This document discusses conditions for performing a chi-square goodness-of-fit test and chi-square test for homogeneity. The chi-square goodness-of-fit test requires that all expected counts be greater than 5. The chi-square test for homogeneity can be used to compare distributions across multiple groups, requires random sampling and independence of observations, and that expected counts be at least 5. The document provides an example comparing wine purchases with different music, calculates the chi-square statistic, and interprets the p-value to determine if distributions differ.

1. Focus Fox
Select the best option:
Which of the following is a condition that must be met in order to
carry out a chi-square goodness-of-fit test?
a. The population must be normally distributed, or the sample size
must be greater than 30
b. The cell counts for our sample have to be approximately the same
as the expected counts
c. All observed cell counts must be greater than 5
d. All expected counts must be greater than 5
e. More than one of these conditions must be met

2. Chi-Square Test Homgeneity
Chi-Square Test for Homogeneity – two way table
If Random – random samples from each population or groups in
randomized experiment
If Large Sample Size – all expected counts are at least 5
If Independent – individual observations & 10% condition
H0: there is no difference in the distribution of a categorical variable
for several populations or treatments
Ha: there is a difference in the distribution of a categorical variable
for several populations or treatments
Find expected counts
df = (number of rows – 1)(number of columns – 1)
P-value is the area to the right of χ2 under the density curve

3. Chi-Square Test Homgeneity
H0: there is no difference in the distributions of wine purchases at
this store when no music, French accordion music, or Italian string
music is played.
Ha: there is a difference in the distributions of wine purchases at this
store when no music, French accordion, or Italian string music is
played.
Previously calculated test statistic χ2 = 18.28
Wine No Music French Italian Totals
French 30 39 30 99
Italian 11 1 19 31
Other 43 35 35 113
Totals 84 75 84 243

4. Chi-Square Test Homgeneity
Previously calculated test statistic χ2 = 18.28
Use Table C to find the P-value.
Now use χ2cdf command on calculator
Wine No Music French Italian Totals
French 30 39 30 99
Italian 11 1 19 31
Other 43 35 35 113
Totals 84 75 84 243

5. Chi-Square Test Homgeneity
Previously calculated test statistic χ2 = 18.28
Interpret the P-value from the calculator in context.
What conclusion would you draw?
Wine No Music French Italian Totals
French 30 39 30 99
Italian 11 1 19 31
Other 43 35 35 113
Totals 84 75 84 243

6. Chi-Square Test Homgeneity
Using Technology:
Enter the observed data in Matrix A
Select χ2 test - STATS – TESTS – χ2 - Test
Observed: [A]
Expected: [B]
Choose calculate then 2nd Enter and Draw
Select and enter Matrix [B]
Tables are commonly referenced by dimensions row by column –
2x3
Pg. 706

7. Chi-Square Test Homgeneity
Follow-Up Analysis
The chi-square test for homogeneity allows us to compare the
distribution of a categorical variable for any number of populations
or treatments.
If we reject the null of no difference, follow-up analysis is required
- Examine which cell in the two-way table show large deviations
between observed and expected counts.
- Look at the which contribute the most to the chi-square statistic
Pg. 709 – minitab output and analysis