ANOVA Assumptions
1. Normality (sampling distribution of the mean)
2. Homogeneity of Variance
3. Independence of Observations
- reason for random assignment
2
3
2
2
2
1ppp
σσσ
= =
31 ppp
µµµ
2
Docsity.com
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Guidelines and tips
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community
Ask the community for help and clear up your study doubts
University Rankings
Discover the best universities in your country according to Docsity users
Free resources
Our save-the-student-ebooks!
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
ANOVA Assumptions - Data Analysis in Psychology - Lecture Slides, Slides of Advanced Data Analysis
Anova Assumptions, Sampling Distribution of the Mean, Homogeneity of Variance, Independence of Observations, Random Assignment, Unbiased Estimator, Assume Homogeneity of Variance, Best Estimate, Equal Sample Sizes, Unequal Sample Sizes. These are the important points of Data Analysis in Psychology.
Typology: Slides
2012/2013
1 / 8
This page cannot be seen from the preview
Don't miss anything!
Related documents
Partial preview of the text
Download ANOVA Assumptions - Data Analysis in Psychology - Lecture Slides and more Slides Advanced Data Analysis in PDF only on Docsity!
ANOVA Assumptions
1. Normality (sampling distribution of the mean)
2. Homogeneity of Variance
3. Independence of Observations
- reason for random assignment
2 3
2 2
2 σ (^) p 1 =σ p = σ p
μ (^) p 1 μ p 2 μ p 3
If we sample from p 1
s 2 is an unbiased estimator of
also, and
Because we assume homogeneity of variance
and,
For the best estimate of
2
1
2 σ (^) p 1 = sp
σ
2
2
2 σ (^) p 2 = sp
2
3
2 σ p 3 = sp
2 3
2 2
2 σ (^) p 1 = σ p = σ p
2 =σ ε
2 1
2 σ (^) e = s p 2 2
2 σ (^) e = s p 2 3
2 σ e = s p
2 σ e
2
e j
= s
2 σ (^) k
s s
j j
∑
2 2
(for equal sample sizes)
2 3 3
2 2 2
2 (^211)
- −
N N N
N s N s N s s
p p p j
(for unequal sample sizes)
Another look,
n n n n k
n s n s n s n s
k
s
j k
j j j k k
- −
∑
... ...
1 2
2 2 2 2 2
2 1 1
2
( 1 ) ( 1 ) ... ( 1 ) ... ( 1 )
) 1
) ... ( 1 )( 1
) ... ( 1 )( 1
) ( 1 )( 1
( 1 )(
1 2
2 2
2
2 2 2 1
2 1
− + − + + − + + −
−
- − −
- − −
− −
−
=
∑ ∑ ∑ ∑
j k
k
k k j
j j
n n n n
n
x n n
x n n
x n n
x n
( 1 ) ( 1 ) ... ( 1 ) ... ( 1 )
... ...
1 2
2 2 2 2
2 1
− + − + + − + + −
=
∑ ∑ ∑ ∑
j k
j k
n n n n
x x x x
( − 1 )
2
j
j i
ij
n
x
( 1 )
( 1 ) ... ( 1 )
( 1 ) ... ( 1 )
( 1 )
( 1 )
( 1 )
( 1 )
... ( 1 )
... ( 1 ) ( 1 )
2
2
1
1
2 2
2
2 2
1
2 1
−
−
−
−
−
−
−
−
−
−
−
−
∑ ∑ ∑ ∑
k
k
j
j
k
k
j
j
n
n
n
n
n
n
n
n
n
x
n
x
n
x
n
x
k
s s sj sk
1 1 ... 1 ... 1
... ...
2 2 2 2
2 1
=
k
∑ s^ j
2
Docsity.com
If we assume not only homogeneity of variance but also that the
null hypothesis is true
then,
and, by rearranging
where
s n thesizeof all samples n
= (^) X (^) p =
2
2
σ ε
2 2 X p
σ ε = ns
2 2
∑
k
X Gn
s
pi Xp
Expected Values
2 E ( MSerror ) =σ ε
2 2 σ ε = s j
2 MS error = s j
2 X
MS treat = ns
2 2 X
σ ε = ns
2 2
E ( MStreat ) =σ ε + n σ e
If the null hypothesis is true,
( ) =σ (^) ε + (φ )
2 E MStreat n
2 =σ ε
= E ( MSerror )
If the null hypothesis is false
E ( MStreat ) =σ (^) e + n ( something >φ )
2
2
σ e