To complete the ANOVA table, we need to understand the relationships between the columns.
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Degrees of Freedom (df):
- For treatments (C), df = k - 1, where k is the number of groups.
- For error (E), df = N - k, where N is the total number of observations.
- For total (B), df = N - 1.
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Mean Square (MS):
- For treatments, \( MS_{\text{Treatment}} = \frac{SS_{\text{Treatment}}}{df_{\text{Treatment}}} \),
- For error, \( MS_{\text{Error}} = \frac{SS_{\text{Error}}}{df_{\text{Error}}} \).
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F-statistic:
- \( F = \frac{MS_{\text{Treatment}}}{MS_{\text{Error}}} \).
Let's denote:
- C = df Treatment
- A = SS Error
- E = df Error
- B = SS Total
- F = F-statistic value for Treatment (we need to calculate this)
- G = table F-value at the critical level.
In the provided ANOVA table, we have certain values:
- SS Treatment = 12,689
- F-statistic = 10.6642
- Total df = 11
To proceed:
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Find the total sum of squares (B): \[ SS_{\text{Total}} = SS_{\text{Treatment}} + SS_{\text{Error}} \quad \Rightarrow \quad B = 12,689 + A. \]
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Determine the degrees of freedom for error (E) and the total degrees of freedom (11): \[ C + E = 11. \]
Given that \( MS_{\text{Error}} = 595.3611 \): \[ MS_{\text{Error}} = \frac{A}{E} \quad \Rightarrow \quad A = 595.3611 \times E. \]
We know \( MS_{\text{Treatment}} \) is calculated as follows: \[ MS_{\text{Treatment}} = \frac{12,689}{C}. \]
And F-statistic can be expressed as: \[ F = \frac{MS_{\text{Treatment}}}{MS_{\text{Error}}}. \] Substituting: \[ 10.6642 = \frac{\frac{12,689}{C}}{595.3611}. \] Solving for \( C \): \[ 10.6642 \times 595.3611 = \frac{12,689}{C}, \] \[ 6355.287537 = \frac{12,689}{C}, \] \[ C = \frac{12,689}{6355.287537} \approx 2. \]
So, the value of C is 2.
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