ChiSquare Test for Homogeneity
Day 116  Lesson 11.2
Learning Targets

State appropriate hypotheses and compute the expected counts and chisquare test statistic for a chisquare test based on data in a twoway table.

State and check the Random, 10%, and Large Counts conditions for a chisquare test based on data in a twoway table.

Calculate the degrees of freedom and Pvalue for a chisquare test based on data in a twoway table.

Perform a chisquare test for homogeneity.
Activity: Does Gummy Bear Brand Matter?
Activity:
Answer Key:
Everyone knows that Haribo gummy bears are the best. Seriously. But we wondered if the distribution of color was the same for the offbrand. So we took a random sample of Haribo gummy bears and a random sample of “Meijer” gummy bears and compared the distribution of color.
Chisquare GOF or Chisquare Test for Homogeneity?
This Activity may look similar to the M&M activity from the previous lesson, but there is a very important distinction. With the M&M activity, we were comparing data from one sample to a claimed distribution of color. With this gummy bear activity we are comparing data from one sample to data from another sample. This is analogous to the difference between a oneproportion ztest and a twoproportion z test.
Chisure Test for Homogeneity or Chisquare Test for Independence?
Tomorrow, we will do a chisquare test for independence. The mechanics of this test are identical to the mechanics for the chisquare test of homogeneity. The difference is that a chisquare test for homogeneity has 2+ populations (Haribo, Meijer) and measures 1 categorical variable (color). For a chisquare test for independence, there is 1 population (senior students) and measures 2 categorical variables (gender, favorite class).