Question

# A random sample of 320 medical doctors showed that 162 had a solo practice. (a) Let...

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 along with the margin of error.Report the margin of error.     Report the confidence interval.Report .

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

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