Question

Reserve Problems Chapter 10 Section 1 Problem 1 Consider the hypothesis test Ho : H1 - H2 = 0 against H : H1 – 12 #0 samples below: I 36 38 32 32 33 30 31 29 39 38 30 37 37 30 39 30 34 40 II 32 30 35 32 31 29 30 3731 34 30 30 30 32 33 34 Variances: 1 = 3.9, 02 = 2.4. Use a = 0.05. (a) Test the hypothesis and find the P-value. Find the test statistic. Round your answers to four decimal places (e.g. 98.7654). 20-0.6450 = P-value 0.5189 (b) Explain how this test could be conducted with a confidence interval. Round your answers to three decimal places (e.g. 98.765). We are 95% confident that value 1 exceeds value 2 by between -2.860 (c) What is the power of the test for the true difference in means of 2? Round your answer to three decimal places (e.g. 98.765). The power of the test is 1-B= Ish-H25 1.444

Answer 1

Soln To test To test given hypothesis Ho! Mi - (M2=0 Vs HiiH-M₂ & 0 we use 2- test statistic for two sample given by 71 - ñ-(HiT) 2 = 72 ~ N(011) 2 + ne Since under null hypothesis Ni-M2=0 - nu 2 - 34. 16667 – 31.875 ✓ (3.912 + (2.412 16 t TR ni n2 18 a) 2= 2.0877 the corresponding p-value is given by b-val = 2 PLZ 7 2.0877) By using stondard normat pdf table calculate it as p-val = 0.0368 we Since Z N NcOil) b thers P(-2d <2&2a1) 21-d or PL-20 < xinh (H1-H2) 2011 ind to ni nu 2 PL- - 2012 OR ri < x-u - Minh t ni m < 2012 =1-9 пL

ar 2 P(21-22-29 <H2 - 14 fr Hz 55-+2 Jan -- = 1.96 Since here l-a = 0.95 d x=0.05 and 2012 - 20.025 so by putting values we get confidence interval as P(0.140 & -h 4.443) = 0.95 c © Here it is given that Hi-12=2 S Power of test is given by 1- R =P(Reject Hol 141-1/2 =2) As we reject Ho if 27202 2 - 2012 Thus Xin <-29, 2 x' 614 or tu in 720 AL f ni h anar 2-2.1515 xinde 7.2.1515 1104 Plen- <- 21515) M-H2=2) + P(xit 72.1515 / M, M2=2) - 2 Plonin > 2.1515 / M 12 =2) a I-B -2 Plz 2-1515-2 + 2 V 2 ni

Notice that here we use the fact that if Mirl2 =1 than - -2 NO.) tru ni ni so 1-R = 2* P(Z > 2.1515) 0.031