# As a condition of employment, Fashion industries applicants must pass a drug test. Of the last...

## Question:

As a condition of employment, Fashion industries applicants must pass a drug test. Of the last {eq}265 {/eq} applicants, {eq}16 {/eq} failed the test.

(a) Develop a {eq}95\% {/eq} confidence interval for the proportion of applicants that fail the test. (Round your answer to 3 decimal places.)

(b) Would it be reasonable to conclude that more than {eq}11\% {/eq} of the applicants fail the test?

## Confidence Interval

The confidence interval is a type of interval estimation. The confidence interval estimation is used to estimate the population parameter. The 95% confidence interval means that there is a 95% probability of true value lie between the lower limit and upper limit.

Given Information:

The total number of applicants is, n = 265

The number of applicants failed in test, X = 16

(a)

Let X is the random variable that denotes the number of applicants failed in test,

The proportion is given as follows,

{eq}\begin{align*} \hat p &= \dfrac{X}{n}\\ & = \dfrac{{16}}{{265}}\\ & = 0.0604 \end{align*}{/eq}

The formula for confidence interval is given as,

{eq}\hat p \pm {z_{\alpha /2}} \cdot \sqrt {\dfrac{{\hat p \cdot \left( {1 - \hat p} \right)}}{n}}{/eq}

Here,

{eq}\hat p{/eq} is probability of success which is estimated by, {eq}\hat p = \dfrac{X}{n}{/eq}.

{eq}{z_{\alpha /2}}{/eq} is the critical value at level of significance {eq}\alpha{/eq} .

The lower limit of 95% confidence interval is given as,

{eq}\begin{align*} \hat p - {z_{\alpha /2}} \cdot \sqrt {\dfrac{{\hat p \cdot \left( {1 - \hat p} \right)}}{n}} & = 0.0604 - 1.96 \cdot \sqrt {\dfrac{{0.0604 \cdot \left( {1 - 0.0604} \right)}}{{265}}} \\ & = 0.032 \end{align*}{/eq}

The upper limit of 95% confidence interval is given as,

{eq}\begin{align*} \hat p + {z_{\alpha /2}} \cdot \sqrt {\dfrac{{\hat p \cdot \left( {1 - \hat p} \right)}}{n}} &= 0.0604 + 1.96 \cdot \sqrt {\dfrac{{0.0604 \cdot \left( {1 - 0.0604} \right)}}{{265}}} \\ & = 0.089 \end{align*}{/eq}

Therefore, the 95% confidence intervals for the population proportion is {eq}\left( {0.032,0.089} \right){/eq}.

(b)

No, it would be not reasonable to conclude that more than 11% applicants failed the test because the upper limit of confidence interval is 8.9% which less than 11%.