How to Perform a 2×2 Chi‑Square Test in SPSS: Step‑by‑Step Guide

Introduction

Chi‑square test (2x2) is a statistical method used to examine whether there is a significant association between two categorical variables, often to see if one variable can predict the other.

In SPSS the test can be performed via the “Crosstabs” module. Choose the two categorical variables, go to the “Statistics” tab, select “Chi‑square”, and click OK. SPSS will generate a cross‑tabulation and display the chi‑square results.

Example

Example: 198 patients with acute myocardial infarction were treated with drug A (11 deaths, 5.56% mortality) and 42 patients with drug B (6 deaths, 14.29% mortality). Question: Is there a difference in mortality between the two groups?

This is a typical binary outcome scenario. The 2 × 2 contingency table (four‑cell table) is used for the chi‑square test.

Data entry

(1) Variable view

(2) Data view

Weight cases: Choose Data → Weight Cases, check “Weight cases by”, place the frequency variable, and click OK. Because each row represents multiple observations, weighting is required.

If each row represents a single observation, weighting is unnecessary.

Analyze → Descriptive Statistics → Crosstabs

Options settings

(1) Main dialog: move the grouping variable (Drug) to the Row(s) box and the outcome variable (Outcome) to the Column(s) box. The variables can be swapped without affecting the result.

(2) Statistics: check “Chi‑square” to perform the chi‑square test for grouped count data.

(3) Cells: under Counts check “Observed” and “Expected”; under Percentages check “Row” to display row percentages.

Result interpretation:

Table 2 shows observed frequencies, percentages, and expected frequencies (row total × column total / grand total). The expected frequencies and total sample size determine which chi‑square result to use.

Conclusions:

If total N ≥ 40 and all expected frequencies ≥ 5, use Pearson Chi‑Square.

If total N ≥ 40 and one expected frequency is between 1 and 5, use Continuity Correction.

If total N ≥ 40 and at least two expected frequencies are between 1 and 5, use Fisher’s Exact Test.

If total N < 40 or any expected frequency < 1, use Fisher’s Exact Test.

In this example N = 240 > 40, with one expected frequency = 3.0 < 5, so the Continuity Correction result is considered: χ² = 2.796, P = 0.095 > 0.05. Therefore the mortality rates (5.6 % vs 14.3 %) differ but not statistically significant.

Reference: Li Yanlong, “Chi‑square test (2x2) – SPSS tutorial”.