A random sample of 334 medical doctors showed that 174 had asolo practice.
(a) Let p represent the proportion of all medical doctors whohave a solo practice. Find a point estimate for p. (Use 3 decimalplaces.)
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(b) Find a 95% confidence interval for p. (Use 3 decimalplaces.)
lower limit
upper limit
Give a brief explanation of the meaning of the interval.
95% of the all confidence intervals would include the trueproportion of physicians with solo practices.
95% of the confidence intervals created using this methodwould include the true proportion of physicians with solopractices.
5% of the all confidence intervals would include the trueproportion of physicians with solo practices.
5% of the confidence intervals created using this method wouldinclude the true proportion of physicians with solopractices.
(c) As a news writer, how would you report the survey resultsregarding the percentage of medical doctors in solo practice?
Report the confidence interval.
Report the margin of error.
Report p̂ along with the margin of error.
Report p̂.
What is the margin of error based on a 95% confidenceinterval? (Use 3 decimal places.)
