The problem asks to test the significance of three chi-square tests given the sample size $N$, number of rows $R$, number of columns $C$, the chi-square statistic value $\chi^2$, and the significance level $\alpha = 0.05$. If the test is significant, calculate the effect size using Cramer's V.

1. Problem Description

The problem asks to test the significance of three chi-square tests given the sample size $N$, number of rows $R$, number of columns $C$, the chi-square statistic value $\chi^2$, and the significance level $\alpha = 0.05$. If the test is significant, calculate the effect size using Cramer's V.

2. Solution Steps

First, we need to determine the degrees of freedom ($df$) for each test. The formula for degrees of freedom in a chi-square test of independence is:

$df = (R - 1)(C - 1)$

Next, we compare the given $\chi^2$ value with the critical $\chi^2$ value at $\alpha = 0.05$ for the corresponding degrees of freedom. If the given $\chi^2$ value is greater than the critical value, we reject the null hypothesis and conclude that the test is significant.

If the test is significant, we calculate Cramer's V, which is a measure of effect size. The formula for Cramer's V is:

$V = \sqrt{\frac{\chi^2}{N \cdot min(R-1, C-1)}}$

1. Test 1: $N = 19$, $R = 3$, $C = 2$, $\chi^2(2) = 7.89$, $\alpha = 0.05$

$df = (3 - 1)(2 - 1) = 2 \cdot 1 = 2$

The critical $\chi^2$ value for $df = 2$ and $\alpha = 0.05$ is 5.

9

9

1. Since $7.89 > 5.991$, the test is significant.

$V = \sqrt{\frac{7.89}{19 \cdot min(3-1, 2-1)}} = \sqrt{\frac{7.89}{19 \cdot 1}} = \sqrt{\frac{7.89}{19}} \approx \sqrt{0.415} \approx 0.644$

2. Test 2: $N = 12$, $R = 2$, $C = 2$, $\chi^2(1) = 3.12$, $\alpha = 0.05$

$df = (2 - 1)(2 - 1) = 1 \cdot 1 = 1$

The critical $\chi^2$ value for $df = 1$ and $\alpha = 0.05$ is 3.

8

4

1. Since $3.12 < 3.841$, the test is not significant.

3. Test 3: $N = 74$, $R = 3$, $C = 3$, $\chi^2(4) = 28.41$, $\alpha = 0.05$

$df = (3 - 1)(3 - 1) = 2 \cdot 2 = 4$

The critical $\chi^2$ value for $df = 4$ and $\alpha = 0.05$ is 9.

4

8

8. Since $28.41 > 9.488$, the test is significant.

$V = \sqrt{\frac{28.41}{74 \cdot min(3-1, 3-1)}} = \sqrt{\frac{28.41}{74 \cdot 2}} = \sqrt{\frac{28.41}{148}} \approx \sqrt{0.192} \approx 0.438$

3. Final Answer

1. Test 1: Significant, $V \approx 0.644$

2. Test 2: Not significant

3. Test 3: Significant, $V \approx 0.438$