The Valentine’s Day Chocolate Case
In the Valentine’s Day chocolate case we are testing H0: μ = 330 versus Ha: μ ≠ 330 by setting α = .05. We have seen that the mean of the reported order quantities of a random sample of n = 100 large retail stores is . Assuming that σ equals 40, it follows that because
is between −z.025 = −1.96 and z.025 = 1.96, we cannot reject H0: μ = 330 by setting α = .05. Since we cannot reject H0, we might have committed a Type II error. Suppose that the candy company decides that failing to reject H0: μ = 330 when μ differs from 330 by as many as 15 valentine boxes (that is, when μ is 315 or 345) is a serious Type II error. Because we have set α equal to .05, β for the alternative value μa = 315 (that is, the probability of not rejecting H0: μ = 330 when μ equals 315) is the area under the standard normal curve to the left of
Here z* = zα/2 = z.05/2 = z.025 since the alternative hypothesis (μ ≠ 330) is two-sided. The area under the standard normal curve to the left of −1.79 is 1 − .9633 = .0377. Therefore, β for the alternative value μa = 315 is .0377. Similarly, it can be verified that β for the alternative value μa = 345 is .0377. It follows, because we cannot reject H0: μ = 330 by setting α = .05, and because we have just shown that there is a reasonably small (.0377) probability that we have failed to detect a serious (that is, a 15 valentine box) deviation of μ from 330, that it is reasonable for the candy company to base this year’s production of valentine boxes on the projected mean order quantity of 330 boxes per large retail store.
In the following box we present (without proof) a formula that tells us the sample size needed to make both the probability of a Type I error and the probability of a Type II error as small as we wish: