A Weak Consignment of Coir Rope: One Tail or Two?
The purchase officer of a coir cooperative buys rope on the specification that the mean breaking strength is 1000 newtons. A consignment arrives, and a laboratory tests 25 randomly chosen rope samples. The sample mean breaking strength is 960 N with a sample standard deviation of 100 N. The officer suspects the consignment is weaker than specified and wants to reject it, but the supplier argues that a sample mean of 960 N could easily arise by chance from a consignment that is exactly on specification. The cooperative's rule is to test at the 5 per cent level of significance. The officer has read that a one-tailed test is used when a direction is suspected, but the supplier insists on the usual two-tailed test. The laboratory also reports a 95 per cent confidence interval for the mean. The board wants a clear recommendation, the reasoning behind the choice of test, and an explanation of what could go wrong with whatever decision is taken.
1. Carry out the two-tailed test at the 5% level and state the conclusion.
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Hypotheses: $H_0:\mu=1000$ N against $H_1:\mu\ne1000$ N.
Test statistic: the population SD is unknown and $n=25$ is small, so use the t-test with $df=24$, assuming the strengths are approximately normal and the sample is random.
$$t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}=\frac{960-1000}{100/\sqrt{25}}=\frac{-40}{20}=-2.00$$
Critical values: at 5% two-tailed with $df=24$, $t_{0.025}=\pm2.064$.
Decision: $|t|=2.00<2.064$, so $H_0$ is not rejected at the 5% level.
The two-tailed p-value is about 0.057, just above 0.05. Conclusion in context: the sample does not provide sufficient evidence at the 5% level to establish that the mean strength differs from 1000 N; a shortfall of 40 N is compatible with sampling fluctuation. Using z instead of t (critical value 1.96) would wrongly tip the decision - with $n=25$ and $\sigma$ estimated, the t-distribution is the correct reference.
2. Explain why a one-tailed test gives a different answer here, and advise which test the cooperative should adopt.
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One-tailed version: $H_1:\mu<1000$ N, all 5% placed in the lower tail. With $df=24$, the critical value is $t_{0.05}=1.711$ (lower tail $-1.711$).
Since $t=-2.00<-1.711$, $H_0$ is rejected at the 5% level. The one-tailed p-value is about 0.028.
Same data, opposite conclusions - the difference is entirely in how the 5% is allocated. A two-tailed 5% test is equivalent to a one-tailed test at 2.5%, so the one-tailed test is more powerful in the suspected direction but blind to over-strength and to any surprise above specification.
Which to adopt: the one-tailed test is defensible only if the cooperative's sole concern was weakness, and if that directional hypothesis was fixed before the sample was drawn. Since the specification is a two-sided quality standard (rope that is far too strong is also off-specification and costs more), and since rejecting a consignment is a serious commercial act, the honest choice is the two-tailed test. Choosing the one-tailed test after seeing that the mean fell below 1000 is not a legitimate way to reach significance.
3. Compute the 95% confidence interval, relate it to the test result, and explain the practical risk in the final decision.
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Interval:
$$\bar{x}\pm t_{0.025,24}\frac{s}{\sqrt{n}}=960\pm2.064\left(\frac{100}{5}\right)=960\pm41.3$$
$$(918.7,\ 1001.3)\ \text{N}$$
Relation to the test: the interval contains 1000 N, which is exactly consistent with not rejecting $H_0$ in the two-tailed test at 5%. This duality ($1-\alpha$ interval, $\alpha$-level two-tailed test) is the point to state explicitly.
Practical risk is not removed by 'not rejected': the interval is compatible with a true mean as low as about 919 N, a 81 N shortfall - commercially serious even if not statistically demonstrated. The non-rejection is largely a statement about sample size: with $n=25$ the test has limited power to detect a 40 N shortfall.
Errors:
- Type I error (5%): rejecting a consignment that is actually on specification - the cooperative loses a supplier and incurs unjustified rejection costs.
- Type II error: accepting a genuinely weak consignment; its probability here is large because of the small sample.
Recommendation: do not accept the consignment outright. Report that the evidence is inconclusive at 5% two-tailed, quote the interval, and resolve the matter by testing more samples (power rises with $n$) or by agreeing a practical tolerance band around 1000 N in the purchase contract, rather than leaning entirely on a borderline $t=-2.00$.