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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

Uploaded on 01/01/2013

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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
pf3
pf4
pf5
pf8

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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

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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

> E ( MSerror )