Chi-Square Tests

The Idea in Plain Language

Measurements have averages and spreads. Categories have counts. When your data are counts in categories (defect types, shifts, suppliers, weekdays), the chi-square test compares the counts you observed with the counts you would expect if a stated idea were true. The more the two differ, the larger the chi-square statistic, and the harder the idea is to believe.

There are two main uses, and the calculation is the same:

When to Use It

How It Works

1. Observed counts go in a table.
2. Expected counts. For association: expected = (row total × column total) / grand total. For goodness of fit: expected = total × the stated proportion.
3. Contribution of each cell = (O − E)² / E. Add them to get χ².
4. Degrees of freedom. Association: (rows − 1)(columns − 1). Goodness of fit: categories − 1 (minus one for each parameter estimated from the data).
5. Compare χ² with the chi-square distribution, and read the p-value from the upper tail.

The statistic says whether there is an association, not how strong it is. For strength use Cramér’s V, which runs from 0 (none) to 1 (perfect): V = √(χ² / (n × (k − 1))), where k is the smaller of the number of rows and columns. As a rough guide for small tables, 0.1 is weak, 0.3 moderate, and 0.5 strong. To see where the association lives, look at the standardized residuals: values beyond about ± 2 mark cells that contribute.

Worked Example 1: Does the Defect Type Depend on the Shift?

A plant recorded 280 defects by type and shift. Is the mix of defect types the same on every shift?

1. Hypotheses. H₀: defect type and shift are independent. H₁: they are associated. α = 0.05.
2. Expected counts. For Scratch on Day: (row total 95 × column total 80) / 280 = 27.14.
3. Check the assumption: the smallest expected count is 25.71, well above 5.
4. Add the contributions to get χ² = 25.41.
5. Degrees of freedom = (3 − 1)(3 − 1) = 4. The critical value at 0.05 is 9.49, and the p-value is 0.00004.
6. Strength. Cramér’s V = 0.21, a weak to moderate association.

Worked Example 2: Goodness of Fit

130 late deliveries were logged over several weeks. If lateness had no connection with the weekday, the five weekdays would each have about 26. Do the counts fit that?

1. Hypotheses. H₀: late deliveries are spread evenly over the five days. H₁: they are not.
2. Expected = 130 × 0.2 = 26 per day. The contributions add to χ² = 11.92.
3. Degrees of freedom = 5 − 1 = 4, and the p-value is 0.0179.

The same test checks whether data follow a distribution. Put the data into bins, calculate the expected count in each bin from the distribution (for example, the normal with the sample mean and standard deviation), and subtract one degree of freedom for each parameter estimated from the data.

Run It in Excel and Minitab

Chi-Square Test for Association: Defect, Shift

Rows: Defect   Columns: Shift

                     Day  Evening    Night     All
Scratch               40       35       20      95
  Expected         27.14    30.54    37.32
  Contribution      6.09     0.65     8.04
Dent                  22       30       38      90
  Expected         25.71    28.93    35.36
  Contribution      0.54     0.04     0.20
Discoloration         18       25       52      95
  Expected         27.14    30.54    37.32
  Contribution      3.08     1.00     5.77
All                   80       90      110     280

Cell Contents: Count, Expected count, Contribution to Chi-square

Chi-Square Test

                       Chi-Square  DF  P-Value
Pearson                    25.412   4    0.000
Likelihood Ratio           26.122   4    0.000

Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: Late

Category  Observed  Test Prop  Expected  Contribution
Mon             31       0.20      26.0        0.9615
Tue             18       0.20      26.0        2.4615
Wed             20       0.20      26.0        1.3846
Thu             22       0.20      26.0        0.6154
Fri             39       0.20      26.0        6.5000

  N  DF  Chi-Sq  P-Value
130   4  11.9231   0.0179

Reading and Reporting

1. Give the table of counts (or the percentages with the totals).
2. Report χ², the degrees of freedom, n, and the p-value, and the effect size (Cramér’s V).
3. Say where the association lives, using residuals or percentages by row.
4. Check and say that the expected counts were large enough.
5. Say what it means in practice, and what you would investigate.

Common Mistakes

Try It Yourself

Customer complaints by region and type: Region A: 30 billing, 20 delivery. Region B: 15 billing, 35 delivery.

• Test whether complaint type depends on region.
• How strong is the association, and what would you check?

Related Pages and Tools

Chi-Square Tests: Frequently Asked Questions

Sources and Further Reading