91Ó°ÊÓ

Margin of error$=0.04$

Confidence level$=99\%$

Educated guess $=0.5$

When the margin of error is $0.04$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

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

$=0.25{\left(\frac{2.575}{0.04}\right)}^2$

$=0.25(4,144.1)$

$=1,036.025$

$≈1,037.$

Margin of error$=0.04$

Confidence level$=99\%$

Educated guess$=0.5$

When the margin of error is $0.04$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

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

$=0.25{\left(\frac{2.575}{0.04}\right)}^2$

$=0.25(4,144.14)$

$=1,036.04$

$≈1,037.$

The sample size obtain in part a is smaller when compared to the sample size obtained in $11.36$. because in part a the educated guess $0.25$is used which provides the more accurate sample, whereas in $11.36$the constant$0.25$ is used which provides the larger sample.