Question

QUESTION 13 4 points Save Answer Given X1 = 20, N2 = 6; X2 = 15, N1 = 5, 57,-iz Our conclusion would be: = 2.5 a=0.05, 2 tail. O since df=9, reject the null O since df = 11, reject the null. O since df = 9, retain the null O since df 11, retain the null

Answer 1

Pago No.: γουν Date: Gliveni X = 20 X2 = 15 n2=5 ni = 6 Sx-X2 = 2.5 x=0.05 2 tail test Degrees of freedom (df) = n, t02-2 = 6+ 5-2 - (1-2 =9 calculate critical t values for two tail test Tedrop- using t table, search now headed df = 9 and column headed two tal area (&)=0.05 -Hence we get two tail df area (2) Foios 2 2.262 Hence awe get & -2.262 critical values are 2.262

Page No.: YOUVA Date: Now calculate t test statistic X - 72 t SH ++ ni nz 2o 15 2.5 * +5 S 2.5 0.60553007 13:30 Hence we get t test statistic (t) = 3.30 Decision rule : retain the null Tift test statistic is lie in unshased region Reject & the is gull, -2.262 2.262 ttest statistic is lie in Shased reglon since, test statistic (t) = 3.30 is lie shaded region, so we will reject the null hypothesis → Reject the null 2.262 3.30 ANS : (A) since df = 9, reject the null