using chi-square test of association/independence to determine association between two variables

1. Comparing Several PropsThe chi-square test for Homogeneity allows us to test if
many populations and/or treatments have the same
proportional distributions
H0: p1 = p2 = p3 … or “no difference in the proportions of
successes for various populations or treatments”
Ha: at least two pi’s are different or “there is a difference in
the proportions of successes for various populations or
treatments”
Example on page 710
The chi-square statistic χ2 of a 2x2 table is equivalent to a
two-sample z test statistic squared z2.

2. Comparing Several PropsIt is also common to take a single random sample of a single
population and classify the data in two categorical variables.
A study followed a random sample of 8474 people with normal
blood pressure for about four years. All the individuals were free
of heart disease at the beginning of the study. Each person took
the Spielberger Trait Anger Scale Test, which measures how prone
a person is to sudden anger. Researchers also recorded whether
each individual developed coronary hear disease (CHD). This
includes people who had heart attacks and those who needed
medical treatment for heart disease.
Low Anger Moderate Anger High Anger Total
CHD 53 110 27 190
NO CHD 3057 4621 606 8284
Total 3110 4731 663 8474

3. Comparing Several PropsRandom sample of 8474 people with normal blood pressure. All
the individuals were free of heart disease at the beginning of the
study. Each person took the Spielberger Trait Anger Scale Test,
which measures how prone a person is to sudden anger. Whether
each individual developed coronary hear disease (CHD, including
people who had heart attacks and those who needed medical
treatment for heart disease.
Calculate appropriate conditional distributions to describe the
relationship between anger and CHD status
Low Anger Moderate Anger High Anger Total
CHD 53 110 27 190
NO CHD 3057 4621 606 8284
Total 3110 4731 663 8474

4. Comparing Several Props
Make a well-labeled graph that compares the conditional
distributions in part A
Write a few sentences describing the relationship between the two
variables.
Do these data provide convincing evidence of an association
between the variables in the larger population? … need new test!!
Low Anger Moderate Anger High Anger Total
CHD 53 110 27 190
NO CHD 3057 4621 606 8284
Total 3110 4731 663 8474

5. Comparing Several PropsChi-Square Test for Association/Independence
Null – no association between categorical variables
Alternative – there is an association between variables
H0: there is no association between anger level and heart
disease in the population of people with normal blood pressure
Ha: there is an association between anger level and heart
disease in the population of people with normal blood pressure
No association means one variable does not occur in common
with the other; the variables are independent
Goal – analyze the relationship between two variables.

6. Comparing Several PropsH0: there is no association between anger level and heart
disease in the population of people with normal blood pressure
Ha: there is an association between anger level and heart
disease in the population of people with normal blood pressure
Similarly:
H0: anger and heart disease are independent in the
population of people with normal blood pressure
Ha: anger and heart disease are not independent in the
population of people with normal blood pressure.
Begin by comparing observed counts with the expected counts
if null is true.

7. Comparing Several PropsFind expected counts
The null hypothesis is that there is no association between
anger level and heart disease in the population of interest. If we
assume that H0 is true, then anger level and heart disease are
independent. We can find the expected cell counts
Low Anger Moderate Anger High Anger Total
CHD 53 110 27 190
NO CHD 3057 4621 606 8284
Total 3110 4731 663 8474

8. Comparing Several Props
Chi-Square Test for Association/Independence
If Large Sample Size, Random, & Independent conditions are met
H0: There is no association between two categorical variables in the
population of interest
Ha: there is an association between two categorical variables in the
population of interest
Or
H0: two categorical variables are independent in the population of
interest
Ha: two categorical variables are not independent in the population of
interest
Find expected counts, then calculate chi-square statistic
Degree of freedom = (number of rows – 1)(number of columns – 1)
P-value area to the right of the statistic

9. Comparing Several Props
Do the data provide convincing evidence of an association between
anger level and heart disease in the population of interest? Carry
out an appropriate test to help answer the question.
Observed counts: Expected Counts:
State:
Plan:
Do:
Conclude:
Pg. 717
Low Mod. High Low Mod. High
CHD 53 110 27 69.73 106.08 14.19
NO CHD 3057 4621 606 3040.27 4624.92 618.81

10. Comparing Several Props
Can we conclude that being prone to anger causes heart disease?
- Observational study
- Lurking variables