Question

A random sample of 320 medical doctors showed that 180 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 all confidence intervals would include the true proportion of physicians with solo practices. O 2% of the all confidence intervals would include the true proportion of physicians with solo practices. • 98% of the confidence intervals created using this method would include the true proportion of physicians with solo practices. O 2% of the confidence intervals created using this method 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 p along with the margin of error, O Report the margin of error O Report the confidence interval. Report What is the margin of error based on a 98% confidence interval? (Use 3 decimal places.)

Answer 1

X = = 0. 563 180 320 b 98% CI for p 0.563 +2.33 Sona) p = x ľ + ²x12 | P (1-2) 0.563 * 0.437 320 0: 530 < 5 < 0.595 Explanation : Oplień 3 c) Report F along with the margin Margin of Errol = zar PCI-P on errol n 0.032 CS Scanned with CamScanner

0.587 = 3278 = = b و (B - X = 3278 0 وعد 5 کا لا97 (ط for pe re 22/2 * ( 1 ) A 0.584 * 3، 576 * 3ا * 0.584 5589 0 : 514 كطك0 : 57۹ = F (وه X /669 = 0.322 5/ 80 99% C. I for p 2/2 £ظ * / (-) 0.322 * 3.576 (322 0.67 5/ 8o 0.321 كط ك0315 0.315 < CS Scanned with CamScanner