To complete the ANOVA table and find the value for (c), we need to use the following relationships:
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Total Sum of Squares (SS_total) is the sum of the Treatment SS, Block SS, and Error SS: \[ SS_{total} = SS_{treatments} + SS_{blocks} + SS_{errors} \]
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The degrees of freedom (df) for each component can be found using:
- df for treatments = k - 1 (where k is the number of treatment groups)
- df for blocks = b - 1 (where b is the number of blocks)
- df for errors = (k - 1)(b - 1)
- Total df = df for treatments + df for blocks + df for errors
Given:
- SS_treatments = 142.67
- SS_blocks = 68.14
- SS_errors = b (unknown, but we have df = 20)
Let's denote:
- SS_errors = b
- Total SS = c
- Total df = e
- df_treatments = d
- df_blocks = 5
- df_errors = 20
To find \( c \): \[ c = 142.67 + 68.14 + b \]
Next, we note that the total df: \[ e = d + 5 + 20 \]
Also, since we know \( MS = \frac{SS}{df} \):
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For Blocks, we can calculate \( MS_{blocks} \): \[ MS_{blocks} = \frac{68.14}{5} = 13.628 \] So, \( f = 13.628 \).
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For Errors, since \( MS_{errors} = 1.9165 \) (given), but we don't have \( SS_{errors} \) yet.
Finding \( SS_{errors} \) using \( MS \):
We have: \[ MS_{errors} = \frac{SS_{errors}}{df_{errors}} \] This gives us: \[ SS_{errors} = MS_{errors} \times df_{errors} = 1.9165 \times 20 = 38.33 \] Now, substitute back: \[ SS_{total} = SS_{treatments} + SS_{blocks} + SS_{errors} = 142.67 + 68.14 + 38.33 \] Calculating: \[ SS_{total} = 249.14 \] So, the value of \( c \) is: \[ c = 249.14 \]
Thus, the final answer is:
(c) = 249.14