MULTIPLE COMPARISONS
Slide 1
Multiple Comparison Procedures
 Suppose that analysis of variance has provided
statistical evidence to reject the null hypothesis of
equal population means.
 Fisher’s least significant difference (LSD) procedure
can be used to determine where the differences
occur.
Slide 2
Fisher’s LSD Procedure
 Hypotheses
H 0 : i 
= j
H a : i   j
Test
Statistic
t
xi  x j
MSE( 1 n  1 n )
i
j
Slide 3
Fisher’s LSD Procedure
 Rejection Rule
p-value Approach:
Reject H0 if p-value < a
Critical Value Approach:
Reject H0 if t < -ta/2 or t > ta/2
where the value of ta/2 is based on a
t distribution with nT - k degrees of freedom.
Slide 4
Fisher’s LSD Procedure_ _
Based on the Test Statistic xi - xj
 Hypotheses
H 0 : i =
 j
H a : i   j
 Test Statistic
xi  x j
 Rejection Rule
where
Reject H0 if xi  x j > LSD
LSD  ta /2 MSE( 1 n  1 n )
i
j
Slide 5
Fisher’s LSD Procedure
Based on the Test Statistic xi - xj
Reed Manufacturing
Recall that Janet Reed wants to know if there is any
significant difference in the mean number of hours
worked per week for the department managers at her
three manufacturing plants.
Analysis of variance has provided statistical
evidence to reject the null hypothesis of equal
population means. Fisher’s least significant difference
(LSD) procedure can be used to determine where the
differences occur.
Slide 6
Fisher’s LSD Procedure
Based on the Test Statistic xi - xj
For a = .05 and nT - k = 15 – 3 = 12
degrees of freedom, t.025 = 2.179
LSD  ta /2 MSE( 1 n  1 n )
i
j
LSD  2. 179 25. 667( 1 5  1 5 )  6. 98
MSE value was
computed earlier
Slide 7
Fisher’s LSD Procedure
Based on the Test Statistic xi - xj
 LSD for Plants 1 and 2
•
Hypotheses (A)
•
Rejection Rule
H 0 : 1 =  2
H a : 1   2
Reject H0 if x1  x2 > 6.98
•
Test Statistic
x1  x2 = |55  68| = 13
•
Conclusion
The mean number of hours worked at Plant 1 is
not equal to the mean number worked at Plant 2.
Slide 8
Fisher’s LSD Procedure
Based on the Test Statistic xi - xj
 LSD for Plants 1 and 3
•
Hypotheses (B)
•
Rejection Rule
H 0 : 1 =  3
H a : 1   3
Reject H0 if x1  x3 > 6.98
•
Test Statistic
x1  x3 = |55  57| = 2
•
Conclusion
There is no significant difference between the mean
number of hours worked at Plant 1 and the mean
number of hours worked at Plant 3.
Slide 9
Fisher’s LSD Procedure
Based on the Test Statistic xi - xj
 LSD for Plants 2 and 3
•
Hypotheses (C)
•
Rejection Rule
H 0 : 2 =
 3
H a : 2  3
Reject H0 if x2  x3 > 6.98
•
Test Statistic
x2  x3 = |68  57| = 11
•
Conclusion
The mean number of hours worked at Plant 2 is
not equal to the mean number worked at Plant 3.
Slide 10
MULTIPLE COMPARISONS
PRACTICE
Slide 11
Perform Multiple Comparisons
Direct
Indirect
Combination
x
17
20.4
25
s2
5.01
6.26
4.01
n
7
7
7
Analysis of Variance
Source
SS
d.f.
MS
F
p-value
Treatment 225.68
2 112.84 22.15 0.000014
Error 91.68
18
5.09
Total 317.36
20
a=0.05
Slide 12
F Distribution
Slide 13
TYPE I ERRORS
Slide 14
Type I Error Rates
 The comparison-wise Type I error rate a indicates
the level of significance associated with a single
pairwise comparison.
 The experiment-wise Type I error rate aEW is the
probability of making a Type I error on at least one of
the (k – 1)! pairwise comparisons.
aEW = 1 – (1 – a)(k – 1)!
 The experiment-wise Type I error rate gets larger for
problems with more populations (larger k).
Slide 15
Slide 16