A random sample of 322 medical doctors showed that 172 had a solo practice. (a) Let p represent the proportion of all medical doctors who have a...
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# A. A random sample of 322 medical doctors showed that 172 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 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.

99% of the confidence intervals created using this method 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 along with the margin of error.

Report the confidence interval.

Report .

Report the margin of error.

What is the margin of error based on a 99% confidence interval?

B. A random sample of 336 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 95% confidence interval for p. (Use 3 decimal places.)

lower limit

upper limit

Give a brief explanation of the meaning of the interval.

5% of the all confidence intervals would include the true proportion of physicians with solo practices.

5% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.

95% of the all confidence intervals would include the true proportion of physicians with solo practices.

95% of the confidence intervals created using this method 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 along with the margin of error.

Report the confidence interval.

Report .

Report the margin of error.

What is the margin of error based on a 95% confidence interval?

A a) p =0.534 b) Lower limit = 0.463 Upper limit = 0.606 99% of the confidence intervals created using this method would... View the full answer

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