Question

A random sample of 1900 workers in a particular city found 494 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: _______   to________ %

Answer 1

Solution :

Given that,

n = 1900

x = 494

= x / n = 494 / 1900= 0.26

1 - = 1 - 0.26 = 0.74

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 = 1.960

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

= 1.960 * (((0.26 * 0.74 ) / 1900 )

= 0.02

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.26 - 0.02 < p < 0.26 + 0.02

0.24 < p < 0.28  

(24%, to 28% )