Supplement 2--One-way ANOVA example (revised)
Data in matrix format:
alfalfa strain (j = 1 to k )
a
b
c
i = 1 to n
1
3.2
4.2
4.8
2
5.4
4.5
4.4
3
2.4
6.0
4.1
4
7.1
3.1
5.9
5
3.2
3.9
5.2
SUM Xj
21.3
21.7
24.4
d
8.5
9.3
5.9
6.4
6.8
1 A. Does the strain of alfalfa affect nitrogen
fixation levels?
B. Ho: μ1 = μ2 = μ3 = μ4
HA: at least one mean is different
TOTAL
2
104.300 Å Sum Xij
36.9
We'll use α = .02, two-tailed ⇒ α/2 = .01 (your
author does not provide an α = .005 table)
dfnum = k - 1 = 4 - 1 = 3
dfdenom = nT - k = 20 - 4 = 16
critical F .01,3,16 = 5.29
meanj
4.26
4.34
4.88
7.38
20.000 Å nT
5.215 Å overall
2
j
3.768
1.133
0.497
2.107
mean
3
F =
j =3
j =4
4
Decision rule: Reject Ho if F > 5.29
nj
s
5
5
5
j =1
j =2
4.5601
3.8281
0.5611 23.4361
TOTAL
32.3855 Å SSTR
j =1
15.0720
j =2
4.5320
j =3
1.9880
TOTAL
30.0200 Å SSE
SSTR
SSE
5
ANOVA table
Sum of
Source of
variation
squares
Treatments
SSTR
Error
SSE
Total
SST
df
k-1
N-k
N-1
Mean
Square F
MSTR
MSE
j =4
8.4280
5.7536
(see work at left)
F = 5.7536
reject
Ho
FTR Ho
F
critical F = 5.29
5
Source of Sum of
variation squares df
Treatments 32.3855
3
Error
30.0200
16
Total
62.4055
19
Mean
Square F
10.7952 5.7536
1.8763
Decision: Reject Ho because F > 5.29
Conclusion: At α = .02, we conclude that the
strain of alfalfa grown significantly
affects the mean level of nitrogen fixation.
Fisher's LSD pairwise t-tests
Ho : μi = μ j
*
*
*
1 v. 2
1 v. 3
1 v. 4
2 v. 3
2 v. 4
3 v. 4
t
‐0.0923
‐0.7157
‐3.6015
‐0.6233
‐3.5091
‐2.8858
ni
5
5
5
5
5
5
nj
5
5
5
5
5
5
xi
xj
4.26
4.26
4.26
4.34
4.34
4.88
4.34
4.88
7.38
4.88
7.38
7.38
MSE
1.87625
* = significant at α /2 = .01
Type I Error Rates (p. 511) and cumulative experimentwise α (αew)
0.88584 ← probability of making no Type I errors in 6 tests = (1 - α )6 = 0.98 6
0.11416 ← probability of making at least one Type I error in 6 tests
= cumulative experimentwise alpha error risk = α ew = 1 - (1 - α )6 = 1 - 0.98 6
0.00333 ← Bonferroni adjusted α = α /6
NOTE: Here, the i and j subscripts do
not refer to rows (observations) and
and columns (treatments), even though
that is how they have been used up to
this point. i refers to the first
treatment mean of the pair tested (μi)
and j to the second (μj).
critical t.01,16 = 2.583
where (df = nT - k) ANOVA in MINITAB Input file: ALFALFANOVA.MTW Download this
file from the
WebCT site
Procedure:
Stat Æ ANOVA Æ One-way… Æ Response: Nitrogen (y)
Factor: Strain (x)
Click Comparisons…
check Fisher’s and
set family error rate to 1*
OK
OK
*this sets α/2 = .01
One-way ANOVA: Nitrogen (y) versus Strain (x)
Source
Strain (x)
Error
Total
DF
3
16
19
S = 1.370
R-Sq = 51.90%
Level
a
b
c
d
N
5
5
5
5
Mean
4.260
4.340
4.880
7.380
SS
32.39
30.02
62.41
StDev
1.941
1.064
0.705
1.452
MS
10.80
1.88
F
5.75
P
0.007
R-Sq(adj) = 42.88%
Individual 98% CIs For Mean Based on
Pooled StDev
---+---------+---------+---------+-----(---------*---------)
(---------*---------)
(--------*---------)
(---------*---------)
---+---------+---------+---------+-----3.2
4.8
6.4
8.0
Pooled StDev = 1.370
Fisher 98% Individual Confidence Intervals
Simultaneous confidence level = 91.58%
All Pairwise Comparisons among Levels of Strain (x)
(I’m not sure how Minitab calculates this. ASW’s method gives 88.584%)
Strain (x) = a subtracted from:
Strain
(x)
Lower Center Upper --------+---------+---------+---------+b
-2.158
0.080 2.318
(------*-------)
c
-1.618
0.620 2.858
(------*-------)
d*
0.882
3.120 5.358
(------*-------)
--------+---------+---------+---------+-3.0
0.0
3.0
6.0
Strain (x) = b subtracted from:
Strain
(x)
Lower Center Upper --------+---------+---------+---------+c
-1.698
0.540 2.778
(-------*------)
d*
0.802
3.040 5.278
(------*-------)
--------+---------+---------+---------+-3.0
0.0
3.0
6.0
Strain (x) = c subtracted from:
Strain
(x)
Lower Center Upper --------+---------+---------+---------+d*
0.262
2.500 4.738
(------*-------)
--------+---------+---------+---------+-3.0
0.0
3.0
6.0
Unhelpfully, Minitab conducts the Fisher’s
LSD test using confidence intervals only.
Reject Ho if 0 does not lie in the interval. If 0 is
in the interval for the difference, then we
cannot be 98% confident that the difference is
significantly different from zero, i.e., that there
is no difference. I’ve marked the significant
results with a * and bold print. They’re the
same results we obtained above.