## Can anyone help me with my issue? Have been trying all day and still no luck with my math.?

This is elementary Statistics question:

An insurance executive asserts that the mean amount paid by his firm for personal injury claims resulting from auto accidents is ,500. An actuary wants to check the accuracy of this claim and so samples randomly 36 cases involving personal injury. The sample mean is ,415 and an all-knowing alien being tells the actuary that σ = ,600. Can we reject the executive’s claim at the α = .05 level?

### One Response to “Can anyone help me with my issue? Have been trying all day and still no luck with my math.?”

1. Keun Says:

This is hypothesis testing for claims about mean with sigma known.

Our original claim is the mu = 18500. Negation of this claim is mu =/ 18500. Since the negation does not contain equality, this will the alternative hypothesis.

Ho: mu = 18500 (original claim)
Ha: mu =/ 18500

To find test statistic, use the formula z = (x bar – mu) / [sigma/sqrt(n)]. The givens are:
x bar = 19415, mu = 18500, sigma = 2600, and n = 36

Plug in then evaluate

z = (19415 – 18500) / [2600 / sqrt(36)] = 2.11

For p-value method, find the area to the right of this z score either by using table or calculator. Since this is two-tailed test (Ha: mu =/ 18500), multiply this area by 2. You should get p-value = 0.0349.

Conclusion: Since p-value = 0.0349 is less than the significance level α = .05, we reject the null hypothesis.

Interpretation: There’s enough evidence to support that the mean amount paid by his firm is not equal to 18500. So we reject the executive’s claim.