Question

For the given data Question in Red ed2 mean = 96.42, st dev = 8.2586, n = 10 Assuming alpha = 0.05 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 4 = 100 Ha: 4 < 100 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided the significance level is a = 0.05, and the critical value for a left- tailed test is te = -1.833. The rejection region for this left-tailed test is R=t:t < -1.833 * Can find the you explain how to p-value? (3) Test Statistics The t-statistic is computed as follows: *I found t= -1.371 but can't X - μo 96.42-100 = -1.371 figure out how they got p= 0.1018 s/n 8.2586/10 using the Tables?? (4) Decision about the null hypothesis Since it is observed that t = -1.371 > t = -1.833, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p = 0.1018, and since p=0.1018 > 0.05, it is concluded that the null hypothesis is not rejected. ?? Thanks!! Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the true average strength is less than 100. at the 0.05 significance level.

Answer 1

Given Hos l=100 Hao [<100 F:-1.371 n-10 . (<-) =(>) df = 10 -1 =9 -1.3 371] = P[ta > 1.37 P-value = P[ta < -1.371 > 1. 371 tesing Excel right distribution formula <TI DIST. RT (1.3719) 0.1018 t df since P value = 0.1018