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?

H0: µ = 80 -->null hypothesis

H1: µ < 80 -->alternative hypothesis

Critical value for a one-tailed test (the alternative hypothesis shows a specific direction) can be found using a z-table at .05 level of significance.

Calculate the test statistic using a one-sample z-test.

With your data:
z = (76 - 80)/(10/√25)

Finish the calculation.

If the test statistic exceeds the critical value from the table, reject the null and conclude µ < 80. If the test statistic does not exceed the critical value from the table, do not reject the null (you cannot conclude a difference in this case).

I hope this will help get you started.

yes you did Thank you...!!!!!

A. The null hypothesis, denoted as H0, for this test would be "The cost-cutting measures have no effect on the average cost to process a claim." The alternative hypothesis, denoted as Ha, would be "The cost-cutting measures have a significant effect on the average cost to process a claim."

B. To find the critical value for this test, we need to refer to the t-distribution table or use statistical software. Since the sample size is 25, we have 24 degrees of freedom (n-1). With a significance level of 5% (or 0.05), a two-tailed test is appropriate. Retrieving the critical value from the t-distribution table or using software, we find t-critical to be approximately ±2.064.

C. The formula to calculate the test statistic in this case is:
t = (sample mean - hypothesized mean) / (sample standard deviation / √sample size)

Given:
Sample mean = $76
Hypothesized mean (from the null hypothesis) = $80
Sample standard deviation = $10
Sample size = 25

Plugging in these values, we can calculate the test statistic as follows:
t = (76 - 80) / (10 / √25)
t = -4 / (10/5)
t = -4 / 2
t = -2

D. To determine if we should reject or fail to reject the null hypothesis, we compare the absolute value of the test statistic (2) to the critical value (±2.064).

Since the absolute value of the test statistic (2) does not exceed the critical value (2.064), we fail to reject the null hypothesis.

E. Based on the results, we conclude that there is not enough evidence to suggest that the cost-cutting measures have a significant effect on the average cost to process claims at the 5% significance level.