Question

10-38. You are given the following results of a paired difference test: d =-4.6 d0.25 n=16

Answer 1

Please don't hesitate to give a "thumbs up" for the answer in case you're satisfied with the answer

dbar = -4.6
sd = .25
n = 16

The 99% confidence interval has a t statistic of 2.95 for df = n-1=15
So, the confidence interval is dbar +/- t*s/sqrt(n)
= -4.6 +/- 2.95*(.25/sqrt(16))
= -4.78 to -4.42

It means that " the true paired difference in mean values lies between -4.78 and -4.42 with a confidence of 99%

The 90% confidence interval has a t statistic of 2.95 for df = n-1=15
So, the confidence interval is dbar +/- t*s/sqrt(n)
= -4.6 +/- 1.75*(.25/sqrt(16))
= -4.71 to -4.49

It means that " the true paired difference in mean values lies between -4.71 and -4.49 with a confidence of 90%