Question

A random sample of 338 medical doctors showed that 168 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.

99% of the confidence intervals created using this method 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.

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.

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

Report the margin of error.

Report p̂ along with the margin of error.

Report p̂. Report the confidence interval.

Answer 1

Solution :

Given that,

n = 338

x = 168

a) Point estimate = sample proportion = = x / n = 168 / 338 = 0.497

1 - = 1 - 0.497 = 0.503

b) At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.497 * 0.503) / 338)

= 0.070

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.497 - 0.070 < p < 0.497 + 0.070

(0.427 < p < 0.567)

lower limit = 0.427

upper limit = 0.567

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.