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To determine if there is a significant relationship between being an athlete and choosing a particular college major, we can perform a Chi-Square Test of Independence. Here is a detailed breakdown of the steps involved in this analysis:
### Step 1: Set Up the Hypotheses
The hypotheses for the Chi-Square Test of Independence are:
- Null Hypothesis ([tex]$H_0$[/tex]): The variables "being an athlete" and "major choice" are independent.
- Alternative Hypothesis ([tex]$H_a$[/tex]): The variables "being an athlete" and "major choice" are dependent.
### Step 2: Create the Contingency Table
We start with the given contingency table:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{} & \text{Arts} & \text{Humanities} & \text{Sciences} & \text{Row Total} \\ \hline \text{Athlete} & 17 & 24 & 15 & 56 \\ \hline \text{Non-Athlete} & 19 & 11 & 9 & 39 \\ \hline \text{Column Total} & 36 & 35 & 24 & 95 \\ \hline \end{array} \][/tex]
### Step 3: Calculate the Expected Frequencies
The expected frequency for each cell in the contingency table is calculated using the formula:
[tex]\[ E_{ij} = \frac{(\text{Row Total for row } i) \times (\text{Column Total for column } j)}{\text{Grand Total}} \][/tex]
Where [tex]\(E_{ij}\)[/tex] represents the expected frequency for the cell in the [tex]\(i\)[/tex]-th row and [tex]\(j\)[/tex]-th column.
1. Arts:
- Athlete: [tex]\(\frac{56 \times 36}{95} \approx 21.2\)[/tex]
- Non-Athlete: [tex]\(\frac{39 \times 36}{95} \approx 14.8\)[/tex]
2. Humanities:
- Athlete: [tex]\(\frac{56 \times 35}{95} \approx 20.6\)[/tex]
- Non-Athlete: [tex]\(\frac{39 \times 35}{95} \approx 14.4\)[/tex]
3. Sciences:
- Athlete: [tex]\(\frac{56 \times 24}{95} \approx 14.1\)[/tex]
- Non-Athlete: [tex]\(\frac{39 \times 24}{95} \approx 9.9\)[/tex]
### Step 4: Compute the Chi-Square Test Statistic
The Chi-Square test statistic is calculated as follows:
[tex]\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
Where [tex]\(O_{ij}\)[/tex] is the observed frequency and [tex]\(E_{ij}\)[/tex] is the expected frequency.
Using the observed and expected frequencies:
1. Arts (Athlete): [tex]\(\frac{(17 - 21.2)^2}{21.2} \approx 0.832\)[/tex]
2. Arts (Non-Athlete): [tex]\(\frac{(19 - 14.8)^2}{14.8} \approx 1.181\)[/tex]
3. Humanities (Athlete): [tex]\(\frac{(24 - 20.6)^2}{20.6} \approx 0.557\)[/tex]
4. Humanities (Non-Athlete): [tex]\(\frac{(11 - 14.4)^2}{14.4} \approx 0.805\)[/tex]
5. Sciences (Athlete): [tex]\(\frac{(15 - 14.1)^2}{14.1} \approx 0.064\)[/tex]
6. Sciences (Non-Athlete): [tex]\(\frac{(9 - 9.9)^2}{9.9} \approx 0.082\)[/tex]
Adding these values together:
[tex]\[ \chi^2 = 0.832 + 1.181 + 0.557 + 0.805 + 0.064 + 0.082 \approx 3.521 \][/tex]
We round the final chi-square test statistic to one decimal place:
[tex]\[ \chi^2 \approx 3.5 \][/tex]
### Step 5: Interpret the Result
The test statistic [tex]\(\chi^2\)[/tex] is 3.5. This value would be compared to the critical value from the Chi-Square distribution table at the 5% significance level and with appropriate degrees of freedom (in this case, [tex]\((2-1) \times (3-1) = 2\)[/tex] degrees of freedom).
If the test statistic is less than the critical value, we do not reject the null hypothesis; hence, there is insufficient evidence to suggest that college major depends on whether a student is an athlete or not.
Given our test statistic of 3.5, we'd look up the critical value for [tex]\(\chi^2\)[/tex] with 2 degrees of freedom at the 5% significance level, which is approximately 5.991. Since 3.5 < 5.991, we fail to reject the null hypothesis.
Hence, we conclude that there is no strong evidence to suggest that being an athlete significantly affects the choice of college major.
### Step 1: Set Up the Hypotheses
The hypotheses for the Chi-Square Test of Independence are:
- Null Hypothesis ([tex]$H_0$[/tex]): The variables "being an athlete" and "major choice" are independent.
- Alternative Hypothesis ([tex]$H_a$[/tex]): The variables "being an athlete" and "major choice" are dependent.
### Step 2: Create the Contingency Table
We start with the given contingency table:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline \text{} & \text{Arts} & \text{Humanities} & \text{Sciences} & \text{Row Total} \\ \hline \text{Athlete} & 17 & 24 & 15 & 56 \\ \hline \text{Non-Athlete} & 19 & 11 & 9 & 39 \\ \hline \text{Column Total} & 36 & 35 & 24 & 95 \\ \hline \end{array} \][/tex]
### Step 3: Calculate the Expected Frequencies
The expected frequency for each cell in the contingency table is calculated using the formula:
[tex]\[ E_{ij} = \frac{(\text{Row Total for row } i) \times (\text{Column Total for column } j)}{\text{Grand Total}} \][/tex]
Where [tex]\(E_{ij}\)[/tex] represents the expected frequency for the cell in the [tex]\(i\)[/tex]-th row and [tex]\(j\)[/tex]-th column.
1. Arts:
- Athlete: [tex]\(\frac{56 \times 36}{95} \approx 21.2\)[/tex]
- Non-Athlete: [tex]\(\frac{39 \times 36}{95} \approx 14.8\)[/tex]
2. Humanities:
- Athlete: [tex]\(\frac{56 \times 35}{95} \approx 20.6\)[/tex]
- Non-Athlete: [tex]\(\frac{39 \times 35}{95} \approx 14.4\)[/tex]
3. Sciences:
- Athlete: [tex]\(\frac{56 \times 24}{95} \approx 14.1\)[/tex]
- Non-Athlete: [tex]\(\frac{39 \times 24}{95} \approx 9.9\)[/tex]
### Step 4: Compute the Chi-Square Test Statistic
The Chi-Square test statistic is calculated as follows:
[tex]\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \][/tex]
Where [tex]\(O_{ij}\)[/tex] is the observed frequency and [tex]\(E_{ij}\)[/tex] is the expected frequency.
Using the observed and expected frequencies:
1. Arts (Athlete): [tex]\(\frac{(17 - 21.2)^2}{21.2} \approx 0.832\)[/tex]
2. Arts (Non-Athlete): [tex]\(\frac{(19 - 14.8)^2}{14.8} \approx 1.181\)[/tex]
3. Humanities (Athlete): [tex]\(\frac{(24 - 20.6)^2}{20.6} \approx 0.557\)[/tex]
4. Humanities (Non-Athlete): [tex]\(\frac{(11 - 14.4)^2}{14.4} \approx 0.805\)[/tex]
5. Sciences (Athlete): [tex]\(\frac{(15 - 14.1)^2}{14.1} \approx 0.064\)[/tex]
6. Sciences (Non-Athlete): [tex]\(\frac{(9 - 9.9)^2}{9.9} \approx 0.082\)[/tex]
Adding these values together:
[tex]\[ \chi^2 = 0.832 + 1.181 + 0.557 + 0.805 + 0.064 + 0.082 \approx 3.521 \][/tex]
We round the final chi-square test statistic to one decimal place:
[tex]\[ \chi^2 \approx 3.5 \][/tex]
### Step 5: Interpret the Result
The test statistic [tex]\(\chi^2\)[/tex] is 3.5. This value would be compared to the critical value from the Chi-Square distribution table at the 5% significance level and with appropriate degrees of freedom (in this case, [tex]\((2-1) \times (3-1) = 2\)[/tex] degrees of freedom).
If the test statistic is less than the critical value, we do not reject the null hypothesis; hence, there is insufficient evidence to suggest that college major depends on whether a student is an athlete or not.
Given our test statistic of 3.5, we'd look up the critical value for [tex]\(\chi^2\)[/tex] with 2 degrees of freedom at the 5% significance level, which is approximately 5.991. Since 3.5 < 5.991, we fail to reject the null hypothesis.
Hence, we conclude that there is no strong evidence to suggest that being an athlete significantly affects the choice of college major.
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