Chi-Square Test: Contingency Tables

Chi-Square Test: Contingency Tables

• Used to determine whether there is an association between two categorical variables.
  • Example: Personality (Introvert, Extrovert) and Colour Preference (Red, Yellow, Green, Blue).

Application

• The Chi-Square test may be used to investigate the association between personality and colour preference.
• Note: The Chi-Square test may be used for ordinal data, but the test will treat the ordinal data as categorical. In R, it is possible to modify the Chi-Square test using the linear-by-linear option to ensure the order is taken into account.

Hypotheses

• Null Hypothesis ($H_0$): There is no association between the variables.
• Alternative Hypothesis ($H_1$): There is an association between the variables.
• The method is based on comparing observed frequencies with the frequencies you would expect to get by chance.

Test Statistic

• For a table with $r$ rows and $c$ columns, the Chi-Square statistic is calculated as:
  $x^2 = \sum<em>{i=1}^{r} \sum</em>{j=1}^{c} \frac{(O<em>{ij} - E</em>{ij})^2}{E_{ij}}$
  where:

  • $O_{ij}$ represents the observed frequency.
  • $E_{ij}$ represents the expected frequency.

• $x^2$ approximately follows a $\chi^2$ distribution with $(r-1)(c-1)$ degrees of freedom.

Expected Frequencies

• The expected frequency $E<em>{ij}$ is calculated as: $E</em>{ij} = \frac{Y<em>{i.} \times Y</em>{.j}}{n}$ where:
  • $Y_{i.}$ gives the row totals.
  • $Y_{.j}$ gives the column totals.
  • $n$ is the total number of observations.

Evaluation

• We evaluate $x^2$ using tables of $\chi^2$ distribution with $(r-1)(c-1)$ degrees of freedom.

Yates' Continuity Correction

• For 2x2 frequency tables (where degrees of freedom, $df = 1 = (r-1)(c-1)$), the Chi-Square test produces overly significant results (rejecting $H_0$ when it is true).
• In such cases, we apply Yates' Continuity Correction to the test statistic:
  $x^2<em>{\text{corrected}} = \sum</em>{i=1}^{r} \sum<em>{j=1}^{c} \frac{(|O</em>{ij} - E<em>{ij}| - 0.5)^2}{E</em>{ij}}$
• Yates' continuity correction is also applied to the $x^2$ goodness-of-fit test when $df = K-1 = 1$.

Effect Size: Strength of Association

• Chi-Square tests do not tell us how strong an association is; therefore, consider effect size measures.

  
  
  
  

  • Phi Coefficient ($\phi$):

• Used for 2x2 tables only.
  $\phi = \sqrt{\frac{x^2}{n}}$
• Guidelines:
  • Small: 0.1
  • Medium: 0.3
  • Large: 0.5

• Can be used with 2 categorical variables when each variable has 2 or more categories.
  $V = \sqrt{\frac{x^2}{n \times df<em>v}}$ where $df</em>v = \min(c-1, r-1)$

Evaluation of Odds Ratio

• OR = 1: Belonging to Group 1 has not affected the odds of Outcome A.
• OR > 1: Belonging to Group 1 has increased the odds of Outcome A.
• OR < 1: Belonging to Group 1 has decreased the odds of Outcome A.

Post Hoc Tests

• If $x^2$ is significant ⇒ association between the variables, but it does not provide any specific information about the association.
• In R, we can use post hoc tests to investigate further. We will use the standardised residuals approach.

The Likelihood Ratio

• An alternative to the $x^2$ test uses a Model based on Maximum-likelihood theory.
  L{x^2} = 2 \left{ \sum{i=1}^{r} \sum{j=1}^{c} y{ij} \ln \left( \frac{y{ij}}{E{ij}} \right) \right}

Evaluation of Likelihood Ratio

• Evaluate $L_{x^2}$ in the same way as $x^2$.
• Example: Following Example 3.6, find $L_{x^2} = 0.42$.
• Again, $\chi^2<em>{0.05, 1} = 3.841 > 0.42 = L</em>{x^2}$. Do not reject \H_0 ⇒ there does not appear to be an association between Education Levels and Department.

Violation of the expected frequency assumption

Fisher's Exact Test