Question

Obtain a sample size that will ensure a margin of error of at most the one specified.

Margin of error$=0.03$

Confidence level$=99\%$

Short answer

The required sample size is$1842.$

Step-by-step solution

Given information

The given data is

Margin of error$=0.03$

Confidence level$=99\%$

Explanation

Margin of error$=0.03$

Confidence level $=99\%$

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 is

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

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

$=0.25(7,367.361)$

$=1,841.84$

$\approx 1842.$