UNIVERSITY
O}'
NORTH CAROLINA
Department of Statistics
Chapel Hill) N. C.
ILLUSTR/,TIOIJ OJ? A TECHNIQUE HHT.CH TESTS WHETHER T'VJO
REGRESSION LINES .AR:E PARALLEL 1iJHL"'N THE VARIANCES ARE UNEQUAL
by
Richard F. Potthoff
February, 1962
Contract No. AF 49(638)-213
It may be desired to test the hypothesis that two regression lines are parallel vnthout assuming that the variances
of the two sets of error terms are necessarily equal. This
paper presents a relatively non-'tecbnical discussion of a
test vlhich can be used for this situation. A numerical illustration is included. The test statistic used is analogous to the well-knO'lm Wilcoxon statistic. This paper is
intended for the practitioner rather than for the theoretician; the more technical aspects of the test are covered in
a separate paper.
This research was supported partially by the Iiathematics Division of the Air
Force Office of Scientific Research and partially by Educational Testing Service.
Institute of Statistics
Mimeo Series No. 320
ILLUSTRATION OF' A TLCHlTIQUE \'JHICH TESTS 1JH:UrHCR THO
REGRESSION LINES ARE PARf-l.LIJEL millN THE VARIANCES P.RE mmQUAL
l
by
Richard F. Potthoff
1.
Suppose
Introduction.
such that, for each
Yi
i,
vTe
have
~
ay
and
~Y
are ulli~nown parameters (regression coefficients), the
e.'s
are specified constants, and the
ulli~novm
variance
... ,
(zn'
J
where
and
~Z
(J 2 .
e
CY.
Z
+
2
m < n.
J
;:;:
J
n pairs
observes the relation
.J
parameters (regression coefficients), the
f.ls
J
H. t S
J
are normal and independent with mean
0
Note that we are alloirlng for the possibility that the variance of the
may be different from the variance of the
~Z·
0
vie will assume (with no loss of generality) that
the hypothesis that the two
~Y
l
~ZH.+f.
are specified constants, and the
and ullimovm variance (Jf'
X. 's
are normal and independent with mean
Suppose also that we have
ulli~novm
are
l
such that, for each
U)
n'
z.;:;:
(1.2)
l
,
e.
where
e.'s
pairs (Y , Xl)' (Yo'
X2 ), ... , (Ym,XmL
, c:.
1
observes the relation
(1.1)
and
ill
In other words, we
~~coefficients
v~nt
f.'s.
J
Suppose we desire to test
are equal, i.e., the hypothesis that
to test the hypothesis that the two regression
lines associated with (1.1) and (1.2) are parallel, and of course lTe 't-.1t1nt to have
a test vThich will be valid regardless of what the values of (J2 and
e
In
this paper we shall present such a test, along with a numerical example which will
illustrate how' the computations are made.
Not only can
1le
test the hJrpothesis
lThis research v~s supported in part by Educational Testing Service, and was
supported in part by the Air Force Office of Scientific Research.
-2-
4If
~y = ~Z'
(~Z - ~y).
but we can also obtain confidence bounds on
A practical situation in which the problem treated by this paper could arise
might be as follows.
Suppose that we have (m + n) classes of students,
which receive curriculum nuraber
culum number 2.
m of
1 and the rewaining n of which receive curri(m + n) classes, we have
Suppose also that, for each of the
available (i) some standard measure of the achievement of the class obtainecl after
completion of the course, and (ii) a concomitant variate representing some standard
measure of the ability of the class obtained before the course started.
the
j.-th class receiving currj.culum nU11'lber 1,
Y.
J.
would be the achievement
measure of this class obtained after the course, and
Xi
would be the ability
measure of the class obtained before the course started; for the
ceiving curriculum number 2,
Zj
Then, for
j-th class re-
vTOuld be the achievement measnre of the class
obtained after the course, and W would be the ability measure of the class obj
tained before the course. It is assumed that the Y.'s are connected 1nth the
J.
Xi's
by a linear relation of the form
with the
(1.1), and that the Zj'S are associated
itT.'s via a linear regression of the form (1.2); the variances assoJ
ciated with the two curriculums may be different.
We suppose that we desire to
compare curriculum number 1 1nth curriculum number 2 to
better.
dete~illine
As a first step in this comparison, we will probably
pothesis that
~Y
= ~Z;
which one is
v~nt
to test the hy-
this is essentially the hypothesis that the difference be-
tween the effects of the two curriculums (if any) is the same irrespective of
ability level.
(If
~y
is different from
~Z'
then that means that the difference
between the effects of the two curriculuras is not constant but rather depends on
the ability level, and might even mean that one curriculum is better for brighter
students while the other is better for duller students.)
The discussion in this paper will be relatively non-technical.
paper covers the technical aspects of the topic of the present taper.
A separate
-3The test.
2.
(i, I)
pair
He nOv1 present the formula for the test statistic.
such that
I < i < I < m ~there are
2I
For every
m (m-l) such pairs alto-
gethe~7, let us define
(2.1)
l:: j
For every pair (j,J) such that
altogetheE7,
C
Now
:5
n fthere are
~
n(n-l) such pairs
let us define
(2.1) will have expected value and median f3 y ' v1h:i.le the expectation and
iI
median of
DjJ (2.2) will be
f3Z'
Thus the expected value and median of
ZJ - Zj
(2.3 )
'nIl be
<J
::::
13 z
-
f3 y '
Vl . .. i'lj
J
f3 y '"' 13 z
Hence, if the null hypothesis
pect about half the
ViIjJ'S
is true, we V10uld ex-
to be positive and half to be negative.
Altogether there will be a total of
4I
I,et
let
'f
be the number of
8
be the proportion which are positive.
(2.4)
'fhis statistic
v1
v1
ron (m-l) Cn-l)
V.. IJ' S
J.J
Then
48
::::
ron ( m-l rcn:Tj
will have expected value
.~ if the null hypothesis
true, and will be approximately normally distributed..
~
the hypothesis
f3 y :::: f3
v1h:i.ch are positive, and
Z
that, regardless of '-That
is true will depend on
~: and ~~ are,
The variance of
is
W
vThen
~; and ~~, but it can be ShOi'1ll
-4··
2m + 5
18m (m-l)
(Remember that
is the smaller of m,n
ill
The test is as follovls:
if
ill
and
n are different).
if vTe want (e.g.) to use a tvTO-tailed test at the
0
5 /0 level to test the hypothesis ~y = ~Z' we can reject the hypothesis if
(2.6)
>/
-i&;---(m-l)
.l
2m+5--
'2
>
and accept otherwise.
This test
(2.6)
will be conservative in the sense that the probabiHty of re-
5 0/0,
jecting the null hypothesis when it is true will not be exactly equal to
but will generally be somewhat less than
5%
•
The reason for this is that
val' (w) (which is unknovm) 'Jill generally be somewhat belo,,, the bound (2.5).
It is possible that some of the ViIjJts (2.3) may be undefined because of a
condition
Xi
= XI
and/or
Wj
= WJ '
It is also possible that some of the
ViIjJ'S, although well-defined, may turn out to be exactly zero, so that they are
neither positive nor negative.
In either of these cases, we can handle the situa-
tion by tallying ~
for each such V
iljJ
pIe, suppose that
= 8,
"
ViljJ ! o.
m
n
:=
Suppose that all the
10.
in the determination of
are different, but that there are tvo
which are alike (vnth the rest all being different).
will be undefined.
S
e
=
exam-
28
of the
W.'s
J
V.I.JIs
J.
J
5 are zero, 711 are posi-
Then we can calculate
711 + ~ (28 + 5)
It is most desirable to have all the
if possible.
Then
Of the remain:tng 1232, suppose that
tive, and 516 are negative.
]'01'
28 x;i·5 ::: 1260 potential
Then there 1'1111 be
Xi'S
S.
=
727.5
X. t s
J.
different and all the
H. Is different,
J
-5The statistic
vi
(2.4) used here bears a certain similarity to the vrell-lmovID
Wilcoxon statistic, whose use with respect to a problem simpler than but somewhat
similar to the one considerecl in this paper is discussed in £1, _~7.
It might be
noted that, in the problem considered in 1~~7 and ~~7, it v~s necessary to
assume only that both sets of error terms followed
vlhereas in this paper we have assumed
distributions,
that both sets of error terms are normal
(although with possibly different variances).
norrnality here is that this assumption
s~mnetrical
v~s
The reason for the assumption of
used in prov:tng the bouno. (2.5).
It is
not knovID whether there are symmetrical distributions besides the normal for which
the bound (2.5) would still hold, but, in view of the results of ~~7, which are
concerned with an analogous situation, it would not be too surprising if
(2.5)
turned out to be valid for a 'tvide class of symmetrical distributions rather than
4It
just for the normal.
If such proved to be the case, we could of course relax the
assumption of normality.
3. Numerical example. We now give a numerical example to illustrate the computation of the test statistic presented in the previous section.
ill ::::
6 and n::: 7,
Yl
Xl
::::
and suppose that our tva samples are
482.9, Y2
:::
, )[2
:::
92
;:
Suppose that
538.7, Y?:
J
:=
102
:::
)
X3
557.1, Y4
108 , J'>.4
"IT
::::
:::
591.2, Y5 ::: 597.1, Y6
112 , X _. 117 , X6
5
::::
650.6
::::
126
ancl
Zl=
511+.0,
v'l::: 92
Z2
::: 527.7, Z ::: 530.0, Z4 ::: 537·3, z5
3
,'tl
:::
sake
or
2
99
,ll
3
:::
100 } w4 ::: 103
,Vl
5
==
538.8,
::: 105
==
550.1, Z7 ::: 553.3
109 , 1;'7 :::: 114
orderliness, vie arranged the first sample so that it is in order
of increasing magnitude of the
X.'s, and we arranged the second sample so that it
~
is in order of increasing magnitude of the H,' s.
J
-6Our first step is to compute the
will be ~ .6(6-1)
==
~
15 CiI's ana. ~ • 7(7 - 1)
Y2 - Yl
x - X
==
538.7 - ~.82.9
102 - 92
==
Y3 - Yl
13 - X
3 Xl
==
557·1 .. 482.9
108
92
::
2
1
-
C
C
::::
C
:::
1lt
15
(2.1) and the D.J's
(2.2). There
J
C'r's
-
591.2 - 482.~
112 - 92
:=
21 DjJ'S. We obtain
==
5·58
4.64
5.42
::::
Also
527.7 - 514.0
99- 92
:::
530.0 - 51).j.• O
100 - 92
1.96
2.00
==
2.12, D
15
::
1.91, D16
==
2.12, D17
==
D2~
-)
2·30, D24
==
2.40, D
25
::
: : 1. 71,
1.85, D26 ::: 2.24, Do,",
c_ (
D
3lt
::::
2.43, D
35
==
1. rr6, D
36
==
2.23, D ::: 1.66,
37
D)+0c
==
1.1~5,
:: 2.82, D ::: 1. 61, D :: 0.64 ..
67
57
Dl ).j.
:::
2.13, D ::
47
D
56
==
1. 79,
D)t=:
+:;
= 0·75,
From this point on, the computation is closely similar to the computation of
the Wilcoxon statistic) which was discussed in
21 DjJtS.
Let us fonn a
row in the chart
per~ains
Cil's
and the
He have
15 CiI's and
21 x 15 chart with 315 entries (see Table I.).
to a
D~T'
J~
both the
13.7.
DjJ'S
and each column to a
C~T'
Each
For convenience,
~~
are arranged in increasing order of nBgnitude.
The 315 entries in the chart represent the 315 Vi1jJ'S (2.3).
-7lIe count up and find that, of the ;15 ViIj/s,
"
maining 296 are negative.
Hence
S'
1'7 ::::
19 are positive and the re-
';15
:::
.060
Thus our test statistic faee (2.6t.7 is
/):S~iT
(vr- .~)
2m+5
2
Y
Since
:::., /i8~F- .060- .500) ;: j~40( - .440)::::5. 636( - .440)::: - 2.48
11
2x +5
(
17
-2.48 is greater in absolute value than 1.96, we reject the null hypothesis
~y ::: ~Z if we are n~king a tvro-tailed test at the
5%
level.
It is of course also possible to make a one-tailed test of the hypothesis
~y :::
f3 Z (against the alternative f3
Z
~z
> f3 y or the alternative
< f3 y ) if this
is vrhat is desired.
If we are interested only in testing the hypothesis
(~Z
interested in getting confidence bounds on
compute the ;15 entries of Table I.
~y
=
~Z
and are nat
- f3 y )' it is not necessary to
It will suffice simply to find out how many
of these extries are positive and how many negative.
To find this out, an alternative computing technique, which operates by rdnking the combined group of ~ m(~l) CiI's
and ~ n(n-l) DjJ'S, is available.
alternative technique, vrhich is described in
v~y
of determining
fidence bounds on
£g7,
S for testing the hypothesis
(f3z -
~y)
This
provides perhaps the easiest
~y
= ~Z'
assuming that no con-
are wanted.
If' confidence bounds are vlanted, hOi-lever, ,V'e can obtain them by utilizing
Table Ie
4. Confidence bounds. The methOd. of obtaining confidence bounds here is
closely analogous to the method described and illustrated in
~
example, that we want to obtain a two-sided 95
0/0
£~7.
Suppose, for
confidence interval for
(~z-f3y)'
-8To do this,
we find that valu:; of
1)
= f3 Z
- f3 y
which, when subtracted from every
entry in Table I., would cause (the resulting new)
significance.
w to be on the threshh01d of
Now w will be on the threshhold of being significantly large if
267 of the 315 entries are positive and 1 entry is zero, since
( 1.960 +
5.636
and
'itT
.500) x 315 _. . 8h8 x 315
will be on the threshho1d of being significantly s:lJ1.a11 if
entries in the table are positive and
(
267.1 ;
of the 315
entry is zero, since
-
Upon looking at Table I., we can
larger than
1
47
=
-3.94 (With
1
detel~ine
267 of its 315 entries are
entry being exactly equal to
the lower end of our confidence interval.
in the table exceed
that
-3.94).
We also find that
47
Thus
-1.66).
Hence
is
of the entries
-1.66 (one entry is equal to -1.66 and the remaining
entries are smaller than
-3.94
267
-1.66 is the upper end of the confidence
interval, and we can state that
0
with confidence coefficient ~ 95 /0 •
In most practical situations the experimenter 'VTould never find out 'Vlhat the
true values Of the parameters were.
Hm"ever, the e:r,a,mple in this paper 'Vlas con··
structed artificially, so that all the parameters are ImO'VID.
ay
= 20,
f3 y
= 5, az
= 330,
When the values of
f3 z
= 2,
m and n
u e2
= 100,
and u f2
= 4•
The values used i"ere
Thus
(f3z - f3 y )
= -3.
are small, the cornputations which are required
for obtaining the test statistic and confidence bounds discussed in this paper are
not too lengthy and can be performed with a desk calculator. However, for a sufficiently heavy volume of data it would probably pay to utilize more automatic methods
to do the calculations.
-9-
REFERIT;NCES
~~7
Richard F. Potthoff, "Use of the Wilcoxon Statistic for a Generalized
Behrens-Fisher Problem)" Institute of Statistics Himeograph Series No. 315, University of North Carolina.
£..7
~
Richard F. Potthoff, "Comparing the Beans of Two Symmetrical Popula tions~'
Institute of Statistics 11imeograph Series No. 316, University of North Carolina.
.i
"
"
.
~
~
....
-
.j!'
e
T,lIBLE I.
C, ;:
C ;:
C ;:
C ;:
C ;:
C ;:
C
-0.54
-0.43
0.27
0.43
0.48
0.53
0.58
0.61
0.67
0.73
0.78
0,.82
0.94
0.94
0·95
1.05
1.06
1.12
1.22
1.25
1.64
-2.43
-2·3:2
-1.6:2
-1.46
-1.41
-1.36
-1.31
-3·25
-3.14
-2.44
-2.28
-2.23
-2.18
-2.13
-2.10
-2.04
-1.98
-1.93
-1.89
-1.77
-1.77
-1·76
-1.66
-1.65
-1·59
-1.49
-1.46
-1.07
-3.60
-3.49
-2·79
-2.63
-2.58
-2·53
-2.48
-2.45
-2.39
-2·33
-2.28
-2.24
-2.12
-2.12
-2.11
-2.01
-2.00
-1.94
-1.84
-1.81
-1.42
-3.80
-3.69
-2·99
-2.83
-2·78
-2·73
-2.68
-2.65
-2·59
-2·53
-2.48
-2.44
-2·32
-2.32
-2·31
-2.21
-2.20
-2.14
-2.04
-2.01
-1.62
-3.93
-3.82
-3·12
-2.96
-2.91
-2.86
-2.81
-2.78
-2.72
-2.66
-2.61
-2.57
-2.45
-2.45
-2.44
-2.34
-2.33
-2.27
-2.17
-2.14
-1.75
-4.00
-3.89
-3·19
-3.03
-2·98
-2·93
-2.88
-2.85
-2·79
-2.73
-2.68
-2.64
-2·52
-2·52
-2·51
-2.41
-2.40
-2.34
-2.24
-2.21
-1.82
i.J.5
1.18
;:O,.64
67
D ;:0.75
45
D =1.45
47
D_,-,=1.61
)(
D =1.66
37
D27=1. 71
D ::=1. 76
35
D ;:1. 79
17
D ::=1.85
25
D ::=1. 91
15
D ::=1.96
12
D ::=2.00
13
D14=2.12
D16=2.12
D46=2.13
D =2.23
36
D2C 2 •24
D =2.30
23
'D =2.40
24
D ::=2.43
34
D5C 2 •82
D
23
3.07
-1.2B
-1.22
-1.16
-1.11
-1.07
-0.95
-0·95
-0.94
-0.8h
-0.83
-0.77
-0.67
-0. 6L~
-0·25
25
3.89
46
4.24
35
4.44
15
4·57
13 = C26= C16= C36= C24= C14;:
4.64 4.66 4.93 5.19 5·25 5·42
-4.02
-3·91
-3·21
-3.05
-3·00
-2·95
-2.90
-2.87
-2.81
-2·75
-2·70
-2.66
-2.54
-2.54
-2·53
-2.43
-2.42
-2.36
-2.26
-2.23
-1.84
-4.29
-4.18
-3,.48
-3.32
-3,.27
-3.22
-3.17
-3.14
-3.08
-3·02
-2.97
-2.93
-2.81
-2.81
-2.80
-2·70
-2.69
-2.63
-2.53
-2·50
-2.11
-4.55 -4.61
-4.44 -4·50
-3.74 -3.80
-3.58 -3.64
-3·53 -3·59
-3.48 -3.54
-3.43 -3.49
-3.40 -3.46
-3.34 -3.40
-3.28 -3.34
-3.23 -3.29
-3·19 -3.25
-3.07 -3.13
-3·07 -3·13
-3.06 -3.12
-2.96 -3·02
-2,.95 -3·01
-2.89 -2·95
-2·79 -2.85
-2.76 -2.82
-2·37 -2.43
-4.78
-4.67
-3·97
-3.81
-3·76
-3·71
-3.66
-3.63
-3·57
-3·51
-3 .~L6
-3.42
-3·30
-3.30
-3·29
-3·19
-3.18
-3·12
-3.02
-2.99
-2.60
C12=
5·58
C56=
5.94
C ;:
34
8.52
-4.94
-4.83
-4.13
-3·97
-3.92
-3.87
-3082
-3·79
-3.73
-3.67
-3.62
-3.58
-3.46
-3.46
-5.30
-5.19
-4.49
-4.33
-4.28
-4.23
-4.18
-4.15
-4.09
-4.03
-3.98
-3.94
-3.82
-3.82
-3.81
-3.71
-3·70
-3.64
-3.54
-3·51
-3.12
-7.88
-7·7'7
-7.07
-6.91
-6.86
-6.81
-6.76
-6.73
-6.67
-6.61
-6.56
-6.52
-6.40
-6.40
-6.39
-6.29
-6.28
-6.22
-6.12
-6.09
-5·70
~3.45
-3.35
-3.34
-3.28
-3.18
-3.15
-2.76