Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.