Question

A random sample of 328 medical doctors showed that 162 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: (Select one)

1% of the confidence intervals created using this method 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.

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.

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

Report p̂ along with the margin of error.

Report the margin of error.

Report p̂.

Report the confidence interval.

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

Answer 1

'

a) point estimate =0.494

b)

lower limit =0.423

upper limit = 0.565

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

c)

Report p̂ along with the margin of error.

margin of error based on a 99% confidence interval =0.071