Question

2 Exercise 11-11 (LO11-2) Ms. Lisa Monnin is the budget director for Nexus Media Inc. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information. points Skipped Sales ($) 144 148 148 166 138 140 Audit ($) 130 102 129 143 125 120 139 At the 0.010 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff? a. State the decision rule. (Round your answer to 3 decimal places.) References b. Compute the pooled estimate of the population variance. (Round your answer to 2 decimal places.)

Answer 1

for sales staff::$\overline{X}=147.33,S=10.01,n=6$

for audit staff:$\overline{X_2}=126.86,S_2=13.48,n_2=7$

The provided sample means are shown below: X1 147.33 X2 = 126.86 Also, the provided sample standard deviations are 81 10.01 $213.48 and the sample sizes are n 6 and n2 7 (1) Null and Aternative Hypotheses The following null and alternative hypotheses need to be tested: This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. Testing for Equality of Variances A F-test is used to test for the equality of variances. The following F-ratio is obtained: Fsi10.012 j13.4870.551 The critical values are F = 0.069 and Fr hypothesis of equal variances is not rejected. 1 1.464, and since F = 0.551, then the null

(2) Rejection Region Based on the information provided, the significance level is α-: 0.01, and the degrees of freedom are df-11 In fact, the degrees of freedom are computed as follows, assuming that the population Variances are equal Hence, it is found that the critical value for this right-tailed test is te df = 11 2.718, for a 0.01 and The rejection region for this right-tailed test is R-t:t > 2.718) Since it is assumed that the population variances are equal, the t-statistic is computed as follows 147.33 - 126.86 0.254 (6-1)10.0147(7-1)13.482(- ) 6+7-2 Since it is observed that t = 0.254 <ち= 2.7 18, it is then concluded that the nul hypothesis is not rejected 0.402, and since p-0.402 > 0.01, it is concluded Using the P-value approach. The p-value is p that the null hypothesis is not rejected It is concluded that the null hypothesis Ho is nof rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is greater than μ2, at the 0.01 significance level The 99% confidence interval is-229.49 < μι-μ2 270.43