1. A random sample of 328 medical doctors showed that 171 have a solo practice.

ID: 3356348 • Letter: 1

Question

a. Let P represent the proportion of all medical doctors who have a solo practice. Find a point estimate for P.

b. Using 80% confidence level, find the critical value that will be used for finding confidence interval for P.

c. Based on the given information, find the standard deviation of the point estimate in part (a).

d. Using 80% confidence level find the margin of error (M.E)

e. Find 80% confidence interval for the proportion all medical doctors who have a solo practice.

f. Interpret the confidence interval you found in part (e)

a) point estimate for P phat =171/328=0.5213

b) for 80% CI ; critical value of z =1.2816

c)std deviation =(p(!-p)/n)1/2 =(0.5213*(1-0.5213)/328)1/2 =0.0276

d) margin of error =z*std error =0.0353

e) 80% confidence interval for the proportion all medical doctors who have a solo practic

=sample proportion -/+ margin of error =0.4860 to 0.5567

f) above interval gives 80% confidence to contain true proportion of all medical doctors who have a solo practice.

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