Question

come from populations (1 point) Test t mean. Assume that the samples are independent simple random samples. Use a significance level of a 0.01 Sample 1: n1 15, 1-28.4, 81-6.07 Sample 2: n2 10, 2 22, 82 8.92 (a) The degree of freedom is (b) The standardized test statistic is (c) The final conclusion is O A. We can reject the null hypothesis that (14-Ha) 0 and accept that (M1-μ2) 0 B. There is not sufficient evidence to reject the null hypothesis that (M1-μ2 0.

Answer 1

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used (2) Rejection Region Based on the information provided, the significance level is α = 0.01, and the degrees of freedom are df 23. 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 two-tailed test is tc 23 2.807, for a 0.01 and df - The rejection region for this two-tailed test is R [t t2.807) (3)_Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows 28.4-22 118.9 2.142 15-16.07 (4) Decision about the null hypothesis Since it is observed that2.142te 2.807, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p0.043, and since p 0.043 0.01, it is concluded that the null hypothesis is not rejected (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ| is different than μ , at the 0.01 significance level The 99% confidence interval is-1.988 < μ1-μ2 < 14.788