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.

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 = t_0.05,44=1.680

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

= 1.680 * (65/ $\sqrt$ 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)

orchestra answered 2 years ago

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