Question

You are investigating the yield stress of a specific alloy to be used in a gear. The yield stress of the alloy is known to be approximately normally distributed with o = 60 psi. A random sample of 12 specimens has a mean yield stress of 3450 psi. (a) If the mean yield stress is 3500 psi, what is the smallest level of significance at which you would be willing to reject the null hypothesis? (b) Calculate Type II error if the true mean is 3470 psi and a=0.01? (c) Suppose that you wanted to reject the null hypothesis with probability at least 0.8 if mean yield stress is 3470 psi. What should be the sample size?

Answer 1

The solution is given as follows.

DATE The 34 50 psi mean yield stress is x = 3450 psi j 12 60 psi - a) Hoy te = = 3500 vs H, : ll 33500 we are willing Ho a = Pł = P .; The small level of significance for which to reject is Type I error P [ Reject Ho Ho is true -PL P[ 1X>3450 % = 3500 EP [13123450 - 3500 60 /52 P. [121 > 2.89 11 L = 0.0019 b) HA 2 M = 3.470 = 0.01 is given. Now

Reject JO Ho if 131> K by Neyman Pearson lemma. af 1% Los K is such that PIRIIK Ho is true Ho is true = 0.01=& P [17]<k in -k < <k Tre = 3500 = 3500 35007 = 0.99 1 A 0.99 -k - 3500 = 0.99 60/512 <Zc K-3500 60/512 2.5758 K - 3500 60 /512 k = 3455.385 P[ Type B - P II error] = P[Accept Hof He is falle P[Accept Hou= 3 470 = [15 < 3455.385 | pe=3470 - P-3455-385-3470 <<<3455-385-3470 60 / 52 60 0/52

DAT 1$ = P [<-0.84379]- P[2<-399.837 0.19939 0.11939 ci = 3470 psi Pl1Z] > Zenit] =1-0.8 = 0.2 PIX-> -0.2 Zcoit = Zoot 1.282 X - 601 Sn 1.282 Sn 1.282 -0.333 - (3.8463) n 14.79 15