Drug Tests There is a 3.9% rate of positive drug test results among workers in the United States (based on data from Quest Diagnostics). Assuming that this statistic is based on a sample of size 2000, construct a 95% confidence interval estimate of the percentage of positive drug test results. Write a brief statement that interprets the confidence interval.

The estimated proportion of positive drug results\(\left( {\hat p} \right)\)is 3.9%.

A 95% confidence interval is to be constructed for the percentage of positive drug test results for a sample size of 2000.

The confidence level is 95% or 0.95.

The following is the formula to compute the 95% confidence interval (CI) of the population proportion:

\(\begin{array}{c}CI = \hat p \pm E\\ = \hat p \pm {z_{\frac{\alpha }{2}}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}} \end{array}\)

Here, E is the margin of error and n is the sample size.

Here, the proportion of positive drug test result is obtainedas shown below:

\(\begin{array}{c}\hat p = 3.9\% \\ = \frac{{3.9}}{{100}}\\ = 0.039\end{array}\)

The sample size (n) is 2000.

Here,

\(\begin{array}{c}100\left( {1 - \alpha } \right) = 95\\1 - \alpha = 0.95\\\alpha = 1 - 0.95\\ = 0.05\end{array}\)

The value of the z-score corresponding to\(\alpha = 0.05\)for the two-tailed test is 1.96.

Thus, the margin of error is

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}} \\ = 1.96\sqrt {\frac{{0.039\left( {1 - 0.039} \right)}}{{2000}}} \\ = 0.008484672.\end{array}\)

And the confidence interval of population proportion becomes

\(\begin{array}{c}CI = \left( {\hat p - E,\hat p + E} \right)\\ = \left( {0.039 - 0.008484672,0.039 + 0.008484672} \right)\\ = \left( {0.0305,0.0474} \right)\\ = \left( {3.1\% ,4.7\% } \right).\end{array}\)

Therefore, the 95% confidence interval estimate of the percentage of positive drug test results is (3.10%,4.7%).

The confidence interval has the following interpretation:

It can be inferred that out of all the samples collected, 95% of them will have a value of positive drug test results that will fall between 3.1% and 4.7%.