For the first sample,, we have
"N1=18"
"\\overline{X1} =87.1"
"S1=6.5"
For the second sample, we have
"N2=12"
"\\overline{X2} =80.8"
"S2=7.4"
Null hypothesis, "H0:\u03bc1=\u03bc2"
Alternate hypothesis: "Ha: \u03bc1\u2260\u03bc2"
Test statistic, "T=(\\overline{X1}-\\overline{X2})\/\\sqrt{\\smash[b]{(S1^2\/N1)+(S2^2\/N2)}}"
Putting all the value, we can get the value of test statistic as , "T=2.395"
Now, we have to find the degree of freedom,
For unequal variance degree of freedom is given by,
"\u03c5=(S1^2\/N1+S2^2\/N2)\/{(S1^2\/N1)^2\/(N1\u22121)+(S2^2\/N2)^2\/(N2\u22121)}"
Putting all the values, the value of v will be "v\\approx23"
So, for "\\alpha=0..01" , the critical region will be "t<-2.807 ,t>2.807"
It can be seen that the test statistics does not fall in the critical region, so the null hypothesis cannot be rejected.
To reject the null hypothesis the test statistic should lie in the critical region.
Comments
Leave a comment