Question

A random sample of 334 medical doctors showed that 174 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. 1% 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. 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. (c) As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice? Report the confidence interval. Report the margin of error. Report p̂. Report p̂ along with the margin of error. What is the margin of error based on a 99% confidence interval? (Use 3 decimal places.)

Answer 1

Solution :

Given that,

n = 334

x = 174

a) Point estimate = sample proportion = = x / n = 174 / 334 = 0.521

1 - = 1-0.521 = 0.479

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

Z/2 = Z_{0.005 = 2.576}

Z/2 = 2.576

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

=2.576 * ((0.521*(0.479) /334 )

= 0.070

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.521 - 0.070< p < 0.521 + 0.070

0.451 < p < 0.591  

( 0.451, 0.591 )

Lower limit = 0.451

Upper limit = 0.591

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

A 99% confidence interval created using this method would include the population proportion p is 0.451 and 0.591.

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

=2.576 * ((0.521*(0.479) /334 )

= 0.070