One-way Analysis of Variance

1 Grouped observations

Suppose we have k groups of observations ( of the same variable ) , where the n _i observations in each group [∑_{i = 1...k} n_i = n] are assumed to have the same distribution, i.e. all the observation in a group have the same mean, and the variance is the same for all the observations:

H₁: y ∼ N ( μ, σ²I ) , ...μ = ( μ_1,1, ⋯ , μ_1,n1, .... . . ...μ_k,1, ⋯ , μ_k,nk ) ...and ...μ_i,j = θ_i. ...So μ ∊ L₁, dim ( L₁ ) = k, and L₁ is spanned by v₁, ⋯ , v_k where:

v₁ = ( 1, ⋯ ,1,0, ⋯ ,0, .... . . ...0, ⋯ , 0 )

....

....

v_i = ( 0, ⋯ ,0, .... . . ...1, ⋯ ,1, .... . . ...0, ⋯ , 0 ) ............[i^th group]

....

....

v_k = ( 0, ⋯ ,0, .... . . ...0, ⋯ ,0,1, ⋯ ,1 )

Now ( y – p₁ ( y ) ) · v_i = 0, ...i = 1, ⋯ k ..., ...i.e.

y · v_i ...= ...p₁ ( y ) · v_i, ............∀ i

but

p₁ ( y ) · v_i ...= ...n_i

∧ θi .

y · v_i ...= ...

ni
∑		xi,j
j = 1

so

∧ θi .

...= ...

	1
	ni

ni
∑		xi,j
j = 1

...= ...y_i ...............the group mean

p₁ ( y ) ...= ...( y₁, ⋯ y₁, ....... . . ......y_i, ⋯ y_i, ....... . . ......y_k, ⋯ y_k )

RSS ...= ...|| y – p₁ ( y ) || ² ...= ...

∑
i

∑		( yi, j – yi ) 2
j

s₁² ...= ...

	RSS
	n – k

2 Testing for uniform mean

Consider the hypothesis H₂: θ₁ = θ₂ = ⋯ = θ_k = θ
This is the same as the hypothesis in " Uniform Normal Distribution ", so the RSS [with n – 1 degrees of freedom] in that model now becomes the TSS in this one:

TSS ...= ...|| y – p₂ ( y ) || ² ...= ...

∑		( yi, j – y ) 2
i, j

...............where y ...= ...

	1
	n

∑ yi, j i , j

s₂² ...= ...

	1
	n – 1

n
∑		( xi – y ) 2
1

Also

ESS ...= ...|| p₁ ( y ) – p₂ ( y ) || ² ...= ...

k
∑		ni ( yi – y ) 2
i = 1

Anova Table:

Sum of Squares		df	MS	r
ESS =	∑ ni ( yi – y ) 2 i	k – 1	ESS k – 1	ESS / k – 1 RSS / n – k
RSS =	∑ i ∑ ( yi, j – yi ) 2 j	n – k	RSS n – k = s12
TSS =	∑ ( yi, j – y ) 2 i, j	n – 1	TSS n – 1 = s22

Now

r ...= ...

	ESS / k – 1
	RSS / n – k

......∼ ......F ( k – 1, n – k )

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