Inference for Tables ChiSquare Tests Ch 11 ChiSquare

Inference for Tables Chi-Square Tests Ch. 11

Chi-Square Test of Independence • The chi-square(c 2) 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. 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 Deployed Not deployed Unintentional Injury Illness Homicide or Suicide 183 784 30 264 11 308 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.

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

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

Expected count formula Observed Deployed Not deployed Expected Deployed Not deployed Unintentional Injury Illness Homicide or Suicide 183 784 30 264 11 308 967 294 319 Unintentional Injury Illness Homicide or Suicide ? ? 41. 68 ? ? ? 224 1356 1580

Expected count formula Observed Deployed Not deployed Expected Deployed Not deployed Unintentional Injury Illness Homicide or Suicide 183 784 967 30 264 294 11 308 319 Unintentional Injury Illness Homicide or Suicide 137. 09 ? 41. 68 ? ? ? 224 1356 1580

Expected count formula Observed Deployed Not deployed Expected Deployed Not deployed Unintentional Injury Illness Homicide or Suicide 183 784 967 30 264 294 11 308 319 Unintentional Injury Illness Homicide or Suicide 137. 09 829. 91 41. 68 252. 32 45. 23 273. 77 224 1356 1580

Chi-square test of independence Ho: Deployment status and cause of death are independent Ha: They are dependent Expected values table: Deployed Not deployed Unintentional Injury Illness Homicide or Suicide 137. 09 829. 91 41. 68 252. 32 45. 23 273. 77

Formula: Deployed Not deployed Unintentional Injury Illness Homicide or Suicide 183 784 30 264 11 308 Unintentional Injury Illness Homicide or Suicide 137. 09 829. 91 41. 68 252. 32 45. 23 273. 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(c 2) 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 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.

Chi-square test of homogeneity Ho: the opinions of consumers and dentists are the same Ha: The opinions are different Expected values table:

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