Question

Save Homework: HW 25: 811.3 through $11.5, inclusive Score: 0.33 of 1 pt W 11.3.3-T 2 of 19 (17 complete) HW Score: 58.44%, 11.1 of 19 pt Question Help Use the given statistics to complete parts (a) and (b). Assume that the populations are normally distributed. (a) Test whether H1 H2 at the c = 0.05 level of significance for the given sample data. (b) Construct a 90% confidence interval about 141-42 n Population 1 24 49.4 5.7 Population 2 19 47.1 13.8 x S (a) Identify the null and alternative hypotheses for this test. OA HOFHH2 H4 H4 12 D. HOW1 - 42 H₂: Hq <H2 B. Ho: H2 HH142 OE HoM1 =12 H:H *12 0 6. Họ: A H2 H: 11 12 OF HO 1112 H: 142 Find the test statistic for this hypothesis test. (Round to two decimal places as needed.) Enter your answer in the answer box and then click Check Answer. 3 parts remaining Clear All Check Answer

Answer 1

The provided sample means are shown below: X1 = 49.4 X2 = 47.1 Also, the provided sample standard deviations are: 81 = 5.7 82 = 13.8 and the sample sizes are ni 24 and 2 19. Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 2 Ha: Hi > 2 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. Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 41. 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 = 1.683, for a = 0.05 and df = 41.

Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= Xi - X2 (11-1)3 +(no-1) 72 -1 -2 11 12 49.4 - 47.1 (24-1)5.72 +19-1)13.89 24-19-2 0.742 (11 + 1) Decision about the null hypothesis Since it is observed that t = 0.742 <te = 1.683, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p=0.2311, and since p=0.2311 > 0.05, it is concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean Mi is greater than 12, at the 0.05 significance level.

Do not reject Ho.there is not sufficient evidence at a= 0.05 l.o,s to conclude that u1> u2

b)

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = n1 + 12 – 2 = 24 +19 – 2 = 41. The critical value for a = 0.1 and df = 41 degrees of freedom is te = t1-a/2:1–1 = 1.683. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: Sp (nı – 1)sî + (n2 – 1)s ni + n2 - 2 (24 – 1) 5.72 + (19 - 1) x 13.82 24 + 19 - 2 = 10.091 Since we assume that the population variances are equal, the standard error is computed as follows: 1 se = 1 8pV ni + 12 10.091 x 1 24 + 1 19 3.099

Now, we finally compute the confidence interval: CI (X1 - X, - te x se, X1 - X, +te x se) (49.4 – 47.1 - 1.683 x 3.099, 49.4 – 47.1 +1.683 x 3.099) (-2.915, 7.515) Therefore, based on the data provided, the 90% confidence interval for the difference between the population means ji – pa is –2.915 <pli – Jl2 <7.515, which indicates that we are 90% confident that the true difference between population means is contained by the interval (-2.915, 7.515).

the 90% confidence interval about u1-u2 is range from a lower bound of -2.915 to upper bound of  7.515