Applied Statistics Using SPSS, STATISTICA, MATLAB and R

5.1.3 The Chi-Square Goodness of Fit Test
5.1 Inference on One Population 179
The previous binomial test applied to a dichotomised population. When there are
more than two categories, one often wishes to assess whether the observed
frequencies of occurrence in each category are in accordance to what should be
expected. Let us start with the random variable 5.4 and square it:
Z
2
2
( P − p)
2 ⎛ 1 1 ⎞ ( X 1 − np)
( X 2 − nq)
= = n(
P − p)
=
+
pq / n
⎜ +
p q
⎟
, 5.5
⎝ ⎠ np nq
where X1 and X2 are the random variables associated with the number of
derivation note that denoting Q = 1 − P we have (nP – np) 2 = (nQ – nq) 2 “successes” and “failures” in the n-sized sample, respectively. In the above
. Formula
5.5 conveniently expresses the fitting of X1 = nP and X2 = nQ to the theoretical
values in terms of square deviations. Square deviation is a popular distance
measure given its many useful properties, and will be extensively used in
Chapter 7.
Let us now consider k categories of events, each one represented by a random
variable Xi, and, furthermore, let us denote by pi the probability of occurrence of
each category. Note that the joint distribution of the Xi is a multinomial
distribution, described in B.1.6. The result 5.5 is generalised for this multinomial
distribution, as follows (see property 5 of B.2.7):
2 ( X np ) 2
k
* 2
i i
= ~ χ k −1
i= 1 npi
∑ −
χ , 5.6
where the number of degrees of freedom, df = k – 1, is imposed by the restriction:
k
∑
i=1
x
i =
n . 5.7
As a matter of fact, the chi-square law is only an approximation for the sampling
distribution of χ ∗2 , given the dependency expressed by 5.7.
In order to test the goodness of fit of the observed counts Oi to the expected
counts Ei, that is, to test whether or not the following null hypothesis is rejected:
H0: The population has absolute frequencies Ei for each of the i =1, .., k
categories,
we then use test the statistic:
* 2
( O − E )
k
2
i i
∑
i= 1 Ei
χ =
, 5.8
2
2