Question

In: Statistics and Probability

A random sample of 45 life insurance policy holders showed that the average premiums paid on...

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.

    1. Construct a 90% confidence interval for the population mean. Make a statement about this in context of the problem.

Solutions

Expert Solution

Solution :

Given that,

= $345

s = $65

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 = t0.05,44=1.680

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

= 1.680 * (65/ 45) = 16.2786

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

- E < < + E

345 - 16.2786 < < 345+ 16.2786

328.7214 < < 361.2786

(328.7214, 361.2786)


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