Question

a sample of 800 persons showed that 120 persons do not have any health insurance. the 95% confidence interval for the population proportion of all persons who do not have any health insurance is close to: (use at least five decimal places during your computation)

Answer 1

The 95% confidence interval for the population proportion of all persons who do not have any health insurance is (0.148, 0.192).

Here is the calculation:

Sample size = 800

Number of persons who do not have any health insurance = 120

Sample proportion = 120 / 800 = 0.15

Confidence level = 95%

Z-critical value for 95% confidence level = 1.96

Confidence interval = (sample proportion - Z-critical value * standard error of the sample proportion, sample proportion + Z-critical value * standard error of the sample proportion)

Standard error of the sample proportion = sqrt(sample proportion * (1 - sample proportion) / sample size)

Standard error of the sample proportion = sqrt(0.15 * (1 - 0.15) / 800) = 0.0128

Confidence interval = (0.15 - 1.96 * 0.0128, 0.15 + 1.96 * 0.0128)

Confidence interval = (0.148, 0.192)

Explanation: