Question

Hint1 Hint In one city, 28.8% of the 7500 students surveyed said they eat at the school cafeteria. Find the 95% confidence interval for the proportion of students that said they eat at the school cafeteria. % to %

Answer 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = p̂ ± Z* sqrt(p̂*(1 – p̂)/n)

Where, p̂ is the sample proportion, Z is critical value, and n is sample size.

We are given

Sample size = n = 7500

Sample proportion = p̂ = 0.288 or 28.8%

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = p̂ ± Z* sqrt(p̂*(1 – p̂)/n)

Confidence Interval = 0.288 ± 1.96* sqrt(0.288*(1 – 0.288)/7500)

Confidence Interval = 0.288 ± 1.96*0.0052

Confidence Interval = 0.288 ± 0.0102

Lower limit = 0.288 - 0.0102 = 0.2778

Upper limit = 0.288 + 0.0102 = 0.2982

Confidence interval = (0.2778, 0.2982)

Confidence interval: 27.78% to 29.82%