A statistician believes that women received more bachelor's degrees than men did last year. To test this claim, she selects coll
eges and universities randomly and compares the number of bachelor's degrees awarded to men and women. Suppose that data were collected for a random sample of 7 colleges and universities, where each difference is calculated by subtracting the number of degrees earned by men from the number of degrees earned by women. Assume that the number of degrees is normally distributed. The statistician uses the alternative hypothesis Ha:μd>0. Using a test statistic of t≈2.968, which has 6 degrees of freedom, determine the range that contains the p-value.
The P-value can be calculated by using excel function T.DIST.RT(2.968,6).
Right tail is used in the function because our alternative hypothesis has greater than sign (>) whereas 2.968 represent test statistic value and 6 represents degree of freedom. The resultant p-value from excel function is 0.0125.
Or if we use t- distribution right tail area table, we see that 6 for degree of freedom the value in table is there are two value 2.44691 and 3.14627 which corresponds to p=0.025 and p=0.01. As 2.44691 is smaller than 2.968 so we take p=0.01 against 3.14627. Thus, in this scenario p-value would be 0.01.
