Question

Q4- Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. It is known that the distribution is normal and a = 02 = 1.0 psi. From a random sample of size n = 10 and n2 = 12, we obtain x1 = 162.5 and X2 = 155. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. (For this question, to decide on the test statistic only check whether the population variances are known or not. Neglect sample sizes.) a) Based on the sample information, should it use plastic 1? Use a =0.05 in reaching a decision. b) Suppose that the true difference in means is really 12 psi. Find the power of the test assuming that a =0.05.

Answer 1

H. Mi-l2 <10 vs Hi 4.-M27, 10 " ,- >c] = x 1 PL (7, -72) - 10 - 10 J = 0.05 Onths) 2 04.) 76 C-LO ) -26-1-64) 0.43 C-10 = 1.64 0.43 210 71 Reject Ho it. F-227 10.71 from the data we got XrX 2 = 16.5-155 = 7.5 710.71 we do not reject Ho at ss level. we do not wre plastic 1. (6 pl xi-x2 > 10.71] - P[ x,-*2-12 -3] 0.43 & (3) - 0.99 8 = power.