Chi-square test of independence
This test checks whether two categorical variables are related. The null says they are independent. Compute the sum of (O - E) squared / E over all cells, with (rows - 1) x (columns - 1) degrees of freedom.
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Analogy
Do people in group A answer yes more often than people in group B?
Example
Observed 30, 20, 20, 30; expected 25 eachEach cell: 25 / 25 = 1Chi-square = 4 x 1 = 4.0df = 1, critical = 3.841p = 0.046: reject independence
In practice
Used for survey results, such as whether preference depends on age group.
Practice
Flashcards
- Degrees of freedom for an r by c table?(r - 1) x (c - 1).
Interview questions
- In a 2 by 2 table the observed counts are 30, 20, 20 and 30. Test whether the two variables are independent at the 5% level.
- Expected count 25 in every cell
- Chi-square = 4.0
- df = 1, critical 3.841
- Reject independence
Questions people ask
How do I find the degrees of freedom for a table?
Multiply the number of rows minus 1 by the number of columns minus 1. A 3 by 4 table has 2 x 3 = 6 degrees of freedom. A 2 by 2 table has 1. This counts how many cells can vary once the totals are fixed.
Does a significant chi-square show that one variable causes the other?
No. It only shows the variables are associated. Other factors may explain the link, and the direction of any cause is unknown. Only a well-designed experiment with random assignment can support a causal claim. A survey can show that two things travel together, but not why they do.