To solve this problem, we need to perform a one-way ANOVA to determine if there's a statistically significant difference in exam scores among the three sections. Here is the step-by-step solution:
1. State the hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): The mean exam scores of all sections are equal.
- Alternative hypothesis ([tex]\(H_1\)[/tex]): At least one section has a different mean exam score.
2. List the data:
[tex]\[
\begin{array}{|c|c|c|}
\hline
\text{Section 1} & \text{Section 2} & \text{Section 3} \\
\hline
88.2 & 45.1 & 74.3 \\
78.8 & 68.1 & 77.4 \\
71.9 & 69.1 & 75.6 \\
64.1 & 87.4 & 84.8 \\
70.8 & 71.9 & 67.4 \\
73.9 & 47.7 & 82.3 \\
82.5 & 83.8 & 78.2 \\
79.7 & 61.8 & 71.1 \\
84.5 & 60.9 & 78.8 \\
\hline
\end{array}
\][/tex]
3. Calculate the ANOVA summary statistics:
We would calculate:
- The mean and variance of each section.
- The overall mean of all the scores.
- Between-group variability (sum of squares between or [tex]\(SSB\)[/tex]).
- Within-group variability (sum of squares within or [tex]\(SSW\)[/tex]).
- The degrees of freedom between ([tex]\(df_b\)[/tex]) and within ([tex]\(df_w\)[/tex]).
- The mean square between ([tex]\(MSB\)[/tex]) and mean square within ([tex]\(MSW\)[/tex]).
- The F-ratio ([tex]\(F\)[/tex]).
4. Compute the F-ratio:
The F-ratio is calculated by dividing the mean square between by the mean square within.
5. Determine the p-value:
Using the F-distribution and the calculated F-ratio and degrees of freedom, we compare to a critical value or use statistical software to determine the p-value.
Given that all calculations yield:
[tex]\[
F = 3.547
\][/tex]
[tex]\[
p = 0.0447
\][/tex]
6. Conclusion:
- Compare the p-value with the significance level ([tex]\(\alpha = 0.05\)[/tex]).
- Since [tex]\(p = 0.0447 < 0.05\)[/tex], we reject the null hypothesis.
Therefore, we have evidence to suggest that there is a statistically significant difference in exam scores among the three sections.
[tex]\[
\begin{array}{l}
F = 3.547 \\
p = 0.0447
\end{array}
\][/tex]
