Are you really being served red snapper? Refer to the Nature (July 15, 2004) study of fish specimens labeled “red snapper,” Exercise 3.75 (p. 196). Recall that federal law prohibits restaurants from serving a cheaper, look-alike variety of fish (e.g., vermillion snapper or lane snapper) to customers who order red snapper. A team of University of North Carolina (UNC) researchers analyzed the meat from each in a sample of 22 “red snapper” fish fillets purchased from vendors across the United States in an effort to estimate the true proportion of fillets that are really red snapper. DNA tests revealed that 17 of the 22 fillets (or 77%) were not red snapper but the cheaper, look-alike variety of fish.

a. Identify the parameter of interest to the UNC researchers.

b. Explain why a large-sample confidence interval is inappropriate to apply in this study.

c. Construct a 95% confidence interval for the parameter of interest using Wilson’s adjustment.

d. Give a practical interpretation of the confidence interval.

Referring to the Nature study of fish specimens labeled red snapper, exercise 3.75, a team of University of North Carolina (UNC) researchers analyzed the meat from each in a sample of 22 “red snapper” fish fillets purchased from vendors across the United States in an estimate of the true proportion of fillets that are really red snapper.

DNA tests revealed that 17 of the 22 fillets were not red snapper but the cheaper look-alike variety of fish.

a.

The parameter of interest to the University of North Carolina (UNC) researchers is the true proportion of fillets that are really red snapper.

b.

Here, n is very low, that is why the large sample confidence interval is inappropriate to apply to the study.

c.

Using Wilson’s adjustment,

The adjusted sample proportion is,

$\begin{array}{rcl}\mathrm{p}\%& =& \frac{\mathrm{x}+2}{\mathrm{n}+4}\\ & =& \frac{17+2}{22+4}\\ & =& \frac{19}{26}\\ & =& 0.7308\end{array}$

The proportion of red snapper is,

$\begin{array}{rcl}\hat{\mathrm{p}}& =& 1-0.7308\\ & =& 0.2692\end{array}$

Using Wilson’s adjustment, the 95% confidence interval for true proportion is,

$\begin{array}{rcl}\hat{\mathrm{p}}\pm {\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\hat{\mathrm{p}}\left(1-\hat{\mathrm{p}}\right)}{\mathrm{n}+4}}& =& 0.2692\pm {\mathrm{z}}_{0.025}\sqrt{\frac{0.2692\left(1-0.2692\right)}{22+4}}\\ & =& 0.2692\pm 1.960\sqrt{\frac{0.1967}{26}}\left[\mathrm{Using}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ & =& 0.2692\pm \left(1.960\times 0.0869\right)\\ & =& 0.2692\pm 0.1703\\ & \approx & 0.27\pm 0.17\end{array}$

Therefore, the 95% confidence interval for the true proportion is $\left(0.27\pm 0.17\right)$.

d.

$\begin{array}{rcl}\left(0.27\pm 0.17\right)& =& \left(0.27+0.17,0.27-0.17\right)\\ & =& (0.1,0.44)\end{array}$

95% confident that true proportion of all fillets that are red snapper is between 0.10 and 0.44