MATH SOLVE

4 months ago

Q:
# An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered by the company. A random sample of 200 claims shows that the insurance company covered 80 accident claims and did not cover 120 claims. Construct a 90 percent confidence interval estimate of the true proportion of claims covered by the insurance company.

Accepted Solution

A:

Answer:0.343 < Pp < 0.457Step-by-step explanation:We are going to denote the population proportion of claims covered as Pp.Ps is going to be the sample proportion of claims covered by the insurance company. Ps = 80/200 = 0.40Qs = 1 - Ps = 0.60n = total number of claims = 200E is going to be the margin of error.[tex]E=z\sqrt{\frac{(Ps)(Qs)}{n} }[/tex]z is the critical value. The z score for a confidence level of 90 % is going to be: 1.645[tex]E=1.645\sqrt{\frac{(0.4)(0.6)}{200} }[/tex]E = 0.057The confidence interval for the population proportion is: (Ps - E, Ps + E)(0.343, 0.457) or 0.343 < Pp < 0.457