Margin of error $=0.01$

Confidence level$=95\%$

Educated guess $=0.3$

(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.

The given values are

Margin of error$=0.01$

Confidence level$=95\%$

Educated guess$=0.3$

Margin of error$=0.01$

Confidence level$=95\%$

Educated guess $=0.3$

When the margin of error is $0.01$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

$n=0.3(1-0.3){\left(\frac{z_{\underset{¯}{a}}}{E}\right)}^2$

$=0.3(1-0.3){\left(\frac{1.96}{0.01}\right)}^2$

$=0.21(38,416)$

$=8,067.36$

$\approx 8,068.$

Margin of error$=0.01$

Confidence level$=95\%$

Educated guess$=0.3$

From the given data

When the margin of error is $0.01$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

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

$=0.25{\left(\frac{1.96}{0.01}\right)}^2$

$=0.25(38,416)$

$=9,604.$

When compared to the sample size obtained in $11.31$, the sample size obtained in part a is smaller. since in part (a) informed guess is used, resulting in a more accurate sample, whereas in $11.31$the constant$0.25$ is used, resulting in a bigger sample.