Math Problem Statement
Solution
The image provided shows a contingency table with observed values for two categorical variables, followed by a hypothesis test to check for independence between the row and column variables. The hypothesis is tested at a significance level of 0.05.
Here’s how we proceed:
Step 1: Hypotheses
- (Null Hypothesis): Row and column variables are independent.
- (Alternative Hypothesis): Row and column variables are not independent.
Step 2: Observed Table
The contingency table is presented as follows:
\hline & C_1 & C_2 & \text{Total} \\ \hline R_1 & 15 & 7 & 22 \\ R_2 & 37 & 43 & 80 \\ \hline \text{Total} & 52 & 50 & 102 \\ \hline \end{array}$$ ### Step 3: Expected Frequencies We calculate the expected values for each cell using the formula: $$E_{ij} = \frac{( \text{Row Total} ) ( \text{Column Total} )}{ \text{Grand Total} }$$ For example, for cell $$(R_1, C_1)$$: $$E_{11} = \frac{22 \times 52}{102} \approx 11.22$$ Similarly, we calculate for other cells: $$E_{12} = \frac{22 \times 50}{102} \approx 10.78$$ $$E_{21} = \frac{80 \times 52}{102} \approx 40.78$$ $$E_{22} = \frac{80 \times 50}{102} \approx 39.22$$ ### Step 4: Chi-Square Test Statistic The chi-square test statistic is calculated using the formula: $$\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$$ Where $$O_{ij}$$ are the observed frequencies and $$E_{ij}$$ are the expected frequencies. Using the observed and expected values: For $$(R_1, C_1)$$: $$\frac{(15 - 11.22)^2}{11.22} \approx 1.29$$ For $$(R_1, C_2)$$: $$\frac{(7 - 10.78)^2}{10.78} \approx 1.32$$ For $$(R_2, C_1)$$: $$\frac{(37 - 40.78)^2}{40.78} \approx 0.35$$ For $$(R_2, C_2)$$: $$\frac{(43 - 39.22)^2}{39.22} \approx 0.36$$ Summing these gives: $$\chi^2 = 1.29 + 1.32 + 0.35 + 0.36 = 3.32$$ ### Step 5: Degrees of Freedom The degrees of freedom $$df$$ are calculated as: $$df = ( \text{Number of Rows} - 1) \times (\text{Number of Columns} - 1) = (2-1) \times (2-1) = 1$$ ### Step 6: Critical Value and Conclusion At a significance level $$\alpha = 0.05$$, with 1 degree of freedom, the critical value from the chi-square distribution table is approximately $$3.841$$. Since the test statistic $$\chi^2 = 3.32$$ is less than the critical value $$3.841$$, we **fail to reject the null hypothesis**. This suggests that there is not enough evidence to conclude that the row and column variables are dependent. ### Final Answer: We fail to reject the null hypothesis, so the row and column variables appear to be independent at the 0.05 significance level. --- Do you want further clarification on any of the steps, or need additional details? Here are 5 related questions for deeper understanding: 1. How are the expected frequencies calculated in a contingency table? 2. What does it mean if we reject or fail to reject the null hypothesis in a chi-square test? 3. How do we choose the significance level for hypothesis testing? 4. How can the degrees of freedom impact the critical value in a chi-square test? 5. What are other types of tests used for categorical data analysis? **Tip**: Always ensure that the sample size is sufficiently large when applying the chi-square test, as small sample sizes can lead to inaccurate results.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Chi-square test
Contingency table
Hypothesis testing
Degrees of freedom
Formulas
Expected frequency formula: Eij = (Row Total * Column Total) / Grand Total
Chi-square test statistic: χ² = Σ((Oij - Eij)² / Eij)
Degrees of freedom: df = (Number of Rows - 1) * (Number of Columns - 1)
Theorems
Chi-square distribution
Null and alternative hypotheses in hypothesis testing
Suitable Grade Level
College Level (Statistics/Econometrics)
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