We are given data on the extroversion levels of students in four different majors: English, History, Psychology, and Math. We want to perform an ANOVA test to determine if there is a significant difference in extroversion levels between the majors. We are given the sample means for each major ($\bar{X}_1 = 3.67, \bar{X}_2 = 3.02, \bar{X}_3 = 5.10, \bar{X}_4 = 3.47$), the sample sizes for each major ($n_1 = 60, n_2 = 44, n_3 = 21, n_4 = 60$), the between-groups sum of squares ($SS_B = 62.92$), the within-groups sum of squares ($SS_W = 1943.06$), and the significance level ($\alpha = 0.05$). We need to calculate the F-statistic and determine if it is significant.

1. Problem Description

We are given data on the extroversion levels of students in four different majors: English, History, Psychology, and Math. We want to perform an ANOVA test to determine if there is a significant difference in extroversion levels between the majors. We are given the sample means for each major ($\bar{X}_1 = 3.67, \bar{X}_2 = 3.02, \bar{X}_3 = 5.10, \bar{X}_4 = 3.47$), the sample sizes for each major ($n_1 = 60, n_2 = 44, n_3 = 21, n_4 = 60$), the between-groups sum of squares ($SS_B = 62.92$), the within-groups sum of squares ($SS_W = 1943.06$), and the significance level ($\alpha = 0.05$). We need to calculate the F-statistic and determine if it is significant.

2. Solution Steps

First, we need to calculate the degrees of freedom.

The degrees of freedom between groups ($df_B$) is the number of groups minus

1.

$df_B = k - 1$, where $k$ is the number of groups.

$df_B = 4 - 1 = 3$

The degrees of freedom within groups ($df_W$) is the total sample size minus the number of groups.

$df_W = N - k$, where $N$ is the total sample size.

$N = n_1 + n_2 + n_3 + n_4 = 60 + 44 + 21 + 60 = 185$

$df_W = 185 - 4 = 181$

Next, we calculate the mean square between groups ($MS_B$) and the mean square within groups ($MS_W$).

$MS_B = \frac{SS_B}{df_B} = \frac{62.92}{3} = 20.9733$

$MS_W = \frac{SS_W}{df_W} = \frac{1943.06}{181} = 10.7351$

Now, we can calculate the F-statistic.

$F = \frac{MS_B}{MS_W} = \frac{20.9733}{10.7351} = 1.9537$

We are given $\alpha = 0.05$. We need to find the critical F-value with $df_B = 3$ and $df_W = 181$. Since 181 is not a common value in F-tables, we can approximate using 120 or 200 degrees of freedom. Using an online F-distribution calculator, the critical F-value for $df_1 = 3, df_2 = 181$, and $\alpha = 0.05$ is approximately $2.65$.

Since our calculated F-statistic ($1.9537$) is less than the critical F-value ($2.65$), we fail to reject the null hypothesis.

3. Final Answer

Fail to reject the null hypothesis.