91Ó°ÊÓ

Here costs associated with conducting interviews for a job opening have skyrocketed over the years. In this problem given that $211$of $502$, senior human resources executives at U.S companies believe that their hiring managers are interviewing too many people to find qualified candidates for the job.

The population of interest in statistical research is the group from which the researcher wishes to draw findings. Hiring managers is the population of interest in this study.

The population parameter of interest $p$is hiring managers. The parameter of interest is a statistical variable that provides additional data on the study sample or group under investigation. These criteria define and characterize a specific study group.

Let’s sample size=$\mathrm{n}$. Then, the sample size$n$=$502$

Then the reliable estimate is given by$\begin{array}{c}\mathrm{p}=\frac{\stackrel{¯}{\mathrm{X}}}{\mathrm{n}}\\ =\frac{211}{502}\\ =0.42\end{array}$

Here sample size is large than a reliable estimate. So, the sample size is large enough to provide a reliable estimate of $p$.

It is given that, n = 502 and x = 211

The formula CI for p is:

$\mathrm{CI}=\stackrel{⌢}{\mathrm{p}}Â\pm {\mathrm{z}}_{\frac{\mathrm{a}}{2}}\frac{\sqrt{\stackrel{⌢}{\mathrm{p}}(1−\stackrel{⌢}{\mathrm{p}})}}{\mathrm{n}}$

The z- critical value at a 0.01 level of significance is 2.33

The sample proportion is,

$\begin{array}{c}\stackrel{⌢}{\mathrm{p}}=\mathrm{x}/\mathrm{h}\\ =211/502\\ =\mathrm{0.4203.}\end{array}$

The 98% CI for p is:

$\begin{array}{c}\mathrm{CI}=\stackrel{⌢}{\mathrm{p}}Â\pm {\mathrm{z}}_{\frac{\mathrm{a}}{2}}\frac{\sqrt{\stackrel{⌢}{\mathrm{p}}(1−\stackrel{⌢}{\mathrm{p}})}}{\mathrm{n}}\\ =0.4203Â\pm 2.33\frac{\sqrt{0.4203(1−0.4203)}}{502}\\ =0.4203Â\pm 2.33\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ =0.4203Â\pm (2.33×0.022)\\ =0.4203Â\pm 0.0513\end{array}$

$\begin{array}{l}=(0.4203−0.0513,0.4203+0.0513)\\ =(0.369,0.472)\end{array}$

Therefore, the 98% CI for p is (0.369, 0.472)

For $90\mathrm{\%}$confidence interval, the interval estimate is given by:

$\begin{array}{c}\mathrm{z}=\stackrel{¯}{\mathrm{X}}Â\pm 1.64×\frac{\mathrm{σ}}{\sqrt{\mathrm{n}}}\\ =211Â\pm 1.64×\frac{1}{\sqrt{502}}\\ =211.07\end{array}$

And $210.93$

$\begin{array}{l}\text{Theproportionis=}\frac{210.93}{211.07}\\ =0.99\end{array}$

The interval estimate is given by $0.14$. It is a narrower value.