Questions: You are interested in investigating whether gender and major are associated at your college. The table below shows the results of a survey. Frequencies of Majors and Gender - Math/Science: Men 92, Women 113 - Arts/Humanities: Men 93, Women 113 - Business/Econ.: Men 99, Women 94 - Other: Men 58, Women 33 What can be concluded at the alpha=0.05 significance level? a. What is the correct statistical test to use? - One sample t-test - Two Sample t-test - Paired t-test - Chi-square Test of Independence b. What are the null and alternative hypotheses? H0: - College major and gender are independent. - The distribution of college major is not the same for each gender. - The distribution of college major is the same for each gender. - College major and gender are dependent.

Solution Steps

We are testing the association between college major and gender. The null and alternative hypotheses are defined as follows:

• \( H_0 \): College major and gender are independent.
• \( H_1 \): College major and gender are dependent.

The observed frequencies from the survey are organized in a contingency table as follows:

\[ \begin{array}{|c|c|c|c|c|} \hline & \text{Math/Science} & \text{Arts/Humanities} & \text{Business/Econ.} & \text{Other} \\ \hline \text{Men} & 92 & 93 & 99 & 58 \\ \hline \text{Women} & 113 & 113 & 94 & 33 \\ \hline \end{array} \]

The expected frequencies for each cell in the contingency table are calculated using the formula:

\[ E = \frac{R_i \times C_j}{N} \]

where \( R_i \) is the total for row \( i \), \( C_j \) is the total for column \( j \), and \( N \) is the grand total. The expected frequencies are:

\[ \begin{array}{|c|c|c|c|c|} \hline & \text{Math/Science} & \text{Arts/Humanities} & \text{Business/Econ.} & \text{Other} \\ \hline \text{Men} & 100.8777 & 101.3698 & 94.9727 & 44.7799 \\ \hline \text{Women} & 104.1223 & 104.6302 & 98.0273 & 46.2201 \\ \hline \end{array} \]

The Chi-Square test statistic \( \chi^2 \) is calculated using the formula:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

where \( O \) is the observed frequency and \( E \) is the expected frequency. The contributions from each cell are:

• For cell (1, 1): \( O = 92, E = 100.8777 \Rightarrow \frac{(92 - 100.8777)^2}{100.8777} = 0.7813 \)
• For cell (1, 2): \( O = 93, E = 101.3698 \Rightarrow \frac{(93 - 101.3698)^2}{101.3698} = 0.6911 \)
• For cell (1, 3): \( O = 99, E = 94.9727 \Rightarrow \frac{(99 - 94.9727)^2}{94.9727} = 0.1708 \)
• For cell (1, 4): \( O = 58, E = 44.7799 \Rightarrow \frac{(58 - 44.7799)^2}{44.7799} = 3.9029 \)
• For cell (2, 1): \( O = 113, E = 104.1223 \Rightarrow \frac{(113 - 104.1223)^2}{104.1223} = 0.7569 \)
• For cell (2, 2): \( O = 113, E = 104.6302 \Rightarrow \frac{(113 - 104.6302)^2}{104.6302} = 0.6695 \)
• For cell (2, 3): \( O = 94, E = 98.0273 \Rightarrow \frac{(94 - 98.0273)^2}{98.0273} = 0.1655 \)
• For cell (2, 4): \( O = 33, E = 46.2201 \Rightarrow \frac{(33 - 46.2201)^2}{46.2201} = 3.7813 \)

Summing these contributions gives:

\[ \chi^2 = 10.9193 \]

The degrees of freedom \( df \) for the Chi-Square test is calculated as:

\[ df = (r - 1)(c - 1) \]

where \( r \) is the number of rows and \( c \) is the number of columns. In this case:

\[ df = (2 - 1)(4 - 1) = 3 \]

The critical value for \( \chi^2 \) at \( \alpha = 0.05 \) with 3 degrees of freedom is:

\[ \chi^2_{\alpha, df} = 7.8147 \]

The p-value associated with the calculated Chi-Square statistic is:

\[ P = P(\chi^2 > 10.9193) = 0.0122 \]

Since the p-value \( 0.0122 \) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis \( H_0 \).

There is sufficient evidence to conclude that college major and gender are dependent.

Final Answer