Answer :
Answer:
To test whether there is any significant association between economic conditions and the level of IQ, we can use the Chi-Square test of independence. Here are the steps:
1. **State the hypotheses:**
- Null hypothesis (\(H_0\)): There is no significant association between economic conditions and the level of IQ.
- Alternative hypothesis (\(H_a\)): There is a significant association between economic conditions and the level of IQ.
2. **Calculate the expected frequencies:**
The expected frequency for each cell in a contingency table is calculated using the formula:
\[
E_{ij} = \frac{(Row \, Total)_i \times (Column \, Total)_j}{Grand \, Total}
\]
Where \( E_{ij} \) is the expected frequency for cell \(i,j\).
3. **Create the observed frequency table:**
| Economic Condition | Rich | Poor | Total |
|--------------------|------|------|-------|
| High | 160 | 140 | 300 |
| Medium | 300 | 100 | 400 |
| Low | 140 | 160 | 300 |
| **Total** | 600 | 400 | 1000 |
4. **Calculate the expected frequencies:**
\[
E_{ij} = \frac{(Row \, Total)_i \times (Column \, Total)_j}{Grand \, Total}
\]
For \( E_{11} \) (High-Rich):
\[
E_{11} = \frac{300 \times 600}{1000} = 180
\]
For \( E_{12} \) (High-Poor):
\[
E_{12} = \frac{300 \times 400}{1000} = 120
\]
For \( E_{21} \) (Medium-Rich):
\[
E_{21} = \frac{400 \times 600}{1000} = 240
\]
For \( E_{22} \) (Medium-Poor):
\[
E_{22} = \frac{400 \times 400}{1000} = 160
\]
For \( E_{31} \) (Low-Rich):
\[
E_{31} = \frac{300 \times 600}{1000} = 180
\]
For \( E_{32} \) (Low-Poor):
\[
E_{32} = \frac{300 \times 400}{1000} = 120
\]
Expected frequency table:
| Economic Condition | Rich | Poor | Total |
|--------------------|------|------|-------|
| High | 180 | 120 | 300 |
| Medium | 240 | 160 | 400 |
| Low | 180 | 120 | 300 |
| **Total** | 600 | 400 | 1000 |
5. **Calculate the Chi-Square statistic:**
\[
\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}
\]
Where \( O_{ij} \) is the observed frequency and \( E_{ij} \) is the expected frequency.
\[
\chi^2 = \frac{(160 - 180)^2}{180} + \frac{(140 - 120)^2}{120} + \frac{(300 - 240)^2}{240} + \frac{(100 - 160)^2}{160} + \frac{(140 - 180)^2}{180} + \frac{(160 - 120)^2}{120}
\]
Let's calculate each term:
\[
\chi^2 = \frac{400}{180} + \frac{400}{120} + \frac{3600}{240} + \frac{3600}{160} + \frac{1600}{180} + \frac{1600}{120}
\]
\[
\chi^2 = 2.222 + 3.333 + 15 + 22.5 + 8.889 + 13.333
\]
\[
\chi^2 \approx 65.277
\]
6. **Compare the Chi-Square statistic with the critical value:**
- Degrees of freedom (df) = (number of rows - 1) \(\times\) (number of columns - 1) = (3-1) \(\times\) (2-1) = 2
- Critical value for \(\chi^2\) at 0.05 significance level for 2 degrees of freedom = 5.99
7. **Decision:**
Since \(\chi^2 \approx 65.277\) is much greater than the critical value 5.99, we reject the null hypothesis.
8. **Conclusion:**
There is a significant association between economic conditions and the level of IQ.