Answer
$n=39$
If the level of confidence decreases, the sample size decreases too.
This is because: a lower level of confidence leads to a lower $t_{\frac{α}{2}}$. And the sample size, $n$, is directly proportional to the square of $t_{\frac{α}{2}}$.
Work Step by Step
We expect to survey a large number of people. In this case, $t_{\frac{α}{2}}\approx z_{\frac{α}{2}}$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.1$
$z_{\frac{α}{2}}=z_{0.05}$
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: $\frac{1.64+1.65}{2}=1.645$
$t_{0.025}\approx z_{0.025}=1.645$
$E=2$ (within 2 hours)
$s=7.5$
$n=(\frac{t_{\frac{α}{2}}.s}{E})^2$
$n=(\frac{1.645\times7.5}{2})^2$
$n=38.05$
Round up:
$n=39$
