Question

A random sample of 328 medical doctors showed that 160 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. 10% of the confidence intervals created using this method would include the true proportion of physicians with solo practices. 90% of the confidence intervals created using this method would include the true proportion of physicians with solo practices. 10% of the all confidence intervals would include the true proportion of physicians with solo practices. 90% 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 90% confidence interval? (Use 3 decimal places.

Answer 1

Solution :

Given that,

n = 328

x = 160

a) Point estimate = sample proportion = = x / n = 160 / 328 = 0.488

1 - = 1 - 0.488 = 0.512

b) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 = 1.645

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

= 1.645 (((0.488 * 0.512) / 328)

= 0.045

A 90% confidence interval for population proportion p is ,

± E

= 0.488 ± 0.045

= ( 0.443, 0.533 )

lower limit = 0.443

upper limit = 0.533

90% 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 = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.488 * 0.512) / 328)

= 0.045