
ChiSquare Test for Homogeneity 
ChiSquare Test for Homogeneity The chisquare test for homogeneity is a test
made to determine whether several populations are similar or equal or homogeneous
in
1. Know what is meant by the test for homegeneity. The test for Homogeneity is evalauting the equality of several
populations of categorical data. The test asked whether 3 or more
The homogeneity chisquare test statistics is computed exactly the same
as the test for independence using contingency
table as
The only difference between the test for independence and the homogeneity test is the stating of the null hypothesis: Homogeneity tests a null hypothesis asserting that various populations
are homogeneous or equal with respect to some charateristics
If is populations ith (categorized) then (the null hypothesis) (note other statistics could be used to compare pairwair combinations
of all possible pairs; however if there is a mistake in judging the
2. Know how to compute the chisquare homegeniety test statistics. Problem: An ads agency wishes
to determine if there are any differences with respect to reader recall
among 3 kinds magazine ads.
Step 1. Make a problems statement: (becomes the hypothesis statement, H_{o} ). The main question to answer is: are there any differences in recalling
property (mnemonic) of the three kinds of advertisements?
Hypothesis: If
represent the proportion rememebering the ith ad (i=1 for Humorous, i=2
for Technical and i=3 for Comparision) then
The ads make no difference in (helping) reader recall. H_{a}: H_{o} is not true. (alternate hypothesis): There is a difference between the ads in terms of reader recall. Step 2. Choose , the significance level of the test. If you want the be 95 % certain that the test is true, then = 0.05 =(10095)/100 The df = (r1)(c1), or (m1)(k1) = (31)(21)=2 So df = 2 Step 3. Look up
from chisquare
table:
For d.f. = 12, Step 4. (4) Determine or compute , The Expected Frequencies: The following table is the Expected Frequencies, E of each cells
in the study:
Step 5. Compute
= 19.02221 from computational table above. Step 6 Perform test chisquare test: , Since , i.e. 19.0221 > 5.99, Then we assume that the null hypothesis of equaly proportions must be rejected Make Conclusion or inference: We conclude that the three ads are not equally easy to remembered. So H_{a} (alternate hypothesis) is favored by this test. 3. Know how to calculate the contingency coefficient, C. The contingency coefficient, C is a statistics that measures
the strength of the relationship between variables or categories in a
It is interpretated like the correlation coefficient. , where is the chisquare statistics obtained for the contingency table and n is the grand total of all frequencies. This statistics is rarely used by practitioners. Workshop Problem (test for Homogeneity) A dietician wants to know if the time of day influences the tendency
to consume coffee. The following data represents the beverage
The null huypothesis is that the proportions of coffee purchases throughout
the day are identical. Using alpha of 0.05, what conclusion
