Question

A random sample of 1600 workers in a particular city found 432 workers who had full health insurance coverage. Find a 95% confidence interval for the true percent of workers in this city who have full health insurance coverage. Express your results to the nearest hundredth of a percent. .

Answer 1

Solution :

Given that,

n = 1600

x = 432

= x / n = 432 / 1600 = 0.270

1 - = 1 - 0.270 = 0.730

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.270 * 0.730) / 1600)

= 0.0218

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.270 - 0.0218 < p < 0.270 + 0.0218

0.2482 < p < 0.2918

(24.82% , 29.18%