Answer
Confidence interval: $6.968\lt x ̅\lt8.388$
Work Step by Step
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
Since 357 respondents is a large sample size, we can use the approximation $t_{\frac{α}{2}}\approx z_{\frac{α}{2}}$
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
$t_{0.025}\approx z_{0.025}=1.96$
$n=357$
$s=6.843$
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=7.678-1.96\times\frac{6.843}{\sqrt {357}}=6.968$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=7.678+1.96\times\frac{6.843}{\sqrt {357}}=8.388$
These results are very close to the results in the problem text. The difference is due to the approximation $t_{\frac{α}{2}}\approx z_{\frac{α}{2}}$. According to table VI:
$t_{0.025}=1.984$ for d.f. = 100 and $t_{0.025}=1.962$ for d.f = 1000
