Question

16. Medical Malpractice In a study of 1228 randomly selected medical malpractice lawsuits,
it was found that 856 of them were dropped or dismissed (based on data from the Physicians
Insurers Association of America). Construct a 95% confidence interval for the proportion of
medical malpractice lawsuits that are dropped or dismissed.

Answer 1

Solution :

Given that,

n = 1228

x = 856

Point estimate = sample proportion = = x / n = 856/1228=0.697

1 -   = 1-0.697 =0.303

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.697*0.303) / 1228)

E = 0.026

A 95% confidence interval is ,

- E < p < + E

0.697-0.026 < p < 0.697+0.026

0.671< p < 0.723