F = (MSB)/(MSW) = (15.83335)/(2.285185) ~= 6.928

$$F = \frac{MSB}{MSW} = \frac{15.83335}{2.285185} \approx 6.928$$

The calculated F-value is $\boxed{6.928}$.

Part iv: State the critical F-value.

To find the critical F-value, we use the F-distribution table with:

• Degrees of freedom for the numerator ($df_1$) = 2
• Degrees of freedom for the denominator ($df_2$) = 27
• Level of significance ($\alpha$) = 0.05

Looking up the F-distribution table for $df_1 = 2$ and $df_2 = 27$ at $\alpha = 0.05$, the critical F-value is approximately 3.35.

• Critical F-value ($F_{crit}$): $\boxed{3.35}$

Part v: State the decision rule.

• Decision Rule: Reject the null hypothesis ($H_0$) if the calculated F-value is greater than the critical F-value ($F_{calc} > F_{crit}$). Otherwise, do not reject the null hypothesis.

Part vi: Make a decision.

• Calculated F-value ($F_{calc}$): 6.928
• Critical F-value ($F_{crit}$): 3.35

Since $6.928 > 3.35$, we reject the null hypothesis.

• Decision: $\boxed{\text{Reject } H_0}$

Part vii: Interpret the result.

Since the null hypothesis is rejected, there is sufficient evidence to conclude that there is a significant difference in the mean scores among students in Agric Education, Business Education, and Industrial Technical Education. This means that at least one of the group means is significantly different from the others.