Question

A random sample of 1600 workers in a particular city found 688 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 = 1600

x = 688

Point estimate = sample proportion = = x / n =688/1600=0.43

1 - = 1 - 0.43=0.57

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

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

= 1.96 (((0.43*0.57) /1600 )

= 0.024

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.43 -0.024 < p <0.43 + 0.024

0.406< p < 0.454

40.6% and 45.4%