Tablet Weight: A Significant Difference That Is Not a Defect
A pharmaceutical unit in Bhubaneswar presses 500 mg paracetamol tablets. The pharmacopoeia tolerates individual tablet weights between 475 and 525 mg. A quality-control officer weighs a random sample of 25 tablets from a batch and obtains a mean of 496.5 mg with a standard deviation of 8.0 mg; tablet weights in this plant are known to be approximately normal. The production manager wants to release the batch. The QC officer replies that the mean is 'significantly below 500 mg' and recommends rejection. Company procedure, written when the plant ran a single high-precision press, prescribes a two-sided test at the 5 per cent level. A consultant observes that the company's second plant routinely samples 100 tablets, and that the same 3.5 mg shortfall there would produce a much larger test statistic. Nobody has yet asked whether the question of interest is really 'is the mean different from 500 mg' or rather 'are the tablets underweight'.
1. Carry out the prescribed test at the 5% level and state your conclusion, with the assumptions on which it rests.
Read the worked answer
H0: μ = 500 mg; H1: μ ≠ 500 mg (two-sided, as the written procedure demands).
Standard error = s/√n = 8.0/5 = 1.6 mg. Degrees of freedom = 25 − 1 = 24.
t = (x̄ − μ0)/SE = (496.5 − 500)/1.6 = −3.5/1.6 = −2.19.
Critical value: t(0.025; 24) = 2.064. Since |−2.19| > 2.064, reject H0 at the 5% level; the two-tailed p-value is about 0.038.
The conclusion is that the batch mean is statistically below 500 mg. Assumptions to state: (i) the 25 tablets are a genuinely random sample of the batch - if they came from the top of a hopper this fails outright; (ii) the weights are approximately normally distributed, which the plant's records support and which matters because n is small and σ is estimated; (iii) the sample standard deviation 8.0 is a usable estimate of the process standard deviation; (iv) tablets are independent - no clustering from a single out-of-adjustment hour of pressing. The test says nothing about individual tablets, only about the mean, and a t-test with n = 25 has limited power to detect small shifts.
2. Construct the 95% confidence interval for the batch mean and explain its relationship to the test.
Read the worked answer
x̄ ± t(0.025; 24) × s/√n = 496.5 ± 2.064 × 1.6 = 496.5 ± 3.30.
95% CI: (493.2 mg, 499.8 mg).
The link to the test is exact: a two-sided test at the 5% level rejects H0 precisely when the 95% confidence interval excludes the hypothesised value. Here 500 lies just outside the upper limit, which is why |t| = 2.19 just exceeds 2.064 - the rejection is marginal, and a 4% level test would have failed to reject. Reporting only 'p = 0.038, significant' conceals this fragility; the interval shows it plainly.
The interval is also more informative for the manager: the data are consistent with a true batch mean anywhere from 3.2 mg to 6.8 mg below the label value. If the tolerable shortfall were, say, 2 mg, the interval would show that it cannot be excluded and the batch would be suspect; if the tolerable shortfall were 10 mg, the interval would clear it. The interval is in the units of the problem, survives the yes/no of a test, and its width (6.6 mg) shows how much the 25-tablet sample actually pins down.
3. The QC officer wants the batch rejected on the strength of the test. Advise the plant, addressing practical significance, the 100-tablet plant, and the choice of one- or two-sided testing.
Read the worked answer
Statistical significance and product quality are different questions here. The pharmacopoeia allows 475–525 mg, and the point estimate 496.5 mg, together with the whole interval 493.2–499.8 mg, sits comfortably inside those limits. The label value of 500 mg is a target, not a specification limit; nothing about the batch is non-conforming. Rejecting it would be a costly Type I error - scrapping a conforming batch. The genuine regulatory worry is the proportion of individual tablets outside tolerance. With mean 496.5 and s = 8, P(W < 475) corresponds to z = (475 − 496.5)/8 = −2.69, about 0.36% of tablets - worth checking against the pharmacopoeial limit on individual content/weight, which is a separate test from the mean.
The 100-tablet plant illustrates that test statistics grow with √n: the same data give z = −3.5/(8/10) = −4.375, p < 0.0001. A larger sample detects ever smaller departures, so 'significant' is not a measure of importance; the effect size (3.5 mg on 500, i.e. 0.7%) and the confidence interval should be reported instead.
On sidedness: the real question is 'are the tablets underweight', so a one-sided test (critical t = 1.711) is the appropriate design and would reject more readily. But the direction must be fixed in the written procedure before the data are seen; choosing the one-sided test after noticing that x̄ fell below 500 is p-hacking and invalidates the p-value. The recommendation: release the batch, report mean, CI and the proportion outside tolerance, and revise the procedure to specify a one-sided test of underweight and a sampling plan with adequate power.