Question

In: Statistics and Probability

A random sample of 2000 workers in a particular city found 820 workers who had full...

A random sample of 2000 workers in a particular city found 820 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.

Solutions

Expert Solution

Solution :

Given that,

n = 2000

x = 820

Point estimate = sample proportion = = x / n = 820/2000=0.41

1 - = 1-0.41=0.59

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )   

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

= 1.96 (((0.41*0.59) / 2000)

= 0.0216

A 95% confidence interval is ,

- E < p < + E

0.41-0.0216 < p < 0.41+0.0216

0.3884<p<0.4316

38.84% , 43.16%


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