Questions: Are A and B independent? Yes, since P(A ∩ B) ≠ 0 Yes, since P(A ∩ B) = P(A)P(B) No, since P(A ∩ B) ≠ P(A)P(B) No, since P(A ∩ B) = 0.10

Solution Steps

To determine the expected frequencies for each cell in the contingency table, we use the formula:

\[ E = \frac{R_i \times C_j}{N} \]

where \(R_i\) is the total for row \(i\), \(C_j\) is the total for column \(j\), and \(N\) is the grand total.

• For cell (1, 1): \[ E = \frac{40 \times 50}{100} = 20.0 \]

• For cell (1, 2): \[ E = \frac{40 \times 50}{100} = 20.0 \]

• For cell (2, 1): \[ E = \frac{60 \times 50}{100} = 30.0 \]

• For cell (2, 2): \[ E = \frac{60 \times 50}{100} = 30.0 \]

Thus, the expected frequencies are: \[ \begin{bmatrix} 20.0 & 20.0 \\ 30.0 & 30.0 \end{bmatrix} \]

The Chi-Square test statistic (\(\chi^2\)) is calculated using the formula:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

where \(O\) is the observed frequency and \(E\) is the expected frequency.

Calculating for each cell:

• For cell 1: \[ O = 30, \quad E = 20.0 \quad \Rightarrow \quad \frac{(30 - 20.0)^2}{20.0} = 5.0 \]

• For cell 2: \[ O = 10, \quad E = 20.0 \quad \Rightarrow \quad \frac{(10 - 20.0)^2}{20.0} = 5.0 \]

• For cell 3: \[ O = 20, \quad E = 30.0 \quad \Rightarrow \quad \frac{(20 - 30.0)^2}{30.0} = 3.3333 \]

• For cell 4: \[ O = 40, \quad E = 30.0 \quad \Rightarrow \quad \frac{(40 - 30.0)^2}{30.0} = 3.3333 \]

Summing these values gives: \[ \chi^2 = 5.0 + 5.0 + 3.3333 + 3.3333 = 15.0417 \]

The critical value for a Chi-Square distribution at \(\alpha = 0.05\) with 1 degree of freedom is: \[ \chi^2_{\alpha, df} = \chi^2_{(0.05, 1)} = 3.8415 \]

The p-value associated with the calculated Chi-Square statistic is: \[ P = P(\chi^2 > 15.0417) = 0.0001 \]

Since the p-value \(0.0001\) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis of independence. Therefore, variables A and B are not independent.

Final Answer