We are given a contingency table showing the number of students from different majors (Psychology, Business, and Math) who go to graduate school (Yes/No). We need to test if there is a relationship between college major and going to graduate school. We can use a Chi-square test for independence to determine if there's a statistically significant association between the two categorical variables.

1. Problem Description

We are given a contingency table showing the number of students from different majors (Psychology, Business, and Math) who go to graduate school (Yes/No). We need to test if there is a relationship between college major and going to graduate school. We can use a Chi-square test for independence to determine if there's a statistically significant association between the two categorical variables.

2. Solution Steps

First, we construct the contingency table.

| | Psychology | Business | Math | Total |

|------------------|------------|----------|------|-------|

| Graduate School (Yes) | 32 | 18 | 36 | 86 |

| No | 12 | 41 | 16 | 69 |

| Total | 44 | 59 | 52 | 155 |

Next, we compute the expected frequencies for each cell in the table. The expected frequency for a cell is calculated as:

$E_{ij} = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}$

So, the expected values are:

$E_{11} = \frac{86 \times 44}{155} \approx 24.42$

$E_{12} = \frac{86 \times 59}{155} \approx 32.76$

$E_{13} = \frac{86 \times 52}{155} \approx 28.81$

$E_{21} = \frac{69 \times 44}{155} \approx 19.58$

$E_{22} = \frac{69 \times 59}{155} \approx 26.24$

$E_{23} = \frac{69 \times 52}{155} \approx 23.19$

Now we have:

| | Psychology | Business | Math | Total |

|------------------|------------|----------|------|-------|

| Graduate School (Yes) | 32 (24.42) | 18 (32.76) | 36 (28.81) | 86 |

| No | 12 (19.58) | 41 (26.24) | 16 (23.19) | 69 |

| Total | 44 | 59 | 52 | 155 |

Next, we 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.

$\chi^2 = \frac{(32 - 24.42)^2}{24.42} + \frac{(18 - 32.76)^2}{32.76} + \frac{(36 - 28.81)^2}{28.81} + \frac{(12 - 19.58)^2}{19.58} + \frac{(41 - 26.24)^2}{26.24} + \frac{(16 - 23.19)^2}{23.19}$

$\chi^2 \approx \frac{(7.58)^2}{24.42} + \frac{(-14.76)^2}{32.76} + \frac{(7.19)^2}{28.81} + \frac{(-7.58)^2}{19.58} + \frac{(14.76)^2}{26.24} + \frac{(-7.19)^2}{23.19}$

$\chi^2 \approx \frac{57.4564}{24.42} + \frac{217.8576}{32.76} + \frac{51.6961}{28.81} + \frac{57.4564}{19.58} + \frac{217.8576}{26.24} + \frac{51.6961}{23.19}$

$\chi^2 \approx 2.35 + 6.65 + 1.79 + 2.93 + 8.30 + 2.23 = 24.25$

The degrees of freedom (df) are calculated as $(r-1)(c-1)$, where $r$ is the number of rows and $c$ is the number of columns. In this case, $df = (2-1)(3-1) = 1 \times 2 = 2$.

We compare our calculated $\chi^2$ value (24.25) with the critical value from the Chi-square distribution with 2 degrees of freedom. For a significance level of $\alpha = 0.05$, the critical value is 5.

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1. Since 24.25 > 5.991, we reject the null hypothesis that there is no association.

3. Final Answer

There is a statistically significant relationship between college major and going to graduate school.