91Ó°ÊÓ

Margin of error$=0.03$

Confidence level$=99\%$

Educated guess$=0.5$

When the margin of error is $0.03$and the confidence level is $99\%$, calculate the sample size.

With a $99\%$confidence level, the required value of $z_{\frac{a}{2}}$from table areas under the standard normal curve is $2.575$.

The sample size is

$n=0.25{\left(\frac{\frac{z_a}{2}}{E}\right)}^2$

$=0.25{\left(\frac{2.575}{0.03}\right)}^2$

$=0.25(7,367.361)$

$=1,841.84$

$≈1,842.$

Margin of error$=0.03$

Confidence level$=99\%$

Educated guess$=0.5$

When the margin of error is $0.03$and the confidence level is $99\%$, calculate the sample size.

With a $99\%$confidence level, the required value of $z_{\frac{a}{2}}$from table areas under the standard normal curve is $2.575$.

The sample size is

$n=0.25{\left(\frac{\frac{z_{\underset{¯}{a}}}{E}}{E}\right)}^2$

$=0.25{\left(\frac{2.575}{0.03}\right)}^2$

$=0.25(7.367.361)$

$=1,841.84$

$≈1,842.$

Because in part a, the educated guess $0.25$is used, which offers a more accurate sample, and in $11.35$, the constant $0.25$ is used, which produces a bigger sample, the sample size achieved in part an is smaller than the sample size obtained in .