In a random sample 500 people eating lunch at a cafeteria on various Fridays, it was found that 160 preferred seafood. Find 99% confidence interval for the actual proportion of the people who eat seafood on Fridays at this cafeteria.Give the upper limit.

Answer 1

To find the 99% confidence interval for the actual proportion of people who eat seafood on Fridays at this cafeteria, we can use the formula:

CI = p̂ ± z * √(p̂(1-p̂)/n)

Where:
CI = Confidence Interval
p̂ = Sample Proportion
z = Z-score corresponding to the desired confidence level
n = Sample Size

First, calculate the sample proportion:
p̂ = 160/500 = 0.32

Next, find the Z-score corresponding to the 99% confidence level. The Z-score for a 99% confidence level is approximately 2.576.

Plug the values into the formula:
CI = 0.32 ± 2.576 * √((0.32*(1-0.32))/500)

Calculating the standard deviation:
√((0.32*(1-0.32))/500) ≈ 0.023

CI = 0.32 ± 2.576 * 0.023

Calculating the confidence interval:
CI = 0.32 ± 0.0593

To find the upper limit of the confidence interval, add the margin of error to the sample proportion:
Upper Limit = 0.32 + 0.0593 = 0.3793

Therefore, the upper limit of the 99% confidence interval for the actual proportion of people who eat seafood on Fridays at this cafeteria is approximately 0.3793.