Chi Square test for homogeneity and Independence

Homogeneity

1. State (Hypothesis Formulation)

• Null Hypothesis ($H_0$): The distributions of the categorical variable are the same across all populations/groups.
• Alternative Hypothesis ($H_a$): The distributions of the categorical variable are not the same across all populations/groups (i.e., at least two groups differ).
• Mathematical Representation:
  • $H_0$: $P_{1,1}=P_{1,2}=...=P_{1,n}$
  • $H_a$: At least two of the probabilities are different.

---

2. Check Conditions

To ensure the validity of the Chi-Square test for homogeneity, we must satisfy these conditions:

1. Randomness: The data must come from independent random samples or a properly randomized experiment.
2. Large Sample Size: Each expected count should be at least 5 in all categories.
3. Independence:
   • Each observation must be independent.
   • If sampling without replacement, the total population must be at least 10 times the sample size (10% condition).

---

3. Calculation

The Chi-Square statistic is calculated using:

$${\chi }^2=\sum _{i=1}^r\sum _{j=1}^c\frac{({\text{Observed}}_{ij}-{\text{Expected}}_{ij}{)}^2}{{\text{Expected}}_{ij}}$$

Where:

• $r$ is the number of rows (groups),
• $c$ is the number of columns (categories),
• Observed${}_{ij}$ represents the actual count in row $i$ and column $j$,
• Expected${}_{ij}$ is calculated as:

Expectedij=(Row Total)i×(Column Total)jGrand Total\text{Expected}_{ij} = \frac{(\text{Row Total})_i \times (\text{Column Total})_j}{\text{Grand Total}}

• Degrees of Freedom (df):

df=(Number of Rows−1)×(Number of Columns−1)df = (\text{Number of Rows} - 1) \times (\text{Number of Columns} - 1)

• P-value Calculation:
  Given ${\chi }^2$ and $df$, the p-value is computed as:

$$p_{val}=P({\chi }^2\ge X)=1-\frac{\gamma \left(\frac{df}{2},\frac{{\chi }^2}{2}\right)}{\Gamma \left(\frac{df}{2}\right)}$$

where:

• $\gamma (s,x)$ is the Lower incomplete gamma function:

$$\gamma (s,x)=\int 0xts-1e-tdt\gamma (s,x)={\int }_0^x{t}^{s-1}{e}^{-t}dt$$

• $\Gamma (s)$ is the Gamma function:

$$\Gamma (s)=\int 0\infty ts-1e-tdt\Gamma (s)={\int }_0^{\infty }{t}^{s-1}{e}^{-t}dt$$

• The Regularized Incomplete Gamma Function:

$$\gamma (s,x)\Gamma (s)\frac{\gamma (s,x)}{\Gamma (s)}$$

which is also known as $P(s,x)$ and implemented in scipy.special.gammainc().

Using scipy.special.gammainc for p-value computation:

import scipy.special as sp
 
def chi2_p_value(chi2_val, df):
    p_val = 1 - sp.gammainc(df / 2, chi2_val / 2)
    return p_val
 
# Example usage
chi2_val = 15.2  # Chi-Square statistic
df = 6           # Degrees of freedom
p_value = chi2_p_value(chi2_val, df)
print(f"P-value: {p_value:.5f}")

---

4. Decision Rule

• If $p_{val}\ge \alpha$ (common choices: $\alpha =0.05,0.01$):
  • Fail to reject $H_0$ → The distributions across groups are not significantly different.
• If $p_{val}\le \alpha$:
  • Reject $H_0$ → At least two groups have significantly different distributions.

Additional Step:
If we reject $H_0$, we can analyze which categories contribute most to the differences by examining:

$$(Observedij-Expectedij)2Expectedij\frac{({\text{Observed}}_{ij}-{\text{Expected}}_{ij}{)}^2}{{\text{Expected}}_{ij}}$$

---

5. Conclusion

Based on the test results:

• If $H_0$ is not rejected, we conclude that there is no significant difference in distributions across the groups.
• If $H_0$ is rejected, we conclude that there is sufficient statistical evidence to suggest that at least two groups have different distributions.

Final conclusion should be written in context:

We have sufficient statistical evidence that [in context].\text{We have sufficient statistical evidence that [in context].}

Independence