Experience in investigating insurance claims shows that the average cost to process a claim is approximately normally distributed with a mean of $80. New cost-cutting measures were started and a sample of 25 claims was tested. The sample mean of the costs to process these claims was %76 and the sample standard deviation of the costs was $10. We would like to test whether the cost-cutting measures seem to be working at the 5% significance level.

Question

Experience in investigating insurance claims shows that the average cost to process a claim is approximately normally distributed with a mean of $80. New cost-cutting measures were started and a sample of 25 claims was tested. The sample mean of the costs to process these claims was %76 and the sample standard deviation of the costs was $10. We would like to test whether the cost-cutting measures seem to be working at the 5% significance level.

A. State the null and alternative hypotheses for this test.
B. find the critical value for this test.
C. Calculate the test statistic.
D. Should we reject or fail to reject the null hypothesis?
E.Based on your answer to a-d, what is your conclusion about the cost cutting measures?

Answer 1

A. Null hypothesis (H0): The cost-cutting measures have no effect and the average cost to process a claim is still $80.

Alternative hypothesis (Ha): The cost-cutting measures have an effect and the average cost to process a claim is less than $80.

B. To find the critical value for this test, we need to determine the significance level, sample size, and the type of test (one-tailed or two-tailed). In this case, the significance level is 5% (0.05), the sample size is 25, and we are conducting a one-tailed test (since we are testing if the average cost is less than $80).

Using a t-distribution table or a statistical software, we find the critical t-value for a one-tailed test with a significance level of 0.05 and 24 degrees of freedom to be approximately -1.711.

C. To calculate the test statistic, we use the formula:

t = (sample mean - population mean) / (sample standard deviation / √sample size)

In this case, the sample mean is $76, the population mean is $80, the sample standard deviation is $10, and the sample size is 25.

t = (76 - 80) / (10 / √25)
t = -4 / (10 / 5)
t = -4 / 2
t = -2

D. To determine whether to reject or fail to reject the null hypothesis, we compare the test statistic (-2) with the critical value (-1.711). If the test statistic is less than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

Since -2 is greater than -1.711, we fail to reject the null hypothesis.

E. Based on the above analysis, we conclude that there is not enough evidence to suggest that the cost-cutting measures have significantly reduced the average cost to process a claim.