Question

A consumer organization collected data on two types of automobile batteries, A and B. They took random samples of size 12 from each battery. The summary statistics are ¯ xA = 36.51, ¯ xB = 34.21, sA = 1.43, and sB = 0.93. Assume that the data are normally distributed with σA = σB.

(a) Is there evidence to support the claim that type A battery mean life exceeds that of type B? Use α = 0.01.

(b) Construct a one-sided 99% confidence bound for the difference in mean battery life (corresponding to the test in part a).

Answer 1

.

.

Answer - given that NA = RB = 12 88 = 34,21 2:001 : XA = 36.51 ; SA = 1.43 : Sp=0.93 The pooled estimate for the variance is Ś = (ha-D så + (M8- 1) sở - (18-1) 1.437 02-1)0.93 HA+ng-2 12+12-2 S* => 14549 .: S= 1.2069 Hypothesis: Ho: MAMB VS H.: MA> Mo The test statistic under Ho ip, t = XA-XB Ana+MB-2 = 4.6707 shat the - 36.51 - 34.21 1.2062 ta t ta critical value : kiable = krat ne-2,2 = 2.8188 = k12+12-2,0.01 Scanned with CamScanner

Decision Rule: Here, Estat = 4.6707 >bon = 2.8188 .: Reject Ho conclusion : to conclude the That of type B. Hence, Ther is type A battery sufficient evidence a mean life exceeds on The 99% confidence Interval for The difference mean battery life is, [TA-To] & KnAMB-2, ap / h the 36.51-34.21 I fiat19-2,0048 12062 tat ta 2.83 2.8188 x 0.4924 2.81 1.388 [0.9123.688]. - The end. Thank's you. Scanned with CamScanner