Question

Acme Products manufactures widgets. It has two machines which produce widgets. It is desirable that both machines produce widgets of the same dimensions. At regular time intervals Acme's quality control tests the widgets from both machines. The last test produced the following results. SM1 = 5.31 S1 = 1.27 n1 = 64 SM2 = 5.69 S2 = 1.56 n2 = 64 Test Ho; p 1 = 2 vs. Ha:p1<>u2 at the a = .05 level of significance. Assume that the two standard deviations are equal (pooled yes). In this exam <> means “not equal". Your answer will be a triple (t-value, p-value, conclusion) where conclusion answers the question as to reject or not reject Ho. O (t = -1.9964, p - ,0480, reject Ho) O (t = -1.9753, p=,0252, reject Ho) o (t = -1.5112, p=0666, do not reject Ho) 0 (t = -1.5112, p - 1332, do not reject Ho)

Answer 1

Jucky Page No.: Date : Solution: 11 We have SMI = 5.31 S = 1.27 SM2= 5-69 52 = 1.56 11=64 h2=64 The hypothesis are, Ho ll=42 Vs .H: 41 242 he test Statistic assuming equal raglance ta SMI-SM2 6.-16:24 13-15, ni th2-2 yo 64 Host how, petting values we get ta 5.31 – 5.69 63x1.272+63 x 1.5721 64764-2 84 54 a ~0.38 0.2514 Scanned with CamScanrler

Jucky Page No.: Date : I 1 titl=1.5112 now Exalue = Pct>1.5112) p-Vallee = 0.13 3 2 ( By using t-table) Here. p-value = 0.1332 0.05 fo-value > 0.05. Hence we donot reject to. CSScanhed with CamScanner

The answer is <d> (t=1:5712, p=0.1332. do not reject Ho) CSScanned with CamScanner