Answer
Confidence interval: $201.5\lt x ̅\lt234.5$
We are 95% confident that the mean total work for the sports-drink treatment is between 201.5 and 234.5 kilojoules.
Work Step by Step
$n=16$, so:
$d.f.=n-1=15$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.131$
(According to Table VI, for d.f. = 15 and area in right tail = 0.025)
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=218-2.131\times\frac{31}{\sqrt {16}}=201.5$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=218+2.131\times\frac{31}{\sqrt {16}}=234.5$
