Certainly! Let's walk through the process of performing an Analysis of Variance (ANOVA) test. We have data on the total Fe content for four types of iron formations: carbonate, silicate, magnetite, and hematite.
### Hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): The means of the total Fe content for all four types of iron formations are equal.
[tex]\[
H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4
\][/tex]
- Alternative hypothesis ([tex]\(H_a\)[/tex]): At least two of the means are different.
[tex]\[
H_a: \text{At least two }\mu_i \text{ are unequal}
\][/tex]
### Building the ANOVA Table:
To carry out the ANOVA, we create an ANOVA table which includes sources of variation, degrees of freedom (df), sum of squares (SS), mean squares (MS), and the F-statistic.
1. Degrees of Freedom (df):
- Between groups (treatments): [tex]\( df_{between} = k - 1 = 4 - 1 = 3 \)[/tex], where [tex]\(k = 4\)[/tex] is the number of groups.
- Within groups (error): [tex]\( df_{within} = N - k = 40 - 4 = 36 \)[/tex], where [tex]\(N = 40\)[/tex] is the total number of observations.
2. Sum of Squares (SS):
- Between groups (SS[tex]\(_{between}\)[/tex]): 522.2
- Within groups (SS[tex]\(_{within}\)[/tex]): 174.00
3. Mean Squares (MS):
- Between groups (MS[tex]\(_{between}\)[/tex]):
[tex]\[
MS_{between} = \frac{SS_{between}}{df_{between}} = \frac{522.2}{3} = 174.07
\][/tex]
- Within groups (MS[tex]\(_{within}\)[/tex]):
[tex]\[
MS_{within} = \frac{SS_{within}}{df_{within}} = \frac{174.00}{36} = 4.83
\][/tex]
4. F-Statistic:
- The F-statistic is the ratio of the mean squares:
[tex]\[
F_{calculated} = \frac{MS_{between}}{MS_{within}} = \frac{174.07}{4.83} = 36.01
\][/tex]
5. p-value:
- Obtained p-value is approximately [tex]\(2.82 \times 10^{-5}\)[/tex].
### ANOVA Summary Table:
```
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
\multirow{2}{*}{Source} & \multicolumn{2}{|c|}{df} & \multicolumn{2}{|c|}{Sum of Squares} & \multicolumn{2}{|c|}{Mean Squares} & \multicolumn{2}{|c|}{F} \\
\hline
& Treatment & Error & Treatment & Error & Treatment & Error & Calculated F & p-value \\
\hline
& 3 & 36 & 522.2 & 174.0 & 174.07 & 4.83 & 36.01 & [tex]\(2.82 \times 10^{-5}\)[/tex] \\
\hline
Total & & & 696.2 & & & & \\
\hline
\end{tabular}
```
### Conclusion:
Since the calculated p-value ([tex]\(2.82 \times 10^{-5}\)[/tex]) is significantly less than the significance level ([tex]\(\alpha = 0.01\)[/tex]), we reject the null hypothesis. This indicates that there are significant differences between the means of the total Fe content for the different types of iron formations. In other words, not all the means are equal, and at least two of the means are significantly different.
