Data Analysis Statistics An Introduction to Statistical Inference and Data Analysis

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2. �2 = 1 � �1 + (-1) � �2 + 0 � �3 and 1 - 1 + 0 = 0.
We usually identify different contrasts by their coefficients, e.g. c =
(1, 1, -2).
Orthogonal Contrasts
" We want to test H0 : � = 0 vs. H1 : � = 0. An unbiased estimator of �
is
k
� �
� = ciXi�;
i=1
�
we will reject H0 if � is observed sufficiently far from zero.
�
" The quantity (�)2 is not a satisfactory measure of departure from H0 :
� = 0 because it depends on the magnitude of the coefficients in the
contrast. Accordingly, we define the sum of squares associated with
the contrast � to be
2
k
�
ciXi�
i=1
SS� = .
k
c2/ni
i=1 i
" Fact: Under H0 : �1 = � � � = �k, SS� is independent of SSW and
SS�/�2
" Fact: Under H0 : �1 = � � � = �k,
SS�
/1
SS�
�2
F (�) = =
SSW
SSW /(N - k)
/(N - k)
�2
220 CHAPTER 11. K-SAMPLE LOCATION PROBLEMS
" The F -test of H0 : � = 0 is to reject if and only if
P = PH0 (F (�) e" f(�)) d" �,
i.e. if and only if
f(�) e" q =qf(1-�,df1=1,df2=N-k),
where f(�) denotes the observed value of F (�).
" Definition 11.2 Two contrasts with coefficient vectors (c1, . . . , ck) and
(d1, . . . , dk) are orthogonal if
k
cidi
= 0.
ni
i=1
" Notice that, if n1 = � � � = nk, then the orthogonality condition simpli-
fies to
k
cidi = 0.
i=1
" In the Heyl (1930) example:
If n1 = n2 = n3, then �1 and �2 are orthogonal because
1 � 1 + 1 � (-1) + (-2) � 0 = 0.
If n1 = 6 and n2 = n3 = 5, then �1 and �2 are not orthogonal
because
1 � 1 1 � (-1) (-2) � 0 1 1
+ + = - = 0.
6 5 5 6 5
However, �1 is orthogonal to �3 = 18�1 - 17�2 - �3 because
1 � 18 1 � (-17) (-2) � (-1)
+ + = 3 - 3.2 + 0.2 = 0.
6 5 5
" One can construct families of up to k-1 mutually orthogonal contrasts.
Such families have several very pleasant properties.
" First, any family of k-1 mutually orthogonal contrasts partitions SSB
into k - 1 separate components,
SSB = SS�1 + � � � + SS�k-1,
each with one degree of freedom.
11.1. THE NORMAL K-SAMPLE LOCATION PROBLEM 221
" For example, Heyl (1930) collected the following data:
Gold 83 81 76 78 79 72
Platinum 61 61 67 67 64
Glass 78 71 75 72 74
This results in the following anova table:
Source SS df MS F P
Between 565.1 2 282.6 26.1 .000028
�1 29.2 1 29.2 2.7 .124793
�3 535.9 1 535.9 49.5 .000009
Within 140.8 13 10.8
Total 705.9 15
" Definition 11.3 Given a family of contrasts, the family rate � of
Type I error is the probability under H0 : �1 = � � � = �k of falsely
rejecting at least one null hypothesis.
" A second pleasant property of mutually orthogonal contrasts is that
the tests of the contrasts are mutually independent. This allows us to
deduce the relation between the significance level(s) of the individual
tests and the family rate of Type I error.
Let Er denote the event that H0 : �r = 0 is falsely rejected. Then
P (Er) = � is the rate of Type I error for an individual test.
Let E denote the event that at least one Type I error is commit-
ted, i.e.
k-1
E = Er.
r=1
The family rate of Type I error is � = P (E).
The event that no Type I errors are committed and
k-1
c
Ec = Er
r=1
and the probability of this event is P (Ec) = 1 - � .
222 CHAPTER 11. K-SAMPLE LOCATION PROBLEMS
By independence,
c c
1 - � = P (Ec) = P (E1) � � � � � P (Ek-1) = (1 - �)k-1;
hence,
� = 1 - (1 - �)k-1.
" Notice that � > �, i.e. the family rate of Type I error is greater than
the rate for an individual test. For example, if k = 3 and � = .05, then
� = 1 - (1 - .05)2 = .0975.
This phenomenon is sometimes called alpha slippage. To protect
against alpha slippage, we usually prefer to specify the family rate of
Type I error that will be tolerated and compute a significance level
that will ensure the specified family rate. For example, if k = 3 and
� = .05, then we solve
.05 = 1 - (1 - �)2
to obtain a significance level of
"
.
� = 1 - .95 = .0253.
Bonferroni t-Tests
" Now suppose that we plan m pairwise comparisons. These comparisons
are defined by contrasts �1, . . . , �m of the form �i - �j, not necessarily
mutually orthogonal. Notice that each H0 : �r = 0 vs. H1 : �r = 0 is a
normal 2-sample location problem with equal variances.
" Fact: Under H0 : �1 = � � � = �k,
� �
Xi� - Xj�
Z =
1 1
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