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).
<h3>z-distribution interval:</h3>In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
- In which z is the z-score that has a p-value of
.
For this problem:
- 1215 samples, hence
.
- 33% was mislabeled or misidentified, hence
.
- 95% confidence level, hence
, z is the value of Z that has a p-value of
, so
.
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 brainly.com/question/25730047