Question

6.8.4. A health planning agency wishes to know, for a certain geographic region, what proportion of patients admitted to hospitals for the treatment of trauma die in the hospital. A 95 percent confidence interval is desired, the width of the interval must be .06, and the population proportion, from other evidence, is estimated to be .20. How large a sample is needed?

Answer 1

600

Step-by-step explanation:

Let p = the estimated L.P.P.

The width of the confidence interval = 0.06

The z-score at 95% CI = 1.96

Therefore,

N = [(1.96)^2 * (p(1-p))] / (0.06^2)

= [(1.96)^2 * (0.20(1-0.20))] / (0.06^2)

= 600