Answer
Confidence interval: $1.152\lt x ̅\lt1.288$
The mean number of family members in jail is between 1.152 and 1.288, with 99% confidence.
Work Step by Step
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
Since 499 respondents is a large sample size, we can use the approximation $t_{\frac{α}{2}}\approx z_{\frac{α}{2}}$
$z_{\frac{α}{2}}=z_{0.005}$
If the area of the standard normal curve to the right of $z_{0.005}$ is 0.005, then the area of the standard normal curve to the left of $z_{0.005}$ is $1−0.005=0.995$
According to Table V, there are 2 z-scores which give the closest value to 0.995: 2.57 and 2.58. So, let's find the mean of these z-scores: $\frac{2.57+2.58}{2}=2.575$
$t_{0.005}\approx z_{0.005}=2.575$
$n=499$
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=1.22-2.575\times\frac{0.59}{\sqrt {499}}=1.152$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=1.22+2.575\times\frac{0.59}{\sqrt {499}}=1.288$
