Question

Question 3 4 pts Acme Products manufactures widgets. It has two machines which produce widgets. It is desirable that both machines produce widgets of the same dimensions. At regular time intervals Acme's quality control tests the widgets from both machines. The last test produced the following results. In this context sM is the sample mean. SM1 = 5.61 S1 = 1.07 n1 = 100 SM2 = 5.89 S2 = 1.26 n2 = 100 Do the machines produce different sized widgets? Test Ho: u1 = u2 vs. Ha:u1<>u2 at the a = .05 level of significance. Assume that the two standard deviations are equal (pooled yes). Your answer is in form of a triple (Z Statistic, P-value, Conclusion) Conclusion is the answer to the question Reject Ho. For this exam,"<>" means not equal. (t = -1.6939, p = .0459, Reject Ho) O (t = 1.8361, p = .0339, reject Ho) O (t = 1.4783, p = .0705, Do Not Reject Ho) O (t = -1.6939, p = .0919, Do Not Reject Ho)

Answer 1

Answer : Last option (t= -1.6939 , p=0.0919, Donot reject H₀)
Solution :

The provided sample means are shown below: X1 = 5.61 X2 = 5.89 Also, the provided sample standard deviations are: S1 = 1.07 S2 = 1.26 and the sample sizes are ni = 100 and n2 = 100. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = H2 Ha: Mi 7 p2 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 a = 0.05, and the degrees of freedom are df = 198. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: = 1.972, for a = 0.05 and df = Hence, it is found that the critical value for this two-tailed test is to 198.


The rejection region for this two-tailed test is R = {t: t > 1.972}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X2 t= (n1 –1) s+(n2-1)s(+ na) ni+ng-2 5.61 – 5.89 (100-1)1.072+(100-1)1.262 100+100-2 (10o + 100 -1.694 (4) Decision about the null hypothesis Since it is observed that t = 1.694 <te = 1.694 <te = 1.972, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.0919, and since p= 0.0919 > 0.05, 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 Mi is different than um, at the 0.05 significance level.