Question

A random sample of 2000 people there were 11% that have diabetes. Determine a 95% confidence interval estimate for the proportion of people with diabetes.

Answer 1

Solution:

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

n = 2000

P = x/n = 0.11 = 11%

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

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

Confidence Interval = 0.11 ± 1.96* sqrt(0.11*(1 – 0.11)/ 2000)

Confidence Interval = 0.11 ± 1.96*0.0070

Confidence Interval = 0.11 ± 0.0137

Lower limit = 0.11 - 0.0137 = 0.0963

Upper limit = 0.11 + 0.0137 = 0.1237

Confidence interval = (0.0963, 0.1237)

We are 95% confident that the population proportion of people with diabetes will lies between 0.0963 and 0.1237.