untitled

The one-factor ANOVAtests the hypothesis that the response variable does not vary

with the level of the factor. The alternative hypothesis is that the response variable

differs according to the level of the factor, either generally increasing or decreasing

with its level, or going up then down or the reverse, or varying in an unsystematic

manner.

Our example (Box 16.1) comprises counts of kangaroos on randomly placed

east–west transects, each 90 km in length, on a single degree block in southwest

Queensland, Australia. The question of particular concern is whether there is an order

effect in days of survey. Did the kangaroos become increasingly disturbed by the

aircraft and therefore seek cover whenever one was heard? Or did they become pro-

gressively habituated to the noise such that more were seen each day as the survey

280 Chapter 16

16.6.1One-factor

ANOVA

Day 1 Day 2 Day 3

96 71 28

38 45 43

80 45 29

35 67 36

50 31 37

55 28 59

38 84 lost

64 70 lost

n 1 = 8 n 2 = 8 n 3 = 6

T 1 = 456 T 2 = 441 T 3 = 232

x ̄ 1 =57.0 x ̄ 2 =55.1 x ̄ 3 =38.7

k=number of classes = 3

N=number of samples =n 1 +n 2 +n 3 = 22

∑Xij= 96 + 38 +...+ 37 + 59 = 1129

∑Xij^2 = 962 + 382 +...+ 372 + 592 =66,251

∑(Ti^2 /ni) = 4562 /8 + 4412 /8 + 2322 /6 =59,273

Main effects sum of squares (SS):

∑(Ti^2 /ni) −(∑Xij)^2 /N=59,273 − 11292 /22 = 1335

Residual SS:

∑Xij^2 −∑(Ti^2 /ni) =66,251 −59,273 = 6978

Total SS:

∑Xij^2 −(∑Xij)^2 /N=66,251 − 11292 /22 = 8313

ANOVA

Source SS d.f. MS F

Main effect 1335 k− 1 = 2 667.5

=1.8

Residual 6978 N−k= 19 367

Total 8313 N− 1 = 21

F=1.8 with 2 d.f. in the numerator and 19 in the denominator. The probability is 0.19, too high to argue for

rejection of the null hypothesis that observable density does not differ by day of survey.

667.5

367

Box 16.1Red

kangaroos counted on

the Cunnamulla degree

block (10,870 km^2 ) in

August 1986. Each

replicate is the number

of kangaroos counted

on a transect measuring

0.4 km by 90 km.

WECC16 18/08/2005 14:47 Page 280