Answer
Confidence interval: $100.417\lt x ̅\lt115.583$
The margin of error has increased.
Work Step by Step
$n=10$, so:
$d.f.=n-1=9$
$level~of~confidence=(1-α).100$%
$96$% $=(1-α).100$%
$0.96=1-α$
$α=0.04$
$t_{\frac{α}{2}}=t_{0.02}=2.398$
(According to Table VI, for d.f. = 9 and area in right tail = 0.02)
$Lower~bound=x ̅ -t_{\frac{α}{2}}.\frac{s}{\sqrt n}=108-2.398\times\frac{10}{\sqrt {10}}=100.417$
$Upper~bound=x ̅ +t_{\frac{α}{2}}.\frac{s}{\sqrt n}=108+2.398\times\frac{10}{\sqrt {10}}=115.583$
