A consumer advocacy group published a study of labeling of seafood sold in three U.S. states. The group purchased 226 pieces of seafood from various kinds of food stores and restaurants in the three states and genetically compared the pieces to standard gene fragments that can identify the species. The study found that 11 of the 26 "red snapper" packages tested were a different kind of fish. Assume that the study used a simple random sample. Complete parts b through c below.
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.)
c) Explain what the confidence interval from part (b) says about "red snapper" sold in these three states. Select the correct choice below and fill in the answer boxes within your choice. (Round to one decimal place as needed.)
A.Ninety-five percent of the time, the true proportion of "red snapper" sold in these three states that is falsely labeled is between ____% and ____%.
B.There is a 95% chance that the probability of any given "red snapper" sold in these three states being actual red snapper is between ____% and _____%.
C.One is 95% confident that between _____% and _____% of all "red snapper" purchased for the study in these three states was not actually red snapper.
D.One is 95% confident that between ____% and _____%of all "red snapper" sold in food stores and restaurants in these three states is not actually red snapper.
b)
n = 26, x = 11, p̅ = x/n = 0.4231
95% Confidence interval for the proportion:
At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960
Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.4231 - 1.96 *√(0.4231*0.5769/26) = 0.233
Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.4231 + 1.96 *√(0.4231*0.5769/26) = 0.613
0.233 < p < 0.613
c)
Answer D.One is 95% confident that between 23.3% and 61.3% of all "red snapper" sold in food stores and restaurants in these three states is not actually red snapper.
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