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Sagot :
To address the given problem, we need to determine if gender and education level are independent or if there is a relationship between them. This requires performing a Chi-Square Test for Independence. Here’s a step-by-step solution:
Step 1: Formulate the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): Gender and education level are independent.
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): Gender and education level are not independent.
Step 2: Construct the Contingency Table
Based on the provided data, the contingency table is:
| | High School | Bachelors | Masters | Ph.D. | Total |
|---------------|-------------|-----------|---------|-------|-------|
| Female | 60 | 54 | 46 | 41 | 201 |
| Male | 40 | 44 | 53 | 57 | 194 |
| Total | 100 | 98 | 99 | 98 | 395 |
Step 3: Calculate Expected Frequencies
The expected frequency for each cell is calculated using:
[tex]\[ E_{ij} = \frac{(row \, total \, of \, cell) \times (column \, total \, of \, cell)}{total \, sample \, size} \][/tex]
For example, the expected frequency for females with High School education:
[tex]\[ E_{11} = \frac{201 \times 100}{395} \approx 50.886 \][/tex]
We perform this calculation for all cells to get the expected frequency table:
| | High School | Bachelors | Masters | Ph.D. |
|-----------|-------------|-----------|---------|--------|
| Female| 50.886 | 49.868 | 50.377 | 49.868 |
| Male | 49.114 | 48.132 | 48.623 | 48.132 |
Step 4: Calculate the Chi-Square Statistic
The Chi-Square statistic is given by:
[tex]\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
Using the observed ([tex]\(O_{ij}\)[/tex]) and expected ([tex]\(E_{ij}\)[/tex]) values:
For each cell:
- Female, High School:
[tex]\[ \frac{(60 - 50.886)^2}{50.886} \approx 1.643 \][/tex]
- Female, Bachelors:
[tex]\[ \frac{(54 - 49.868)^2}{49.868} \approx 0.344 \][/tex]
- Female, Masters:
[tex]\[ \frac{(46 - 50.377)^2}{50.377} \approx 0.380 \][/tex]
- Female, Ph.D.:
[tex]\[ \frac{(41 - 49.868)^2}{49.868} \approx 1.577 \][/tex]
- Male, High School:
[tex]\[ \frac{(40 - 49.114)^2}{49.114} \approx 1.690 \][/tex]
- Male, Bachelors:
[tex]\[ \frac{(44 - 48.132)^2}{48.132} \approx 0.355 \][/tex]
- Male, Masters:
[tex]\[ \frac{(53 - 48.623)^2}{48.623} \approx 0.385 \][/tex]
- Male, Ph.D.:
[tex]\[ \frac{(57 - 48.132)^2}{48.132} \approx 1.632 \][/tex]
Adding these values gives us the Chi-Square statistic:
[tex]\[ \chi^2 \approx 1.643 + 0.344 + 0.380 + 1.577 + 1.690 + 0.355 + 0.385 + 1.632 = 8.006 \][/tex]
Step 5: Determine the Degrees of Freedom
The degrees of freedom (dof) for the test is calculated as:
[tex]\[ \text{dof} = (number \, of \, rows - 1) \times (number \, of \, columns - 1) \][/tex]
In this case:
[tex]\[ \text{dof} = (2 - 1) \times (4 - 1) = 3 \][/tex]
Step 6: Determine the p-value
The p-value corresponds to the calculated Chi-Square statistic with the given degrees of freedom.
For [tex]\(\chi^2 = 8.006\)[/tex] and [tex]\(dof = 3\)[/tex], the p-value is approximately [tex]\(0.0459\)[/tex].
Step 7: Compare the p-value to the Significance Level
The significance level [tex]\(\alpha\)[/tex] is [tex]\(0.05\)[/tex].
Since [tex]\( p \approx 0.0459 < 0.05\)[/tex]:
- We reject the null hypothesis.
Conclusion:
At the 5% significance level, there is sufficient evidence to conclude that gender and education level are not independent; there is a significant relationship between gender and education level.
Step 1: Formulate the Hypotheses
- Null Hypothesis ([tex]\(H_0\)[/tex]): Gender and education level are independent.
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): Gender and education level are not independent.
Step 2: Construct the Contingency Table
Based on the provided data, the contingency table is:
| | High School | Bachelors | Masters | Ph.D. | Total |
|---------------|-------------|-----------|---------|-------|-------|
| Female | 60 | 54 | 46 | 41 | 201 |
| Male | 40 | 44 | 53 | 57 | 194 |
| Total | 100 | 98 | 99 | 98 | 395 |
Step 3: Calculate Expected Frequencies
The expected frequency for each cell is calculated using:
[tex]\[ E_{ij} = \frac{(row \, total \, of \, cell) \times (column \, total \, of \, cell)}{total \, sample \, size} \][/tex]
For example, the expected frequency for females with High School education:
[tex]\[ E_{11} = \frac{201 \times 100}{395} \approx 50.886 \][/tex]
We perform this calculation for all cells to get the expected frequency table:
| | High School | Bachelors | Masters | Ph.D. |
|-----------|-------------|-----------|---------|--------|
| Female| 50.886 | 49.868 | 50.377 | 49.868 |
| Male | 49.114 | 48.132 | 48.623 | 48.132 |
Step 4: Calculate the Chi-Square Statistic
The Chi-Square statistic is given by:
[tex]\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
Using the observed ([tex]\(O_{ij}\)[/tex]) and expected ([tex]\(E_{ij}\)[/tex]) values:
For each cell:
- Female, High School:
[tex]\[ \frac{(60 - 50.886)^2}{50.886} \approx 1.643 \][/tex]
- Female, Bachelors:
[tex]\[ \frac{(54 - 49.868)^2}{49.868} \approx 0.344 \][/tex]
- Female, Masters:
[tex]\[ \frac{(46 - 50.377)^2}{50.377} \approx 0.380 \][/tex]
- Female, Ph.D.:
[tex]\[ \frac{(41 - 49.868)^2}{49.868} \approx 1.577 \][/tex]
- Male, High School:
[tex]\[ \frac{(40 - 49.114)^2}{49.114} \approx 1.690 \][/tex]
- Male, Bachelors:
[tex]\[ \frac{(44 - 48.132)^2}{48.132} \approx 0.355 \][/tex]
- Male, Masters:
[tex]\[ \frac{(53 - 48.623)^2}{48.623} \approx 0.385 \][/tex]
- Male, Ph.D.:
[tex]\[ \frac{(57 - 48.132)^2}{48.132} \approx 1.632 \][/tex]
Adding these values gives us the Chi-Square statistic:
[tex]\[ \chi^2 \approx 1.643 + 0.344 + 0.380 + 1.577 + 1.690 + 0.355 + 0.385 + 1.632 = 8.006 \][/tex]
Step 5: Determine the Degrees of Freedom
The degrees of freedom (dof) for the test is calculated as:
[tex]\[ \text{dof} = (number \, of \, rows - 1) \times (number \, of \, columns - 1) \][/tex]
In this case:
[tex]\[ \text{dof} = (2 - 1) \times (4 - 1) = 3 \][/tex]
Step 6: Determine the p-value
The p-value corresponds to the calculated Chi-Square statistic with the given degrees of freedom.
For [tex]\(\chi^2 = 8.006\)[/tex] and [tex]\(dof = 3\)[/tex], the p-value is approximately [tex]\(0.0459\)[/tex].
Step 7: Compare the p-value to the Significance Level
The significance level [tex]\(\alpha\)[/tex] is [tex]\(0.05\)[/tex].
Since [tex]\( p \approx 0.0459 < 0.05\)[/tex]:
- We reject the null hypothesis.
Conclusion:
At the 5% significance level, there is sufficient evidence to conclude that gender and education level are not independent; there is a significant relationship between gender and education level.
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