Questions: A researcher investigated the effect of guilt emotion on how a decision maker focuses on the problem. A total of 171 volunteer students participated in the experiment, where each was assigned to one of three emotional states (guilt, anger, or neutral) through a reading/writing task. Immediately after the task, students were presented with a decision problem where the stated option has predominantly negative features (for example, spending money on repairing a very old car). The results (number responding in each category) are summarized in the accompanying table. Is there sufficient evidence (at α=0.01) to claim that the option choice depends on emotional state? Specify the null and alternative hypotheses. Choose the correct answer below. A. H0: The classifications choose and do not choose are independent. Ha: The classifications choose and do not choose are dependent. B. H0: The classifications emotional state and option choice are dependent. Ha: The classifications emotional state and option choice are independent. C. H0: The classifications choose and do not choose are dependent. Ha: The classifications choose and do not choose are independent. D. H0: The classifications emotional state and option choice are independent. Ha: The classifications emotional state and option choice are dependent. Contingency Table Emotional State Option Choice Totals Choose Stated Option Do Not Choose Stated Option Guilt 44 10 54 Anger 7 50 57 Neutral 9 51 60 Totals 60 111 171

Solution Steps

We are testing whether the option choice depends on the emotional state of the participants. The null and alternative hypotheses are defined as follows:

• \( H_0 \): The classifications emotional state and option choice are independent.
• \( H_a \): The classifications emotional state and option choice are dependent.

The expected frequencies for each cell in the contingency table are calculated based on the row and column totals. The expected frequencies are as follows:

• For cell (1, 1): \[ E = \frac{R_1 \times C_1}{N} = \frac{54 \times 60}{171} = 18.95 \]
• For cell (1, 2): \[ E = \frac{R_1 \times C_2}{N} = \frac{54 \times 111}{171} = 35.05 \]
• For cell (2, 1): \[ E = \frac{R_2 \times C_1}{N} = \frac{57 \times 60}{171} = 20.0 \]
• For cell (2, 2): \[ E = \frac{R_2 \times C_2}{N} = \frac{57 \times 111}{171} = 37.0 \]
• For cell (3, 1): \[ E = \frac{R_3 \times C_1}{N} = \frac{60 \times 60}{171} = 21.05 \]
• For cell (3, 2): \[ E = \frac{R_3 \times C_2}{N} = \frac{60 \times 111}{171} = 38.95 \]

The expected frequencies are: \[ \begin{bmatrix} 18.95 & 35.05 \\ 20.00 & 37.00 \\ 21.05 & 38.95 \end{bmatrix} \]

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. The calculations for each cell are as follows:

• For cell (1, 1): \[ O = 44, E = 18.95, \frac{(O - E)^2}{E} = \frac{(44 - 18.95)^2}{18.95} = 33.13 \]
• For cell (1, 2): \[ O = 10, E = 35.05, \frac{(O - E)^2}{E} = \frac{(10 - 35.05)^2}{35.05} = 17.91 \]
• For cell (2, 1): \[ O = 7, E = 20.0, \frac{(O - E)^2}{E} = \frac{(7 - 20.0)^2}{20.0} = 8.45 \]
• For cell (2, 2): \[ O = 50, E = 37.0, \frac{(O - E)^2}{E} = \frac{(50 - 37.0)^2}{37.0} = 4.57 \]
• For cell (3, 1): \[ O = 9, E = 21.05, \frac{(O - E)^2}{E} = \frac{(9 - 21.05)^2}{21.05} = 6.90 \]
• For cell (3, 2): \[ O = 51, E = 38.95, \frac{(O - E)^2}{E} = \frac{(51 - 38.95)^2}{38.95} = 3.73 \]

Summing these values gives: \[ \chi^2 = 33.13 + 17.91 + 8.45 + 4.57 + 6.90 + 3.73 = 74.68 \]

The critical value for a Chi-Square distribution with 2 degrees of freedom at \( \alpha = 0.01 \) is: \[ \chi^2_{\alpha, df} = \chi^2_{(0.01, 2)} = 9.21 \]

The p-value associated with the test statistic \( \chi^2 = 74.68 \) is: \[ P = P(\chi^2 > 74.68) = 0.0 \]

Since the calculated Chi-Square test statistic \( 74.68 \) is greater than the critical value \( 9.21 \) and the p-value \( 0.0 \) is less than \( \alpha = 0.01 \), we reject the null hypothesis.

Final Answer