Question

Chapter 10 Section 1 Additional Problem 1 Consider the hypothesis test Ho : M1 = up against H1 : M1 < u2 with known variances 01 = 10 and 02 = - 6. Suppose that sample sizes ni = 11 and n2 = 14 and that ži = 14.3 and 12 = 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 ui is 4 units less than u2? (c) Assuming equal sample sizes, what sample size should be used to obtain ß = 0.05 if u1 is 4 units less than 42 ? Assume that a = 0.05. (a) The null hypothesis (e.g. 98.7654). rejected. The P-value is Round your answer to four decimal places (b) The power is Round your answer to two decimal places (e.g. 98.76). (c) ni = n2 Round your answer up to the nearest integer. Statistical Tables and Charts

Answer 1

Solection : and sample Lets to. To test Ho: M. Pe against #r: ,<p3 with 6 = 10 and 6+6(knoon) sample sizes on, ass and ny = 14 ple mean à = 14-3 and 2 = 19.6. To remain clear,' we will assume normal distribution without loss of generality. try find the 'test statistie Seeppose mi(+11) id samples ples are drawn from NCM36 - 10).> independend and 2 (= 14) ind samples are doawon from a (M2'; 6 = 6) therefore, obviously, Tim N(Ms, %) independently and X a do ( 6 ) X2 NCHI-Hz, +2) Coeproductive property) (31-7z) - (11-12) Nou) can be resed as 67 52² test statistic and render Ho: Hi-fly=0 Z Nos) +02 Decision Rule. Reject Ho at level de ef. z<-Toe otherwise accept where To - repper-x point of (0, 1) disth (a) calecelation: observed 7 =- (2, - Z) (under Ho) => 7 - ni nz N nz it Gi n na 14-3 - 19.6 - 5.3 = -1.5519. 3.415 100 11 + 36 14 and at given a = 0·05, - 70.05 Sinee observed 71 Ho = -1055197 - To.05 = -49657 we can't rejeeti Ho at x = 0.05 the null hypothesis is not rejected. þ.Value = Po(7* < -1-5519) where zi NO,1) $(-1-5519) = 1 - $ (1-5519) - 1-0.9394 = 0·0606 (from table)

(b) power, P = P(dejeet Ho) - P. (22-To-06) PH (P <-3-65) (from Hable) (X.-X2) - (Ni-M2) 1•65+ (M-H2) < nz Hila-4 Ho: Mo-M2. = o against 4: M.-M.-4 power, p = $ 1-65 - 4 Too + TA $ (0-6384) - 0-75. ( from table) - (0.202 PAT Given that, in this case. (e) assume, n = n2 = n (say) x=0.05 and Power, B- X; -X2 - (M-H2) 6² +62² and Mi= M2-4 100 + 36 render this condition. B=0.05 $(0-2 $(0• 202 In -0.05 N = 1:645 2 1.65 +(H1-12) Joi G²+62² n n (N-23:35 E- & (0.20250) 2 - 66-317 na 67. -X