# Help

The traditional cell-mean (or structural) model of one-way within-subject
(repeated-measures) ANOVA is

_{ij}= μ + α

_{i}+ β

_{j}+ ε

_{ij}(I)

Y

_{ij}independent variable;

_{i}constants subject to Σα

_{i}= 0 – simple effect of factor A at level

*i*,

*i*= 1, 2, ...,

*a*;

_{j}independent

*N*(0, σ

_{p}

^{2}) – random effect of subject

*j*,

*j*= 1, 2, ...,

*b*(σ

_{p}

^{2}- population variance);

_{ij}independent

*N*(0, σ

^{2}) – random error or within-subject variability or interaction between the factor of interest and subject (σ

^{2}- variance of sampling error).

E(Y

_{ij}) = μ + α

_{i}, Var(Y

_{ij}) = σ

_{p}

^{2}+ σ

^{2}, Cov(Y

_{ij}, Y

_{i'j}) = σ

_{p}

^{2}(i ~= i'), Cov(Y

_{ij},Y

_{i'j'}) = 0 (j ~ = j');

_{p}

^{2}/(σ

_{p}

^{2}+ σ

^{2}).

Suppose we want to model a new one-way within-subject
(repeated-measures) ANOVA:

Y_{ij} = μ_{i}+ β_{j} + ε_{ij}
(II)

where all terms bear the same meaning as in model (I) except for factor
effects (constants) {μ_{i}, *i* = 1, 2, ..., *a*}. And the new assumptions are the same
as before except E(Y_{ij}) = μ_{i}.

The absence of a constant in model (II) is because we don't want a common mean removed from each effect and would like to test the following null hypothesis

H_{0}: μ_{1} = 0, μ_{2} = 0, ..., and μ* _{a}*
= 0 (III)

instead of the main effect test of factor A in model (I),

H_{0}: α_{1} = 0, α_{2} = 0, ..., and α* _{a}*
= 0 (no difference among factor levels)

(III) is an extension of one-sample

*t*test to a whole set of factor effects instead of one specific simple effect μ

_{i}.

I failed to find any discussion about model (II) and (III) in any text books I could access to, so I tried the following myself. Solving (II) as a general linear model is very straightforwad by coding factor A with a dummy variable, and we get an F test with F = {[(residuals SS of the reduced model) - (residuals SS of full)]/DF1}/[(residuals SS of full)/DF2]. The nice thing about this test is that there is only one F and the degrees of freedom are accurate.

However I would like
to analyze it in a more computationally economical way by calculating
sums of squares as in the traditional ANOVA approach.

First I started with the reduced model under null hypothesis (III)

Y_{ij} = β_{j} + ε_{ij}
(IV)

The errors under model (II) and (IV) are respectively,

ε_{ij} = Y_{ij} - μ_{i }- β_{j}

ε_{ij} = Y_{ij} - β_{j}

But I could not go any further from here, thus failed to work out a
formula for hypothesis (III). Any suggestions about this approach?

Then I tried the following method.

Intuitively we would pick SSA* = bΣ_{i}Y_{i}.^{2}^{}
(thereafter a dot in the place of an index indicates the term as the
mean over that index) as the sums of squares for hypothesis (III). It seems SSA* is not really a perfect candidate for hypothesis (III) since

Y_{i.} = (1/b)Σ_{j}Y_{ij}= μ_{i}+ β_{.} + ε_{i.},

which still contains subject effect β_{.}_{}, but I failed to come up with a better solution with some combination of Y_{i.}, Y_{.j}, Y_{ij}, and Y_{..}. The expected value of MSA* can be derived as the following

E(SSA*)

= bE(Σ_{i}Y_{i}.^{2})

= bE[Σ_{i}(μ_{i}+ β_{.} + ε_{i.})^{2}]

= bE[Σ_{i}(μ_{i}^{2}+ β_{.}^{2} + ε_{i.}^{2}+2μ_{i}β_{.}
+ 2μ_{i}ε_{i.}+ 2β_{.}ε_{i.})]

= bΣ_{i}[μ_{i}^{2}+ Eβ_{.}^{2} + Eε_{i.}^{2}+2E(μ_{i}β_{.})
+ 2E(μ_{i}ε_{i.}) + 2E(β_{.}ε_{i.})]

= bΣ_{i}(μ_{i}^{2}+ σ_{p}^{2}/b + σ^{2}/b)

= bΣ_{i}μ_{i}^{2}+ a(σ_{p}^{2} + σ^{2})

As the degrees of freedom for SSA* is *a*, we have

E(MSA*) = (bΣ_{i}μ_{i}^{2})/a+ (σ_{p}^{2} + σ^{2})

Similar derivation can be applied to MSS (mean squares for Subject) and
MSAS (mean squares for interaction term between factor A and Subject),
which are the same as in traditional one-way within-subject ANOVA,

E(MSS) = aσ_{p}^{2} + σ^{2},

^{2}

Based on the above 3 terms, we can construct the following F tests for
(III),

F_{1} = aMSA*/[MSS+(*a*-1)MSAS]

F_{2} = (MSA* - MSS)/[MSAS/(1-*a*)]

F_{2} is not a good candidate since the numerator might be negative. With a quasi-F statistic F_{1} with a composite mean square term in the denominator, MSS+(*a*-1)MSAS, the degrees of freedom for the denominator can be approximated as

df_{denom} = [MSS+(*a*-1)MSAS]^{2}/{MSS^{2}/(b-1)+(*a*-1)^{2}MSAS^{2}/[(a-1)(b-1)]} (IV)

And the degrees of freedom for F_{1} is F(*a*, df_{denom}).

Questions:

(1) Is everything above correct?

(2) I am not so sure about the multiplier (*a*-1)^{2} in (IV): Is it correct?

(3) One thing I don't feel
comfortable with this approach is that there might have multiple choices of F,
and I have to approximate the degrees of freedom for those composite sums of squares. And
I am not so sure whether this approach would be equivalent to the one with general linear model.

(4) Any better choices of F statistic for hypoethesis (III)?

(5) Hypothesis
(III) would be denied (rejected) if either

(i) the main effect of factor A in model (I) is significant
(because this denies that the α_{i} are equal at all), or

(ii) the hypothesis that the grand mean μ in model (I) is zero is rejected?

This seems very nice because it is much simpler
than the above approach with composite sums of squares, but my sub-questions are:

A. Yes, hypothesis (III) would
be denied (rejected) if either (i) or (ii) is
rejected, but is (III) equivalent to the following 2 hypotheses in the traditional ANOVA combined? Or in
other words, is it true that rejecting (III) would lead to at least one
of the following 2 hypotheses for model (I) being rejected?

H_{0} : the main effect of factor A is 0, or *α** _{i}* = 0,

*i*= 1, 2, ...,

*a*

and

H_{0} : μ = 0

B. One annoying part of this new approach is, I will have to have 2 separate F tests, lack of the nice feature of only one statistic for the hypothesis.