A consumer advocacy group published a study on seafood labeling in three U.S. states. The study found that 10 out of the 20 “red snapper” packages tested were mislabeled and contained a different kind of fish. Assume that the study used a simple random sample. Complete parts a through c below.---**a) Are the conditions for creating a confidence interval satisfied? Explain.**- **A.** Yes, because the sample is a simple random sample, the sample proportion is between 10% and 90%, and there are at least 20 “successes” and 20 “failures.” - **B.** No, because the sample is a simple random sample, the sample proportion is between 10% and 90%, and there are at least 20 “successes” and 20 “failures.” - **C.** Yes, because the sample is a simple random sample, the sample is less than 10% of the population, and there are at least 10 “successes” and 10 “failures.” [Selected Answer] - **D.** No, because the sample is a simple random sample, the sample is less than 10% of the population, and there are at least 10 “successes” and 10 “failures.”**b) Construct a 95% confidence interval for the proportion of “red snapper” packages that were a different kind of fish.**(Round to three decimal places as needed.)

a) Suppose p is the true proportion of “red snapper” packages that were a different kind of fish.According to the question, among a random sample of 20 “red snapper” 10 were a different kind of fish. The sample size is n = 20 and the number of favorable cases is x = 10. The sample proportion is p̂ = x/n = 10/20 = 0.5. It is assumed that the population is large enough and the sample is less than 10% of the population. The number of success is, np̂ = (20)(0.5) = 10 and the number of failures is, n(1 – p̂) = (20)(1 – 0.5) = 10. Hence, there are at least 10 “successes” and 10 “failures”. In addition, the sample is a simple random sample. Hence, yes, the conditions for creating a confidence interval are satisfied. The correct option is C. b) The confidence level is 95%. The level of significance is α = 1 – 0.95 = 0.05. Now, α/2 = 0.025 and 1 – α/2 = 0.975. Using Excel formula: =NORM.S.INV(0.975), or from standard normal table corresponding to row of 1.9 and column of 0.06, it is obtained that the critical value is, zα/2 = z0.025 ≈ 1.96. The excel output is, The 95% confidence interval is, The 95% confidence interval for the proportion of ““red snapper” packages that were a different kind of fish is (0.281, 0.719).