Transcript PPT 14

Presentation 14
ANOVA
ANOVA (Chapter 16.1)
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ANOVA is an extension of the 2-sample ttest or test of 2 means. It is an extension
because it allows us to test the equality of
3 or more means.
We use it when we have a quantitative
response variable, AND a categorical
predictor variable with MORE than two
levels.
ANOVA Details
Hypotheses:
H0: All means are equal between the groups.
Ha: At least one mean is different.
(Is called F-test because is based on F-distribution.)
Conditions:
1. Samples are random and independent.
2. Response is normal for each group (population).
3. All populations have the same standard deviation (they
may have different means).
ANOVA Example
We wanted to test 3 fertilizer treatments on corn
yield. The three treatments were each applied
to 50 lots of corn plants in a randomized design.
The yield in each lot was recorded.
H0: The mean yield is the same for all
fertilizers.
Ha: At least one fertilizer is different than the
rest in terms of mean yield.
MINITAB Output For ANOVA
One-way ANOVA: Yield versus Fertilizer
Analysis of Variance for Yield
Source
DF
SS
MS
F
P
Fertiliz 2 21547.3 10773.7 130.07 0.000
p-value of the test!
Error
147 12176.4
82.8
Total
149 33723.7
Individual 95% CIs For Mean
Based on Pooled StDev
Level
N
Mean
StDev -----+---------+---------+---------+Fert1
50 68.586 10.115
(--*-)
Fert2
50 76.293
7.840
(-*--)
Fert3
50 47.906
9.205 (--*-)
-----+---------+---------+---------+Pooled StDev = 9.101
50
60
70
80
ANOVA - Conclusions
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The p-value is less than .05 so we reject the null
hypothesis and conclude that at least one treatment is
different than the rest.
Looking at the CI’s, it appears that the mean yield for
fertilizer 2 is the highest, fertilizer 1 is second best, and
fertilizer 3 is clearly the worst in terms of mean yield.