Contrasts: Lubricant Example
(a)
Source
d.f .
SS
MS
F
Lubricant
3
47.84375
15.9479
5.5557
Error
28
80.37500
2.8705
C. Total
31
128.21875
Test
H 0 : µ=
µ=
µ=
µ4
1
2
3
H a : at least
at α = .05 .
vs.
Reject the null hypothesis
(b)
The sample means are :
y1 = 9.75 , y2 = 8.0 , y3 = 7.124 , y4 = 6.5
LSD=
2.048 × 2.8705 × 2 / 8 = 1.735
Blu-Tek
Z907
L43
SP51
6.5
7.125
8.0
------------------
9.75
From JMP:
LSMeans Differences Student's t
α=0.050
Level
1
2
3
4
t=2.04841
A
B
B
B
Least Sq Mean
9.7500000
8.0000000
7.1250000
6.5000000
Levels not connected by same letter are significantly different.
(c)
W=
3.87 × 2.8705 × 1/ 8 = 2.32
Blu-Tek
Z907
L43
SP51
6.5
7.125
8.0 9.75
---------------------------
p-value
<.0001
one inequality. Since p-value<.05,
From JMP:
LSMeans Differences Tukey HSD
α= 0.050 Q= 2.73031 <<<<<<<Note :This Q value is different from the value from the table. i.e. 3.87/ 2 =2.736
Level
1
2
3
4
A
A
B
B
B
Least Sq Mean
9.7500000
8.0000000
7.1250000
6.5000000
Levels not connected by same letter are significantly different.
(d)
Hypothesis
H 0 :1/ 2( µ1 + µ2 ) − 1/ 2( µ3 + µ4 ) =
0
Contrast :
=
1 1 −1 −1
1
Estimate:
ˆ 1 = 9.75 + 8.0 − 7.125 − 6.5 = 4.125
12 + 12 + 12 + 12
4
2.8705 = (1.694255)
=
  1.198
8
8
Standard Error
=
=
V (ˆ 1 )
Thus the t-statistic is:
=
tc
4.125
= 3.4432 and the percentile from the t-table is t.025,28 = 2.048
1.198
Thus R.R. is t > 2.048 ; Thus the null hypothesis is rejected at α = .05 since
3.4434 is in the rejection region.
From JMP output:
t-statistic = 3.4432
p-value = .0018
Reject null hypothesis at α = .05
Hand computations for the other three comparisons are not shown here but those
computations are similar to above and are summarized below.
Contrast
SP51
L43
Z907
Blu-Tek
Est
s.e.
t
p-value
i)
1
1
-1
-1
4.125
1.198
3.4432
.0018
ii)
1
-1
1
-1
2.375
1.198
1.98
.0573
iii)
1
-1
0
0
1.75
0.8471
2.0658
.0482
iv)
0
0
1
-1
0.625
0.8471
0.7378
.4668
The results from JMP are shown below.
Contrast
Test Detail
1
2
3
4
Estimate
Std
Error
t Ratio
Prob>|t|
SS
0.5
0.5
1
0
0.5
-0.5
-1
0
-0.5
0.5
0
1
-0.5
-0.5
0
-1
2.0625 1.1875
1.75 0.625
0.599 0.599 0.8471 0.8471
3.4432 1.9824 2.0658 0.7378
0.0018 0.0573 0.0482 0.4668
34.031 11.281 12.25 1.5625
Summary Statement
The data show that there is a significant difference between the average effects of petroleum distillate
based lubricants from those of synthetic oil based lubricants on the wear of the machine part. The
average wear appear to be lower for the synthetic oil based lubricants. There is no significant difference
between the average effects of silicon additives vs. the hi-tech detergent additives on machine part
wear. There is a significant difference between the average effects of silicon additive vs. the hi-tech
detergent additive when used with petroleum distillate base as opposed to synthetic oil base, when the
two additive effects were not significantly different.