Margin of error$=0.02$

Confidence level$=90\%$

Educated guess$=0.1$

(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$=0.02$

Confidence level$=90\%$

Educated guess$=0.1$

From the given values

When the margin of error is $0.02$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

$n=0.1(1-0.1){\left(\frac{z_{\underset{¯}{a}}}{E}\right)}^2$

$=0.1(1-0.1){\left(\frac{1.645}{0.02}\right)}^2$

$=0.09(6,765.06)$

$=608.8$

$\approx 609.$

Margin of error$=0.02$

Confidence level$=90\%$

Educated guess$=0.1$

When the margin of error is $0.02$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

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

$=0.25{\left(\frac{1.645}{0.02}\right)}^2$

$=0.25(6,765.0625)$

$=1,691.266$

$\approx 1,692.$

When compared to the sample size obtained in $11.33$, 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 $11.33$ the constant $0.25$ is used, resulting in a bigger sample.