Topics: Using Fit Y by X for Chi-Square test of homogeneity and independence.
Category: Basic Inference - Proportions and Means
JMP features: Analyze > Fit Y by X.
Open One-Page Guide (PDF)
How can I test if type and marital status are independent of each other in the Chi-Square test under contingency analysis that is mentioned in one page guide. How can I consider both variables as Y? If I chose both type and marital status for Y (response), what should I chose for X (factor) to test for independence in the lunch platform? pls help.
You can choose either variable to be the X, and the other variable to be the Y.
Even though this makes it sound like you are hypothesizing that one of the variables is a predictor and the other is a response, the test itself doesn't do this.
The Chi-Square Test of independence is eqivalent to testing if the proportions in one variable (Y) differ based on the groups in the other variable (X). Or vice versa: the Chi-Square Test of independence is eqivalent to testing if the proportions in one variable (X) differ based on the groups in the other variable (Y). So you can put either variable in the Y box, and put the other variable in the X box, and you will get a Chi-Square test of independence between those two variables.