Posted by gina on .
Experience in investigating insurance claims shows that the average cost to process a claim is approximately normally distributed with a mean of $80. New costcutting 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 costcutting 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 ad, what is your conclusion about the cost cutting measures?

statistics please 
MathGuru,
H0: µ = 80 >null hypothesis
H1: µ < 80 >alternative hypothesis
Critical value for a onetailed test (the alternative hypothesis shows a specific direction) can be found using a ztable at .05 level of significance.
Calculate the test statistic using a onesample ztest.
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. 
statistics please 
gina,
yes you did Thank you...!!!!!