Question

Chapter 10 Section 1 Additional Problem 1 Consider the hypothesis test H:41 = 112 against H :< u2 with known variances (j = 9 and 02 = 5. Suppose that sample sizes nj = 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 2? (C) Assuming equal sample sizes, what sample size should be used to obtain B = 0.05 if Mi is 4 units less than uz ? 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

The hypotheses are H : 14 = uy vs H7:44 <M. The known population standard deviations are 01 = 9 and 52=5. The sample sizes are n = 9 and n2 =15. The sample means are a = 14.3 and 72 = 19.5. The given level of significance is a = 0.05. Z + (a) The population variances are known; therefore, z-test is applicable in this case. The test statistic is calculated as follows: - t lol ni n2 14.3-19.5 192 52 V9 15 z=-1.59 + The P-value of left tailed test is calculated as follows: P-value = P(Z <-1.59) P-value = 0.0559 The decision rule is that reject the null hypothesis if P-value is less than the level of significance. Since 0.0559 > 0.05, therefore, the null hypothesis is not rejected, and it can be concluded that there is not sufficient evidence against the null hypothesis. Hence, the null hypothesis is not rejected. The P-value is 0.0559.

(b) The test is rejected if calculated Z is less than Ze, that is, Z<Zc. The power of the test is calculated as follows: Power=1-B = 1-P(Type II error) = 1-P(Fail to reject H. H) = P(Reject H, H, is true) = P(ā- 5-5.2 M-H2 =-4) x1 - 22 +4 =P -5.2+4 192 32 15 + + ni n2 = P(Z <-0.37) Power = 0.3557 The power is 0.3557. (c) Assume the sample sizes are equal, that is, n = n2 =n.

The value of n is calculated as follows: B = 0.05 = P(Type II error) = 0.05 = P(Fail to reject H, H2) = 0.05 =P(51 - >-5.24 - Hy = -4)= 0.05 11 - 12 +4 P -5.2 +4 19252 = 0.05 + + n n n n -1.2 PZ> =0.05 192 32 + n n -1.2 1-PZ< = 0.05 n n

-1.2 PZ< = 0.95 92 52 + n 1 0-(0.95) -1.2 192 52 + n n -1.2 11 =1.645 52 -1.2 >> + n n 1.645 52 + -1.2 1.645 n n 81+25 -1.2 1.645 n 1 n 106 -1.2 1.645 106 n= (-1.2/1.645) n=199.2 >n 200 Hence, n = n2 = 200