Clinical Biochemistry of Domestic Animals (Sixth Edition) - UMK ...

14
Chapter | 1 Concepts of Normality in Clinical Biochemistry
for comparing two responses involves using each subject
as its own control. For example, a pretreatment response
in an individual might be compared with a posttreatment
response. Clearly in this design, the pre- and posttreatment
responses may not be and most likely are not independent
so the procedure given earlier for comparing two
groups would not be appropriate. Rather, the differences in
response (pretreatment minus posttreatment response) are
formed and the population of interest is the single population
of differences having as one of its parameters the mean
difference, μ d . The quantity μ d is estimated by the mean
difference for the sample of n differences, d − Σ d i / n , and
an interval estimate is formed by d − ( t 1 α /2; n 1 s d )/ √ n
where s d is an estimate of σ d . If the interval so constructed
covers zero, then it may be that there is no difference
between the mean pre- and posttreatment values.
F . Comparing the Mean Response of
Three or More Populations Using
Independent Samples
1 . By Extension of the 2-Sample t-Test
Comparison of more than two groups is the natural progression
from the methodology discussed to this point. Consider
the comparison of three independent groups. The approach
that immediately comes to mind is that of estimating the
three groups ’ means and standard deviations and then constructing
three sets of confidence intervals (the first group
versus the second group, the first group versus the third
group, and the second group versus the third group) using
the approach described earlier for two independent groups.
However, some modifications need to be made. First, because
we are assuming that all three groups have equal variances,
pooling of the variances for the three groups provides a better
estimate of the common variance than does pooling of
just the variances for the two groups being compared. The
form of the pooled variance is the natural extension of that
for two groups, namely s p
2
[( n 1 1) s 1
2
( n 2 1) s 2
2
( n 3 1) s 3
2
]/(n 1 n 2 n 3 3). When the group sample
sizes are equal—that is, n 1 n 2 n 3 —s p
2
Σ s i
2
/3. The
quantity s p
2
is used for all three interval estimates.
Second, the error rate of each comparison has to be
adjusted so that the error rate over all three comparisons will
be α . This is required because theoretically it turns out that
the error rate over all three comparisons is larger than that for
a single comparison. Several approaches are suggested in the
literature for circumventing this problem. One such approach,
based on the Bonferroni inequality ( Neter et al. , 1996 ,
Stevens, 2002 ), is called the Bonferroni/Dunn procedure
( Zar, 1999 ). The Bonferroni/Dunn procedure involves making
each single comparison in a “ family ” of comparisons
with an error rate of α /m , where m is the total number of
comparisons to be made. This approach gives an error rate
of α over all comparisons—that is, over all members of the
family of comparisons. In the context of comparing k groups
in a pairwise manner, the form of the interval estimate is
( x − i x − j ) [( t 1 α /2 m ; n 1 n 2 … nk k )( s p )(1/ n i 1 / n j ) 1/2 ],
where m is the total number of comparisons to be made and
k is the total number of groups, which in the present example
is three. Intervals covering zero would indicate no difference
in the central value of the groups being compared.
2 . By One-Way Analysis of Variance
The process of deciding whether or not there are differences
between groups in the central value of the response being
evaluated can also be approached using the method of analysis
of variance (ANOVA). ANOVA involves decomposing
the total variability in a given set of data into parts reflective
of the amount of variability attributable to various sources.
One source of variability is that within the group. Because
the groups are assumed to have the same spread, this source
of variability is estimated as the pooling of the estimated
variances for the k groups considered and is equal to s p
2
defined earlier. The second source of variability results from
the variability among groups means. If there is no difference
in the groups ’ means, then the k samples can be thought of
as being k independent samples from a common population
and the k means, therefore, represent a sample of size k from
the sampling distribution of x − . Their variance represents
TABLE 1-5 Analysis of Variance Table for Classification of Responses on Basis of One Factor, Equal Responses
for Each Class a
Source of variation Degrees of freedom Sum of squares Mean square F value
Among group means k 1 SS A ( k 1) MS A
Within group k ( n 1) SS W k ( n 1) MS W
MS
MS
A
k
∑
n [ x ( x / k)]
i1
k
s2
∑ i
i
W
1
k
i
k
∑
i1
k 1
i
2
MS A / MS W
Total kn 1 SS T SS A SS W
a
k is the number of classes of groups, n is the number of responses per group, constant over all groups, s i 2 is the variability of the responses within the ith group.