Basic Statistics

APPROXIMATE TTEST 121

computing the variance. When unequal variances are found, the first thing that should
be examined is why they are different. Is it the case of a few outliers, or are the two
distributions different? The use of box plots is highly recommended, or histograms
can be examined for each sample. Note that some computer programs do provide
tests for equal variances and they all compute the standard deviation, so the variance
can easily be obtained.

9.3 APPROXIMATE t TEST

In the first F test of the hemoglobin levels of children with cyanotic and acyanotic
heart disease, we were not able to reject the null hypothesis of equal variances.
Similarly, we were not able to reject the null hypothesis in the diet example. Thus, in
both cases it was possible that the population variances were equal. In Section 7.5.3,
when we computed the confidence interval for the difference in the mean between
two independent groups using the t distribution, we stated that one of the assumptions
was equal population variances in the two groups. The same assumption was made in
Section 8.2.2 when we tested whether the means of two independent populations were
equal. The F tests that we performed in this chapter give us additional confidence that
the t test of Section 8.2.2 was appropriate since we did not find significant differences
in the variances.
But what if the variances had been significantly different? If the pooled-variance
t test given in Chapter 8 is used when the population variances are not equal, the P
value from the test may be untrustworthy. The amount of the error depends on how
unequal the variances are and also on how unequal the sample sizes are. For nearly
equal sample sizes, the error tends to be small.
One course of action that is commonly taken when we wish to perform a test
concerning the means of two independent groups is to use an approximate t test
(see van Belle et al. [2004]). If the observations are independent and are simple
random samples from populations that are normally distributed, then after making an
adjustment to the d.f., we can use the following test statistic:

which has an approximate t distribution. Suppose now in the hemoglobin example
given in Table 8.2 that si = 10 and s: = 1.0167. If we perform the F test, equal
variance would be rejected at the Q = .05 level. Then our computed approximate t
sta tistic is
-2.71

• 
  
  
  
  • = -2.88
    13.03 - 15.74
    t=
    J1.0167/19 + 10/12 &%%
    The d.f.’s are approximately