Getting a Job. The National Association of Colleges and Employers sponsors the Graduating Student and Alumni Survey. Part of the survey gauges student optimism in landing a job after graduation. According to one year's survey results, published in American Demographics, among the $1218$respondents, $733$said that they expected difficulty finding a job. Note: In this problem and the next, round your proportion answers to four decimal places.

a. Determine a sample size that will ensure a margin of error of at most $0.02$for a $95\%$confidence interval without making a guess for the observed value of $\hat{p}$.

b. Find a $95\%$confidence interval for $p$if, for a sample of the size determined in part (a), $58.7\%$of those surveyed say that they expect difficulty finding a job.

c. Determine the margin of error for the estimate in part (b), and compare it to the required margin of error specified in part (a).

d. Repeat parts (a)-(c) if you can reasonably presume that the percentage of those surveyed who say that they expect difficulty finding a job will be at least $56\%$.

e. Compare the results obtained in parts (a)-(c) with those obtained in part (d).

According to a survey, $733$ out of $1218$ graduates said they had trouble obtaining work following graduation.

For a population with a margin of error of at most $E$, $A(1-\alpha )$level confidence interval is calculated by

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

$\alpha =0.05$for a $95\%$confidence interval.

As a result, ${\mathrm{z}}_{\raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\mathrm{z}}_{0.025}=1.96$.

$\mathrm{E}=0.02$

As a result, the needed sample size $n$is determined as follows:

$\begin{array}{l}n=0.25{\left(\frac{1.96}{0.02}\right)}^2\\ \Rightarrow n=2401\end{array}$

As a result, $n=2401$ is the sample size necessary to keep the margin of error to a maximum of $0.02$.

The sample proportion is $0.587$and the sample size is $2401$.

$\alpha =0.05$for a $95\%$confidence level.

So,$$\begin{gathered} z_{a/2}=z_{0.025} \\ =1.96 \end{gathered}$$.

As a result, the $95\%$confidence interval will be

$\begin{array}{l}\mathrm{CI}=\widehat{p}\pm z_{\alpha /2}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ \Rightarrow \mathrm{CI}=0.587\pm 1.96\sqrt{\frac{0.587(1-0.587)}{2401}}\\ \Rightarrow \mathrm{CI}=0.587\pm 0.0197\\ \Rightarrow \mathrm{CI}=0.5673\text{to}0.6067\end{array}$

The sample proportion is $0.587$and the sample size is $2401$.

The $95\%$confidence interval for part $\left(c\right)$is $0.5673$to $0.6067$.

Divide the breadth of the confidence interval by two to get the margin of error.

As a result,

$$\begin{gathered} E=\frac{0.6067-0.5673}{2} \\ =0.0197 \end{gathered}$$

This margin of error is $0.0197$, which is nearly equal to the $0.02$margin of error given in part (b).

According to a survey, $733$out of $1218$graduates said they had trouble obtaining work following graduation.

For a population with a margin of error of at most $E$, $A(1-\alpha )$level confidence interval is calculated by

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

For a $95\%$ confidence level, $\alpha =0.05$.

So, $$\begin{gathered} z_{a/2}=z_{0.025} \\ =1.96 \end{gathered}$$.

$E=0.02$

As a result, the needed sample size $n$is determined as follows:

$$\begin{gathered} n=0.25{\left(\frac{1.96}{0.02}\right)}^2 \\ =2401 \end{gathered}$$

As a result, $n=2401$is the sample size necessary to keep the margin of error to a maximum of $0.02$.

$n=2401,\hat{p}=0.56$is given.

$\alpha =0.05$for a $95\%$confidence level.

As a result, $$\begin{gathered} z_{\alpha /2}=z_{0.025} \\ =1.96 \end{gathered}$$.

As a result, the $95\%$confidence interval will be written as

$\begin{array}{l}\mathrm{CI}=\widehat{p}\pm z_{\alpha /2}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ \Rightarrow \mathrm{CI}=0.56\pm 1.96\sqrt{\frac{0.56(1-0.56)}{2401}}\\ \Rightarrow \mathrm{CI}=0.56\pm 0.02\\ \Rightarrow \mathrm{CI}=0.54\text{to}0.58\end{array}$

As a result, the $90\%$confidence interval is $0.54$to $0.58$.

Divide the breadth of the confidence interval by two to get the margin of error.

$$\begin{gathered} E=\frac{0.58-0.54}{2} \\ =0.02 \end{gathered}$$

According to a survey,$733$ out of$1218$ graduates said they had trouble obtaining work following graduation.

Part $\left(a\right)-\left(c\right)$results:$n=2401,90\mathrm{\%}\mathrm{CI}=0.54$to $0.58,E=0.02$

Part$\left(d\right)$results:$n=2401,90\mathrm{\%}\mathrm{CI}=0.54$to$0.58,E=0.02$.