Question

A random sample of 200 students was drawn from a large university. It was found that 73 students in this sample live in off-campus apartment. Construct a 90% confidence interval for the proportion of students in this university that live in off-campus apartment. Use normal approximation in this question. You can also take a look at Wilson score interval.

Answer 1

Confidence interval for Population Proportion is given as below:

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

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

We are given

x = 73

n = 200

P = x/n = 73/200 = 0.365

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

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

Confidence Interval = 0.365 ± 1.6449* sqrt(0.365*(1 – 0.365)/200)

Confidence Interval = 0.365 ± 1.6449* 0.0340

Confidence Interval = 0.365 ± 0.0560

Lower limit = 0.365 - 0.0560 = 0.3090

Upper limit = 0.365 + 0.0560 = 0.4210

Confidence interval = (0.3090, 0.4210)