Confidence Intervals Practice

Question

Q1: In December 2011, Consumer Reports published their study of labeling of seafood sold in New York, New Jersey, and Connecticut. They purchased a random sample of seafood from various kinds of seafood stores and restaurants and found that 12 of the 22 "red snapper" packages tested were actually a different kind of fish. Are the conditions satisfied for creating a confidence interval?

Q2: Construct a 95% confidence interval for the proportion of all red snapper packages that are mislabeled. It goes from to (Round to two decimal places)

Q3: Explain what your confidence interval says about "red snapper" sold in these three states.

Q4: Explain what it means to be 95% confident in this context.

Answer 1

Since 12 are mislabeled and 10 are not mislabeled, there are at least 10 successes and failures.

The sample is random and the sample is less than 10% of the population so all conditions are met.

12/22 = 0.55 are mislabeled (phat)

CI = phat +/- z-critical * sqrt(phat*(1-phat)/n), where n = sample size and z-critical for 95% CI is 1.96

= 0.55 +/- 1.96 * sqrt(0.55 * 0.45/22) = 0.34, 0.76

The plausible values for the proportion of mislabeled fish ranges between 0.34 and 0.76 at the 95 percent confidence level.

There is a 95% probability that a sampling of 95% confidence intervals will contain the true population proportion.