## Related questions with answers

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. 172). 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.

Construct a $95 \%$ confidence interval for the parameter of interest using Wilson's adjustment.

Solution

VerifiedGiven:

$\begin{aligned} n&=\text{Sample size}=22 \\ x&=\text{Number of successes}=17 \\ \tilde{p}&=\text{Sample proportion}=77\%=0.77 \\ c&=\text{Confidence level}=95\%=0.95 \end{aligned}$

Wilson's adjustment adds 2 successes and 2 failures to the sample.

The sample proportion is the number of successes divided by the sample size:

$\hat{p}=\dfrac{x+2}{n+4}=\dfrac{17+2}{22+2}=\dfrac{19}{26}\approx 0.7308$

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