Question

Consider the following partially completed ANOVA and give the value of (e). / Beskou die onvoltooide ANOVA en gee die waarde van (e).


Source SS df MS Fstat Ftab
Treatments 142.6 (d) 35.6675 (g) (i)

Blocks 68.14 5 (f) (h) (j)
Errors (b) 20 1.9165
Total (c) (e)

Answer 1

To fill in the values for the ANOVA table, we need to first understand some key relationships in ANOVA. Let's go through it step by step:

1. **Degrees of Freedom (df)**:
- For Treatments (df\_treatments): Typically, df is one less than the number of treatment groups. Therefore, if we let \( d \) be the number of treatment groups, then:
\[
\text{df treating} = d - 1
\]
- For Blocks: Already given as 5, which means there are 6 blocks (block df = 5).
- For Errors: df for errors is the total df minus the sum of df for treatments and blocks.
- Total df (df\_total) is the total number of observations minus 1.

2. **Mean Squares (MS)**:
- Mean Square for Treatments (MS\_treatments) is given as:
\[
\text{MS Treatments} = \frac{SS Treatments}{df Treatments}
\]
\[
f = \frac{SS treatments}{d} = 35.6675
\]
- For Blocks:
\[
\text{MS Blocks} = \frac{SS Blocks}{df Blocks} = \frac{68.14}{5} = 13.628
\]
- For Errors:
\[
\text{MS Errors} = \frac{SS Errors}{df Errors} = 1.9165
\]

3. **Sum of Squares (SS)**:
- Total SS is the sum of Treatments, Blocks, and Errors:
\[
\text{Total SS} = SS Treatments + SS Blocks + SS Errors
\]
- From given data, we can denote:
\[
c = 142.6 + 68.14 + b
\]

4. **Calculating SS Errors**:
You have \( df\_errors = 20 \). Since MS for errors is given as:
\[
\text{MS Errors} = \frac{SS Errors}{df Errors} = 1.9165
\]
Then:
\[
b = MS Errors \times df Errors = 1.9165 \times 20 = 38.33
\]

5. **Finding Values**:
- **Value of (b)**: Now that we have \( b \):
\[
b = 38.33
\]
- **Finding Total SS (c)**:
\[
c = 142.6 + 68.14 + 38.33 = 249.07
\]
- **Finding Total df (e)**:
Total df = Total number of samples (N) - 1. Assuming that N is the total number of observations across treatments and blocks.

Given that:
- If Treatments have \( d \) groups and Blocks have 6, then if we have:

For the formula:
\[
N = df\_treatments + df\_blocks + df\_errors + 1 = (d - 1) + 5 + 20 + 1
\]
This would mean:
\[
N = d + 25
\]
Thus, the total df (e) for the ANOVA table would be:
\[
e = N - 1 = (d + 25) - 1 = d + 24
\]

Assuming there are 6 treatment groups:

If \( d = 6 \):
\[
e = 6 + 24 = 30
\]

Using these relations, we can fill out the missing values in the ANOVA table.

In conclusion:
- The value of (e) as calculated is:
**e = 30**