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 n1-1, n2-1, where n1 and n2 are the sample sizes.
Let
x1(bar) = mean of sample ofsize n1 frompopulation P1
s1 = standard deviation calculated from the sample
x2(bar) = mean of sample of size n2 from populationP2
s2 = standard deviation calculated from the sample
m1 = unknown mean of populationP1
m2 = unknown mean of populationP2
To test the null hypothesis m1 = m2, we calculate the tstatistic
t = [x1(bar)-x2(bar)]/[sqrt(s12/n1 + s22/n2)],
with df = the smaller of n1-1, n2-1.
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
(x1(bar)-x2(bar)) plus/minust*[sqrt(s12/n1 + s22/n2)].
Example:
Here are hypothetical results of a computerprogramming aptitude test that was administered to random samples of51 men and 60 women.
| Men | nm = 51 | x(bar)m = 82 | sm = 5 |
| Women | nw = 60 | x(bar)w = 84 | sw = 7 |
Null hypothesis H0: mm = mw
Alternate hypothesis Ha: mm is not equal to mw
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 tcdf(-1E99,-1.75,50) =.0431266224, which also indicates that we are not in the "outer 2.5%"in a 2-tail test.
We could construct a 95% confidence interval formm - mw. 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.