Question

Consider the hypothesis test Ho: N1-N2 = 0 against Hi: 11-42 + 0 samples below: 1 37 38 31 33 33 30 31 29 39 38 30 38 37 30 39 31 35 38 II 32 29 35 32 32 29 30 37 31 34 30 30 30 32 33 34 Variances: 01 = 4.0, 02 = -0.5. Use 0 = 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). 2.5264 202 P-value = 0.0115 (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 0.5387 (c) What is the power of the test for the true difference in means of 3? Round your answer to three decimal places (e.g. 98.765). The power of the test is 1 - B = 4.2669 SW1-W23

Answer 1

4 Aim: To out the test carry compute confidence interval and the power of the test: Given: 0 f 82= -0.5 O-005 Observations of two pamples. Parameters; Let Me & Mé' be the population meano. Hypothesis; Ho: The two population means do not differ significantly thi The two population means differ significantly Ho: M1 =12 Here as population standard deviations r2 are given, we we 2-test. vs. Mathematically vis Hl: Me # Ma. Jest statistic Lcale (XI-X2) -(41–42) N NCoid) ri 2 022 + ni na Here ti f xã are the sample means of both the samples. Ixli 617 34:2778 = 34.278 ni 18 X2 510 ΣΧ2 n2 31.875 16 bo under Ho, M1-42 =0. Ical= (xi 812 622 ni 12 +

Zeal 11 34-278 - 31.875 (4)2 + (-0.52 18 16

2-403 0-9511 Zeal 2- 5261 Reject Ho it Izcall> 2x12. 12 =0.025 . Decision : Here = 0.05 bo 20.025 here Izcall - 1.96. - 2.5 262 > Z</2 = 1.96. so we zeject to Also reject Hoit p value Here p-value -oo sejection conclusion: _The population mean may differ significantly - 0:0115 < =0:05

ce 95-)• confidence interval can be computed as- 1-0 = p [ 1Zcal i < 2x121 CXI -X2) - (H1-H2) < 2212 012 + 122 Di n2 =p/lexi - X2) = (41-42) I < 2x12 8012 +022 ni 1 2 =plcXI – X2) 022 < (HI- M2) < Z212 0,2 ni + n2 (XI-X2) + 2x12 x 012 (22 + OL n2 Lower bound - (x1 - x2 ) 2212012 ni + 022 Az 2-403 -1.96 x 0.9511 0.538

opper bound a ( XL - X2) + 2212 x 012 22 + ni n2 2.403 + 1.96 X 0.9511 4.267 The 95.1, confidenec interval for Mi- M2 is Mi-M2 <6.2672 ( 0.538 < Using confidence interval we accept Ho when only o is included in the obtained interval, because under Mo, Mi = 42 Mi 42 = 0. otherwise we Ho. Now, hese as the confidence interval does not contain 0. teject Ho. we seject Me doo one can notice from the obtained confidence interval that the difference beth Mi 4 He io positive as both lower 4 upper bounds ase positive That mean So we conclude that the population means differ significantly futher more that ML >M2. ML - 42 so Mis M2

C. difference in meano = M1-42 = 3. 1-B = Power of the test P[ conreat decision] EL [ Rejecting Ho when He jo true) Reject to I Hiio true pſ Ical > 2012 f H1: H1-42 = 3 3). A : 2 - ال let A-3 sp (-x2)-(-H2) 2212 Mi-M2) 012 022 622 + 02 + ni ni n2 n2 - P Zcal > 2212 A r12 2 न ni n2 012 0-025 Here = 0.05 20.025 = 1: 96 PL Zeal > 1.96 3 (4)2 18 + (-0.52 16 M =plzcol 1.96 – 3.1544) P [ zcal > - 1.1944] = p [ zcal < 1.1944) - Normal jo a symmetric dioth. 1- p [ zcal > 1.1944) - 1-pc Zcal ? 1:19] -1- 0.11702 - 0.883

Anowers: a). || Test otatistic Zo Zcal = 2.5267 p-value = 0.0115 b). we are 95.4. confident that value I eacceeds value 2 beth s 41 - 42 4-267. by 0.538 c) The power of the test = 1-8 0.883,