Getting a Job. The National Association of Colleges and Employers sponsors the Graduating Student and Alumni Survey. Part f the survey gauges student optimism in landing a job after graduation. According to one year's survey results, published in American Demographics, 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. Use these data to find and interpret a $95\%$ confidence interval for the proportion of students who expect difficulty finding a job.

b. Find the margin of error for the estimate of $p$.

c. Express the confidence interval in the form point estimate $\pm$ margin of error.

According to a poll, $733$ out of $1218$ graduates expect trouble finding work following graduation.

Sample size is $n=1218$.

The number of successes is $x=733$.

Sample proportion would be $\hat{p}=\frac{x}{n}$

$=\frac{733}{1218}$

$=0.6018$

A confidence interval of

$1-\alpha$, find $z_{\alpha /2}$using table II.

Confidence level is $95\%$.

i.e. $\alpha =0.05$.

Using table II,

$z_{\alpha /2}=z_{0.05/2}$

$=z_{0.025}$

$=1.96$

The confidence interval for $p$is given as-

$\hat{p}\pm z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$\Rightarrow 0.6018\pm 1.96\sqrt{\frac{0.6018(1-0.6018)}{1218}}$

$\Rightarrow 0.6018\pm 0.0275$

$\Rightarrow 0.5743\text{to}0.6293$

$\Rightarrow 57.43\%\text{to}62.93\%$

As a result, one can be confident that between $57.43\%$and $62.93\%$of U.S. grads will have trouble finding work following graduation.

The sample size is $n=1218$.

The number of successes is $x=733$.

Sample proportion would be $\hat{p}=\frac{x}{n}$

$=\frac{733}{1218}$

$=0.6018$

A confidence interval of$1-\alpha$,

To find $z_{\alpha /2}$using table II.

Confidence level is $95\%$. i.e. $\alpha =0.05$.

Using table II,

$z_{\alpha /2}=z_{0.05/2}$

$=z_{0.025}$

$=1.96$

The margin of error for the estimate of $p$is given as-

$E=z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$\Rightarrow E=1.96\sqrt{\frac{0.6018(1-0.6018)}{1218}}$

$\Rightarrow E=0.0275$

$\Rightarrow E=2.75\%$

The estimated $p$ has a margin of error of $0.0275$ ($2.75\%$).

Sample size is $n=1218$.

The number of success is $x=733$.

Sample proportion would be role="math" localid="1651412370640" $\hat{p}=\frac{x}{n}$

$=\frac{733}{1218}$

$=0.6018$

Confidence interval

$\hat{p}\pm z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$\Rightarrow CI=0.6018\pm 0.0275$

As a result, there is a $95\%$ chance that between $0.6018\pm 0.0275\%$ of U.S. grads expect trouble finding work following graduation.