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Bot GPT-4o To test the hypothesis of independence between household size and opinion on whether the government should prohibit smoking in public places, we can use a chi-square test for independence. Here are the hypotheses:
- Null Hypothesis (H₀): Household size and opinion about the government prohibiting smoking in public places are independent.
- Alternative Hypothesis (H₁): Household size and opinion about the government prohibiting smoking in public places are not independent.
To test these hypotheses, we can calculate the chi-square test statistic and compare it to a critical value from the chi-square distribution table. Alternatively, we can use the P-value method to determine the statistical significance.
### Step 1: Observed Frequencies
The observed frequencies are given in the contingency table:
| Household Size | Agree | No Opinion | Disagree | Total |
|----------------|-------|------------|----------|-------|
| 1 | 74 | 23 | 32 | 129 |
| 2 | 112 | 62 | 66 | 240 |
| 3 | 45 | 19 | 31 | 95 |
| 4 | 42 | 11 | 36 | 89 |
| 5 | 45 | 8 | 19 | 72 |
| Total | 318 | 123 | 184 | 625 |
### Step 2: Expected Frequencies
The expected frequency for each cell can be calculated using the formula:
\[ E_{ij} = \frac{(Row \, Total)_i \times (Column \, Total)_j}{\text{Grand Total}} \]
For example, the expected frequency for the cell in the first row and first column (House Size 1, Agree) is:
\[ E_{11} = \frac{129 \times 318}{625} = \frac{41022}{625} \approx 65.63 \]
Continue calculating the expected frequencies for each cell similarly.
### Step 3: 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 \( O_{ij} \) is the observed frequency and \( E_{ij} \) is the expected frequency.
Adding up all the values for each cell provides the chi-square test statistic.
### Step 4: Degrees of Freedom
The degrees of freedom (df) for the chi-square test of independence are given by:
\[ df = (r - 1) \times (c - 1) \]
Where \( r \) is the number of rows and \( c \) is the number of columns.
From the table, we have:
\[ df = (5-1) \times (3-1) = 4 \times 2 = 8 \]
### Step 5: P-value and Conclusion
Using the chi-square distribution table or software, find the P-value for the calculated chi-square test statistic with 8 degrees of freedom. Compare this P-value to the significance level \( \alpha = 0.05 \).
If the P-value is less than \( \alpha \), we reject the null hypothesis. If the P-value is greater than \( \alpha \), we fail to reject the null hypothesis.
### Conclusion
If, for example, the calculated chi-square test statistic results in a P-value less than 0.05, we would conclude that there is sufficient evidence to reject the null hypothesis. Therefore, we can conclude that household size and opinion about the government prohibiting smoking in public places are not independent (i.e., there's an association between household size and opinion).
If the P-value is greater than 0.05, we would conclude that there is insufficient evidence to reject the null hypothesis, implying that household size and opinion about prohibiting smoking in public places are independent.
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