Question

You are told that a random sample of 150 people from Iowa has been given cholesterol tests, and 60 of these people had levels over the “safe” count of 200. Construct a 95% confidence interval for the population proportion of people in Iowa with cholesterol levels over 200. Place your LOWER limit, rounded to 3 decimal places, in the first blank. For example, .678 would be a legitimate entry. Place your UPPER limit, rounded to 3 decimal places, in the second blank. For example, .789 would be a legitimate entry. Lower Limit Upper Limit

Answer 1

$\dpi{120} \widehat p$ = 60/150 = 0.4

At 95% confidence level, the critical value is z_0.025 = 1.96

$\dpi{120} \widehat p$ +/- z_0.025 * sqrt($\dpi{120} \widehat p$(1 - $\dpi{120} \widehat p$ )/n)

= 0.4 +/- 1.96 * sqrt((0.4 * 0.6)/150)

= 0.4 +/- 0.0784

= 0.3216, 0.4784

= 0.322, 0.478

Lower limit = 0.322

Upper limit = 0.478