# The human resources department of a very large organisation is trying to determine the proportion...

## Question:

The human resources department of a very large organisation is trying to determine the proportion of all employees that are satisfied with their current position. They randomly select {eq}81 {/eq} employees and ask them: "Are you satisfied with your current position?" {eq}68 {/eq} replied yes they were. Construct a {eq}95\% {/eq} confidence interval to estimate the true proportion of all employees at this workplace who were satisfied with their position.

## Confidence Interval:

Confidence Interval gives the interval estimate of population parameter it gives two values such that population parameter lies in between these two values with some certain proportion of times also called confidence coefficient.

## Answer and Explanation:

We need to calculate confidence Interval of population parameter,

{eq}P( p - z_{\frac{\alpha}{2}} \sqrt{\frac{p(1-p)}{n}} \leq P \leq p + z_{\frac{\alpha}{2}} \sqrt{\frac{p(1-p)}{n}} )=1-\alpha {/eq}

We are given alpha =0.05

So p={eq}\frac{68}{81}=0.8395 {/eq}

Now {eq}P( 0.8395-1.96 \sqrt{\frac{0.8396(1-0.8396)}{81}} \leq P \leq 0.8395+1.96 \sqrt{\frac{0.8395(1-0.8395)}{81}}) =0.95 \\ P(0.8102 \leq P \leq 0.8688)=0.95 {/eq}