Answer
Confidence interval: $17.12\lt x ̅\lt19.68$
The margin of error has decreased.
Work Step by Step
$n=50$, so:
$d.f.=n-1=49$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.009$
(According to Table VI, for d.f. = 50 and area in right tail = 0.025. Notice that there are no available values of $t$ for d.f. = 49, but the given value of $t$ is pretty close)
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=18.4-2.009\times\frac{4.5}{\sqrt {50}}=17.12$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=18.4+2.009\times\frac{4.5}{\sqrt {50}}=19.68$
