AP Statistics 8.6 Carrying Out a Chi-Square Test for Homogeneity or Independence Study Notes
AP Statistics 8.6 Carrying Out a Chi-Square Test for Homogeneity or Independence Study Notes- New syllabus
AP Statistics 8.6 Carrying Out a Chi-Square Test for Homogeneity or Independence Study Notes -As per latest AP Statistics Syllabus.
LEARNING OBJECTIVE
- The chi-square distribution may be used to model variation.
Key Concepts:
- Chi-Square Test Statistic for Homogeneity or Independence’
- Determine the P-Value for a Chi-Square Significance Test for Independence or Homogeneity
- Interpret the P-Value for the Chi-Square Test for Homogeneity or Independence
- Justifying a Claim Based on a Chi-Square Test for Homogeneity or Independence
Chi-Square Test Statistic for Homogeneity or Independence
Chi-Square Test Statistic for Homogeneity or Independence
The chi-square statistic measures the overall difference between observed counts (\(O_{ij}\)) and expected counts (\(E_{ij}\)) in a two-way table. It quantifies how much the observed data deviate from the counts expected if the null hypothesis were true.
Formula:
\(\displaystyle \chi^2 = \sum_{i=1}^{r} \sum_{j=1}^{c} \dfrac{(O_{ij} – E_{ij})^2}{E_{ij}}\)
- \(O_{ij}\) = observed count in row \(i\), column \(j\)
- \(E_{ij}\) = expected count in row \(i\), column \(j\)
- \(r\) = number of rows, \(c\) = number of columns
Notes:
- Expected counts: \(\displaystyle E_{ij} = \frac{(\text{row total}_i)(\text{column total}_j)}{\text{grand total}}\)
- Degrees of freedom: \(df = (r-1)(c-1)\)
- The chi-square statistic is always positive and larger values indicate greater deviation from independence or homogeneity.
Example
Observed counts of snack preference by gender among 100 students:
| Chips | Candy | Row Total | |
|---|---|---|---|
| Male | 30 | 20 | 50 |
| Female | 10 | 40 | 50 |
| Column Total | 40 | 60 | 100 |
Calculate the chi-square statistic for testing independence of snack preference and gender.
▶️ Answer / Explanation
Step 1 — Calculate expected counts:
- Male & Chips: \(E = \frac{50 \times 40}{100} = 20\)
- Male & Candy: \(E = \frac{50 \times 60}{100} = 30\)
- Female & Chips: \(E = \frac{50 \times 40}{100} = 20\)
- Female & Candy: \(E = \frac{50 \times 60}{100} = 30\)
Step 2 — Compute chi-square contributions for each cell:
- Male & Chips: \(\frac{(30-20)^2}{20} = \frac{100}{20} = 5\)
- Male & Candy: \(\frac{(20-30)^2}{30} = \frac{100}{30} \approx 3.33\)
- Female & Chips: \(\frac{(10-20)^2}{20} = \frac{100}{20} = 5\)
- Female & Candy: \(\frac{(40-30)^2}{30} = \frac{100}{30} \approx 3.33\)
Step 3 — Sum contributions:
\(\chi^2 = 5 + 3.33 + 5 + 3.33 \approx 16.66\)
Conclusion: The chi-square statistic is approximately 16.66. This value will be compared to a chi-square distribution with \(df = (2-1)(2-1) = 1\) to determine the p-value.
Determine the P-Value for a Chi-Square Significance Test for Independence or Homogeneity
Determine the P-Value for a Chi-Square Significance Test for Independence or Homogeneity
The p-value represents the probability of obtaining a chi-square statistic as extreme or more extreme than the observed value, assuming the null hypothesis is true.
- The p-value is found using the chi-square distribution with the appropriate degrees of freedom.
- It tells us how surprising the observed data are if the null hypothesis \( H_0 \) (no association or no difference) were true.
Steps to Determine the P-Value:
- Calculate the Chi-Square Statistic:
\( \displaystyle \chi^2 = \sum \dfrac{(O_{ij} – E_{ij})^2}{E_{ij}} \) - Determine Degrees of Freedom:
\( df = (\text{number of rows} – 1)(\text{number of columns} – 1) \) - Find the P-Value:
Use a chi-square distribution table or statistical software to find the probability \( P(\chi^2 \ge \text{observed value}) \).
Key Notes:
- The p-value assumes \( H_0 \) is true.
- Larger values of \( \chi^2 \) → smaller p-values (stronger evidence against \( H_0 \)).
Example:
A study of snack preference by gender among 100 students produced a chi-square statistic of \( \chi^2 = 16.66 \) with \( df = 1 \).
Determine the p-value.
▶️ Answer / Explanation
Step 1 — Find P-Value:
Using the chi-square distribution with \( df = 1 \), \( P(\chi^2 \ge 16.66) ≈ 0.00004 \).
Step 2 — Interpretation:
The probability of observing such an extreme chi-square value if \( H_0 \) is true is extremely small (0.00004).
Conclusion:
This very small p-value indicates that the observed data are highly inconsistent with the null hypothesis.
Interpret the P-Value for the Chi-Square Test for Homogeneity or Independence
Interpret the P-Value for the Chi-Square Test for Homogeneity or Independence
Interpreting the p-value helps us decide whether the observed differences or associations are statistically significant.
- A small p-value (< \( \alpha \)) → Reject \( H_0 \) → evidence of association or difference.
- A large p-value (> \( \alpha \)) → Fail to reject \( H_0 \) → insufficient evidence of association or difference.
Interpretation in Context:
- If \( H_0 \) is true, the p-value is the probability of obtaining a chi-square statistic at least as large as the observed one.
- Always interpret the p-value in the context of the problem (i.e., what the two variables represent).
Decision Guidelines:
| P-Value | Decision | Interpretation |
|---|---|---|
| p < α | Reject \( H_0 \) | Evidence of an association / difference |
| p > α | Fail to Reject \( H_0 \) | No convincing evidence of association / difference |
Example:
From the previous study, \( \chi^2 = 16.66 \), \( df = 1 \), and the p-value = 0.00004. Significance level \( \alpha = 0.05 \).
Interpret the p-value.
▶️ Answer / Explanation
- Since \( p = 0.00004 < 0.05 \), we reject \( H_0 \).
- There is strong statistical evidence that snack preference and gender are not independent.
- In context, this means that the distribution of snack preference differs by gender in the population.
Conclusion: A very small p-value indicates a significant association between the two categorical variables.
Justifying a Claim Based on a Chi-Square Test for Homogeneity or Independence
Justifying a Claim Based on a Chi-Square Test for Homogeneity or Independence
The purpose of this to use the results of a chi-square test to draw conclusions about the population(s) from which the data were collected.
Steps for Justifying a Claim:
1.Compare the p-value to the significance level (\(\alpha\)):
- If p-value < \(\alpha\) → reject \(H_0\)
- If p-value ≥ \(\alpha\) → fail to reject \(H_0\)
2.Make a conclusion in context:
- Reject \(H_0\): There is sufficient evidence that the variables are not independent (association exists) or that distributions differ between populations.
- Fail to reject \(H_0\): There is insufficient evidence of association or difference; we cannot conclude that the null hypothesis is true.
3.Relate to the research question:
- For independence: draw conclusions about the population from which the data were sampled.
- For homogeneity: draw conclusions about the populations being compared.
Notes:
- Always state the conclusion in terms of the context of the variables and the population(s).
- A chi-square test does not “prove” anything; it only provides statistical evidence supporting or failing to support a claim.
Example
A survey of 100 students records snack preference (Chips, Candy) and gender (Male, Female). Chi-square test for independence gives \(\chi^2 = 16.66\), p-value ≈ 0.00004, \(\alpha = 0.05\).
Justify a claim about the population based on the test results.
▶️ Answer / Explanation
Step 1 — Compare p-value to \(\alpha\):
p-value = 0.00004 < 0.05 → reject \(H_0\)
Step 2 — State conclusion in context:
There is strong statistical evidence that snack preference is associated with gender in the population of students surveyed. The distribution of snack preference differs between males and females.
Step 3 — Relate to research question:
The chi-square test supports the claim that gender and snack preference are not independent among students. This conclusion is based on the sample and applies to the population from which the sample was drawn.