Question

Chapter 10 Section 1 Additional Problem 1 Consider the hypothesis test H:41 = M2 against H :Mi < u2 with known variances (j = 10 and 62 = 6. Suppose that sample sizes nj = 11 and n2 = 14 and that Ij = 14.3 and I2 = 19.6. 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 Mi is 4 units less than 12? (c) Assuming equal sample sizes, what sample size should be used to obtain B = 0.05 if H1 is 4 units less than 2? Assume that a = 0.05. (a) The null hypothesis is not 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

=100+14 Page given by The variance is na 102 11 + 14 = 9.090+2.571 = 1.661 The test statistic is 2= x - Xe 14.3 - 19.6 (11.661 -5-3/3.4146 - -1.55 p-value (Ze-list 0.0606 Thues at level L=0.05 we tail to regect the null hypthesis 6 under Ho and at the level 2 = 0.05 we reject the null hypthesis is 2<-1645. thus the power is given by. P(x) = 1.645 P ( n - 22 +4 <-1.645 + 4 JL. 667 M.667 P (25=1:645 + 4 = p (2<-0.473) 0.3192 111.661)

Page we assumed equal sample size then the pooled variance is n Thus power of the test at a = 0.05 is t-sha - 1.645) 12 M-1 +4 -1.645 + 4 T125/2 P( it to 125 125 at we need B=0.05 = power = 0.95 thus we need - 1.645+ 42 = 1:645 1125 - n = 28 1 645/125 n a = 9.196 84.56 This nearest integer is n=85 @ P. value = 0.0606. The null hypothesis is not rejected. (6) The power is @n, na -0.32 85