidence interval for the mean fill and explain why it is consistent with the test result.
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For df = 19, t(0.025) = 2.093.
95% CI = 496.5 ± 2.093 × 1.297 = 496.5 ± 2.71,
giving (493.8, 499.2) ml.
The value 500 ml lies above the whole interval. This is consistent with rejecting the null hypothesis that the mean is 500 ml in favour of a lower mean; the data suggest the true mean is below 500 ml.
3. The colleague says a 99% confidence interval includes 500 ml, so the inspector should not reject the claim. Evaluate this argument.
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The colleague is confusing a two-sided 99% confidence interval with a one-tailed 1% test. The 99% two-sided interval uses t = 2.861 and gives
496.5 ± 2.861 × 1.297 = 496.5 ± 3.71,
so the interval is (492.8, 500.2) ml, which does include 500. That corresponds to a two-tailed test at the 1% level, where the critical values are ±2.861 and t = −2.70 does not reject.
The inspector’s question was directional: is the mean below 500? If that direction was specified before seeing the data, the one-tailed test is appropriate and the 1% rejection stands. If the direction was chosen only after noticing that the sample mean was low, the two-tailed conclusion is safer. The small sample also relies on approximate normality; a strong outlier or skewness would weaken both the interval and the test.