Question

An insurance company is reviewing its current policy rates. When originally setting the rates
they believed that the average claim amount was $1,800. They are concerned that the true mean
does not equal this because they could potentially lose a lot of money. They randomly select 40
claims, and calculate the sample mean of $1,950. Assuming that the standard deviation of claims
follows a Normal distribution with known standard deviation $500, perform a hypothesis testing
to see if the insurance company should be concerned.

Suppose the significance level α is set to 0.01, what is the “acceptance” region?

What is the calculated value for the Z-statistic?

Answer 1

Solution :

= 1800

= 1950

= 500

n = 40

This is the two tailed test .

The null and alternative hypothesis is

H₀ :   = 1800

H_a :    1800

Test statistic = z

= ( - ) / / n

= (1950 - 1800) /500 / 40

= 1.897

p(Z >1.897 ) = 1-P (Z <1.897 ) =0.0578

P-value = 0.0578

= 0.01

p=0.0578 ≥ 0.01

Fail to reject the null hypothesis .