Question

Suppose the national average dollar amount for an automobile insurance claim is $580.659. You work for an agency in Michigan and you are interested in whether or not the state average is greater than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≤ 580.659, Alternative Hypothesis: μ > 580.659. A random sample of 29 claims shows an average amount of $584.228 with a standard deviation of $67.9408. What is the test statistic and p-value for this test?

Question 13 options:

1)	Test Statistic: 0.283, P-Value: 0.7794

2)	Test Statistic: -0.283, P-Value: 0.6103

3)	Test Statistic: 0.283, P-Value: 0.3897

4)	Test Statistic: 0.283, P-Value: 0.6103

5)	Test Statistic: -0.283, P-Value: 0.3897

Answer 1

Solution :

Given that ,

= 580.659  

= 584.228

= 67.9408

n = 29

The null and alternative hypothesis is ,

H₀ :   580.659

H_a : > 580.659

This is the right tailed test .

Test statistic = z

= ( - ) / / n

= ( 584.228 - 580.659) / 67.9408 / 29

= 0.283

The test statistic = 0.283

P - value = P(Z > 0.283 ) = 1 - P (Z < 0.283 )

= 1 - 0.6114

= 0.3886

The P-value = 0.3886