"A knowledge of statistics is like aknowledge of foreign languages or of algebra; it may prove of use atany time under any circumstances."

11.2 COMPARING TWO MEANS (Pages 617 -633)OVERVIEW: One can use thet-distribution on two samples that are not matched pairs. (Hencethere would be no difference column.) Basically, one tests the nullhypothesis that the two samples came from populations with equalmeans. The degrees of freedom is the smaller of n

_{1}-1, n_{2}-1, where n_{1}and n_{2}are the sample sizes.Let

x

_{1}(bar) = mean of sample ofsize n_{1}frompopulation P_{1}s

_{1}= standard deviation calculated from the sample x

_{2}(bar) = mean of sample of size n _{2}from populationP_{2}s

_{2}= standard deviation calculated from the sample m

_{1}= unknown mean of populationP_{1}m

_{2}= unknown mean of populationP_{2}To test the null hypothesis m

_{1}= m _{2}, we calculate the tstatistict = [x

_{1}(bar)-x_{2}(bar)]/[sqrt(s _{1}^{2}/n _{1}+ s _{2}^{2}/n_{2})],with df = the smaller of n

_{1}-1, n_{2}-1.The reader is reminded that, in this situation,"variances add" (see page 400). The expression under the radical inthe denominator is an approximation for the variance of the samplingdistribution of mean differences. Hence, the square root is anapproximation for the standard deviation of the meandifferences.

A confidence interval for the difference ofpopulation means has the form

(x

_{1}(bar)-x_{2}(bar)) plus/minust*[sqrt(s_{1}^{2}/n_{1}+ s_{2}^{2}/n _{2})]. Example:

Here are hypothetical results of a computerprogramming aptitude test that was administered to random samples of51 men and 60 women.

Men

n

_{m}= 51x(bar)

_{m}= 82s

_{m}= 5Women

n

_{w}= 60x(bar)

_{w}= 84s

_{w}= 7Null hypothesis H

_{0}: m_{m}= m _{w}Alternate hypothesis H

_{a}: m_{m}is not equal to m _{w}Degrees of freedom = 51-1 = 50.

t = (82-84)/sqrt(25/51 + 49/60) = -1.75

If we test Ho at the 5% level of significance(2-tail), the critical region is t < -2.009 or t > 2.009. Sincethe calculated t is not in the critical region, the difference 82-84is not significant at this level.

We can also note that

-1E99,-1.75,50) =.0431266224, which also indicates that we are not in the "outer 2.5%"in a 2-tail test.tcdf( We could construct a 95% confidence interval form

_{m}- m_{w}. This interval is

(82-84) plus/minus (2.009)sqrt(25/51 + 49/60) = -2 plus/minus 2.30.This is the interval [-4.30, 0.30].

Note that this interval contains 0, supporting thepreviously obtained conclusion that the mean difference of -2 is notsignificant at the 5% level.