ETX5900 · Business Statistics
Chi-Square Test for Independence
Module 6 (Week 6) of ETX5900 Business Statistics at Monash University introduces the chi-square (χ²) test of independence — the formal hypothesis test for whether two categorical variables are related. You lay the counts in an r×c contingency table, work out the frequency expected in each cell if the variables were independent, and measure how far the observed counts stray. It maps to Berenson, Levine, Szabat & Stephan, Basic Business Statistics (Australian 5th ed.), Chapter 15, Section 15.3, and is a core task in the individual assignment and a regular fixture of the final exam.
What this chapter covers
- 01When to use it: two categorical variables cross-classified as counts in an r×c contingency table
- 02Hypotheses: H₀ = the variables are independent (no relationship); H₁ = they are dependent
- 03Expected frequency: fₑ = (row total × column total) / n, computed for every cell assuming H₀ is true
- 04Test statistic: χ² = Σ (fₒ − fₑ)² / fₑ, summed over all cells (always ≥ 0)
- 05Degrees of freedom: df = (r − 1)(c − 1) — set by the table size, not the sample size
- 06Validity condition: every expected count fₑ ≥ 5; if not, combine adjacent categories and recompute
- 07Decision rule: it is an upper-tailed test — reject H₀ if χ²calc > χ²crit(α, df), equivalently if p-value < α
- 08Reading tables / Excel: chi-square table entered by df and upper-tail α; in Excel p = CHISQ.TEST(observed, expected), critical value = CHISQ.INV.RT(α, df)
- 09Conclusion: state the verdict in context (dependent / no evidence of a relationship) and name the possible error (Type I if you reject)
Department vs preference for remote work (2×2 test)
- +1Step 1 — Hypotheses. H₀: department and remote-work preference are independent; H₁: they are dependent. Significance level α = 0.05.
- +1Read the margins. Row totals: Sales = 100, Operations = 100. Column totals: Yes = 90, No = 110. Grand total n = 200. Table is 2 rows × 2 columns.
- +1Step 2a — Expected counts, fₑ = (row × col) / n. Yes cells: 100×90/200 = 45; No cells: 100×110/200 = 55. So each row expects 45 Yes and 55 No. Every fₑ ≥ 5, so the test is valid.
- +1Step 2b — Test statistic. χ² = (60−45)²/45 + (40−55)²/55 + (30−45)²/45 + (70−55)²/55 = 5.000 + 4.091 + 5.000 + 4.091 = 18.182.
- +1Step 3 — Critical value. df = (r−1)(c−1) = (2−1)(2−1) = 1. From the chi-square table (upper tail, α = 0.05), χ²crit = 3.841.
- +1Steps 4 & 5 — Decision and conclusion. Since χ²calc = 18.182 > 3.841 = χ²crit (p-value < 0.05), reject H₀. There is sufficient evidence at the 5% level that department and remote-work preference are dependent (related).
Key terms
- Contingency (cross-classification) table
- An r×c table of counts formed by cross-classifying observations on two categorical variables; its row, column and grand totals are the margins that drive the expected counts.
- Observed frequency (fₒ)
- The actual count in a cell of the contingency table — a frequency, not a percentage.
- Expected frequency (fₑ)
- The count expected in a cell if the two variables were independent: fₑ = (row total × column total) / n. The test is valid only when every fₑ ≥ 5.
- Chi-square statistic (χ²)
- χ² = Σ (fₒ − fₑ)² / fₑ over all cells. It is never negative; larger values mean the observed counts are further from what independence predicts.
- Degrees of freedom (df)
- For a test of independence, df = (r − 1)(c − 1), where r and c are the numbers of rows and columns. It fixes which chi-square curve and critical value to use.
- Upper-tailed test
- A test whose rejection region is entirely in the right tail. Because χ² is never negative, the whole significance level α sits in one (upper) tail — there is no α/2.
- Null / alternative hypothesis
- H₀: the two variables are independent (no relationship); H₁: the two variables are dependent (there is a relationship). Expected counts are built assuming H₀ is true.
- Type I error
- Rejecting H₀ when it is actually true — here, concluding the variables are related when in truth they are independent. Its probability is the significance level α.
Chi-Square Test for Independence FAQ
Why is the chi-square test of independence always upper-tailed, and how many degrees of freedom does it have?
The statistic χ² = Σ(fₒ−fₑ)²/fₑ is a sum of squared, non-negative terms, so it can never be negative and evidence against independence only ever pushes it larger. That means the whole significance level α sits in the right (upper) tail — you reject H₀ when χ²calc exceeds the critical value, and you never split α into two tails. The degrees of freedom come from the table shape: df = (r−1)(c−1), so a 2×2 table has df = 1 and a 3×2 table has df = 2 — it does not depend on the sample size n.
What do I do if some expected frequencies are less than 5?
The chi-square distribution is only an approximation, and it becomes unreliable when an expected count fₑ falls below 5 (check the expected counts, not the observed ones). The standard fix is to combine adjacent or related categories so the merged cells each clear 5, then recompute the expected frequencies, the statistic and the degrees of freedom for the smaller table. Always state that you checked the fₑ ≥ 5 condition — it is an easy mark to earn or lose.
Can AI help me with the chi-square test of independence in ETX5900?
Yes — as a study aid. Sia (AskSia) explains the method step by step: how to set up the contingency table, why H₀ is "independent", how to compute each expected frequency as (row × column) / n, how to build χ² = Σ(fₒ−fₑ)²/fₑ, how to find df = (r−1)(c−1), and how to reach the reject / fail-to-reject decision against the critical value or p-value. It also generates fresh practice tables and checks your own working step by step. It never hands over answers, sits an assessment in your place, or guarantees a grade — always confirm the exam format and academic-integrity rules on Moodle.
Exam move
Treat the chi-square test of independence as a five-step drill you can reproduce under pressure: (1) state H₀ independent vs H₁ dependent, (2) compute every expected frequency fₑ = (row×col)/n and check each is ≥ 5, (3) build χ² = Σ(fₒ−fₑ)²/fₑ, (4) find df = (r−1)(c−1) and read the critical value from the chi-square table in the upper tail, and (5) compare, decide, and write the verdict in context. The exam provides a formula sheet and the statistical tables, so marks come from choosing this test (both variables categorical), laying out all five steps, and reading the table correctly — not from memorising formulae. Since the exam duration is not stated in the unit materials, plan by marks rather than minutes: spend time in proportion to each question's marks and leave a slice to check the fₑ ≥ 5 condition and that your decision matches the comparison. Confirm the exam length and format on Moodle.
Working through Chi-Square Test for Independence in ETX5900? Sia is AskSia’s AI Statistics tutor — ask any ETX5900 Chi-Square Test for Independence question and get a clear, step-by-step explanation grounded in how ETX5900 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
