Question

A random sample of 328 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 90% confidence interval for p. (Use 3 decimal places.)

lower limit
upper limit

Give a brief explanation of the meaning of the interval.

90% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.

10% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.  

  90% of the all confidence intervals would include the true proportion of physicians with solo practices.

10% 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 the confidence interval.

Report p̂.    

Report p̂ along with the margin of error.

Report the margin of error.

What is the margin of error based on a 90% confidence interval? (Use 3 decimal places.)

Answer 1

Solution :

Given that,

n = 328

x = 172

Point estimate = sample proportion = = x / n = 172/328=0.5244

1 - = 1-0.5244=0.4756

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645} ( Using z table )

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.5244*0.4756) /328 )

E = 0.045

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.5244 - 0.045< p <0.5244 + 0.045

0.479< p < 0.569

The 90% confidence interval for the population proportion p is : lower limit =0.479 , upper limit 0.569

brief explanation =90% of the all confidence intervals would include the true proportion of physicians with solo practices.