Transcript Basic principles of probability theory

Basics of ANOVA
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Why ANOVA
Assumptions used in ANOVA
Various forms of ANOVA
Simple ANOVA tables
Interpretation of values in the table
Exercises
Why ANOVA
If we have two samples then under mild conditions we can use t-test to test if difference between
means is significant. When there are more than two sample then using t-test might become
unreliable.
ANalysis Of VAriances – ANOVA is designed to test differences between means in many sample
cases.
Examples of ANOVA: Suppose that we want test effect of various exercises on weight loss. We
want to test 5 different exercises. We recruit 20 men and assign for each exercises four of
them. After few weeks we record weight loss. Let us denote i=1,2,3,4,5 as exercise number
and j=1,2,3,4 person’s number. Then Yij is weight loss for jth person on the ith exercise
programme. It is one-way balanced ANOVA. One way because we have only one category
(exercise programme). Balanced because we have exactly same number of men on each
exercise programme.
Another example: Now we want to subdivide each exercises into 4 subcategories. For each
subcategory of the exercise we recruit four men. We measure weight loss after few weeks.
i – exercise category
j – exercise subcategory
k – kth men.
Then Yijk is weight loss for kth men in the jth subcategory of ith category. Number of observations
is 5x4x4 = 80. It is two-fold nested ANOVA.
We want to test: a) There is no significant differences between categories; b) there is no
significant difference between different subcategories
Examples of ANOVA
One more example: We have 5 categories of exercises and 4 categories of diets. We hire for each
exercise and category 4 persons. There will be 5x4x4=80 men. It is two way crossed
ANOVA. Two-way because we have categorised men in two ways: exercises and diets.
This model is also balanced: we have exactly same number of men for each exercise-diet.
i – exercise number
j – diet number
k – kth person
Yijk – kth person in the ith exercise and jth diet.
In this case we can have two different type of hypothesis testing. Assume that mean for each
exercise-diet combination is ij. If we assume that model is additive, i.e. effect of exercise
and diet adds up then we have: ij = i+j. i is the effect of ith exercise and j the effect if
diet. Then we want to test following hypotheses: a) ij does not depend on exercise and b)
ij does not depend on diet.
Sometimes we do not want to assume additivity. Then we want to test one more hypothesis:
model is additive. If model is not additive then there might be some problems of
interpretations with other hypotheses. In this case it might be useful to use transformation
to make the model additive.
Models used for ANOVA can be made more and more complicated. We can design three, four
ways crossed models or nested models. We can combine nested and crossed models
together. Number of possible ANOVA models is very large.
Assumptions
ANOVA models are special cases of the linear models. We can write the model as:
Y με
Where Y is the observation vector,  -is vector of the means composed of the treatement
means and  is the error vector. Basic assumptions in ANOVA models are:
1.
Expected values of the errors are 0
2.
Variance of all errors are equal to each other
3.
Errors are independent
4.
Errors are normally distributed
All ANOVA treatments are very sensitive to assumptions 1)-3). F-tests meant to be robust
against the assumption 4). If assumptions 1)-3) are valid then 4) will always be valid at
least asymptotically. I.e. for large number of the observations
ANOVA tables
Standard ANOVA tables look like
effect
df
SSh
MS
F
prob
v1
d1
SS1
MS1=SS1/d1
MS1/MSe
pr1
...
...
...
...
vp
dp
SSp
MSp=SSp/dp
MSp/MSe
prp
error
de
SSe
MSe=SSe/de
total
N
SSt
Where v1,,,vp are values we want to test if they are 0. df is degrees of freedom corresponding to
this value. SSh is sum of the squares corresponding to this value (h denotes hypothesis). F
is F-value we want to test. Its degrees of freedom is (di,de). Prob is corresponding
probability. If probability is very low then we reject hypothesis that this value is 0. If the
value for prob is high enough then we do not reject null-hypothesis that this value is 0.
These values are calculated using likelihood ratio test. Let us say we want to test hypothesis:
H0: vi=0 vs H1:vi0
Then we maximise likelihood under null hypothesis find corresponding variance then we
maximise the likelihood under alternative hypothesis and find corresponding variance.
Then we calculate sum of the squares for null and alternative hypothesis find F-statistics
LR test for ANOVA
Suppose variances are:
ˆˆ 2 for null hypothesis ˆ 2 for thealternative hypothesis
Then mean sum of the squares for the null and alternative hypotheses as:
SSh 
SSe 
ˆˆ 2  ˆ 2
dfh
and for thealternative hypothesis
ˆ 2
dfe
Since first sum of squares is 2 with degrees of freedom dfh and the second sum of squares is 2
with degrees of freedom dfe and they are independent then their ratio has F-distribution
with degrees of freedom (dfh,dfe). Degrees of freedom of hypothesis is found using
number of elements in the category-1 in the simplest case.
Using this type of ANOVA tables we can only tell if there is significant differences between
means. It does not tell which one is significantly different.
This ratio has F distribution if null-hypothesis is true. Otherwise it has non-central F-distribution.
Example: Two way ANOVA
Let us consider an example taken from Box, Hunter and Hunter. Experiment was done
on animals. Survival times of the animals for various poisons and treatment was
tested. Table is:
treatment
A
poisons
I
B
C
D
0.31
0.45
0.46
0.43
0.82
1.10
0.88
0.72
0.43
0.45
0.63
0.76
0.45
0.71
0.66
0.62
II
0.36
0.29
0.40
0.23
0.92
0.61
0.49
1.24
0.44
0.35
0.31
0.40
0.56
1.02
0.71
0.38
III
0.22
0.21
0.18
0.23
0.30
0.37
0.38
0.29
0.23
0.25
0.24
0.22
0.30
0.36
0.31
0.33
ANOVA table
ANOVA table produced by R:
Df
Sum Sq
pois
2
1.03828
treat
3
0.92569
pois:treat 6
0.25580
Residuals 36 0.83013
Mean Sq
0.51914
0.30856
0.04263
0.02306
F value
22.5135
13.3814
1.8489
Pr(>F)
4.551e-07 ***
5.057e-06 ***
0.1170
Most important values are F and Pr(>F). In SPSS Pr is replaced by sig.
In this table we have tests for pois. and treat. Moreover we have “interaction” between these
two categories. Interaction means that it would be difficult to separate effects of these
two categories. They should be considered simultaneously. Pr. for interaction is not very
small and it is not large enough to discard interaction effects. In these situations
transformation of the variables might help. Let us consider ANOVA table for the
transformed observations. Let us use transformation 1/y. Now ANOVA table looks like:
Df
Sum Sq Mean Sq
F value
Pr(>F)
pois
2
34.903
17.452
72.2347
2.501e-13 ***
treat
3
20.449
6.816
28.2131
1.457e-09 ***
pois:treat 6
1.579
0.263
1.0892
0.3874
Residuals 36
8.697
0.242
ANOVA table
According to this table Pr. corresponding to the interaction term is high. It means that
interaction for the transformed variables is not significant. We could reject interaction
terms. We can build the ANOVA table without the interactions. It will look like:
Df Sum Sq
pois
2 34.903
treat
3 20.449
Residuals 42 10.276
Mean Sq
17.452
6.816
0.245
F value
71.326
27.858
Pr(>F)
3.124e-14 ***
4.456e-10 ***
Now we can say that there is significant differences between poisons and treatments.
Sometimes it is wise to use transformation to reduce effect of interactions. For this several
different transformations (inverse, inverse square, log) could be used. For each of them
ANOVA tables could be built. Then by inspection you can decide which transformation
gives better results. Following argument could be used to justify transformation. If
effects of two different categories is multiplicative then log of them will have additive
effect. It is easier to interpret additive effects than others.
Use of SPSS
Let us use SPSS for one data set taken from Box, Hunter and Hunter.
That is result from randomised block design on penicillum manufacture
treatment
A
B
C
D
blend of corn
steep liquour
1
89
88
97
94
2
84
77
92
79
3
81
87
87
85
4
87
92
89
84
5
79
81
80
88
Book on SPSS: Brace, N., Kemp, R. & Snelgar, R. (2003) SPSS for Psychologists. A
Guide to Data Analysis using SPSS for Windows. Second edition.
Exercise 3.
a) Data on effect of various dietary supplements on chick weights. It is one way
ANOVA. Use SPSS or other packages and analyse these data. Take the data
from the web page:
http://www.ysbl.york.ac.uk/~garib/mres_course/exercise_3a.html
What do you think about the differences.
b) Another exercise will be ready by the Wednesday, 5th of November.
http://www.ysbl.york.ac.uk/~garib/mres_course/exercise_3b.html
Solutions and small reports should be ready by 17th of November.
All questions after 13th of November.