Chi Square test of Independence

• Assumptions

• Method of simple random sampling is used
• variable is measured on nominal scale of measurement
• expected value of each of cell is at least 5

• 5 Step Hypothesis

• Hypothesis

Ho: each population has the same proportion of observations. That is

P(level_1) population 1 = P(level_1) population 2 = . . . = P(level_1) population r

P(level_2) population 1 = P(level_2) population 2 = . . . = P(level_2) population r

.

.

P(level_c) population 1 = P(level_c) population 2 = . . . = P(level_c) population r

 

Ha: at least one of population has the different proportion of observations. That is

P(level_1) population 1 ≠ P(level_1) population 2 ≠ . . . ≠ P(level_1) population r

P(level_2) population 1 ≠ P(level_2) population 2 ≠ . . . ≠ P(level_2) population r

.

.

P(level_c) population 1 ≠ P(level_c) population 2 ≠ . . . ≠ P(level_c) population r

• Test Statistic

• Critical value

chisq(a,df)  

• P-value

P(CHISQ≥Chisq)  

• Decision rule

Reject Ho if chisq > chisq(a,df)  or p-value < alpha