Question

In: Statistics and Probability

A random sample of 330 medical doctors showed that 164 had a solo practice. (a) Let...

A random sample of 330 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.

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

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.


(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 .     

Report along with the margin of error.

Report the margin of error.



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

Solutions

Expert Solution

Solution :

Given that,

n = 330

x = 164

a) Point estimate = sample proportion = = x / n = 164 / 330 = 0.497

1 - = 1 - 0.497 = 0.503

b) At 90% confidence level

= 1 - 90%

=1 - 0.90 = 0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

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

= 1.645 (((0.497 * 0.503) / 330)

= 0.045

A 90% confidence interval for population proportion p is ,

± E

= 0.497 ± 0.045

= ( 0.452, 0.542 )

lower limit = 0.452

upper limit = 0.542

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

c) Report along with the margin of error

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

= 1.645 (((0.497 * 0.503) / 330)

= 0.045


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