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Mislabeled seafood In 2013 the environmental group Oceana (usa.oceana.org) analyzed 1215 samples of seafood purchased across the United States and genetically compared the pieces to standard gene fragments that can identify the species. Laboratory results indicated that 33% of the seafood was mislabeled according to U.S. Food and Drug Administration guidelines.

Construct a 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified.

Sagot :

Using the z-distribution, it is found that the 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

z-distribution interval:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

  • In which z is the z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].

For this problem:

  • 1215 samples, hence [tex]n = 1215[/tex].
  • 33% was mislabeled or misidentified, hence [tex]p = 0.33[/tex].
  • 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 - 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3036[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.33 + 1.96\sqrt{\frac{0.33(0.67)}{1215}} = 0.3564[/tex]

The 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).

You can learn more about the use of the z-distribution to build a confidence interval at https://brainly.com/question/25730047

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