Question of Confidence interval Mislabeled seafood In December 2011, Consumer Re

Question

Question of Confidence interval

Mislabeled seafood In December 2011, Consumer Reports published their study of labeling of seafood sold in New York, New Jersey, and Connecticut. They purchased 190 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. Laboratory results indicated that 22% of these packages of seafood were mislabeled, incompletely labeled, or misidentified by store or restaurant employees.

a) Construct a 95% confidence interval for the proportion of all seafood packages in those three states that are mislabeled or misidentified.

b) Explain what your confidence interval says about seafood sold in these three states.

c) A 2009 report by the Government Accountability Board says that the Food and Drug Administration has spent very little time recently looking for sea- food fraud. Suppose an official said, “That’s only 190 packages out of the billions of pieces of seafood sold in a year. With the small number tested, I don’t know that one would want to change one’s buying habits.” (An official was quoted similarly in a dif- ferent but similar context). Is this argument valid? Explain.

Explanation / Answer

TRADITIONAL METHOD
given that,
possibile chances (x)=41.8
sample size(n)=190
success rate ( p )= x/n = 0.22
I.
sample proportion = 0.22
standard error = Sqrt ( (0.22*0.78) /190) )
= 0.03
II.B9
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.03
= 0.059
III.
CI = [ p ± margin of error ]
confidence interval = [0.22 ± 0.059]
= [ 0.161 , 0.279]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=41.8
sample size(n)=190
success rate ( p )= x/n = 0.22
CI = confidence interval
confidence interval = [ 0.22 ± 1.96 * Sqrt ( (0.22*0.78) /190) ) ]
= [0.22 - 1.96 * Sqrt ( (0.22*0.78) /190) , 0.22 + 1.96 * Sqrt ( (0.22*0.78) /190) ]
= [0.161 , 0.279]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.161 , 0.279] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

c.
argument valid,
They purchased 190 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. Laboratory results indicated that 22% of these packages of seafood were mislabeled, incompletely labeled, or misidentified by store or restaurant employees.
A 2009 report by the Government Accountability Board says that the Food and Drug Administration has spent very little time recently looking for sea- food fraud,an official said, That’s only 190 packages out of the billions of pieces of seafood sold in a year. With the small number tested, I don’t know that one would want to change one’s buying habits.

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