Late-delivery rate and hypothesis test
An online retailer promises next-day delivery on at least 95% of orders. In a random sample of 200 orders, 16 arrived late. The operations team will redesign the warehouse only if there is significant evidence at the 5% level that the late-delivery rate exceeds 5%. Let X be the number of late orders in the sample; assume orders are independent and use a normal approximation to the binomial distribution with continuity correction. The manager is also worried about the test's ability to detect a true late rate of 8%, because missing such a rate would leave customers dissatisfied. Before making a decision, the team wants the hypotheses, the critical region, the p-value, the probability of a type II error at p=0.08, and a short comment on whether the observed count is practically important rather than only statistically significant.
1. State the hypotheses and determine the critical region for the test at the 5% significance level, using the normal approximation with continuity correction.
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Let p be the true proportion of late orders. The hypotheses are
H0: p≤0.05 versus H1: p>0.05.
Under H0 at the boundary p=0.05, X is approximately normal with mean np=200(0.05)=10 and variance np(1-p)=200(0.05)(0.95)=9.5, so the standard deviation is sqrt(9.5)=3.082. For an upper-tail test at α=0.05, the critical z-value is about 1.645. With continuity correction, reject H0 when
(X-0.5-10)/3.082 ≥1.645.
This gives X-10.5≥5.07, so X≥15.57. Since X is an integer, the critical region is X≥16. The actual significance level is approximately P(Z≥1.784)≈0.037, slightly below 5% because the discrete distribution cannot achieve exactly 5% with this sample size.
2. Compute the p-value for the observed result and state the decision. Then calculate the probability of a type II error at p=0.08 and interpret it.
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The observed value is X=16. With continuity correction,
z=(15.5-10)/3.082≈1.784,
so the p-value is approximately P(Z≥1.784)≈0.037. Since 0.037<0.05, reject H0. There is significant evidence at the 5% level that the late rate exceeds 5%.
For a type II error at p=0.08, X has mean 200(0.08)=16 and variance 200(0.08)(0.92)=14.72, so σ=3.837. A type II error occurs when H0 is not rejected, i.e. X≤15. With continuity correction,
P(X≤15 | p=0.08)≈P(Z≤(15.5-16)/3.837)=P(Z≤-0.130)≈0.448.
Thus the power at p=0.08 is only about 55%. There is roughly a 45% chance that a true 8% late rate would not be detected by this test. The observed 16 late orders out of 200 is 8%, well above the 5% promise, but the test has limited power at that alternative; a larger sample would be needed for a reliable decision.
3. The team asks whether 16 late orders is practically important, not only statistically significant. Give a reasoned comment using the observed proportion and the test's limitations.
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The observed late proportion is 16/200=0.08, which is 8%, or 3 percentage points above the promised maximum of 5%. In relative terms, the observed late rate is 60% higher than the promise. That is operationally meaningful because it represents more missed delivery commitments than advertised, and 16 affected customers in a sample of 200 is not a trivial count.
However, statistical significance at the 5% level does not by itself prove that the long-run rate is far above 5%. The normal approximation gave a p-value near 0.037, which is close to the threshold, and the power at a true rate of 8% is only about 55%. A confidence interval for p would still be fairly wide. The practical decision should therefore combine the observed 8% rate, the cost of redesign, and the consequence of missing the promise. If the cost of redesign is modest, the evidence is strong enough to act; if redesign is expensive, the team should collect a larger sample before committing.