Question

random sample of 328 medical doctors showed that 178 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 98% confidence interval for p. (Use 3 decimal places.)

lower limit
upper limit

Give a brief explanation of the meaning of the interval.

2% of the all confidence intervals would include the true proportion of physicians with solo practices.2% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.    98% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.98% of the all confidence intervals 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 p̂ along with the margin of error.Report the confidence interval.    Report the margin of error.Report p̂.

What is the margin of error based on a 98% confidence interval? (Use 3 decimal places.)

Answer 1