In: Statistics and Probability
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.
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%