A random sample of 336 medical doctors showed that 174 had a

A random sample of 336 medical doctors showed that 174 had a solo practice. (a) Let p represent the proportion of all medical doctors who have a solo practice. Find a point estimate for p. (Use 3 decimal places.)


(b) Find a 99% confidence interval for p. (Use 3 decimal places.)

Give a brief explanation of the meaning of the interval.

(c) As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice?

What is the margin of error based on a 99% confidence interval? (Use 3 decimal places.)

Solution

a)
Point of estimate = Sample proportion = x/n =0.518

Confidence Interval For Proportion
CI = p