Chi square test - 15 marks MPT level answer of this question
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Chi-Square Test (χ²)
Definition
The Chi-square (χ²) test is a non-parametric statistical test used to determine whether there is a significant association between two categorical variables, or whether an observed frequency distribution differs significantly from a theoretically expected distribution. It was introduced by Karl Pearson in 1900.
Types of Chi-Square Tests
| Type | Purpose |
|---|---|
| 1. Chi-square test of independence (association) | Tests association between two categorical variables |
| 2. Chi-square test of goodness of fit | Tests whether observed frequencies fit an expected distribution |
| 3. Yates' correction (corrected χ²) | Applied to 2×2 contingency tables when any expected frequency is 5-40 |
| 4. Mantel-Haenszel χ² | Tests association after stratification for confounders |
Assumptions / Pre-requisites
- Data must be in frequencies (counts), not percentages or means
- Sample size: Total n should be ≥ 40
- Expected frequency: Each cell's expected frequency should be ≥ 5 (in a 2×2 table)
- In larger tables (r×c), not more than 20% of cells should have expected frequency < 5, and no cell should have expected frequency < 1
- Observations must be independent
- Random sampling from the population
- Data should be nominal or ordinal (categorical)
If expected frequency in any cell of a 2×2 table is < 5 - use Fisher's Exact Test If n is < 40 - use Fisher's Exact Test If expected frequency is 5–40 - use Yates' correction If expected frequency is > 40 - use Standard χ² test
Formula
$$\chi^2 = \sum \frac{(O - E)^2}{E}$$
Where:
- O = Observed frequency
- E = Expected frequency
- Σ = Sum over all cells
Expected Frequency Calculation:
$$E = \frac{\text{Row total} \times \text{Column total}}{\text{Grand total}}$$
Yates' Correction for Continuity (2×2 table)
When expected frequency in any cell is between 5 and 40:
$$\chi^2_{corrected} = \sum \frac{(|O - E| - 0.5)^2}{E}$$
Steps in Calculation
Step 1: State the Null hypothesis (H₀): There is no significant association between the two variables.
Step 2: Prepare the contingency table with observed frequencies.
Step 3: Calculate expected frequencies for each cell using: E = (Row total × Column total) / Grand total
Step 4: Apply the formula: χ² = Σ(O-E)²/E
Step 5: Calculate degrees of freedom (df): df = (r-1)(c-1), where r = rows, c = columns
Step 6: Find the tabulated/critical value of χ² at the desired significance level (usually p = 0.05) from χ² distribution table
Step 7: Compare calculated χ² with tabulated χ²
- If calculated χ² > tabulated χ² → Reject H₀ → Association is statistically significant
- If calculated χ² < tabulated χ² → Accept H₀ → Association is NOT significant
Worked Example
Problem: In a study, 200 individuals were assessed for smoking habit and lung cancer:
| Lung Cancer (+) | Lung Cancer (-) | Total | |
|---|---|---|---|
| Smoker | 40 | 60 | 100 |
| Non-smoker | 10 | 90 | 100 |
| Total | 50 | 150 | 200 |
Step 1 - H₀: No association between smoking and lung cancer.
Step 2 - Calculate Expected Frequencies:
| Cell | Formula | E |
|---|---|---|
| Smoker + Cancer | (100×50)/200 | 25 |
| Smoker + No Cancer | (100×150)/200 | 75 |
| Non-smoker + Cancer | (100×50)/200 | 25 |
| Non-smoker + No Cancer | (100×150)/200 | 75 |
All expected frequencies > 40, so standard χ² applies.
Step 3 - Calculate χ²:
| Cell | O | E | (O-E) | (O-E)² | (O-E)²/E |
|---|---|---|---|---|---|
| a | 40 | 25 | 15 | 225 | 9.0 |
| b | 60 | 75 | -15 | 225 | 3.0 |
| c | 10 | 25 | -15 | 225 | 9.0 |
| d | 90 | 75 | 15 | 225 | 3.0 |
χ² = 9.0 + 3.0 + 9.0 + 3.0 = 24.0
Step 4 - Degrees of freedom: df = (2-1)(2-1) = 1
Step 5 - Tabulated value: χ² at df=1, p=0.05 = 3.84
Step 6 - Conclusion: Calculated χ² (24.0) > Tabulated χ² (3.84)
→ Reject H₀. There is a statistically significant association between smoking and lung cancer (p < 0.05).
Degrees of Freedom
| Table | df |
|---|---|
| 2×2 | 1 |
| 2×3 | 2 |
| 3×3 | 4 |
| r×c | (r-1)(c-1) |
Shortcut Formula for 2×2 Table
For a 2×2 contingency table (a, b, c, d):
$$\chi^2 = \frac{N(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}$$
χ² Critical Values Table (commonly used)
| df | p = 0.05 | p = 0.01 |
|---|---|---|
| 1 | 3.84 | 6.64 |
| 2 | 5.99 | 9.21 |
| 3 | 7.81 | 11.34 |
| 4 | 9.49 | 13.28 |
Applications in Community Medicine / Public Health
- Case-control studies - Association between exposure and disease
- Cross-sectional surveys - Association between variables (e.g., diet and BMI category)
- Cohort studies - Comparing proportions of outcomes
- Vaccine efficacy studies - Comparing attack rates between vaccinated and unvaccinated
- Audit and evaluation - Whether outcome proportions differ between groups
- Goodness of fit - Checking if a distribution follows Mendelian ratios (genetics)
Advantages
- Simple to calculate
- Applicable to nominal and ordinal data
- Does not assume normal distribution (non-parametric)
- Applicable to both small and large samples (with modifications)
Limitations
- Cannot be used with continuous data directly
- Sensitive to small expected frequencies
- Does not indicate strength of association (use Cramer's V or Phi coefficient for that)
- Does not indicate direction of association
- Not applicable when expected frequency < 5 in any cell of a 2×2 table
Alternatives When χ² Cannot Be Applied
| Situation | Alternative Test |
|---|---|
| n < 40 OR any E < 5 in 2×2 table | Fisher's Exact Test |
| 5 ≤ E ≤ 40 in 2×2 table | Yates' corrected χ² |
| Paired proportions | McNemar's test |
| Ordinal data, trend | Mantel-Haenszel test |
Summary Points to Remember
- χ² is used for categorical data
- Formula: Σ(O-E)²/E
- df = (r-1)(c-1)
- Minimum expected frequency ≥ 5 in 2×2 table
- If calculated χ² > table value → significant association
- Introduced by Karl Pearson, 1900
- For 2×2 table with E between 5-40: use Yates' correction
- For 2×2 table with E < 5 or n < 40: use Fisher's Exact Test
Explain me in simple language
Chi-Square Test - Simple Explanation 😊
What is it in ONE line?
"Chi-square test checks if two things are RELATED or just happened by CHANCE."
Real Life Analogy
Imagine you're a doctor and you want to know:
"Do smokers get lung cancer MORE than non-smokers?"
You collect data from 200 people. Now you have two questions:
- Is the difference you see REAL?
- Or did it happen just by LUCK/CHANCE?
Chi-square test answers exactly this.
The Core Idea - Super Simple
You compare two things:
| What you ACTUALLY saw | What you EXPECTED to see (if no relation) | |
|---|---|---|
| Name | Observed (O) | Expected (E) |
- If O and E are very close → No relationship → Just chance
- If O and E are very different → There IS a relationship!
The Formula (Don't fear it!)
$$\chi^2 = \sum \frac{(O - E)^2}{E}$$
In plain English:
- For each box in your table: subtract Expected from Observed
- Square it (to make negatives positive)
- Divide by Expected
- Add all of them up
Bigger the number = Bigger the difference = More likely there's a real association
Step-by-Step With a Story
Story: You want to know - "Is smoking linked to lung cancer?"
You study 200 people:
Got Cancer No Cancer TOTAL
Smokers 40 60 100
Non-smokers 10 90 100
TOTAL 50 150 200
Step 1 - Ask: "What SHOULD the numbers be if smoking had NO effect?"
Use this simple formula:
$$E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}$$
For Smokers + Cancer:
$$E = \frac{100 \times 50}{200} = 25$$
So if smoking had NO effect, only 25 smokers should get cancer. But we saw 40! That's a BIG difference.
Step 2 - Calculate χ²
| O | E | (O-E)² / E | |
|---|---|---|---|
| Smoker + Cancer | 40 | 25 | 9.0 |
| Smoker + No Cancer | 60 | 75 | 3.0 |
| Non-smoker + Cancer | 10 | 25 | 9.0 |
| Non-smoker + No Cancer | 90 | 75 | 3.0 |
χ² = 9 + 3 + 9 + 3 = 24
Step 3 - Compare with a "cut-off value"
- The standard cut-off (at p=0.05, df=1) is 3.84
- Our value = 24
- 24 > 3.84 → YES! There is a real association!
The Decision Rule (Simplest Version)
Calculated χ² > Table value (3.84)
↓
SIGNIFICANT
(Real relationship exists)
Calculated χ² < Table value (3.84)
↓
NOT SIGNIFICANT
(Could be just chance)
When Can You Use It?
✅ Data is in counts/numbers (not averages)
✅ You are comparing groups (smoker vs non-smoker, male vs female)
✅ Each person fits in only ONE box
✅ Sample size is at least 40
✅ Every box has at least 5 expected cases
When Can You NOT Use It?
| Problem | Use This Instead |
|---|---|
| Sample < 40, or any box has < 5 expected | Fisher's Exact Test |
| Box has 5-40 expected | Yates' Corrected χ² |
| Same person measured twice | McNemar's Test |
One Trick to Remember the Formula
Think of it as measuring "surprise":
- How surprised are you that O ≠ E?
- The more surprised = The bigger the χ² value
- Big χ² = Something real is going on!
Summary in 5 Points
- Chi-square = comparison of Observed vs Expected frequencies
- Formula = Σ(O-E)²/E
- df = (rows - 1) × (columns - 1)
- If χ² calculated > 3.84 (at df=1) → significant association
- Works only with categorical data and sample ≥ 40 with all expected frequencies ≥ 5
That's it! The whole test is just asking: "Is what I saw different enough from what I expected?" If yes - the association is real. If no - it's just chance.
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