n 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.
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 for a confidence interval for a proportion.
First, find the sample proportion, which is the number of people who preferred seafood divided by the sample size:
p̂ = 160/500 = 0.32
Next, calculate the standard error of the proportion:
SE = √(p̂(1-p̂)/n)
SE = √(0.32(1-0.32)/500)
SE ≈ 0.021
To find the margin of error, we can use the z-value for a 99% confidence level, which corresponds to a 0.995 cumulative probability. From a standard normal distribution table, this z-value is approximately 2.58.
Margin of error = z * SE = 2.58 * 0.021 ≈ 0.054
Finally, calculate the confidence interval by subtracting and adding the margin of error to the sample proportion:
Lower limit = p̂ - margin of error = 0.32 - 0.054 ≈ 0.266
Upper limit = p̂ + margin of error = 0.32 + 0.054 ≈ 0.374
Therefore, the 99% confidence interval for the actual proportion of people who eat seafood on Fridays at this cafeteria is (0.266, 0.374). The upper limit is 0.374.
To find the 99% confidence interval for the actual proportion of people who eat seafood on Fridays at the cafeteria, you can use the formula:
Upper limit = sample proportion + z * √(sample proportion * (1 - sample proportion) / n),
where:
- sample proportion is the proportion of people in the sample who preferred seafood (160/500 = 0.32),
- z is the z-score corresponding to a 99% confidence level (z = 2.576),
- n is the size of the sample (500).
Plugging in the values, we have:
Upper limit = 0.32 + 2.576 * √(0.32 * (1 - 0.32) / 500) = 0.32 + 2.576 * √(0.21728 / 500) = 0.32 + 2.576 * 0.013096 = 0.32 + 0.03371 ≈ 0.3537.
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 0.3537.