00:01
To solve this question, first of all we will let here, mu1 be the mean of preferred hand, preferred hand.
00:19
And mu 2 be the mean value, mean value or non preferred hand.
00:35
So now the simple test statistics for the sample stratetics will be as when n is equal to 5.
00:53
This is n1 is equal to 5, n2 is equal to 5.
00:59
So summation of all values of xi will be as 53 and submission of all values of yi is 43.
01:08
So we can find out value for mean that is fire x as a as a summission of all values of yi is 4.
01:13
Summation of x i divided by n so we will get here 53 divided by 5 which is 10 .5 x bar and similarly our values of y bar will be as submission of y i divided by n to this would be an one so we will get 43 divided by 5 which is equal to um 80 .8 .5 so now now our value for submission of xi squared will be as 58 and then for xy i squared is equal to 375.
02:10
So now our sample variance sample variance s1 square will be as summation of xi square minus n.
02:25
X bar square divided by n1 minus 1 and when we calculate r value will be as for s 1 square is 5 .3 and similarly sample variance s 2 we can calculate s 2 also by using formula submission of y i s square minus n 2 y bar square then n2 minus 1 and when we calculate r value for s2 square will be as 1 .3.
03:03
So then we assume here the sample variance.
03:09
Let's take sample variance as sigma 1 square and sigma 2 square are equal r equal then are pooled variance variance s square will be as n1 minus 1 s 1 square minus n 2 minus 1 s 2 divided by n1 plus n2 minus 2 so let's put down our value this will be as 5 minus 1 will be equal to 4 multiplied by 5 .3 to the value square and 4 multiplied by 1 .3 divided by n1 plus into will be as 5 plus 5 minus 2 so when we calculate r value for pool variance s square will be 3 .3 so therefore our pooled standard deviation which is square root of s will be equal to square root of 3 .3 which is 1 .8166 so after this we will check to test we will set our hypothesis so our null hypothesis will be ho says that mu 1 is equal to mu 2 and our alternative hypothesis will be h a this to mu 1 does not equal to mu 2 this will be the hypothesis now for the solution of first part of the question we are asked what kind of test we will use here.
05:25
So the answer is here we use two -tailed sample test which is t test.
05:46
So mu -1 is equal to mu -2 or you can say mu -1 minus mu -2 is equal to 0.
05:55
This is a two -tailed test...