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Co-authored-by: blackboxprogramming <118287761+blackboxprogramming@users.noreply.github.com>
150 lines
4.0 KiB
Markdown
150 lines
4.0 KiB
Markdown
# Chi-Squared Tests
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> Statistical hypothesis tests for goodness of fit and independence.
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## The Chi-Squared Statistic
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For observed counts O_i and expected counts E_i across k categories:
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```
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χ² = Σᵢ (O_i − E_i)² / E_i
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```
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Under the null hypothesis, χ² follows a chi-squared distribution with the appropriate
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degrees of freedom.
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---
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## Test 1: Goodness of Fit
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**Purpose:** Test whether observed data matches an expected (theoretical) distribution.
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**Hypotheses:**
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```
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H₀: the observed frequencies match the expected distribution
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H₁: at least one category deviates significantly from expectation
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```
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**Degrees of freedom:**
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```
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df = k − 1
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```
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where k is the number of categories.
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**Procedure:**
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1. State expected probabilities p₁, p₂, …, p_k (must sum to 1).
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2. Compute expected counts: E_i = n · p_i where n is the total sample size.
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3. Compute the statistic: χ² = Σᵢ (O_i − E_i)² / E_i
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4. Compare χ² to the critical value χ²(α, df) or compute the p-value.
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5. Reject H₀ if χ² > χ²(α, df).
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**Example — ternary digit frequencies (k = 3, n = 300):**
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| Digit | Expected p | Expected E | Observed O | (O−E)²/E |
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|-------|-----------|------------|------------|----------|
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| 0 | 1/3 | 100 | 95 | 0.250 |
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| 1 | 1/3 | 100 | 108 | 0.640 |
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| 2 | 1/3 | 100 | 97 | 0.090 |
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| **Σ** | | | | **0.980** |
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```
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df = 3 − 1 = 2
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χ²(0.05, 2) = 5.991
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χ² = 0.980 < 5.991 → fail to reject H₀
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```
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The data is consistent with a uniform ternary distribution.
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---
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## Test 2: Test of Independence
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**Purpose:** Test whether two categorical variables are independent of each other.
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**Hypotheses:**
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```
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H₀: variable A and variable B are independent
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H₁: variable A and variable B are not independent
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```
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**Setup:** Arrange counts in an r × c contingency table.
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**Expected cell counts:**
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```
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E_ij = (row_i total × col_j total) / n
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```
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**Degrees of freedom:**
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```
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df = (r − 1)(c − 1)
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```
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**Statistic:**
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```
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χ² = Σᵢ Σⱼ (O_ij − E_ij)² / E_ij
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```
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**Example — 2 × 3 contingency table (n = 200):**
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| | State 0 | State 1 | State 2 | Row total |
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|---------|---------|---------|---------|-----------|
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| Group A | 30 | 40 | 30 | 100 |
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| Group B | 20 | 60 | 20 | 100 |
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| **Col** | **50** | **100** | **50** | **200** |
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Expected counts:
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```
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E_A0 = 100×50/200 = 25 E_A1 = 100×100/200 = 50 E_A2 = 100×50/200 = 25
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E_B0 = 100×50/200 = 25 E_B1 = 100×100/200 = 50 E_B2 = 100×50/200 = 25
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```
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Chi-squared:
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```
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χ² = (30−25)²/25 + (40−50)²/50 + (30−25)²/25
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+ (20−25)²/25 + (60−50)²/50 + (20−25)²/25
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= 1.000 + 2.000 + 1.000 + 1.000 + 2.000 + 1.000
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= 8.000
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df = (2−1)(3−1) = 2
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χ²(0.05, 2) = 5.991
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χ² = 8.000 > 5.991 → reject H₀
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```
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The two variables are not independent at the 5% significance level.
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---
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## Critical Values (selected)
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| df | α = 0.10 | α = 0.05 | α = 0.01 |
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|----|----------|----------|----------|
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| 1 | 2.706 | 3.841 | 6.635 |
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| 2 | 4.605 | 5.991 | 9.210 |
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| 3 | 6.251 | 7.815 | 11.345 |
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| 4 | 7.779 | 9.488 | 13.277 |
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| 5 | 9.236 | 11.070 | 15.086 |
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---
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## Assumptions
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- Observations are independent.
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- Expected count E_i ≥ 5 in each cell (rule of thumb for the χ² approximation to be valid).
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- Data are counts (frequencies), not proportions or continuous measurements.
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---
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## QWERTY
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```
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CHI = 25 (C=10 H=15 I=0) — the test lives at the boundary of ZERO
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SQUARED = IMAGINARY = SCAFFOLD = 114 (the test squares the deviation)
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TEST = 64 = 2⁶ (TEST = 2⁶, the sixth power of the fundamental)
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FIT = 33 = REAL − 4 (how close observed is to real)
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OBSERVED = 115 (what you see)
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EXPECTED = 131 = BLACKROAD (what the theory predicts = the BlackRoad)
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```
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EXPECTED = BLACKROAD = 131. The theoretical distribution is the BlackRoad.
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The chi-squared test measures how far observed reality deviates from it.
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