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ONE-WAY ANOVA AND TWO-SAMPLE COMPARISON
1) The basic procedure.
ANOVA is an acronym for Analysis Of Variance. It is a methodology for testing
hypotheses regarding population means in various contexts. One-way ANOVA refers to
the methodology for testing the equality of two or more population means, based on
independent samples from each of the populations. Thus, if mu1, mu2, … ,muk denote
the means of k populations, one-way ANOVA is a methodology for testing
H_0 : mu1= mu2= … =muk,
against the alternative that not all are equal.
Here we will demonstrate the use of Minitab for carrying out this test procedure using
the data from www.stat.psu.edu/~mga/401/labs/05/lab6/anova.fe.data.txt. The data are
about total Fe for four types of iron formation (1= carbonate, 2= silicate, 3= magnetite,
4= hematite).
If the data from the different populations (also called factor levels in the ANOVA jargon)
are given in different columns, then use the sequence of commands
Stat >ANOVA>One-way (Unstacked)>Enter C1-C4 for Response, 95 for confidence
level>OK.
[If the data from all factor levels are stored in one column, there must also be a second
column which indicates the group membership of each observation in the first column.
Call this second column “formation”. The sequence of commands in this case are:
Stat>ANOVA>One-way>Enter Fe as response and formation as Factor>OK ]
The output that Minitab produces (other software packages produce similar outputs) is:
One-way ANOVA: C1, C2, C3, C4
Source
Factor
Error
Total
DF
3
36
39
S = 3.955
Level
C1
C2
C3
C4
N
10
10
10
10
SS
509.1
563.1
1072.3
MS
169.7
15.6
R-Sq = 47.48%
Mean
26.080
24.690
29.950
33.840
StDev
3.391
4.425
2.854
4.831
Pooled StDev = 3.955
F
10.85
P
0.000
R-Sq(adj) = 43.10%
Individual 95% CIs For Mean Based on
Pooled StDev
-----+---------+---------+---------+---(-----*------)
(------*-----)
(-----*-----)
(------*-----)
-----+---------+---------+---------+---24.0
28.0
32.0
36.0
The ANOVA table gives the decomposition of the total sum of squares into a
sum of squares due to the population differences (Factor) and a sum of squares due to the
intrinsic error. Thus, 509.1 + 563.1 = 1072.3 (not really, due to rounding). MS = SS/DF,
and the F statistic is the ratio of the MS for Factor over MS for error. Typically, statistics
books also have F-tables, where the value of the F statistic can be looked up. We do will
not learn how to do that because Minitab produces the p-value, and does so in much
greater accuracy than what we could do from the F-tables. [The F-distribution is
characterized by two degrees of freedom (here the degrees of freedom of the F-statistic
are 3 and 36) and thus contain only selected percentiles.] Because the p-value is small,
the hypothesis of equality of the population means is rejected. Following the ANOVA
table there is information about the estimate of the standard deviation, which is assumed
to be the same in all populations (here the estimate is S=3.955, and it is also given in the
last line of the output), and the R-Sq, which has the same significance as explained in the
activity for regression. The individual sample means, estimated standard errors, and 95%
CI for each population mean are also given.
2) Multiple Comparisons for One-Way ANOVA.
When the null hypothesis is rejected it means that, the data strongly suggest that, at least
one of the population means is different from the others. When k>2, additional testing
needs to be done to identify which means appear to be different. This additional testing is
called multiple comparisons. It involves performing all pair-wise comparisons (i.e.
testing the null hypothesis of equality of each possible pairs of means) in such a way that
the probability of committing a type I error for any of these pair-wise test procedures
does not exceed the designated level of significance alpha. One of the ways of doing
multiple comparisons, is to perform the aforementioned ANOVA test for each pair-wise
comparison, at an adjusted level of significance. The adjusted level equals the designated
alpha divided by the total number of pair-wise comparisons. This is called the Bonferroni
method.
Here we will demonstrate a different method, called the Tukey method, for doing
pairwise comparisons, in such a way that the overall level of significance is alpha. .
Stat >ANOVA>One-way (Unstacked)>Enter C1-C4 for Response, 95 for confidence
level>Click Comparisons select Tukey’s, enter family error rate (5 for overall level
of significance 0.05)>OK>OK
The additional Minitab output, with my comments in brackets, is:
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons
Individual confidence level = 98.93%
C1 subtracted from:
C2
C3
C4
Lower
-6.155
-0.895
2.995
Center
-1.390
3.870
7.760
Upper
3.375
8.635
12.525
+---------+---------+---------+--------(------*------)
(------*-----)
(------*------)
+---------+---------+---------+---------14.0
-7.0
0.0
7.0
[These are simultaneous CI for the differences mu1-mu2, mu1-mu3, and mu1-mu4.
If a CI does not contain 0, the two means are declared significantly different.
Thus, mu1 is significantly different from mu4, but not significantly different
from mu2 or from mu3.]
C2 subtracted from:
C3
C4
Lower
0.495
4.385
Center
5.260
9.150
Upper
10.025
13.915
+---------+---------+---------+--------(------*-----)
(------*------)
+---------+---------+---------+---------14.0
-7.0
0.0
7.0
[These are simultaneous CI for mu2-mu3 and mu2-mu4. None of these CI contains
zero, and thus mu2 is significantly different from both mu3 and mu4.]
C3 subtracted from:
C4
Lower
-0.875
Center
3.890
Upper
8.655
+---------+---------+---------+--------(------*-----)
+---------+---------+---------+---------14.0
-7.0
0.0
7.0
[This is a simultaneous CI for mu3-mu4. The CI contains zero, and thus mu3 is
not significantly different from mu4.]
3) A Nonparametric Test for Comparing k Means
The ANOVA methodology is exact (i.e. the F-statistic has the F-distribution) only if the
population distributions are normal and have the same variance (they are homoscedastic),
but it is approximately valid if the sample sizes are large without the normality
assumption, provided the populations are homoscedastic. Moreover, the ANOVA
methodology has more power (i.e rejects the null hypothesis, when it is not true, with
higher probability) than any other test only when the k population distributions are
normal and homoscedastic. An alternative test procedure, which is nearly as powerful as
ANOVA under normality and homoscedasticty, but can be much more powerful than
ANOVA when the population distributions are non-normal, is the Kruskal-Wallis test.
Roughly speaking, the Kruskal-Wallis procedure consists of combining the data from the
k populations and ranking the combined data set from smallest to largest. Each of the
original observations is then replaced by its rank, and the ranks are used instead of the
original observations in the ANOVA test statistic. [The actual Kruskal-Wallis test
statistic is somewhat different than the procedure just described, but the difference gets
smaller as the sample sizes increase.]
In this activity we will learn how to implement the Kruskal-Wallis test with Minitab.
We will use the data www.stat.psu.edu/~mga/401/labs/05/lab6/k-w.cortisol.data.txt.
To use the Kruskal-Wallis procedure, the data must be stacked.
Data>Stack>Columns, enter C1-C3 under Stack the Following Columns, click on
Column of Current Worksheet and enter C4, enter C5 in Store Subscripts in>OK
Then use the sequence of commands:
Stat>Nonparametrics>Kruskal-Wallis, enter C4 for Response, and C5 for
Factor>OK
The Minitab output is:
\
Kruskal-Wallis Test: C4 versus C5
Kruskal-Wallis Test on C4
C19
C15
C16
C17
Overall
H = 9.23
N
10
6
6
22
Median
305.5
460.0
729.5
DF = 2
Ave Rank
6.9
15.0
15.7
11.5
P = 0.010
Z
-3.03
1.55
1.84