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.)

Lower limit

Upper limit

Give a brief explanation of the meaning of the interval.

1% of the confidence intervals created using this method would include the true proportion of physicians with solo practices. 99% of the all confidence intervals would include the true proportion of physicians with solo practices. 1% of the all confidence intervals would include the true proportion of physicians with solo practices. 99% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.

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

Report the confidence interval. Report the margin of error. Report p?. Report p? along with the margin of error.

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

Answer rating: 100% (QA)

a Given The sample size n 336 X 174 Sample proportion p 174 336 051786 ... View the full answer