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5.From a sample of 1,000 college students, 28% watch the Super Bowl game each year. Find the margin of error and the 95% confidence interval for the proportion of all college students who watch the Super Bowl.

5.From a sample of 1,000 college students, 28% watch the Super Bowl game each year. Find the margin of error and the 95% confidence interval for the proportion of all college students who watch the Super Bowl.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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**Problem Statement** From a sample of 1,000 college students, 28% watch the Super Bowl game each year. **Task** Find the margin of error and the 95% confidence interval for the proportion of all college students who watch the Super Bowl. **Solution Steps** 1. **Identify sample proportion (p-hat):** - \( p\hat{} = \frac{28}{100} = 0.28 \) 2. **Sample size (n):** - \( n = 1000 \) 3. **Calculate standard error (SE):** - \( SE = \sqrt{\frac{p\hat{} (1 - p\hat{})}{n}} = \sqrt{\frac{0.28 \times 0.72}{1000}} \) 4. **Find z-score for 95% confidence level:** - Generally, \( z = 1.96 \) for 95% confidence 5. **Calculate margin of error (ME):** - \( ME = z \times SE \) 6. **Determine the confidence interval:** - Lower limit = \( p\hat{} - ME \) - Upper limit = \( p\hat{} + ME \) The solution will give the 95% confidence interval for the proportion of college students who watch the Super Bowl.
Transcribed Image Text:**Problem Statement** From a sample of 1,000 college students, 28% watch the Super Bowl game each year. **Task** Find the margin of error and the 95% confidence interval for the proportion of all college students who watch the Super Bowl. **Solution Steps** 1. **Identify sample proportion (p-hat):** - \( p\hat{} = \frac{28}{100} = 0.28 \) 2. **Sample size (n):** - \( n = 1000 \) 3. **Calculate standard error (SE):** - \( SE = \sqrt{\frac{p\hat{} (1 - p\hat{})}{n}} = \sqrt{\frac{0.28 \times 0.72}{1000}} \) 4. **Find z-score for 95% confidence level:** - Generally, \( z = 1.96 \) for 95% confidence 5. **Calculate margin of error (ME):** - \( ME = z \times SE \) 6. **Determine the confidence interval:** - Lower limit = \( p\hat{} - ME \) - Upper limit = \( p\hat{} + ME \) The solution will give the 95% confidence interval for the proportion of college students who watch the Super Bowl.
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