The document describes a study conducted in a supermarket to examine the effect of background music on wine purchases. Customers were randomly assigned to one of three conditions: no music, French accordion music, or Italian string music. The number of bottles of French, Italian, and other wines purchased under each condition were recorded. A chi-square test is proposed to determine if the distribution of wine purchases differs across the three music treatments.

1. Focus FoxA statistically minded toll collector wonders if drivers are equally
likely to choose each of the three lanes at his toll booth. He selects a
random sample from all the cars that approach the booth when all
three lanes are empty, so that the driver’s choice isn’t influenced by
the cars already at the booth.
Which of the following is the correct expression for the chi-square
goodness-of-fit test in this situation?
a.
Lane Left Center right
Number of drivers 137 159 169

2. Inference for Relationships
What if we want to compare a single categorical variable across
several populations or treatments? - we need a new test…
- Determine whether the distribution of the categorical variable
is the same for each population
- Examine related test to see if there is an association between
the variable and populations
Recall:
Two-Way Tables, conditional probabilities

3. Inference for Relationships
Market researchers suspect that background music may affect the
mood and buying behavior of customers. One study in a supermarket
compared three randomly assigned treatments: no music, French
accordion music, and Italian string music. Under each condition, the
researchers recorded the numbers of bottles of French, Italian, and
other wine purchased.
a. Calculate the conditional distribution of the type of wine sold for
each treatment.
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. Inference for Relationships
b. Make an appropriate graph for comparing the conditional
distributions you found.
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. Inference for Relationships
c. Are the distribution of wine purchases under the three music
treatments similar or different? Reference evidence found in parts
a & b.
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. Inference for Relationships
In the wine example, if we use a one sample z test, we could select a
comparison that is significant or isn’t significant.
Individual comparisons don’t tell us whether the three distributions of
the categorical variable are significantly different.
We need to make multiple comparisons
- An overall test to see if there is any differences in parameters
- Detailed follow-up analysis to decide which of the parameters differ and
to estimate how large the differences are
We compare the observed counts in the a two-way table with the
counts we would expect if H0 is true.

7. Inference for Relationships
The null hypothesis in the wine and music experiment is that there is
no difference in the distribution of wine purchases in the store when
no music, French accordion music, or Italian string music is played.
To find the expected counts we start by assuming the H0 is true. We
can see from the two-way table that 99 of the 243 bottles of wine
bought during the study were French wines.
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

8. Inference for Relationships
If the specific type of music that’s playing has no effect on wine
purchases, the proportion of French wine sold under each music
condition should be 99/243 = 0.407.
There are 84 bottles of wine bought when no music is playing, so
0.407•84 = 34.22 bottles of French wine on average.
There are 75 bottles of bought when French music is playing, so
0.407•75 = 30.56 bottles of French wine on average.
There are 84 bottles of wine bought when Italian music is playing, so
0.407•84 = 34.22 bottles of French wine on average.
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

9. Inference for Relationships
Repeat the process for each type of wine using the proportion of total
bottles sold against each type of wine sold.
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
Wine No Music French Italian Totals
French 34.22 30.56 34.22 99
Italian 31
Other 113
Totals 84 75 84 243

10. Inference for Relationships
There is a general formula for the expected count in any cell of a two-
way table:
row total • column total
table total
99 • 84
243
Notice that all the
expected counts in the wine
study are at least 5.
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
Wine No Music French Italian Totals
French 34.22 30.56 34.22 99
Italian 10.72 9.57 10.72 31
Other 39.06 34.88 39.06 113
Totals 84 75 84 243

11. Inference for Relationships
Finding the chi-square statistic χ2 = ∑ (observed – expected)2
Expected
Calculate the chi-square
statistic for the observed
and expected counts of
wine and music.
(30-34.22)2 + (39-30.56)2 +….
34.22 30.56
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
Wine No Music French Italian Totals
French 34.22 30.56 34.22 99
Italian 10.72 9.57 10.72 31
Other 39.06 34.88 39.06 113
Totals 84 75 84 243

12. Inference for Relationships
Think of the chi-square statistic χ2 as a measure of how much the
observed counts deviate from the expected counts.
Large values of χ2 are evidence against the null, and the P-value
measures the strength of the evidence.
We will use Table C, but our df are a little different
df = (number of rows – 1)(number of columns – 1)