Single Factor Analysis of Variance
Using MINITAB
EXAMPLE Some investigators think that the concentration (mg/ml) of a particular antigen in supernatant fluids could be related to
the onset of meningitis in infants. The accompanying data is typical of that given in plots appearing in the paper “Type-Specific
Capsular Antigen is Associated with virulence in Late-Onset Group B Streptococcal Type III Disease” (Infection and
Immunity(1984):124-129)
Asymptomatic infants
1.56
1.06
.87
1.39
.71
.87
.95
1.51
Infants with late onset sephis
1.51
1.78
1.45
1.13
1.87
1.89
1.07
1.72
Infants with late onset meningitis
1.21
1.34
1.95
2.00
2.27
.88
1.67
2.57
INPUTTING DATA:
GROUP
1
1
.
.
3
CONCENTRATION
1.56
1.06
2.57
COMMANDS IN MINITAB:
STAT > ANOVA > One-Way > RESPONSE > CONCENTRATION > FACTOR > GROUP
GRAPHS > Histogram of Residuals, Normal Plot of Residuals > OK
Comparisons > TUKEY’S > OK > OK.
STAT > ANOVA > Test for Equal Variances > RESPONSE > CONCENTRATION > FACTOR > GROUP > OK.
Check Assumptions:
1. Samples are independent
2. Assume samples are random samples
3. Assessing the reasonableness of the normality assumption:
Histogram
Normal Probability Plot
(response is concentration)
(response is concentration)
99
5
95
4
90
Percent
Frequency
80
3
2
70
60
50
40
30
20
10
1
0
5
-0.8
-0.4
0.0
Residual
0.4
0.8
1
-1.0
-0.5
0.0
Residual
Based on the normality plot and the histogram, the normality assumption is reasonable.
4. Assessing the reasonableness of the equal variance assumption:
H0:  12 =  22 =  32
Ha: At least two of the population variances differ
Minitab Output:
Bartlett's Test (Assumes each sample comes from a Normal Distribution)
Test statistic = 3.03, p-value = 0.219
Levene's Test (Assumes each sample comes from Any Continuous Distribution)
Test statistic = 2.01, p-value = 0.159
We fail to reject Ho. The equal variance assumption is reasonable
0.5
1.0
Since assumptions are reasonable, we can perform the Analysis of Variance
One-way Analysis of Variance
Ho: 1= 2 = 3
Ha: At least two of the population means are different.
=.05.
Minitab Ouput:
One-way ANOVA: Concentration versus Group
Source
Group
Error
Total
DF
2
21
23
SS
1.630
3.729
5.359
MS
0.815
0.178
F
4.59
P
0.022
p-value < , thus reject H0. The data provide sufficient evidence to conclude that at least two of the population means are different.
Now Perform Tukey’s Multiple Comparisons to locate where the differences are.
Minitab Output:
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of Group
Individual confidence level = 98.00%
Group = 1 subtracted from:
Group
2
3
Lower
-0.0929
0.0909
Center
0.4375
0.6213
Upper
0.9679
1.1516
----+---------+---------+---------+----(----------*---------)
(---------*----------)
----+---------+---------+---------+-----0.50
0.00
0.50
1.00
Group = 2 subtracted from:
Group
3
Lower
-0.3466
Center
0.1838
Upper
0.7141
----+---------+---------+---------+----(----------*---------)
----+---------+---------+---------+-----0.50
0.00
0.50
1.00
If interval does not include 0, the corresponding population means are declared different. Thus, only 1 and 3 are declared
different.
Interval for 2-1: ( -.0929 , .9679)
Interval for 3-1: (.0909 , 1.1516)*
Interval for 3-2: ( -.3466 , .7141)