8.3 The Chi-Square Test for Homogeneity

• The Chi-Square Test for Independence explicitly compares different groups.

  False
• What does the alternative hypothesis (Hₐ) state in the Chi-Square Test for Homogeneity?

  Distributions differ significantly
• What is the description of the null hypothesis (H₀) in the Chi-Square Test for Homogeneity?

  Distribution is the same
• Why is random sampling necessary for the Chi-Square Test for Homogeneity?

  Ensures unbiased representation
• What is the purpose of the Chi-Square Test for Homogeneity?

  Compares categorical variable distributions
• All expected counts in a Chi-Square Test for Homogeneity must be at least 5 to ensure the Chi-Square distribution approximates the true distribution well.


  True
• The chi-square test statistic measures the difference between observed and expected frequencies.


  True
• Steps to determine whether to reject the null hypothesis using the critical value approach

  1️⃣ Determine the significance level (\(\alpha\))

  2️⃣ Find the degrees of freedom (\(df\))

  3️⃣ Consult the Chi-Square Distribution Table

  4️⃣ Compare the Chi-Square Test Statistic (\(\chi^2\)) with the critical value
• What is the null hypothesis in a Chi-Square Test for Homogeneity?

  Distributions are the same
• The alternative hypothesis (Hₐ) for the Chi-Square Test for Homogeneity states that the distribution of the categorical variable differs across at least two populations.


  True
• What is the formula for calculating expected frequencies under homogeneity?

  $E_{ij} =$$\frac{(\text{Row } i \text{ Total}) \times (\text{Column } j \text{ Total})}{\text{Grand Total}}$
• What does the chi-square test statistic measure?

  Observed versus expected frequencies
• What is the calculated chi-square test statistic for the given observed and expected frequencies?

  2.0834
• What are the degrees of freedom for a contingency table with 3 rows and 2 columns?

  2
• Match the degrees of freedom with the critical value at \(\alpha = 0.05\):

  1 ↔️ 3.841

  2 ↔️ 5.991

  3 ↔️ 7.815
• If \(\chi^2 = 6.5\) with \( df = 2 \) and \( \alpha = 0.05 \), you would reject the null hypothesis
• What is the purpose of the Chi-Square Test for Homogeneity?

  Compares categorical variable distributions
• What does the null hypothesis (H₀) state in the Chi-Square Test for Homogeneity?

  Distributions are the same
• Match the Chi-Square test with its purpose:

  Chi-Square Test for Homogeneity ↔️ Compares categorical variable distributions across populations

  Chi-Square Test for Independence ↔️ Tests if two categorical variables are independent
• One of the assumptions for the Chi-Square Test for Homogeneity is that all expected counts must be at least 5.
• Steps to calculate expected frequencies in a Chi-Square Test for Homogeneity:

  1️⃣ Calculate Row and Column Totals

  2️⃣ Multiply Row Total by Column Total

  3️⃣ Divide by Grand Total

  4️⃣ Record Expected Frequency
• The alternative hypothesis in the Chi-Square Test for Homogeneity states that the proportions of categories differ significantly across at least two populations.
• What are the three assumptions that must be met in a Chi-Square Test for Homogeneity?

  Random sampling, independence, expected counts
• What is the formula to calculate expected frequencies in a Chi-Square Test for Homogeneity?

  $E_{ij} =$$\frac{(\text{Row } i \text{ Total}) \times (\text{Column } j \text{ Total})}{\text{Grand Total}}$
• How are degrees of freedom calculated in a Chi-Square Test for Homogeneity?

  $df =$$(r - 1) \times (c - 1)$
• The Chi-Square Test for Homogeneity is used to determine whether the distribution of a categorical variable is the same across different groups.
• The null hypothesis (H₀) for the Chi-Square Test for Homogeneity states that the distribution of the categorical variable is the same across all populations
• All expected counts in the Chi-Square Test for Homogeneity must be at least 5
• What is the value of the Chi-Square Test Statistic for the example data in the material?

  $\chi^{2} =$$2.0834$
• The symbol \( E_{ij} \) in the chi-square formula represents the expected frequency.

  True
• The formula for calculating degrees of freedom in the chi-square test is \( df = (r - 1) \times (c - 1) \).


  True
• The significance level (\(\alpha\)) in hypothesis testing represents the probability of rejecting a true null hypothesis
• In the p-value approach, you reject the null hypothesis if the p-value is greater than the significance level.

  False
• What does it mean if you fail to reject the null hypothesis in a chi-square test for homogeneity?

  Distributions are consistent
• The research question for the Chi-Square Test for Homogeneity is: "Are the proportions of categories similar across populations?"
• The Chi-Square Test for Homogeneity is used to determine whether the distribution of a categorical variable is the same across different populations.
• What is the description of the alternative hypothesis (Hₐ) in the Chi-Square Test for Homogeneity?

  Distribution differs significantly
• Observations within each sample must be independent in the Chi-Square Test for Homogeneity.

  True
• The null hypothesis in the Chi-Square Test for Homogeneity states that the proportions of categories differ across populations.

  False
• Match the assumption with its importance in the Chi-Square Test for Homogeneity:

  Random Sampling ↔️ Ensures unbiased representation

  Independence ↔️ Prevents artificial inflation

  Expected Counts ≥ 5 ↔️ Ensures Chi-Square approximation