Question

Consider the following sample data drawn independently from normally distributed populations with equal population variances. Use Table 2. Sample 1 12.7 11.7 7.8 11.6 10.8 10.4 94 10.7 Sample 2 8.7 10.8 13.5 11.8 11.5 95 10.8 11.8 Click here for the Excel Data File a. Construct the relevant hypotheses to test if the mean of the second population is greater than the mean of the first population. O Ho: Ni - M2 = 0; HAV1 -20 O Ho: Mi - W220;HAM - W250 Ho: P1 - 120; HAN: - H2> 0 b-1. Calculate the value of the test statistic. (Negative values should be indicated by a minus sign. Round all intermediate calculations to at least 2 decimal places and final answer to 2 decimal places.) Test statistic b-2. Calculate the critical value at the 5% level of significance. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.) Critical value b-3. Using the critical value approach, can we reject the null hypothesis at the 5% significance level? Yes, since the value of the test statistic is less than the critical value of -1.761. Yos, since the value of the test statistic is less than the critical value of -2.145. No, since the value of the test statistic is not less than the critical value of -2.145. No, since the value of the test statistic is not less than the critical value of -1.761. c. Using the critical value approach, can we reject the null hypothesis at the 10% level? No, since the value of the test statistic is not less than the critical value of -1.345. Yes, since the value of the test statistic is less than the critical value of -1.761. Yes, since the value of the test statistic is less than the critical value of -1.345. No, since the value of the test statistic is not less than the critical value of -1.761

Answer 1

Each part is solved with full explanation as follows:

1.

Sample ! 12.7 11.7 Sample 2 8.7 10.8 13.5 7.8 11.6 11.8 11.5 9.5 10.8 10.4 9.4 10.7 - Samples are drawn independently Both the sampled populations are normally distributed o variances of both the populations are equal 10.8 11.8 (a) Hypothesis for mean of the second population is greater than the mean of the first population correct answer is b. Ho: M, M2 30 HA: M-450 For mean of the second population to be greater than mean of first population, ing should be greater than li, which is stated in alternate hypothesis and from this we derived the null hypothesis Cbr1) with above stated three conditions about sample & population, we will calculate pooled variance for these samples: Ś - (Mo-1) s} + {^2-1) 52 Mit M2-2 Mi = No. of observations of the sample Si a variance of the samples

2.

Samble : 21,0 non ha Calculating 52. Mean( ) = 12.7+ 11.7+7.8 + 11.6+ 10.8 + 10.47 7.8 + 11.6+ 10.8 + 10.4+ 9.4 +10.7 - 35,1 = 10.6375 2.10.64 Ex = (12-7)+ (11.7)2+. 17,8)2+ (11.6)+ (10.8) + (ory ley 49:49+(10712 - 16).29 + 136.89 + 60.84 + 134.56 + 116.61 = 921.23 4+ 134.56 + 116.64+ 108.16 +88.36 + 114.49 s2 = ? n(at)2 서 = 921.23-8(10.6375) - 85633ulys 8-1 = 921.23-905.25 = 2.28 Similarly for sample 2: She dy = 8.7+ (0,8+13.5+ 11.8+115795+10.8+11.8 - 38,4 11.05 a to do wii ) Ex? (.1)2+(10.8)*+ (13.5)+(1.3)+(11.5) + (9.5)+(10.894+ (11.012 - 75,69 + 116.64+ 182.25+ 139.24 +132.25+ 90.25+ 116,64+ 139.24 =992.2 M= s?= 20.0-b.. {xt- n(52)? 992.2-8 (11.05.)? Med 18-1 Best i to mistra ud =992.2-976.82 = 2.19 70..- ww Le nu

3.

Pooled mian = m = (m, -19.5++ (924)5; Mit Mq-2 (8-1) (2.28) + (8-1)(2.19). 18+8-2 = 15,96+15.33 - 2.235 14 As our sample size 16) < 30, me af will me t= ( 5-72) - (M1-42) PUM38.231 ze 18)< 30, me te will use t statest = (2.28-2.19)-0 E 0.09 Ves 2.235(+1) 70.555875 - 0.09 0.747 = 0.12 in Test Statistic (t) = 0.12 2.0 2.0 2.2 USD 0 6-2) buitical value at 5). significance level It is a one taited test. Degue of greedom- nitN2-2 ) +1801) +18) - = 8+8–2 = 16-20 +22 214 SCOP - .: 8-0.05 df = 14 , one tailed test e on-3 By looking at t table 18 Air critical value =-1076)

4.

U-3) 220.05 A see t statistics = 0.12 Critical value=-1.761 AM to 12 -1.76) does not exceed a The calculated value of the critical value. t at statistic Hence we the mean os y Correct answer mwl not reject the mill hype second population is not greatu than he null hypothesis and population peatu than mean of sürst Popula is not less No, since the value of the test statistic than the critical value of -1.76). -).345 (c) At 10% level. t-statistics = 0.12 critical value= -1.3 less than critical Again ou test statistics 10.121 is not value of -1.345. Hene arrect answer is a. No, since the value of test the statistics is not less than cutical value of -1.345.