Answer
Confidence interval: $22.091\lt x ̅\lt22.209$
Work Step by Step
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
Since 26540 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.
$t_{0.025}\approx z_{0.025}=1.96$
$n=26540$
$s=4.885$
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=22.150-1.96\times\frac{4.885}{\sqrt {26540}}=22.091$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=22.150+1.96\times\frac{4.885}{\sqrt {26540}}=22.209$
