Chi-Square Test of Independence

The chi-square test of independence (also called a chi-square test of association) determines whether two categorical variables are statistically independent of each other.

In Six Sigma, this is extremely useful for analyzing contingency tables. For example, you might want to see if defect occurrence is independent of the shift, or if customer satisfaction ratings are independent of location. It helps answer questions like, “Are these two factors related, or are any differences just random variation?”

When to use: Use a chi-square test of independence in the Analyze phase when you have counts of outcomes classified by two categorical factors and you suspect there may be a relationship. For instance, you collected data on defects categorized by machine and defect type – is the pattern of defect types the same for each machine (independence), or does one machine have disproportionately more of certain defects (dependence)? Another example: a call center manager might check if call resolution outcome is independent of the agent’s experience level. Essentially, whenever you have an R×C table of counts (R rows = categories of variable1, C columns = categories of variable2), this test can tell you if there’s an association.

Hypotheses:

• H₀: The two categorical variables are independent (no association). In other words, the distribution of one variable is the same for all categories of the other variable.
• H₁: The two variables are not independent (there is an association). At least some categories of one variable are linked to differences in the distribution of the other.

Assumptions: Similar to the GOF test, the observations should be independent of each other, and sample size should be sufficiently large (expected counts ideally ≥ 5 in each cell of the table, or at least in the vast majority of cells). If any cell has an expected count too low, results may not be reliable (you might combine sparse categories or use Fisher’s exact test for a 2×2 table if necessary).

The data for an independence test are typically arranged in a contingency tablesixsigmastudyguide.com. Each cell of the table contains the count of observations that fall into the corresponding row category and column category combination. For example, rows might be “Machine A vs Machine B” and columns “Defect Type X, Y, Z,” with each cell showing how many defects of each type occurred on each machine.

Steps to Perform a Chi-Square Test of Independence

1. Set up hypotheses:
   • Null (H₀): The two variables are independent (no association between them).
   • Alternative (H₁): The two variables are associated (the distribution of one depends on the other).
2. Choose significance level (α): Often 0.05.
3. Create the contingency table of observed counts: Organize your data into a table of r rows and c columns. Calculate the row totals, column totals, and the grand total n. These totals are needed to compute expected counts.
4. Calculate expected counts for each cell: If H₀ is true (independence), the expected frequency in a cell is:
   Eij=(Row i total)×(Column j total)n.E_{ij} = frac{(text{Row i total}) times (text{Column j total})}{n}.Eij​=n(Row i total)×(Column j total)​.
   This comes from probability theory – if events are independent, P(row i and col j) = P(row i)*P(col j), so expected count = (row proportion * column proportion * n). You do this for every cellsixsigmastudyguide.com. For example, if in a survey 60% of responses were from Region A (row) and 30% preferred Product X (column), then under independence you’d expect 0.6 * 0.3 = 18% of total respondents to fall in (Region A, Product X) cell.
5. Compute the chi-square statistic: Sum over all cells:
   χ2=∑i=1r∑j=1c(Oij−Eij)2Eij.chi^2 = sum_{i=1}^{r}sum_{j=1}^{c} frac{(O_{ij} – E_{ij})^2}{E_{ij}}.χ2=∑i=1r​∑j=1c​Eij​(Oij​−Eij​)2​.
   Each cell’s contribution is calculated similarly to the GOF test, comparing observed vs expectedsimplilearn.comsimplilearn.com.
6. Degrees of freedom: df = (number of rows – 1) × (number of columns – 1)simplilearn.com. This formula reflects that once you know all the row and column totals, only that many cells are free to vary independently; the rest are determined by totals.
7. Find the critical value (right-tailed): Use the χ² table for the given df at your α. This test is right-tailed – large χ² means observed counts diverge from expected (indicating dependence). Alternatively, calculate the p-value corresponding to the χ² statistic and df.
8. Conclusion: If χ²_stat > χ²_critical (p < α), reject H₀. Conclude there is a significant association between the two variables – the differences in the contingency table are unlikely due to chance. If χ²_stat is not beyond critical (p ≥ α), fail to reject H₀, meaning you do not have evidence of an association (the variables could be independent).

Independence Test Example: Process Defects by Shift

As a Six Sigma example, suppose a manufacturing team wants to know if the occurrence of defect types is independent of the production shift (Day shift vs. Night shift). They categorize defects into three types: Type A, Type B, Type C. Over a month, they collect the following data:

• Day Shift: 16 defects of Type A, 4 of Type B, 10 of Type C (Total = 30 defects in day shift).
• Night Shift: 8 defects of Type A, 12 of Type B, 10 of Type C (Total = 30 defects in night shift).
  (Total defects observed = 60.)

The contingency table of observed counts would be:

We’ll test at 95% confidence (α = 0.05) if shift and defect type are independent.

Step 1 – Hypotheses:
H₀: Defect type is independent of shift (no relationship between shift and defect distribution).
H₁: Defect type depends on shift (the pattern of defect types differs between day and night shifts).

Step 2 – Significance: α = 0.05.

Step 3 – Observed data: (as in the table above).

Step 4 – Expected counts: Compute for each cell using E = (row total * col total) / grand total. For example:

• Expected in Day Shift, Type A = (Day total 30 * Type A total 24) / 60 = 12.
• Expected in Day, Type B = (30 * 16) / 60 = 8.
• Expected in Day, Type C = (30 * 20) / 60 = 10.
• Expected in Night, Type A = (30 * 24) / 60 = 12.
• Expected in Night, Type B = (30 * 16) / 60 = 8.
• Expected in Night, Type C = (30 * 20) / 60 = 10.

So the expected table (under H₀) would be:

Step 5 – χ² calculation: Sum (O−E)²/E for all 6 cells:

• Day/A: (16−12)²/12 = (4)²/12 = 16/12 = 1.33
• Day/B: (4−8)²/8 = (−4)²/8 = 16/8 = 2.00
• Day/C: (10−10)²/10 = 0
• Night/A: (8−12)²/12 = (−4)²/12 = 16/12 = 1.33
• Night/B: (12−8)²/8 = (4)²/8 = 16/8 = 2.00
• Night/C: (10−10)²/10 = 0

χ²_stat = 1.33 + 2.00 + 0 + 1.33 + 2.00 + 0 = 6.67 (approximately).

Step 6 – Degrees of freedom: df = (2–1)*(3–1) = 1 * 2 = 2.

Step 7 – Critical value: For df = 2 and α = 0.05 (right-tailed), χ²_critical = 5.991sixsigmastudyguide.com.

Step 8 – Conclusion: Our χ²_stat = 6.67 which is greater than 5.991. That means p-value < 0.05. We reject H₀ and conclude there is a significant association between shift and defect type. In practical terms, the mix of defect types is different on day vs. night shift. Looking at the observed vs expected, Day shift had more Type A than expected and fewer Type B, whereas Night shift had more Type B and fewer Type A than expected. This insight suggests that shift (perhaps the conditions or personnel on each shift) has an impact on what kinds of defects occur. That information can guide further root cause analysis (maybe day shift has an issue causing Type A defects, while night shift struggles with Type B).

(Note: If our χ²_stat had been less than 5.991, we’d conclude there’s no evidence of a difference – the defect pattern could be essentially the same on both shifts.)

This example demonstrates how the chi-square test of independence helps identify relationships between categorical factors in a process. In a Six Sigma project, such a finding would prompt investigation into why the association exists (e.g., do day and night shifts follow different procedures or have different skill levels leading to certain defects?). It’s a powerful Analyze phase tool for validating or disproving suspected factor-effect relationships.

(In this test, as with GOF, we used the right tail of the χ² distribution. A large χ² (here ~6.67) beyond the cutoff signaled a significant difference. Independence tests are always right-tailed because only greater divergence (not a deficiency) indicates dependence.)

A Second Example: When the Test Says “No Association”

The example above found a real association (we rejected H₀). It’s just as important to see what it looks like when the test says the opposite: the variables really do look independent. Exam questions test both outcomes, not just the “reject” case.

Suppose a plant wants to know whether meeting spec is independent of which of two production lines made the part. Over a week, they collect:

H₀: Meeting spec is independent of production line. H₁: Meeting spec depends on which line made the part. α = 0.05.

Expected counts (row total × column total ÷ grand total): Line 1/Meets = (100×156)/200 = 78. Line 1/Not = (100×44)/200 = 22. Line 2/Meets = 78. Line 2/Not = 22. All four expected counts are comfortably above the minimum of 5 needed for this test to be reliable.

Chi-square statistic: (76−78)²/78 + (24−22)²/22 + (80−78)²/78 + (20−22)²/22 = 0.05 + 0.18 + 0.05 + 0.18 ≈ 0.47.

Degrees of freedom: (2−1)×(2−1) = 1. Critical value at df = 1, α = 0.05: 3.841.

Conclusion: 0.47 is far below 3.841, so we fail to reject H₀. There’s no evidence that meeting spec depends on which line made the part; the small differences between the two lines are consistent with normal random variation. Notice this is a 2×2 table, the most common shape you’ll actually see on an exam question.

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