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Homework Problems 3
- Ex 6.17 a, b and c based on the Number of Parity Data in Table 6.8.
- Ex. 6.19 a, b and c based on data in Table 6.10 (Milk Transportation Data)
6.17(a), p.
Test for treatment effects using a repeated measures design. Set =.05.
Null hypothesis,
O
H : there are NO treatment effects.
Alternate hypothesis,
A
H : there ARE treatment effects.
3 4 1 2
( + ) - ( + )= Number Format contrast, representing the difference between the presence and
absence of Arabic format numbers
1 3 2 4
= Parity Type contrast, representing the difference between the presence and absence
of “different parity”
1 4 2 3
= Contrast representing the influence of Arabic format on parity differences (i.e.
Format-Parity interaction
With
’=[
1
2
3
4
], the contrast matrix C is
C
The data (see Appendix 6.17) give:
x
and
S
Stat 8108: Multivariate Dr. Sarkar
C x
C x
CSC
36178.35( 1)+25650.31( 1)+17707.06(1)+14637.81(1)36178.
25650.31( 1) 22302.16( 1) 13759.3(1) 13367.56(1)
17707.06( 1) 13759.3( 1) 17865.71(1) 11027.05(1)
14637.81( 1) 13367.56( 1) 11027.05(1) 12182.03(1)
35(1)+25650.31( 1)+17707.06(1)+14637.81( 1 )36178.35(1)+
25650.31(1) 22302.16( 1) 13759.3(1) 13367.56( 1)
17707.06(1) 13759.3( 1) 17865.71(1) 11027.05( 1)
14637.81(1) 13367.56( 1) 11027.05(1) 12182.03( 1)
25650.31( 1)+17707.06( 1)+14637.81(1)
25650.31(1) 22302.16( 1) 13759.3( 1) 13367.56(1)
17707.06(1) 13759.3( 1) 17865.71( 1) 11027.05(1)
14637.81(1) 13367.56( 1) 11027.05( 1) 12182.03(1)
-36178.35-25650.31+17707.06+14637.81 36178.35-25650.31+17707.06-14637.
-25650.31-22302.16+13759.3 13367.56 25650.31-22302.16 13759
-17707.06-13759.3 17865.71 11027.
-14637.81-13367.56 11027.05 12182.
36178.35-25650.31 17707.06+14637.
.3 13367.56 25650.31-22302.16-13759.3 13367.
17707.06 13759.3 17865.71 11027.05 17707.06-13759.3-17865.71 11027.
14637.81 13367.56 11027.05-12182.0314637.81-13367.
-11027.05 12182.
Stat 8108: Multivariate Dr. Sarkar
6.17(b), p.
Construct 95% (simultaneous) confidence intervals for the contrasts representing the number
format effect, the parity type effect and the interaction effect. Interpret the resulting intervals.
To see which of the contrasts are responsible for the rejection of the null hypothesis, we construct 95%
simultaneous confidence intervals.
Here 1
c ' is the 1
st
row of the contrast matrix C.
The contrast 1 3 4 1 2
c ' ( + ) - ( + ) = Number Format influence; is estimated by the confidence
interval:
( x 3 + x 4 ) - ( x 1 + x 2 )
2 1 1
c 'Sc
c *
n
1 1
3 4 1 2
1, 1
( 1)( 1) c 'Sc
q n q
n q
x x x x F
n q n
which = -411.4998 to -186.8802, as calculated by SAS (see Appendix 6.17). This can be
expressed as -299.
112.3098. (-299.19 is the average of -411.4998 and -186.8802, and
112.3098 is half the difference between -411.4998 and -186.8802) Since this entire range is
negative, we conclude that there is a Number Format effect in that the presence of an Arabic
number format reduces the judgment time.
The contrast 1 3 2 4
=Parity Type influence; is estimated by the confidence interval:
2 2 2
1 3 2 4
c 'Sc
( x + x ) - ( x + x ) c *
n
2 2
1 3 2 4
1, 1
( 1)( 1) c 'Sc
q n q
n q
x x x x F
n q n
which = 125.01524 to 280.32476 (see Appendix 6.17). This can be expressed as 202.67
77.6548, which is an entirely positive interval, so we conclude that there is a Parity Type effect
in that the presence of different parity prolongs the judgment time.
The contrast 1 4 2 3
=Format-Parity interaction; is estimated by the confidence
interval:
2 3 3
1 4 2 3
c 'Sc
( x + x ) - ( x + x ) c *
n
Stat 8108: Multivariate Dr. Sarkar
3 3
1 4 2 3
1, 1
( 1)( 1) c 'Sc
q n q
n q
x x x x F
n q n
which = 75.03463 to 32.37463 (see Appendix 6.17) i.e. -21.
53.7046, which includes the
value within the interval i.e. this interval is not significantly different from zero, so we conclude
that there is an absence of interaction.
6.17(c), p.
The absence of interaction supports the M model of numerical cognition, while the presence of
interaction supports the C and C model of numerical cognition. Which model is supported in this
experiment?
The M model of numerical cognition is supported since there is an absence of interaction.
Appendix 6.
N
32
P
4
XBAR VAR F C S
967.56 36178.35 2.93 9.4 36178.35 25650.31 17707.06 14637.
876.89 22302.16 25650.31 22302.16 13759.3 13367.
828.63 17865.71 17707.06 13759.3 17865.71 11027.
716.63 12182.03 14637.81 13367.56 11027.05 12182.
T
HOTELLING
Reject Null Hypothesis
Stat 8108: Multivariate Dr. Sarkar
pooled
S
as calculated by SAS (Appendix 6.19).
1 2
2 1 2
, 1
1 2
p n n p
n n p
c F
n n p
3,36 23 3 1
F
3,
F
2
c
=12.93, as calculated by SAS (Appendix 6.19).
The likelihood ratios test (Johnson, p.285) of 0 1 2 0
H
is based on the square of the statistical
distance,
2
T
. We Reject
0
H
if
1
2 2
1 2 1 2
0 0
1 2
pooled
T x x S x x c
n n
For our purposes, let 0
=0.
We Reject
0
H
if
1
2 2
1 2 1 2
1 2
pooled
T x x S x x c
n n
Is
1
2
T
Stat 8108: Multivariate Dr. Sarkar
Is
1
2
T
…………………………
6.19(b), p.
If the hypothesis of equal cost vectors is rejected in Part (a), find the linear combination of mean
components most responsible for the rejection.
Since 1 2
( x x )=
=
, we can visually see that the Capital cost component on
the surface appears to have the biggest difference (between gasoline and diesel trucks). However, we do
not know if this Capital cost component difference is a statistically significant difference, and/or if Fuel is
also showing a statistically significant difference, and/or if Repair is also showing a statistically
significant difference, and/or if the sum of all three costs would show a statistically significant difference.
For testing 0 1 2
H : 0
, the linear combination â '( x 1 x 2 )
, with coefficient vector
1
1 2
â ( )
pooled
S x x
, quantifies the largest population difference. I.e., if
2
T
rejects
0
H
, then
1 2
â '( x x )
will have a nonzero mean. If the absolute value
1 2
1 2
a '( x x ) a '( )
is
â( )
pooled
c S a
then mathematically it must follow that
1 2 1 2
1 2
a '( ) â( ) a '( ) a '( ) â( )
pooled pooled
x x c S a x x c S a
Referring to the answer to question 6.19(c) below, we see that the Capital cost component is the only
statistically significant difference in cost vectors between gasoline and diesel trucks. Knowing this, we
can conclude that we should choose a vector â which will isolate the Capital cost component when â' is
multiplied by ( x 1 x 2 )
. This is obviously
â'= 0 0 1
. Thus the linear combination is â '( x 1 x 2 )
=
.
6.19(c), p.
Stat 8108: Multivariate Dr. Sarkar
13.5287 Capital 3.
The confidence interval for the difference in the costs of the Capital component does not include zero.
Capital is quite a different cost between gasoline and diesel trucks.
Appendix 6.
N1 P1 XBAR1 S
36 3 12.22 23.01 12.37 2.
8.11 12.37 17.54 4.
9.59 2.91 4.77 13.
N2 P2 XBAR2 S
23 3 10.11 4.36 0.76 2.
10.76 0.76 25.85 7.
18.17 2.36 7.69 46.
SPOOL
15.81 7.89 2.
7.89 20.75 5.
2.7 5.9 26.
C
MEANDIFFERENCE INTERVAL
Stat 8108: Multivariate Dr. Sarkar
u11-u12 -1.71 5.
MEANDIFFERENCE INTERVAL
u11-u12 -7.02 1.
MEANDIFFERENCE INTERVAL
u11-u12 -13.53 -3.
EVAL EVEC
32.955088 0.3837284 0.4924704 0.
20.294232 0.5910466 0.5189794 -0.
9.8906808 0.7095184 -0.698665 0.
99% confidence ellipse:
INTERV
Stat 8108: Multivariate Dr. Sarkar
