WK11: Statistical inference: Two categorical variables: Chi-square test for goodness of fit

Chi-Square Goodness of Fit Test

• Used to test if a single categorical variable has a specific distribution.
• Follows the standard hypothesis testing steps:
  • State null and alternative hypotheses & level of significance.
  • Check conditions.
  • Calculate test statistic.
  • Find p-value.
  • Make a decision and conclude in context.

Hypotheses

• Distribution of a Categorical Variable: Lists categories and their proportions.
• If there are K categories: $P<em>1, P</em>2, …, P_K$ represent the proportions for each category.
• $\sum<em>{i=1}^{K} P</em>i = 1$ (sum of proportions equals 1).
• Null Hypothesis ($H_0$): Specifies proportions for all categories.
  • $P<em>1 = P</em>{10}, P<em>2 = P</em>{20}, …, P<em>K = P</em>{K0}$
• Alternative Hypothesis ($H_a$): At least one proportion is different from what the null hypothesis states.
  • Not all $P<em>i$ are equal to $P</em>{i0}$.

Example: Births and Days of the Week

• Claim: Births are not evenly distributed across days of the week.
• Null Hypothesis ($H_0$): Births are equally likely on all days of the week.
  • $P<em>1 = P</em>2 = … = P_7 = \frac{1}{7}$
• Alternative Hypothesis ($H_a$): Births are not equally likely on all days of the week.
  • Not all $P_i$ are equal to $\frac{1}{7}$.
  • This doesn't specify which days have different proportions.

Conditions for Chi-Square Goodness of Fit Test

• Random Sample.
• Expected counts under the null hypothesis must be:
  • At least 1 for each cell.
  • At least 80% of cells should have expected counts of at least 5.
• Expected Counts Calculation: Multiply the proportion specified in the null hypothesis by the total sample size.

Example: Births and Days of the Week (Conditions)

• Sample size = 700
• Expected count for each day = $\frac{1}{7} \times 700 = 100$
• Since 100 > 5 for all days, the conditions are met.

Test Statistic

• Chi-Square Statistic: Measures the difference between observed and expected counts.
• Formula: $\chi^2 = \sum<em>{i=1}^{K} \frac{(O</em>i - E<em>i)^2}{E</em>i}$
  • $O_i$ = Observed count for category i
  • $E_i$ = Expected count for category i
• Degrees of Freedom: Number of categories minus one (K - 1).

Example: Births and Days of the Week (Test Statistic)

• $\chi^2 = 19.12$
• Degrees of freedom = 7 - 1 = 6

P-Value and Conclusion

• P-value: Probability of observing a test statistic as extreme or more extreme than the one calculated, assuming the null hypothesis is true.
• If p-value < level of significance ($\alpha$): Reject the null hypothesis.
  • The sample gives statistically significant evidence supporting the alternative.
• If p-value > level of significance ($\alpha$): Fail to reject the null hypothesis.
  • The sample does not give statistically significant evidence to support the alternative.

Example: Births and Days of the Week (Conclusion)

• P-value is between 0.0025 and 0.005.
• Level of significance ($\alpha$) = 0.05
• Since p-value < 0.05, reject the null hypothesis.
• Conclusion: At the 5% level of significance, the data gives statistically significant evidence that local births are not equally likely on all days of the week.

Equivalence with Test for Proportion of Successes

• When a categorical variable has only two categories, the chi-square goodness-of-fit test is equivalent to the test for the proportion of successes.
• Both tests will provide the same decision and conclusion.

Example: Spinning a Coin

• Data: 168 heads, 232 tails in 400 spins.
• Test if this contradicts a 50/50 distribution at \alpha = 0.05.

Using Proportion Test

• Success = Landing heads.
• $H_0: P = 0.5$
• $H_a: P \neq 0.5$
• Sample proportion: $\hat{p} = \frac{168}{400} = 0.42$
• Test statistic: $Z = \frac{\hat{p} - P}{\sqrt{\frac{P(1-P)}{n}}} = \frac{0.42 - 0.5}{\sqrt{\frac{0.5(1-0.5)}{400}}} = -3.2$
• P-value: 2 * 0.0007 = 0.0014
• Conclusion: P-value < 0.05, so reject the null hypothesis. The sample gives statistically significant evidence contradicting the 50/50 distribution.

Using Chi-Square Goodness of Fit Test

• Categories: Heads (H) and Tails (T)
• $H<em>0: P</em>H = 0.5, P_T = 0.5$
• $H_a$: At least one is different from 0.5.
• Expected counts: 200 for each category
• Test statistic: $\chi^2 = \sum \frac{(O<em>i - E</em>i)^2}{E_i} = \frac{(168-200)^2}{200} + \frac{(232-200)^2}{200}= 10.24$
• Degrees of freedom: 2 - 1 = 1
• P-value: Between 0.001 and 0.002
• Conclusion: P-value < 0.05, so reject the null hypothesis. The sample gives statistically significant evidence that $P_H \neq 0.5$, which contradicts the 50/50 distribution.

Relationship between Z and Chi-Square

• For two categorical variables with two categories each, $\chi^2 = Z^2$.