To calculate a 99% confidence interval for the proportion of people who prefer seafood at the cafeteria, we can use the formula for the confidence interval for a proportion:
\[ \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]
Where:
- \(\hat{p}\) is the sample proportion of people who prefer seafood.
- \(Z\) is the Z-score corresponding to the desired confidence level (for 99% confidence, \(Z \approx 2.576\)).
- \(n\) is the sample size.
Step 1: Calculate the sample proportion (\(\hat{p}\))
In this case, the number of people who prefer seafood (successes) is 160, and the total sample size \(n\) is 500.
\[ \hat{p} = \frac{160}{500} = 0.32 \]
Step 2: Calculate the standard error (SE)
Next, we calculate the standard error:
\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.32(1 - 0.32)}{500}} = \sqrt{\frac{0.32 \times 0.68}{500}} = \sqrt{\frac{0.2176}{500}} \approx \sqrt{0.0004352} \approx 0.0208 \]
Step 3: Calculate the margin of error
Now we find the margin of error using the Z-score for 99% confidence:
\[ \text{Margin of error} = Z \times SE = 2.576 \times 0.0208 \approx 0.0536 \]
Step 4: Determine the confidence interval
Now we can calculate the confidence interval:
\[ \text{Lower limit} = \hat{p} - \text{Margin of error} = 0.32 - 0.0536 = 0.2664 \] \[ \text{Upper limit} = \hat{p} + \text{Margin of error} = 0.32 + 0.0536 = 0.3736 \]
Step 5: Final result
Therefore, the 99% confidence interval for the actual proportion of people who prefer seafood on Fridays at this cafeteria is:
\[ \boxed{(0.2664, 0.3736)} \]
This means we are 99% confident that the true proportion of people who prefer seafood is between approximately 26.64% and 37.36%.