Margin of error$=0.01$

Confidence level$=90\%$

Educated guess$=0.9$

(a) Obtain a sample size that will ensure a margin of error of at most the one specified (provided of course that the observed value of the sample proportion is further from $0.5$that of the educated guess.

(b). Compare your answer to the corresponding one and explain the reason for the difference, if any.

Margin of error$=0.01$

Confidence level$=90\%$

Educated guess$=0.9$

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 $1.645$.

The sample size is

$n=0.9(1-0.9){\left(\frac{z_{\underset{¯}{a}}}{E}\right)}^2$

$=0.9(1-0.9){\left(\frac{1.645}{0.01}\right)}^2$

$=0.01(27,060.25)$

$=2,435.4225$

$\approx 2,436.$

Margin of error$=0.01$

Confidence level$90\%$

Educated guess$=0.9$

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 $1.645$.

The sample size is

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

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

$=0.25(27,060.25)$

$=6,765.06$

$\approx 6,766.$

When compared to the sample size obtained in $11.34$, the sample size obtained in component an is smaller. since in part an informed guess is used, resulting in a more accurate sample, whereas in $11.34$ the constant $0.25$ is used, resulting in a bigger sample.