STAT3012 Lecture Notes - Lecture 15: Box Plot, Null Hypothesis, Test Statistic
Lecture 15 – More on 1-way ANOVA
New concepts
✷Partitioning ANOVA sums of squares
✷The ANOVA table
✷Treatment sum of squares and its distribution
✷Independent contrasts
Applied Linear Models: Lecture 15 1
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Theory – Structure of the ANOVA table for a single treatment model
For the 1-way model
M1:Yij =µ+αi+ǫij (i= 1, . . . , t, j = 1, . . . , ni)
The ANOVA table seen in the previous lecture with anova() is:
Source of Variation Df Sum Sq Mean Sq F value P value
Treatment (t−1) S1−S0(S1−S0)
(t−1) =a f =a/b p
Residual (N−t)S0S0
(N−t)=b
Total N−1S1
Remark: Often factor is used synonymously for treatment.
Applied Linear Models: Lecture 15 2
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Example – Caffeine: Numbers for the ANOVA table
We have already seen the ANOVA table using R. Recall that,
S1−S0=
t
X
i=1
ni(Yi•−Y••)2
= 10[(244.8−246.5)2+ (246.4−246.5)2+ (248.3−246.5)2] = 61.4
S0= 134.1
N= 30
t= 3,
which gives the table by ‘hand’.
Applied Linear Models: Lecture 15 3
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Document Summary
Treatment sum of squares and its distribution. Theory structure of the anova table for a single treatment model. M1 : yij = + i + ij (i = 1, . The anova table seen in the previous lecture with anova() is: Remark: often factor is used synonymously for treatment. Example ca eine: numbers for the anova table. We have already seen the anova table using r. recall that, S1 s0 = ni(y i y )2 txi=1. = 10[(244. 8 246. 5)2 + (246. 4 246. 5)2 + (248. 3 246. 5)2] = 61. 4. N = 30 t = 3, which gives the table by hand". Theory distribution of treatment sum of squares (tss) normal random variables we can show that i=1 ci i. Using basic properties of e, var and independent. Given a constraintec =pt i /ni)(cid:17) ciy i n(cid:16)ec , 2 txi=1 txi=1 i=1 ci = 0 pt.