To complete the partially finished ANOVA table, we need to determine the missing values for \( (A) \), \( (B) \), \( (C) \), \( (D) \), \( (E) \), and \( (F) \).
From the information provided, we can identify a few key components:
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The degrees of freedom (df) for the total is given as 11. If we assume that there are \( k \) treatments, then the degrees of freedom for the treatments is \( k - 1 \), and for error, it's \( N - k \). The total degrees of freedom is \( N - 1 \), where \( N \) is the total number of observations.
Given that: \[ df_{total} = df_{treatment} + df_{error} = 11 \]
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The mean square for treatments (MS Treatment) can be computed as: \[ MS_{Treatment} = \frac{SS_{Treatment}}{df_{Treatment}} = \frac{12,689}{C} \] The F-statistic is given as: \[ F = \frac{MS_{Treatment}}{MS_{Error}} = \frac{12,689 / C}{595.3611} = 10.6642 \]
Rearranging gives: \[ 10.6642 \times 595.3611 = \frac{12,689}{C} \] Let’s compute \( C \): \[ 10.6642 \times 595.3611 = 6,343.113332 \] \[ 12,689 = C \times 6,343.113332 \] \[ C = \frac{12,689}{6,343.113332} \approx 2.000 \] (This indicates that there are approximately 3 levels of treatment, as \( df_{Treatment} = k - 1 \) implies \( k = C + 1 = 3 \)).
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For the error sum of squares (SS Error) and total (SS Total): We can find \( (A) \) using the formula: \[ SS_{Total} = SS_{Treatment} + SS_{Error} = B \] Hence, \[ B = A + 12,689 \] Given \( df_{Error} (D) = N - k \) is determined:
- If \( C = 2 \), df for treatments is 2, thus \( df_{error} = 11 - 2 = 9 \).
The mean square for error (MS Error) can therefore be calculated: \[ MS_{Error} = 595.3611 \quad \Rightarrow \quad A = MS_{Error} \cdot df_{Error} = 595.3611 \cdot 9 = 5,358.25 \]
The total sum of squares would now be: \[ B = SS_{Treatment} + SS_{Error} = 12,689 + 5,358.25 = 18,047.25 \]
Now we can fill in the ANOVA table values:
- \( (A) = 5,358.25 \)
- \( (B) = 18,047.25 \)
- \( (C) = 2 \)
- \( (D) = 9 \)
- \( (E) = 10.6642 \times 595.3611 = 6,343.113332 \)
- \( (F) = \) Lookup F critical value in F-distribution table, using \( df_{Treatment} = 2 \) and \( df_{Error} = 9 \) at \( \alpha = 0.05 \).
To find \( F_{tab} \):
Using F-distribution tables, we find that: \[ F_{(2, 9, 0.05)} \approx 4.26 \]
Thus, the complete values in the ANOVA table will be approximately: \[ \begin{align*} & (A) = 5,358.25 \ & (B) = 18,047.25 \ & (C) = 2 \ & (D) = 9 \ & (E) = 6,343.113332 \ & (F) = 4.26 \text{ (based on the lookup)} \end{align*} \]
Final Answers:
- The value of \( F \) at \( \alpha = 0.05 \) is approximately 4.26.
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