Answer
99% confidence: $n=298$
95% confidence: $n=173$
If the level of confidence decreases, the sample size decreases too.
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: $\frac{2.57+2.58}{2}=2.575$
$t_{0.005}\approx z_{0.005}=2.575$
$E=2$ (within 2 points)
$s=13.4$
$n=(\frac{t_{\frac{α}{2}}.s}{E})^2$
$n=(\frac{2.575\times13.4}{2})^2$
$n=297.65$
Round up:
$n=298$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
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.
$t_{0.025}\approx z_{0.025}=1.96$
$E=2$ (within 2 points)
$s=13.4$
$n=(\frac{t_{\frac{α}{2}}.s}{E})^2$
$n=(\frac{1.96\times13.4}{2})^2$
$n=172.45$
Round up:
$n=173$
