R(J +1) such that every point in the subspace

j=1

minimizes the sum of squared errors. We select

and the solution of interest by imposing the constraint

J

j=1 a j = 0. These solutions are given by

J

1 1 1

h =1+ ’ . µ = Y—¦—¦

3(J ’ 1) j=1 (n j ’ 1) (N ’ J )

a j = Y—¦ j ’ Y—¦—¦ , for j = 1, . . . , J.

Statistic B is approximately distributed χ 2 (J ’ 1) It is easily shown that the regression sum

under H0 . Large values of B are interpreted as of squares is equal to the treatment sum of

evidence that H0 is false. Therefore the test is squares derived above, and that the test of the

conducted at the (1 ’ p) signi¬cance level by

˜ null hypothesis that there is not a regression

comparing the realized value of B against the relationship [8.4.8] is equivalent to the test that

p-quantiles of χ 2 (J ’ 1) (see Appendix E).

˜ there is not a treatment effect [9.2.4].

9.3: Two Way Analysis of Variance 181

9.3 Two Way Analysis of Variance we again assume to be independent and normally

distributed.

We now extend the model discussed in Section 9.2 A ¬xed effects two way model with interaction

so that it is possible to account for the effects of is given by

two treatments (if a completely randomized design

has been used) or the effects of a treatment and Yi jl = µ + ai + b j + ci j + Ei jl . (9.15)

a block (if a randomized block design has been

The parameters are subject to the constraints

used).

The example we wish to keep in mind is I J

ai = b j = 0, and

the CCCma AMIP experiment. Recall that we

i=1 j=1

have nine DJF seasons in each simulation, each

of which is subjected to a different ˜treatment™ J I

ci j = ci j = 0 for all i and j,

(i.e., the sea-surface temperature and sea-ice

j=1 i=1

regime). The experiment is replicated six times

in six different simulations started from randomly

and the errors are assumed to be iid N (0, σE ). 2

chosen initial conditions. We can think of the six

In both experiments with and without replica-

simulations as blocks.

tion, it is possible to construct models with some

or all of the effects treated as random effects. As

9.3.1 The Two Way ANOVA Model” in [9.2.8], the test statistics used to test for block

Introduction. Suppose an experiment was and treatment effects are identical to the ¬xed

conducted that resulted in one outcome per effects case, but the interpretation of the tests is

treatment or treatment/block combination. (The quite different. There are also differences in the

language we use refers to treatments and blocks calculation of variance proportions.

because that coincides most closely with our We do not discuss random effects models in this

example.) Suppose that I different treatments section, but a two way model with a combination

were used, and that these were applied in random of ¬xed and random effects is discussed in detail in

order to I experimental units in J blocks. Section 9.4 in the context of this chapter™s working

We represent the resulting IJ outcomes of example.

the experiment with random variables Yi j , for

i = 1, . . . , I , and j = 1, . . . , J , which we assume

9.3.2 Two Way Model Without Interaction.

to be independent and normally distributed.

In the setup of (9.14), the total sum of squares

A ¬xed effects model for data of this sort is the

is partitioned into treatment, block, and sum of

two way model without interaction given by

squared errors as

Yi j = µ + ai + b j + Ei j . (9.14)

SST = SSA + SSB + SSE,

The parameters are subject to the constraints

where

I J

ai = 0 and b j = 0. I J

(Yi j ’ Y—¦—¦ )2 ,

SST =

i=1 j=1

i=1 j=1

N (0, σE ).

2

The errors are assumed to be iid

I

An important, and limiting, aspect of this model

is that the treatment and block effects are assumed SSA = J (Yi—¦ ’ Y—¦—¦ )2 ,

i=1

to be additive. This assumption may not be correct,

but we can not determine this with the limited J

SSB = I (Y—¦ j ’ Y—¦—¦ )2 ,

number of data that are available. (9.16)

To test the additivity assumption it is necessary j=1

to have data from a replicated experiment. If

IJ

a completely randomized design is used, every

treatment combination must be used more than SSE = (Yi j ’ Yi—¦ ’ Y—¦ j + Y—¦—¦ )2.

i=1 j=1

once. If a blocked design is used, each treatment

must be used within each block more than once. Using methods similar to those in [9.2.4], the

The outcome of a replicated experiment is following can be shown.

represented by random variables Yi jl , for i =

I

1, . . . , I , j = 1, . . . , J , and l = 1, . . . , n i j , which 1 E(SSA) = J i=1 ai2 + (I ’ 1)σE . 2

9: Analysis of Variance

182

2 If H0 : a1 = · · · = a I = 0 is true, then estimates a number greater than σE . The effect is

2

SSA/σE ∼ χ 2 (I ’ 1).

2 to reduce the power of the tests described above.

The linear contrast methodology described in

3 E(SSB) = I J b2 + (J ’ 1)σE . 2

[9.2.12] and [9.2.13] naturally extends to the

j=1 j

two way case and is not detailed here. Both

4 If H0 : b1 = · · · = b J = 0 is true, then

the treatment and block sums of squares can

SSB/σE ∼ χ 2 (J ’ 1).

2

be partitioned into independent components if

5 E(SSE) = (I ’ 1)(J ’ 1)σE . needed.

2

Diagnostic opportunities for the two way model

2 ∼ χ 2 ((I ’ 1)(J ’ 1)).

6 SSE/σE without interaction are relatively limited because

of the relatively large number of ¬tted parameters

7 SSA, SSB, and SSE are independent. compared with the number of degrees of freedom

It follows from items 1, 2, and 5“7 that the null available for error. None the less, scatter plots of

estimated errors, plotted by treatment and block,

hypothesis of no treatment effect, that is,

can be useful for identifying observations with

H0 : a1 = · · · = a I = 0, large in¬‚uence.

can be tested against the alternative hypothesis that

9.3.3 Two Way ANOVA of the CCCma Multiple

there is a treatment effect by comparing

AMIP Experiment. We now use the two way

SSA/(I ’ 1) model with I = 9 treatments and J = 6

F=

SSE/((I ’ 1)(J ’ 1)) blocks. Because there is only one replication per

treatment/block combination, there are (I ’1)(J ’

with F(I ’ 1, (I ’ 1)(J ’ 1)) critical values (see

1) = 40 df for error.

Appendix G). Similarly, items 3“7 are used to

The F-ratios for the boundary forced effect

show that the no block effect null hypothesis, that

on 850 hPa DJF temperature (not shown) are

is,

very similar to those computed using the one

H0 : b1 = · · · = b J = 0 (9.17) way model. The small reduction in the number

of degrees of freedom available for error results

can be tested against the alternative hypothesis that in a test that is slightly less powerful than in

there is a block effect by comparing the one way case. However, the estimate of error