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