Question

QUESTION 35 Given 71 = 20, N4 = 6; X2 Our conclusion would be: 15, N1 5, Sã,-82 = 2.5 a=0.05, 2 tail. since df=9, reject the null since df = 11, reject the null. since df = 9, retain the null since df =11, retain the null

Answer 1

ANSWER

GIVEN THAT

$\bar{X_1}=20\, \, \, \, \bar{X_2}=15$

nu = 6 n2 = 5

$S_{\bar{X_1}-\bar{X_2}}=2.5$

g= 0,05

2 tail test

Degree of freedom (df)=

ni + n2 - 2

=6+5-2

=11-2

=9

Calculate critical , t values for two tail test

using t table , search row neaded df=9 and column headed

two tail area $(\alpha )=0.05$

Hence we get

05, و نه ده ن ل و 52

Hence we get critical values are -2.262 & 2.262

Now, calculate t test statisitc

$t=\frac{\bar{X_1}-\bar{X_2}}{S*\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

  $=\frac{20-15}{2.5*\sqrt{\frac{1}{6}+\frac{1}{5}}}$

$=\frac{5}{2.5*0.60553007}$

=3.30

Hence we get t test statistic (t)=3.30

Decision rule :

t test statist? retain the null it is lie in unshsed region Refect 1 t test Start 2.262 es lie la shared

Since, test statistic (t)=3.30 is lie in shaded region , so we will reject the null hypothesis

Ans ; A) Since df=9, reject the null

Reject the null 2.262