Question

(1 point) Test the claim that the two samples described below come from populations with the same mean. Assume that the samples are independent simple random samples. Use a significance level of a = 0.05 Sample 1: n = 6, 11 = 25, $1 = 5.29 Sample 2: n2 = 17, I2 = 21.1, S2 = 5.84 (a) The degree of freedom is (b) The test statistic is (c) The final conclusion is A. We can reject the null hypothesis that (41 - H2) = 0 and accept that (u1 - U2) +0. OB. There is not sufficient evidence to reject the null hypothesis that (H1-H2) = 0.

Answer 1

Page Me Given that, 12=17 = 25 6 = 2 21.1 Si=5,29 S2 = 5.84 Null and Alternative hypothesis - Hoi (41-420 Ha: (41-42) to This corresponds to a two-tailed test for which is a t-test for two population means Rejection Region - Rol Based on the information provided the significance level of a=QO5 and gdegree of freedom [df) (Sin t he CO) (+0-1) ( (51292 + 51843) 17 - df = 9,668 N 10 Hence it is found that the critical value for this two-tailed test is te = 2.239 For a=000s and def = 9.668 | Rejection Region - ht: It>2.239

6 Test statistics - the t-statistie ir computed as follow. 25-21 I t=los Test statistice=list Decision about the nell hypotheere- since it pe observed that It=1.51 <te= 2,239 it is then concluded that the hull hypothes Thypothests Te not reject. i Osing p-value approach - P-value = 01163 - since, p value = 0116370.05 it to concluded that the null hypothesis r not rejected,

Pege No O Conclusion - It is concluded that the null hypothers Ho is not rejected. Therefore B There r not slefficient evidence to reject the null hypotheels (4,-42-o.