Question

Of 565 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared to standard gene fragments that can identify the​ species, 45​% were mislabeled.

​a) Construct a 95​% confidence interval for the proportion of all seafood sold in the country that is mislabeled or misidentified.

The 95​% confidence interval is from

​b) What does the confidence interval say about seafood sold in the​ country?

A. In 95​% of samples of seafood sold in the​ country, the proportion that is mislabeled will be in the interval.

B.We are 95​% confident that the interval captures the true proportion of all seafood sold in the country that is mislabeled.

C.There is a 95​% chance that the true proportion of mislabeled seafood is in the interval.

​c) Is the government​ spokesperson's criticism​ valid?

A.​No, as long as the necessary assumptions and conditions were​ met, the results can be generalized.

B.​Yes, the sample size must be at least​ 10% of the​ population, or the results cannot be generalized.

C.No comma until another study produces a different intervalNo, until another study produces a different interval​, the results should be generalized.

Answer 1

a0

sample proportion, = 0.45
sample size, n = 565
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.45 * (1 - 0.45)/565) = 0.0209

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0209
ME = 0.041

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.45 - 1.96 * 0.0209 , 0.45 + 1.96 * 0.0209)
CI = (0.409 , 0.491)

b)

B.We are 95​% confident that the interval captures the true proportion of all seafood sold in the country that is mislabeled.

c)

B.​Yes, the sample size must be at least​ 10% of the​population, or the results cannot be generalized.