91Ó°ÊÓ

Margin of error$=0.02$

Confidence level$=95\%$

Educated guess$=0.6$

From the given values

When the margin of error is $0.02$and the confidence level is $95\%$, calculate the sample size.

With a $95\%$confidence level, the required value of $z_{\frac{a}{2}}$from table areas under the standard normal curve is $1.96$.

The sample size is

$n=0.6(1-0.6){\left(\frac{z_{\frac{\mathrm{Î\pm }}{2}}}{E}\right)}^2$

$=0.6(1-0.6){\left(\frac{1.96}{0.02}\right)}^2$

$=0.24(9,604)$

$=2,305.96$

$≈2,305.$

Margin of error$=0.02$

Confidence level $=95\%$

Educated guess $=0.6$

From the given data

When the margin of error is $0.02$and the confidence level is $95\%$, calculate the sample size.

With a $95\%$confidence level, the required value of $z_{\frac{a}{2}}$from table areas under the standard normal curve is $1.96$.

The sample size is

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

$=0.25{\left(\frac{1.96}{0.02}\right)}^2$

$=0.25(9,604)$

$=2,401.$

When compared to the sample size obtained in $11.32$, the sample size obtained in part an is less. because the educated guess is utilised in part, which produces a more accurate sample, whereas the constant $0.25$ is used in , which provides a bigger sample.