Question

Consider the hypothesis test Ho: Mi = u2 against Hui <Hz with known variances j = 9 and 02 = 5. Suppose that sample sizes n = 9 and n2 = 15 and that j = 14.3 and 12 = 19.5. Use a = 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) if u is 4 units less than 12? (c) Assuming equal sample sizes, what sample size should be used to obtain B = 0.05 if u1 is 4 units less than 42? Assume that a= 0.05. (a) The null hypothesis rejected. The P-value is Round your answer to four decimal places (e.g. 98.7654). (b) The power is Round your answer to two decimal places (e.g. 98.76). (c) n = n2 = Round your answer up to the nearest integer.

Answer 1

Solution Giuen C-9 2=5 n =15 R 14.9 1, - 18.5 L = 0.05 Null hypothesis Hoi ll, a la Altenative hypothely Hi all, el Hypothesis test and p-value: Test statistic Zo 2 n 14.3 -19.5 92 52 9 15 -5.2 1181 9. 25 15 -5.2. 520.11 20.1111-1.6666 -5.2 4.2947 -1.2107 i. Test stalistic : -1.2107

The alternative hypothesis is left tailed. PCZL Zo) P(Z2-1.2107) 0.1130 .: P-value = 0.1130 P-value 88 greater than significance level Cl). So, we fail to reject null hypothche : The null hypothesis is not rregedted. The null hypothers is not- refelled. The Po value is 0.1130 6 Power The Cummulatine Standard nolmal distribution to find the Probability of type " end B for the One sided alternative with A --4 A-A B = 1-0 -22 2 ng ZA 70.05 1.64 - 4-0 B :1-6 -1.64 - 9 9 52 15 ~A 1-% -1.64 4.2947

8:1-06-164 +0.9313] 11-06-0.7087) 1-0.2419 0.7581 Power = 1-B 51-0.7581 0.2419 = 0.24 :. Power :0.24 One sided alternative A-4 hypothesis with K=B=0.05 : Za = ZB = 70.05 21.64 n = n2 (72+Gi++) (A-2019 (9² +9²) (1.64 .64 +1.64 (-4-0²

n = (3.28)*(81+25) (-4) 10.7584 * 106 16 1140.3904 16 : 71.2744 ܙܒ ܐ n71 in, -nean 71 Please give upvete. Thank you