4119 of 40 (6 complete) A c compared the pieces to standard gene fragments that
ID: 3320378 • Letter: 4
Question
4119 of 40 (6 complete) A c compared the pieces to standard gene fragments that can identify the species. The study found that 11 of the 23 red snapper packages tes Complete parts a through c below y of labeling of seafood sold in three U. S states. The group purchased 266 pieces of seafood from various kinds of a) Are the conditions for creating a confidence interval satisfied? Explain This Test: 40 pts p A B. C. No, because the sample is a simple Yes because the sample is a simple ran food stores and restaurants in the three states and genetically Yes, because the sample is a simple random sample the sample proportion is between 10% and 90% and there are at least 20 m sample the sample is less than 10% of th e population and there are at least 10 expected "successes" and 10 D. No, because the sample is a simple random sample the sample proportion is between 10% and 90%, and there are at least 2 b) Construct a 95% confidence interval for the proportion of red sn random sample, the sample is less than 10% of the population, and there are at least 10 exp expected "successes" and 20 expected Yailures. expected "successes" and 10 expected failures expected "failures apper" packages that were a difierent kind of fish 0 expected "successes. and 20 expected-failures. e Explain what the confidence interval from part (b) says about 'red snapper sold in these t 0 A, One is 95% confident that between. hree states Select the correct choice below and fil in the answer boxes within your choice % and. % of all "red snapper purchased for the study in these three states was not actually red snapper time, the true proportion of 'red snapper" sold in these three states that is falsely labeled is between Cick to select your answer(s) % and i % 0 Type here to searc t1:05 AMExplanation / Answer
Solution:- p = 11/23 = 0.478 , q = 1 - 0.478 = 0.522 , n = 23
95% confidence level for Z is 1.96
=> 0.478 +/- 1.96*sqrt(0.478*0.522/23)
= (0.273 , 0.682)
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