Answer
$n=115$
If the level of confidence increases, the sample size increases too.
This is because: a higher level of confidence leads to a higher $t_{\frac{α}{2}}$. 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 adult Americans. In this case, $t_{\frac{α}{2}}\approx z_{\frac{α}{2}}$
$level~of~confidence=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$z_{\frac{α}{2}}=z_{0.005}$
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: $\frac{2.57+2.58}{2}=2.575$
$t_{0.025}\approx z_{0.025}=2.575$
$E=4$ (within four books)
$s=16.6$
$n=(\frac{t_{\frac{α}{2}}.s}{E})^2$
$n=(\frac{2.575\times16.6}{4})^2$
$n=114.20$
Round up:
$n=115$
