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 ```The McFaland Insurance Company claims department reports the mean cost to process a claim is \$60. An industry comparison showed this amount to be larger than most other insurance companies, so they instituted cost-cutting measures. To evaluate the effect of the cost-cutting measures, the supervisor of the claims department selected a random sample of 26 claims processed last month. The sample information is reported below: 45 49 62 40 43 61 48 53 67 63 78 64 48 54 51 56 63 69 58 51 58 59 56 57 38 76 At 95% confidence level is it reasonable to conclude that mean cost to process a claim is now less than \$60...Why?```
 ```Mutiny58 ? 1. Anything in that range that matches our existing condition for mean costs is the hypothesis H0 (also called the null hypothesis) ? that the efforts DO NOT change costs. 2. Only those values outside the range would support hypothesis H1 ? the alternate hypothesis ? that mean costs are now lower. Statistics Glossary ?Hypothesis Testing? http://www.cas.lancs.ac.uk/glossary_v1.1/hyptest.html To calculate the values in the statistical range, we?ll need the mean and standard deviation, easily done in Excel or many programmable calculators: Mean = \$56.42 S.D. = \$10.04 3. It is effective to use the Student?s T-distribution for small sample sizes, as it provides a statistical significance related to the sample size. Our sample size is 26 ? but the Student?s T-distribution uses ?degrees of freedom? ? which is N-1 or 25. What?s the t-critical value? That depends on the confidence level that we?re trying to get ? and whether we?re dealing with a two-tailed or one-tailed T-distribution. Here it?s distributed about the mean, so it?s two-tailed (sometimes samples can vary only one way). The critical values are determined from the following table: Surfstat Statistical Tables ?Student?s T Table? (1997) http://math.uc.edu/~brycw/classes/148/tables.htm 95% confidence: 2.06 4. You?ll accept the H0 hypothesis based on whether or not the same falls in the range of: mean +/- T-crtical * SD In our case this is: \$56.04 +/- 2.06 * \$10.04 = \$34.58 to \$76.72 With \$60 well within that range, there?s no statistical reason to believe that cost-control efforts changed anything, so you?re accept H0 ? no change. BMJ.com ?Statements of probability and confidence intervals? (undated) http://bmj.bmjjournals.com/collections/statsbk/4.shtml 5. What can you do to enhance your conclusions? You can expand your sample size, though your critical T never drops below 1.96 ? and you?d have to get a much tighter standard deviation to judge the cost-control efforts to be effective. Google search strategy t-distribution + critical t-distribution + ?hypothesis testing? Best regards, Omnivorous-GA```