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6.4 CONFIDENCE INTERVAL FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS 181
EXAMPLE 6.4.3
The purpose of a study by Granholm et al. (A-7) was to determine the effectiveness of
an integrated outpatient dual-diagnosis treatment program for mentally ill subjects. The
authors were addressing the problem of substance abuse issues among people with severe
mental disorders. A retrospective chart review was carried out on 50 consecutive patient
referrals to the Substance Abuse/Mental Illness program at the VA San Diego Healthcare
System. One of the outcome variables examined was the number of inpatient treatment
days for psychiatric disorder during the year following the end of the program.
Among 18 subjects with schizophrenia, the mean number of treatment days was 4.7 with
a standard deviation of 9.3. For 10 subjects with bipolar disorder, the mean number of
psychiatric disorder treatment days was 8.8 with a standard deviation of 11.5. We wish
to construct a 95 percent confidence interval for the difference between the means of the
populations represented by these two samples.
Solution:
First we use Equation 6.4.2 to compute the pooled estimate of the common
population variance.
s 2 p = 118 - 1219.32 2 + 110 - 12111.52 2
18 + 10 - 2
= 102.33
When we enter Appendix Table E with 18 + 10 - 2 = 26 degrees of freedom
and a desired confidence level of .95, we find that the reliability factor
is 2.0555. By Expression 6.4.4 we compute the 95 percent confidence interval
for the difference between population means as follows:
102.33
14.7 - 8.82 ; 2.0555 A 18
-4.1 ; 8.20
-12.3, 4.10
+ 102.33
10
We are 95 percent confident that the difference between population means
is somewhere between -12.3 and 4.10. We can say this because we know
that if we were to repeat the study many, many times, and compute confidence
intervals in the same way, about 95 percent of the intervals would
include the difference between the population means.
Since the interval includes zero, we conclude that the population
means may be equal.
■
Population Variances Not Equal When one is unable to conclude that the
variances of two populations of interest are equal, even though the two populations may
be assumed to be normally distributed, it is not proper to use the t distribution as just
outlined in constructing confidence intervals.
A solution to the problem of unequal variances was proposed by Behrens (3) and
later was verified and generalized by Fisher (4, 5). Solutions have also been proposed by
Neyman (6), Scheffé (7, 8), and Welch (9, 10). The problem is discussed in detail by
Cochran (11).