0:00
With our solution.
00:01
So, we are given that x1 bar is equal to 4 .20, x2 bar is equal to 1 .71, s1 is equal to 2 .20, s2 is equal to 0 .72, n1 is equal to 22 and n2 is equal to 22.
00:22
Right, so our null hypothesis will be that is h0 is equal to mu1 is equal to mu2 and our alternative hypothesis will be mu1 is not equal to mu2.
00:36
Right, so on the basis of the information that has been given to us that is alpha value is equal to 0 .05.
00:44
Now degree of freedom will be equal to n1 plus n2 minus 2, so that is equal to 22 plus 22 minus 2, so that is equal to 42.
00:56
Now we have degree of freedom and alpha, so we can find out a critical value that is g of c from the table of t, so that will be 2 .018.
01:09
Right, we have used the table and the alpha value and degree of freedom.
01:15
Right, so the rejection region for this two tail is r is equal to t should be greater than 2 .018.
01:36
Right, okay, now for this we need to find sp...