## Related questions with answers

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 seafood 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 different but similar context). Is this argument valid? Explain.

Solution

VerifiedGiven:

$n=190$

$\hat{p}=22\%=0.22$

$c=95\%=0.95$

(a) For confidence level $1-\alpha=0.95$, determine $z_{\alpha/2}=z_{0.025}$ using the normal probability table in the appendix (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):

$z_{\alpha/2}=1.96$

The margin of error is then:

$E=z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=1.96\times \sqrt{\dfrac{0.22(1-0.22)}{190}}\approx 0.0589$

The boundaries of the confidence interval are then:

$\hat{p}-E=0.22-0.0589=0.1611=16.11\%$

$\hat{p}+E=0.22+0.0589=0.2789=27.89\%$

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