Question

How many Americans should be surveyed in order to be 98 percent confident that the sample mean will not differ from the population mean by more than​ 3%

Answer 1

Solution,

Given that,

$\hat p$ = 1 - $\hat p$ = 0.5

margin of error = E = 0.03

At 98% confidence level

$\alpha$ = 1 - 98%

$\alpha$ = 1 - 0.98 =0.02

$\alpha$/2 = 0.01

Z_$\alpha$/2 = Z_0.01 = 2.326

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (2.326 / 0.03)² * 0.5 * 0.5

= 1502.85

sample size = n = 1503