Question

I report the proportion of people who have health insurance at PCC with the following confidence interval, based on a survey of 200 randomly selected PCC students: 0.753 < p < 0.847.

a. How many people surveyed reported that they had health insurance?

b. At what confidence level was this interval found?

Answer 1

Solution(a)
Number of Sample (n)= 200
Lower confidence interval = 0.753
Upper confidence interval = 0.847
Sample proportion can be calculated as
Sample proportion = (Lower confidence interval + Upper confidence interval)/2 = (0.753+0.847)/2 = 1.6/2 = 0.8
80% of people surveyed reported that they had health insurance. So the number of people surveyed reported that they had health insurance = 0.8*200 = 160
So 160 people surveyed reported that they had health insurance.
Solution(b)
The margin of error can be calculated as
Margin of error(E) = (Upper confidence - Lower Confidence)/2 = (0.847-0.753)/2 = 0.047
Also, Margin of error(E) can be calculated as
The margin of error(E) = Zalpha/2 * sqrt(p*(1-p)/n)
Zalpha/2 can be calculated as
Zalpha/2 = E/sqrt(p*(1-p)/n) = 0.047/sqrt(0.8*(1-0.8)/200) = 0.047/0.0283 = 1.6617
From Z table we found alpha/2 = 0.0483
alpha = 0.0483*2 = 0.0966 or 0.1
So Confidence level = 1 - alpha = 1 - 0.1 = 0.9
So At 90% confidence level was this interval found.