Answer
Confidence interval: $8.86\lt x ̅\lt11.94$
The mean number of e-mails sent in a day is between 8.86 and 11.94, with 95% confidence.
Work Step by Step
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.9=1-α$
$α=0.1$
Since 928 respondents is a large sample size, we can use the approximation $t_{\frac{α}{2}}\approx z_{\frac{α}{2}}$
$z_{\frac{α}{2}}=z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
$t_{0.05}\approx z_{0.05}=1.645$
$n=928$
$Lower~bound=x ̅-t_{\frac{α}{2}}.\frac{s}{\sqrt n}=10.4-1.645\times\frac{28.5}{\sqrt {928}}=8.86$
$Upper~bound=x ̅+t_{\frac{α}{2}}.\frac{s}{\sqrt n}=10.4+1.645\times\frac{28.5}{\sqrt {928}}=11.94$
