Question

Warren County Telephone Company claims in its annual report that "the typical customer spends $60 per month on local and long-distance service." A sample of 12 subscribers revealed the following amounts spent last month. $64 $65 $64 $68 $60 $65 $68 $62 $64 $59 $53 $68

Answer 1

Given Data: 64, 65, 64, 68, 60, 65, 68, 62, 64, 59, 53, 68 We calculate sample mean and std deviation from given data Sample Mean, r - - Sample Mean, 2 D(r) 760 Sample Variance, s2 Sample std dev, s - s2v19.1515154.376244 Test: μ 60 with t dist" Hypothesis test 12 Σ(1-2)2 210.666667 19.151515 (Null Hypothesis) (Alternative Hypothesis, also called H1 This is two tailed test T 63.333333 s 4.376244 n 12 Significance level, α-0.05 (If no value is given, we take level of 0.05) Degrees of freedom df-n-1 11 T-63.333333-60 Test statistic,d4.376244/V12 Test statistic 2.639 Using t-table, we can observe that the p-value is between 0.02 and 0.05 Using excel we find p valueP(t 2.639) 2*t.dist(2.639, 11,1) - - 2.639 n 4.376244/V12

0.0230521603 tcritical-ta/2.df-t0.025.df-1,-2.201 (from t-table, two-tails, d.f. = 11 ) Critical Value, te2.201 Rejection criteria: Rejection region is in direction of Alternative hypothesis, since Ha : μメ60, rejection rule is: Reject Ho if Ito > te 2.201 i.e, we reject if t<-2.201 or t> 2.201 equivalently if p-value < α Rejection region has been shaded: 0.025 0.025 2.201 -2.201 Decision : Since tol > te or equivalently since p-value < α, we rejec the null hypothesis. Hence at 5% significance level, we have sufficient evidence to conclude that μ is different from 60 We reject the claim of the telephone company oaThank youes