Inference for Tables Chi-Square Tests Ch. 11

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

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.

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…

So, if there are 1580 troops, how many would be Unintentional Injury Illness Homicide or Suicide Deployed 183 30 11 Not deployed 784 264 308 967 294 319 224 1356 1580 So, if there are 1580 troops, how many would be expected to be both deployed and die because of illness?

Expected count formula Unintentional Injury Illness Homicide or Suicide Deployed 183 30 11 Not deployed 784 264 308 967 294 319 224 1356 1580

Expected count formula Observed Unintentional Injury Illness Homicide or Suicide Deployed 183 30 11 Not deployed 784 264 308 967 294 319 224 1356 1580 Expected Unintentional Injury Illness Homicide or Suicide Deployed ? 41.68 Not deployed

Expected count formula Observed Unintentional Injury Illness Homicide or Suicide Deployed 183 30 11 Not deployed 784 264 308 967 294 319 224 1356 1580 Expected Unintentional Injury Illness Homicide or Suicide Deployed 137.09 41.68 ? Not deployed

Expected count formula Observed Unintentional Injury Illness Homicide or Suicide Deployed 183 30 11 Not deployed 784 264 308 967 294 319 224 1356 1580 Expected Unintentional Injury Illness Homicide or Suicide Deployed 137.09 41.68 45.23 Not deployed 829.91 252.32 273.77

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.

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

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.

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

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.

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.

Formula: df=(2 – 1)(5 – 1)= 4

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.

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