Answer :
Let's determine if there's an association between sex and character type for smoking movie characters using the chi-square test of independence with a significance level of [tex]\(\alpha = 0.05\)[/tex].
### Step-by-Step Solution
1. State the Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): There is no association between sex and character type for smoking movie characters.
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): There is an association between sex and character type for smoking movie characters.
2. Construct the Contingency Table:
[tex]\[ \begin{tabular}{|l|c|c|c|} \hline & \multicolumn{3}{|c|}{Character Type} \\ \hline Sex & Positive & Negative & Neutral \\ \hline Male & 257 & 107 & 131 \\ \hline Female & 85 & 13 & 51 \\ \hline \end{tabular} \][/tex]
- Total counts for each category:
- Males: [tex]\(257 + 107 + 131 = 495\)[/tex]
- Females: [tex]\(85 + 13 + 51 = 149\)[/tex]
- Positive: [tex]\(257 + 85 = 342\)[/tex]
- Negative: [tex]\(107 + 13 = 120\)[/tex]
- Neutral: [tex]\(131 + 51 = 182\)[/tex]
- Overall total: [tex]\(495 + 149 = 644\)[/tex]
3. Calculate the Expected Counts:
The expected count for each cell in the contingency table is computed using the formula:
[tex]\[ E_{ij} = \frac{(row \; total_i) \times (column \; total_j)}{grand \; total} \][/tex]
Applying this formula, we get:
- Expected count for Males, Positive:
[tex]\[ \frac{495 \times 342}{644} \approx 262.87 \][/tex]
- Expected count for Males, Negative:
[tex]\[ \frac{495 \times 120}{644} \approx 92.24 \][/tex]
- Expected count for Males, Neutral:
[tex]\[ \frac{495 \times 182}{644} \approx 139.89 \][/tex]
- Expected count for Females, Positive:
[tex]\[ \frac{149 \times 342}{644} \approx 79.13 \][/tex]
- Expected count for Females, Negative:
[tex]\[ \frac{149 \times 120}{644} \approx 27.76 \][/tex]
- Expected count for Females, Neutral:
[tex]\[ \frac{149 \times 182}{644} \approx 42.11 \][/tex]
The expected counts matrix:
[tex]\[ \begin{array}{|c|c|c|c|} \hline & \text{Positive} & \text{Negative} & \text{Neutral} \\ \hline \text{Male} & 262.87 & 92.24 & 139.89 \\ \hline \text{Female} & 79.13 & 27.76 & 42.11 \\ \hline \end{array} \][/tex]
4. Compute the Chi-Square Test Statistic:
[tex]\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
Adding the contributions for each cell:
[tex]\[ \begin{aligned} \chi^2 &= \frac{(257 - 262.87)^2}{262.87} + \frac{(107 - 92.24)^2}{92.24} + \frac{(131 - 139.89)^2}{139.89} \\ &+ \frac{(85 - 79.13)^2}{79.13} + \frac{(13 - 27.76)^2}{27.76} + \frac{(51 - 42.11)^2}{42.11} \\ &= 0.13 + 2.24 + 0.57 + 0.44 + 7.34 + 2.50 \\ &= 13.22 \text{ (rounded to two decimal places)} \end{aligned} \][/tex]
5. Determine the Degrees of Freedom:
The degrees of freedom for the chi-square test of independence is given by:
[tex]\[ \text{degrees of freedom} = (rows - 1) \times (columns - 1) \][/tex]
Here, we have 2 rows (Males and Females) and 3 columns (Positive, Negative, Neutral):
[tex]\[ \text{degrees of freedom} = (2 - 1) \times (3 - 1) = 2 \][/tex]
6. Interpret the Results:
The chi-square test statistic is [tex]\(\chi^2 = 13.22\)[/tex] with 2 degrees of freedom. Comparing this to the critical value for chi-square distribution with 2 degrees of freedom at [tex]\(\alpha = 0.05\)[/tex] or considering the p-value [tex]\(p = 0.0013\)[/tex] which is less than 0.05, we reject the null hypothesis.
### Conclusion:
There is sufficient evidence to suggest that there is an association between sex and character type for movie characters who smoke at the [tex]\(\alpha = 0.05\)[/tex] significance level.
### Step-by-Step Solution
1. State the Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): There is no association between sex and character type for smoking movie characters.
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): There is an association between sex and character type for smoking movie characters.
2. Construct the Contingency Table:
[tex]\[ \begin{tabular}{|l|c|c|c|} \hline & \multicolumn{3}{|c|}{Character Type} \\ \hline Sex & Positive & Negative & Neutral \\ \hline Male & 257 & 107 & 131 \\ \hline Female & 85 & 13 & 51 \\ \hline \end{tabular} \][/tex]
- Total counts for each category:
- Males: [tex]\(257 + 107 + 131 = 495\)[/tex]
- Females: [tex]\(85 + 13 + 51 = 149\)[/tex]
- Positive: [tex]\(257 + 85 = 342\)[/tex]
- Negative: [tex]\(107 + 13 = 120\)[/tex]
- Neutral: [tex]\(131 + 51 = 182\)[/tex]
- Overall total: [tex]\(495 + 149 = 644\)[/tex]
3. Calculate the Expected Counts:
The expected count for each cell in the contingency table is computed using the formula:
[tex]\[ E_{ij} = \frac{(row \; total_i) \times (column \; total_j)}{grand \; total} \][/tex]
Applying this formula, we get:
- Expected count for Males, Positive:
[tex]\[ \frac{495 \times 342}{644} \approx 262.87 \][/tex]
- Expected count for Males, Negative:
[tex]\[ \frac{495 \times 120}{644} \approx 92.24 \][/tex]
- Expected count for Males, Neutral:
[tex]\[ \frac{495 \times 182}{644} \approx 139.89 \][/tex]
- Expected count for Females, Positive:
[tex]\[ \frac{149 \times 342}{644} \approx 79.13 \][/tex]
- Expected count for Females, Negative:
[tex]\[ \frac{149 \times 120}{644} \approx 27.76 \][/tex]
- Expected count for Females, Neutral:
[tex]\[ \frac{149 \times 182}{644} \approx 42.11 \][/tex]
The expected counts matrix:
[tex]\[ \begin{array}{|c|c|c|c|} \hline & \text{Positive} & \text{Negative} & \text{Neutral} \\ \hline \text{Male} & 262.87 & 92.24 & 139.89 \\ \hline \text{Female} & 79.13 & 27.76 & 42.11 \\ \hline \end{array} \][/tex]
4. Compute the Chi-Square Test Statistic:
[tex]\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
Adding the contributions for each cell:
[tex]\[ \begin{aligned} \chi^2 &= \frac{(257 - 262.87)^2}{262.87} + \frac{(107 - 92.24)^2}{92.24} + \frac{(131 - 139.89)^2}{139.89} \\ &+ \frac{(85 - 79.13)^2}{79.13} + \frac{(13 - 27.76)^2}{27.76} + \frac{(51 - 42.11)^2}{42.11} \\ &= 0.13 + 2.24 + 0.57 + 0.44 + 7.34 + 2.50 \\ &= 13.22 \text{ (rounded to two decimal places)} \end{aligned} \][/tex]
5. Determine the Degrees of Freedom:
The degrees of freedom for the chi-square test of independence is given by:
[tex]\[ \text{degrees of freedom} = (rows - 1) \times (columns - 1) \][/tex]
Here, we have 2 rows (Males and Females) and 3 columns (Positive, Negative, Neutral):
[tex]\[ \text{degrees of freedom} = (2 - 1) \times (3 - 1) = 2 \][/tex]
6. Interpret the Results:
The chi-square test statistic is [tex]\(\chi^2 = 13.22\)[/tex] with 2 degrees of freedom. Comparing this to the critical value for chi-square distribution with 2 degrees of freedom at [tex]\(\alpha = 0.05\)[/tex] or considering the p-value [tex]\(p = 0.0013\)[/tex] which is less than 0.05, we reject the null hypothesis.
### Conclusion:
There is sufficient evidence to suggest that there is an association between sex and character type for movie characters who smoke at the [tex]\(\alpha = 0.05\)[/tex] significance level.
