9.1 i. (8-24) This is a matched-sample problem. First form differences (new - old), D. The sample mean difference is -4, the sample variance is 21.5. A 95% confidence interval for delta, the population difference is -4 +/- 2.78 * 2.07, or -4 +/- 5.75 (t=2.78, df=4). The hypothesized value, 0, falls within the interval, so the difference is not statistically discernible at the 5% error level.

ii. (8-25) This is an unmatched-sample problem. For the irregulars, we have a mean of 61 and variance of 236.5 and for the regulars we have a mean of 74 and variance of 192.5. Use equation 8-20 to form the confidence interval for the difference (regulars - irregulars), it is (74-61) +/- 2.31 * 14.65 * 0.63, or 13 +/- 21.32.. Here, too, the hypothesized value, 0, falls within the interval, so the difference is not statistically discernible at the 5% error level.

iii. (8-26) A 95% confidence interval for the difference (1963 - 1968) is (0.8 - 0.35) +/- 1.96*0.0440, or 0.45 +/- 0.09. The hypothesized value, 0, is outside the interval, so the difference is statistically discernible at the 5% error level.

9.2 (a.) Use Formula 8-24 (this is a matched sample), Dbar=10, and S_D²=17. A 95% confidence interval for delta, the population difference is 10 +/- 2.78 * 1.83, or 10 +/- 5.09 (t=2.78, df=4). Point estimate is 10, sampling allowance is 5.09.

(b.) The point estimator (10) exceed the sampling allowance (5.09).

(c.) Yes, the value associated with no difference (0) falls outside the 95% confidence interval, so the difference is statistically discernible at the 5% error level.

(d.) Yes.

(e.) H₀: delta = 0, there is no difference between the less experienced appraiser and the other. Yes, this hypothesis can be rejected at the 5% error level.

9.4 (a.) The null hypothesis is that the training programs give equally high productivity on average, but this is written in terms of population means, not sample means (mu2 - mu1 = 0, not Xbar2-Xbar1 =0).

(b.) True.

(c.) The calculated p-value is correct, but is evidence against the hypothesis not in favor of it (it is a small value)!

(d.) True.

(e.) False. The sample should be randomized to avoid bias.

9.7 (a.) Type I, probability alpha.

(b.) Type II, probability beta..

(c.) alpha and beta.

9.11 The new sample should be 25 times larger. To just detect a 0.5mm difference w/ a sample of size 100, 0.5 = 1.96*(sigma/10); to just detect a 0.1mm difference, set 0.1 equal to the allowance for error: 0.1 = 1.96*(sigma/sqrt(n)). The square root of n, sqrt(n) must be 5 times larger than it was (10), so n must be 25 times larger than 100, or 2500.