Question

Consider the hypothesis test Ho: Mi = U2 against HL : M1 <H2 with known variances a = 9 and 62 = 5. Suppose that sample sizes nj = 9 and n2 = 15 and that I = 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 H1 is 4 units less than 2? (c) Assuming equal sample sizes, what sample size should be used to obtain B = 0.05 if yli is 4 units less than 12? 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 (e.g. 98.76). Round your answer to two decimal places (c) ni = n2 = Round your answer up to the nearest integer.

Answer 1

501 تو را hypothesis are, suloy 1 Standa Ho ill all silt Autototesse ei sulor sitzitata H, : 4,6% 44 ) ദാ 21ന് te sorple sizes are -4243 รับการ From the given information, the null and alternative The variances are 0,=9 = 5 Quote Xp2t4.3, 7, +19.5 Level of Signilicance 220.05 X=0.05 nord The quentity of interest is the difference in means, Mylly. Note that Ao=o. Use the following formula to compute the test statistic. X, -,-1 Zo را از او را تر افراد نے ان تا کار را در ساری Sample means are or 31+zifalz fud antra per @! 4 st stets ny T to resistor rotat nodig at mort 14.3-1905-0 Ne nu Filicedog sat od + 45 pistoals at 20 101 qut -5.2 10.66667 -52 3.2659. I - 1.59 From the standard normal table, the critical value corres- ponding to b.05 level of significance is 1.96. PZA.

ME Decision Tule: co Rebect the null hypothests, il the test statiste value is less than the critical value - 1.96 or the test Statistic value is greater than the critical Value, 1.96 Conclusion: Since, afts -1.59 -1.96, that means, the test Statistic value is less than the critical value, -1-96 or the test statistic value is greater than the critical value, 1.96, fail to reject the null hypothesis. compute the p-value. f since, the test is two tailed, then the p-value is p-value= a[1-d1zo] -20-$1.59] = 2[1-09441) - To.lll8 b From the given information, the true difference of means is D=4 Use the following formula to compute the probability of a typell error, Bi A-Ao A-do wallot at so B = 0 74 ni ni & 4.96 4-0 gy is 4 IVE IS 3:2659]-o [-1962 y 3,2659

22.22 Assume that samorizzati 9200 0 (0.73) - $ (-3.18) 0.7673 - 0.0007 = 0.7666 19-1 tap-1) power of test is = 1-3-1-0.7666 = 0.2334 From the given information, the true difference of means is A=4. Level of significake =0.05 B=0.05 Assumei that sample sizes are equal.n,=n, an from the standard normal table, the critical value corresponding to 0.05 level of significance for the two tail test is 1.96. From the standard normal table, the critical value cones- ponding to 0.05 level of significance for the two- tail test is 1.96. From the standard normal table, the critical value corresponding to 0.05 level of significance for the one-tail test is 1.64

Use the following formala to compute the sample size: (24/₂ + Zp)"(ta rons Dog (1.96+1.645 (arts (4-oji nov .:47 on finger for Sos 86 85.86 86 Increase in the sample size results na decrear in the probability of a type ll enoy therefore, to ensure that B is not greater than 5% the sampleste is n=841.2003 apos lorofildod loomaat oor loyal Zoo of otrog