Question

Of 535 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, 75% were mislabeled. ​

a) Construct a 90​% confidence interval for the proportion of all seafood sold in the country that is mislabeled or misidentified. ​The 90​% confidence interval is from____% to ___%

b) Explain what your confidence interval says about seafood sold in the country.

A.We are90​%confident that the interval captures the true proportion of all seafood sold in the country that is mislabeled.

B.In90​%of samples of seafood sold in the​ country, the proportion that is mislabeled will be in the interval.

C.There is a 90​% chance that the true proportion of mislabeled seafood is in the interval.

​c) A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

A.Yes, until another study produces a similar intervalYes, until another study produces a similar interval​, the results cannot be generalized.

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

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

Answer 1

Sample proportion () = 0.75
Sample size (n) = 535
a) Confidence interval(in %) = 90
z @ 90% = 1.645
Since we know that

Required confidence interval = (0.75-0.0308, 0.75+0.0308)
Required confidence interval = (0.7192, 0.7808)
b) A.We are 90​%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.
Please hit thumps up if the answer helped you.