Question

Suppose the national average dollar amount for an automobile insurance claim is $565.412. You work for an agency in Michigan and you are interested in whether or not the state average is less than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≥ 565.412, Alternative Hypothesis: μ < 565.412. A random sample of 46 claims shows an average amount of $582.985 with a standard deviation of $91.4882. What is the test statistic and p-value for this test?

Answer 1

Given, sample mean = = $582.985

Standard deviation = s = $91.4882

Sample size n = 46

We have to use 'one sample t test' because population standard deviation is not known.

Hypothesis :

( claim )

Left tailed test.

Test statistic :

P-value :

P-value for this left tailed test is ,

P-value = P( t < 1.30275 )

Using excel function , =T.DIST( t , df , 1 )   , df = n - 1 = 46 - 1 = 45

P-value = T.DIST( 1.30275 , 45 , 1 ) = 0.9004

P-value = 0.9004

Decision about null hypothesis :

Rule : Reject null hypothesis if p-value less than significance level

Suppose Significance level = 0.05

Here P-value = 0.9004 is greater than = 0.05 . So fail to reject null hypothesis.

So, there is not sufficient evidence to conclude that state average is less than the national average