Problem 8.42 In a study comparing the breaking strengths (measured in psi) of two types of fiber, random samples of sizes n1 = n2 = 9 of both types of fiber were tested, and the following summary statistics were obtained: x1 = 40.1, s2 1 = 2.5; x2 = 41.5, s2 2 = 2.9. Assume both samples come from normal distributions. Although fiber 2 is cheaper to manufacture it will not be used unless its breaking strength is found that to be greater than that of fiber 1. Thus, we want to test H0 : ∆= µ1 − µ2 = 0 against H1 : ∆= µ1 − µ2 < 0. Observe that the alternative hypothesis as formulated here asserts that the breaking strength of fiber 2 is greater than fiber 1. Controlling the type I error here protects you from using an inferior, although cheaper, product.

(a) Use the given data to test the null hypothesis at the levels: α = 0.05; α = 0.01.

(b) Test the null hypothesis by constructing an appropriate one sided 100(1−α)% confidence interval, where α = 0.01.