CHI SQUARE TESTS for goodness of fit and independence (Chapter 12)
The Chi square statistic can be used for tests on distributions — but must be used with frequency counts,[i.e. the
number of observations that fall into certain categories]. We use fito represent the actual frequency for category i
(number of observations — in the actual data — that are in categoryi) and eito represent the expected frequency if H0
is true (number of observations for category ipredicted by H0for a sample of this size).
Our test statistic is (in all cases) χ2=X
I
(fi−ei)2
ei
OR χ2=X
i,j
(fij −eij)2
eij
(Total, over all categories, of (actual minus expected) squared over expected — categories may be based on one variable
– first formula – or two variables – second formula)
NOTE: Expected cell frequency must be at least 5 in order to use the chi-square distribution (rows or columns may be
combined to accomplish this)
Goodness of Fit [One variable — one row of categories]
The issue is to determine whether a particular probability distribution might reasonably describe the population from
which the sample was drawn. Our test is always
H0: The data come from a population with the distribution stated
Ha: The data come from a population which does not fit that distribution
The test statistic is given by: sample χ2=X
I
(fi−ei)2
ei
with df = #categories−1−(number of parameters estimated from data)
In general, the expected frequency for category iis P(X=i)×n(n= sample size) — and is not rounded to a whole
number. (P(X=i) comes from the distribution we are testing for)
Critical values for the distribution are given in table 3 on p.923 [same as used for inference on σ2] but we are only interested
in small areas [columns further to the right].
Decision method: We will reject H0and conclude the proposed distribution does not fit if our sample χ2> χ2
αwith df =
#categories −1−(number of parameters estimated from data)
Independence Test and Contingency Tables [Two variables or two populations making a table of categories]
Events Aand Bare independent if P(A|B) = P(A), [which is equivalent to P(Aand B) = P(A)P(B)] Two variables
are independent if knowing the value for one does not change the probability distribution for the other. (All events that
can be described with one are independent of all events that can be described with the other)
In the contingency table (laying out all the possible combinations of values for the variables — all “contingencies”),
independence means that the probability of any cell can be found as the product of marginal probabilities (P(X=
Aand Y=B) = P(X=A)×P(Y=B)) That is, the probability of column one is the same for every row, probability of
column two is the same for every row, etc. and probability of row 1 is the same for every column, etc. Thus the expected
count eij for the cell in row i, column jis given by
eij =P(row i)×P(column j)×sample size = # row i
sample size ×# column j
sample size ×sample size = # row i×# column j
sample size
The issue is to determine whether the two variables (determining the rows and columns, respectively) are independent.
Test is always
H0: The two variables are independent
Ha: The two variables are not independent
The test statistic is χ2=X
i,j
(fij −eij)2
eij
df = (#rows −1) ×(#columns −1)
Decision method: We will reject H0and conclude the variables are not independent if our sample χ2> χ2
αwith df =
(#rows −1) ×(#columns −1). That is we reject the null hypothesis only if the test statistic is “big”.
MINITAB: [for contingency table] Enter the observed frequencies in adjacent columns, keeping the entries in order (so
you copy the table of observed values). Choose Stat>Tables then choose Chi-Square Test (Table in Worksheet) enter the
appropriate columns (containing the table) in the Columns Containing Table box
Equality of proportions
The chi square test for equality of several proportions (which is the extension of the two-sample test on proportions) is
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