91影视

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 鈥渞ed snapper鈥� 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鈥檚 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鈥檚 adjustment, the 95% confidence interval for true proportion is,

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

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

d.

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

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