2 the mean response to the 1982/83 and 1986/87

F= , boundary conditions are shown in Figure 9.3. F

c2

SSE J j

is signi¬cantly greater than 1 over 34.9% of the

N ’J j=1 n j

globe. The diagram shows that there are substantial

and that H0 should be rejected when F is unusually differences in the atmospheric response to the two

large. Next we show that wc and SSE are warm events in the tropical Paci¬c, the North

independent. Then we argue that wc is normal Paci¬c, and the South Atlantic. On the other hand,

because it is a linear combination of normal the response to the two warm events is similar

random variables. Also, the mean of wc is over Africa and the Indian Ocean during DJF.

zero under the null hypothesis, and therefore Larger differences evolve in subsequent seasons

the numerator of F, when properly scaled, is re¬‚ecting the difference in the phasing of these two

distributed χ 2 (1) under H0 . Finally, we conclude events.4

that F ∼ F(1, N ’ J ) under H0 . Thus the test

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

˜

˜ 9.2.15 Diagnostics. We have not concerned

comparing the computed f with the p-quantile of

F(1, N ’ J ) (see Appendix G). ourselves much, to this point, with diagnostics of

the ¬tted model. Many of the diagnostics discussed

Note that the test of the linear contrast adapts

in connection with regression models (see [8.3.13]

itself correctly to account for unequal sample

and [8.3.14]) are useful here as well. In particular,

sizes, but the test for the treatment effect does not.

scatter plots of the residuals as a function of

Note also that if two contrasts, say c j and d j , for

j = 1, . . . , J , are orthogonal, meaning that the treatment are useful for detecting outliers

4 The ¬ve-month running mean SO index reached a

J

cjdj

= 0, minimum in January of 1983 and again in March or April of

nj 1987.

j=1

9: Analysis of Variance

180

9.2.17 Equivalent Representation of a One Way

and changes in error variance between treatments.

ANOVA Model as a Regression Model. It may

Changes in variability from one treatment to the

be useful at this point to make the connection

next can also be conveniently tested with Bartlett™s

between ANOVA and regression models. We

test.

can write model Yi j = µ + a j + Ei j from

(9.1) in matrix vector form as follows. Let Y

9.2.16 Bartlett™s Test. Suppose we have J be the N -dimensional random vector constructed

samples (or treatments) of possibly unequal sizes by concatenating the J n -dimensional vectors

j

n 1 , . . . , n J and we wish to test the null hypothesis

(Y1, j , Y2, j , . . . , Yn j , j )T, and de¬ne E similarly.

that all errors, either in the ¬xed or random effects

models, have the same variance. The alternative is Let A be the (k + 1)-dimensional vector of

that at least one sample or treatment has a variance parameters (µ, a1 , . . . , a J ) . Then, (9.1) can be

T

expressed as

that is different. That is, we wish to test

Y = X A + E,

H0 : σE j = σE for all j = 1, . . . , J

2 2

where X is the N — (J + 1) design matrix given

by

«

against the alternative that the variances are not

0 ... 0

1 1

all equal. Here we use σE j to denote the variance

2

. . . .

¬ n 1 rows ·

. . . .

¬ ·

. . . .

of the random variables that represent sample or

¬ ·

¬ ·

treatment j. Let S1 , . . . , SJ be the corresponding

2 2

0 ... 0

11

¬ ·

¬ ·

1 ... 0

sample variances of the errors; that is, 10

¬ ·

¬ ·

.. . .

.. . .

¬ n 2 rows ·

.. . .

X =¬ ·

nj

¬ ·

1 ... 0

= (Yi j ’ Y—¦ j )2 /(n j ’ 1),

S2 10

¬ ·

j

¬ ·

.. . .

.. . .

¬ ·

i=1

.. . .

¬ ·

¬ ·

0 ... 1

1 0

¬ ·

and let S 2 be the pooled estimate of the variances ¬ ·

. . . .

p

. . . .

n k rows

. . . .

given by

0 ... 1

1 0

J

j=1 (n j ’ 1)S 2 SSE The normal equations that provide the least

j

S2 = = ,

p

N’J N’J squares estimators of A are given by

X TX A = X Y. (9.13)

J

where N = j=1 n j .

With this notation, Bartlett™s statistic is given by These equations have solutions given by

a = (X TX )’ X T Y,

Q

where (X TX )’ denotes the generalized inverse of

B= ,

h

X TX (see Graybill [148]). The generalized inverse

is required because X TX is a non-invertible matrix

where

(the ¬rst column of X is the sum of the remaining

J columns). In fact, solution (9.13) de¬nes a

J

Q = (N ’ J ) ln(S 2 ) ’ (n j ’ 1) ln(S 2 ), one-dimensional subspace of the parameter space