Fish contaminated by a plant’s discharge. Refer (Example 1.5, p. 38) to the U.S. Army Corps of Engineers data on a sample of 144 contaminated fish collected from the river adjacent to a chemical plant. Estimate the proportion of contaminated fish that are of the channel catfish species. Use a 90% confidence interval and interpret the result.

The U.S army Corps of Engineers data on a sample of 144 contaminated fish collected from the river adjacent to a chemical plant. That is, the sample size is n=144.Let X be the number of catfish species, that is x=96.

Let,

Proportion of contaminated fish that are of the chemical catfish species is

$\begin{array}{rcl}\hat{p}& =& \frac{x}{n}\\ & =& \frac{96}{144}\\ & =& 0.667\\ & & \end{array}$

To ensure that sample size is sufficiently large that the normal distribution provides a reasonable approximation for the sampling distribution of, it require both $n\hat{p}$ and $n\hat{q}$to be at least 15. That is, $n\hat{p}⩾15$and $n\hat{q}⩾15$.

Here $n\hat{p}$and $n\hat{q}$are the number of successes and number of failures, respectively, in the sample.

Now,

$\begin{array}{rcl}n\hat{p}& =& 144\times 0.667\\ & =& 96.048\\ & & \end{array}$

$\begin{array}{rcl}n\hat{q}& =& 144\times \left(1-0.667\right)\\ & =& 47.952\\ & & \end{array}$

$n\hat{p}$and $n\hat{q}$values are at least 15, it may assume that the normal approximation is reasonable and can conclude that the sample size is sufficiently large.

The 90% confidence intervals for true proportion is,

$\begin{array}{rcl}\hat{p}\pm 1.645\sqrt{\frac{\hat{p}\hat{q}}{n}}& =& 0.667\pm 1.645\left(\sqrt{\frac{0.667\left(1-0.667\right)}{144}}\right)\\ & =& 0.667\pm 0.064\\ & =& \left(0.603,0.731\right)\\ & & \end{array}$

Hence, it is 90% confident that, the true proportion of contaminated fish that are of the channel catfish species is between 0.603 and 0.731.