Question

Chapter 10 Section 1 Additional Problem 1 Consider the hypothesis test Ho : M = M2 against Hui <H2 with known variances j = 9 and 02 = 5. Suppose that sample sizes n = 9 and n2 = 15 and that is = 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 fli is 4 units less than 2? (c) Assuming equal sample sizes, what sample size should be used to obtain B = 0.05 if Hi is 4 units less than 112? Assume that a = 0.05. (a) The null hypothesis is v 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) ni = n2 = Round your answer up to the nearest integer. Statistical Tal and

Answer 1

Given that sol- pe Conside the hypothesis text Doa Giho : M = M2 HIM, LM, > variances o, - 9 og a5 for to this >> Sample size in')=9 vel ronn =15 . x SX= 14,3 1, 19.5 2 0:05 0 a The null hypothesis is not rejected Then the pratue is Po 00537 extColors P=0.0606 The power is 0.32 o n - 72 there is the solution a) The pooled rouance is given by 2 02 + 81 +25 9 12 15 the con - 9+1.667 =10.667 OL The test statistic is given by Zi X X2

-5.2 = -1.592 14.3 - 19.5 10 667 3.266 -1.69 b) Then the p value P(22-1189) = 0.0506 Thus at level 2=0.05 we fail to reject the null hypothes's under to and at the level 220.05 w reject the nulle hypothesis it z <-15647 the powy given by . PCA 1.645 is that 10.67 PC & -f ty <- 1.645t y ✓ 10.67 √1067 baton P(Z < -1.645 + ſ1067 1064) -10:32 c) If we assume equal Sample size then the pooded variance is loo n. This power of the text at 2=0.05 is X X1 +4 -1.605 + Evo slob/n T, X2 <-1.605) - P 106 n PC Toon è l.but If we need ß=0.05 => powy 0.95 they we need – logus urn Fool l ſ106 in- 2xl.bus/106 8.46 4 n=18.467 n=72 rn - 846 n-71.5