Question

A random sample of 45 life insurance policy holders showed that the average premiums paid on their life insurance policies was $345 per year with a sample standard deviation of $65. Construct a 90% confidence interval for the population mean. Make a statement about this in context of the problem.

Answer 1

Solution :

Given that,

= 345

= 65

n = 45

at 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645* ( 65/ 45)

= 15.9

At 90% confidence interval estimate of the population mean is,

- E < < + E

345-15.9< < 345+15.9

329.1< < 360.9