1. Inference for Tables
Chi-Square Tests
Ch. 11

2. Chi-Square Test of Independence
The chi-square(c2) test for independence/association test if the variables in a two-way table are related (single random sample)
Ho: the variables are independent
Ha: the variables are dependent

3. A study compared noncombat mortality rates for U. S
A study compared noncombat mortality rates for U.S. military personnel who were deployed in combat situations to those not deployed. The results of a random sample of 1580 military personnel:
Cause of death
Unintentional Injury
Illness
Homicide or Suicide
Deployed
183 30 11
Not deployed
784
264
308
At the 0.05 significance level, test the claim that the cause of a noncombat death is independent of whether the military person was deployed in a combat zone.

4. Ho: Deployment status and cause of death are independent
Chi-square test of independence
Ho: Deployment status and cause of death are independent
Ha: They are dependent
Cause of death
Unintentional Injury
Illness
Homicide or Suicide
Deployed
183 30 11
Not deployed
784
264
308
Expected values table?
What are we expecting to be true about the table?
So, if deployment status and cause of death are independent, then…

5. So, if there are 1580 troops, how many would be
Unintentional Injury
Illness
Homicide or Suicide
Deployed
183 30 11
Not deployed
784
264
308
224
1356
1580
So, if there are 1580 troops, how many would be
expected to be both deployed and die because of illness?

6. Expected count formula
Unintentional Injury
Illness
Homicide or Suicide
Deployed
183 30 11
Not deployed
784
264
308
224
1356
1580

7. Expected count formula
Observed
Unintentional Injury
Illness
Homicide or Suicide
Deployed
183 30 11
Not deployed
784
264
308
224
1356
1580
Expected
Unintentional Injury
Illness
Homicide or Suicide
Deployed ?
41.68
Not deployed

8. Expected count formula
Observed
Unintentional Injury
Illness
Homicide or Suicide
Deployed
183 30 11
Not deployed
784
264
308
224
1356
1580
Expected
Unintentional Injury
Illness
Homicide or Suicide
Deployed
137.09
41.68 ?
Not deployed

9. Expected count formula
Observed
Unintentional Injury
Illness
Homicide or Suicide
Deployed
183 30 11
Not deployed
784
264
308
224
1356
1580
Expected
Unintentional Injury
Illness
Homicide or Suicide
Deployed
137.09
41.68
45.23
Not deployed
829.91
252.32
273.77

10. Ho: Deployment status and cause of death are independent
Chi-square test of independence
Ho: Deployment status and cause of death are independent
Ha: They are dependent
Expected values table:
Unintentional Injury
Illness
Homicide or Suicide
Deployed
137.09
41.68
45.23
Not deployed
829.91
252.32
273.77
Given random sample.
Exp. counts > 5 ∴ large sample
At least 15,800 military persons in pop.

11. Formula: df? df=(2 – 1)(3 – 1)= 2 Unintentional Injury Illness
Homicide or Suicide
Deployed
183 30 11
Not deployed
784
264
308
df?
Unintentional Injury
Illness
Homicide or Suicide
Deployed
137.09
41.68
45.23
Not deployed
829.91
252.32
273.77
df=(2 – 1)(3 – 1)= 2

12. Chi-Square Table
df=2
We reject Ho, since p-value<a there is enough evidence to believe that cause of death depends on deployment status.

13. Chi-Square Test of Homogeneity
The chi-square(c2) test for homogeneity allows the observer to test if the populations within a two-way table are the same (multiple random samples)
Ho: all the populations are the same for the given variable
Ha: all the populations are different

14. In the past, a number of professions were prohibited from advertising
In the past, a number of professions were prohibited from advertising. In 1977, the U.S. Supreme Court ruled that prohibiting doctors and lawyers from advertising violated their right to free speech. The article “Should Dentists Advertise?” compared the attitudes of consumers and dentists toward the advertising of dental services. Separate random samples of 101 consumers and 124 dentists were asked to respond to the following statement “I favor the use of advertising by dentists to attract new patients.” The data is presented below:
The authors of the article were interested in determining whether the two groups differed in their attitudes toward advertising.

15. Ho: the opinions of consumers and dentists are the same
Chi-square test of homogeneity
Ho: the opinions of consumers and dentists are the same
Ha: The opinions are different
Expected values table:
Given random samples.
Exp. counts > 5 ∴ large samples
At least 1240 consumers/dentists in pops.

16. Formula:
df=(2 – 1)(5 – 1)= 4

17. Chi-Square Table
df=4
We reject Ho, since p-value<a there is enough evidence to believe the opinions of consumers is different than the opinions of dentists.

18. Summary of All Chi-Square Tests
Test for Goodness of Fit
Comparing the distribution of a variable to what is expected (given % or equally likely)
Chi-Square test for homogeneity of populations
Two (or more) populations comparing one variable
Stratified sampling
Chi-Square test of association/independence
One population comparing two variables