Question

please show all work

A random sample of 45 life insurance policy holders showed that the average premiums paid on their life insurance policies was $340 per year with a sample standard deviation of $62. Construct a 90% confidence interval for the population mean. Make a statement about this in context of the problem.

Answer 1

Solution :

Given that,

= 340

s = 62

n = 45

Degrees of freedom = df = n - 1 = 45 - 1 = 44

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,44 = 1.680

Margin of error = E = t_/2,df * (s /n)

= 1.680 * (62 / 45)

= 15.53

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

- E < < + E

340- 15.53 < < 340 + 15.53

324.47 < < 355.53

(324.47, 355.53 )