Question

A random sample of 320 medical doctors showed that 162 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 98% confidence interval for *p*. (Use 3 decimal
places.)

lower limit | |

upper limit |

Give a brief explanation of the meaning of the interval.

98% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.2% of the all confidence intervals would include the true proportion of physicians with solo practices. 2% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.98% 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
margin of error. Report the confidence
interval.Report *p̂*.

What is the margin of error based on a 98% confidence interval?
(Use 3 decimal places.)

Answer #1

a)

The point estimate for the population proportion is equal to the sample proportion. This is computed here as:

p = 162 / 320 = 0.506

Therefore 0.506 is the point estimate here.

b)

From standard normal tables, we get:

P( -2.576 < Z < 2.576 ) = 0.99

Therefore the confidence interval here is computed as:

(0.434, 0.578)

Therefore the lower limit there is 0.434 and the upper limit is 0.578

c)

Now lastly the margin of error above is already computed as: 0.072

Therefore 0.072 is the margin of error for 99% confidence interval

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