Question 620853
This is a matched pair t -test.
So d"bar" is 61.86-65.43 = -3.57
the standard deviation is sqrt(70.4762+170.2857)=15.52
n = 7

So the t test statistic is -3.57/(15.52/sqrt(7))=-0.608591

Ho = "mu"d = 0
H1 = "mu"d is not = 0

test statistic is less than the critical values so we cannot reject the null, thus there is no evidence of a difference between the two methods