Question

Question Help Of 515515 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, 6262​% were mislabeled. ​a) Construct a 9090​% confidence interval for the proportion of all seafood sold in the country that is mislabeled or misidentified. ​b) Explain what your confidence interval says about seafood sold in the country. ​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?

Answer 1

Answer:

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.

Step-by-step explanation:

Sample size, n = 51

p = 0.62

1 - p = 1 - 0.62 = 0.38

n = 515

Confidence level = 90% = Zcritical at 90% = 1.645

Confidence interval = (p ± margin of error)

Margin of Error = Zcritical * sqrt[(p(1-p))/n]

Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]

Margin of Error = 1.645 * 0.0214

Margin of Error = 0.035203

Lower boundary = (0.62 - 0.035203) = 0.584797

Upper boundary = (0.62 + 0.035203) = 0.655203

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.