#### Answer

Confidence interval: -4.9831 to -2.4169

#### Work Step by Step

df = n - 1 = 10 - 1 = 9
$t_{α/2} = 0.05$
From student's t distribution table, we have:
$t_{0.05} = 1.833$
$E = t. \frac{s_{d}}{n} = 1.833 . \frac{2.2136}{\sqrt 10}$
$ = 1.833 . 0.7 = 1.283$1$
Confidence interval will be:
$\bar d - E = -3.7 - 1.2831 = -4.9831$
$\bar d - E = -3.7 + 1.2831 = -2.4169$