Answer
90% confidence: $n=188$
95% confidence: $n=267$
If the level of confidence increases, the sample size increases too.
Work Step by Step
We expect to survey a large number of adult Americans. In this case, $t_{\frac{α}{2}}\approx z_{\frac{α}{2}}$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.1$
$z_{\frac{α}{2}}=z_{0.05}$
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: $\frac{1.64+1.65}{2}=1.645$
$t_{0.05}\approx z_{0.05}=1.645$
$E=1.5$ (within 1.5 points)
$s=12.5$
$n=(\frac{t_{\frac{α}{2}}.s}{E})^2$
$n=(\frac{1.645\times12.5}{1.5})^2$
$n=187.92$
Round up:
$n=188$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
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.
$t_{0.025}\approx z_{0.025}=1.96$
$E=1.5$ (within 1.5 points)
$s=12.5$
$n=(\frac{t_{\frac{α}{2}}.s}{E})^2$
$n=(\frac{1.96\times12.5}{1.5})^2$
$n=266.78$
Round up:
$n=267$
