Question

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

Margin of error$=0.01$

Confidence level$=90\%$

Short answer

The required sample size is$6766.$

Step-by-step solution

Given information

The given data is

Margin of error $=0.01$

Confidence level $=90\%$

Explanation

The given data is

Margin of error $=0.01$

Confidence level $=90\%$

When the margin of error is $0.01$and the confidence level is $90\%$, calculate the sample size.

With a $90\%$confidence level, the required value of $z_{\frac{a}{2}}$from table areas under the standard normal curve is $1.645$.

Then the sample size is

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

$=0.25{\left(\frac{1.645}{0.01}\right)}^2$

$=0.25(27,060.25)$

$=6,765.06$

$\approx 6766.$