Question

A random sample of 340 medical doctors showed that 164 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 90% confidence interval for p. (Use 3 decimal places.)

lower limit
upper limit

Give a brief explanation of the meaning of the interval.

A) 90% of the all confidence intervals would include the true proportion of physicians with solo practices.

B) 10% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.

C) 90% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.

D) 10% 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?

A) Report p̂.

B) Report p̂ along with the margin of error.

C) Report the confidence interval.

D) Report the margin of error.

What is the margin of error based on a 90% confidence interval? (Use 3 decimal places.)

Answer 1

Solution :

Given that,

n = 340

x = 164

a) Point estimate = sample proportion = = x / n = 164 / 340 = 0.482

1 - = 1 - 0.482 = 0.518

b) At 90% confidence level the z is,

= 1 - 90%

= 1 - 0.90 = 0.10

/2 = 0.05

Z_/2 = 1.645

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

= 1.645 * (((0.482 * 0.518 ) / 340)

= 0.045

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.482 - 0.045 < p < 0.482 + 0.045

(0.437 < p < 0.527)

lower limit = 0.437

upper limit = 0.527

correct option is = C.

90% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.

c) correct option is = B.

Report along with the margin of error.

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

= 1.645 * (((0.482 * 0.518 ) / 340)

= 0.045