Answer
We can say with 99% confidence that the mean difference is between -5.85 and 3.7.
Work Step by Step
Here we have α = 0.01 and n = 8. df = 8 - 1 = 7
α/2 = 0.005
Using Student's t distribution table, we have:
$t^{*} = 3.499$
$E = t^{*} . s_{d}/\sqrt n$
$ = 3.499 \times 3.833/\sqrt 8 = 3.499 \times 1.355 = 4.742$
$\bar d - E = -1.075 - 4.742 = -5.8$
$\bar d = E = -1.075 + 4.742 = 3.7$
