Question
Consider the following partially completed analysis of variance table:
Source SS df MS F Ftab
Factor A 467.2837 B 467.2837 F H
Factor B 1547.8312 3 D G I
Residuals 503.26 C E
Total A 7
Complete the analysis of variance table and test whether for the difference amongst the treatment (Factor A) means and block (Factor B) means. Test at 10% significance level.
Source SS df MS F Ftab
Factor A 467.2837 B 467.2837 F H
Factor B 1547.8312 3 D G I
Residuals 503.26 C E
Total A 7
Complete the analysis of variance table and test whether for the difference amongst the treatment (Factor A) means and block (Factor B) means. Test at 10% significance level.
Answers
GPT-4o mini
To complete the analysis of variance (ANOVA) table, we need to fill in the degrees of freedom (df), mean squares (MS), F-values, and any other missing values based on given information. Here we have data for Factor A, Factor B, Residuals, and Total.
### Step 1: Define the given information
- **Factor A**:
- SS_A = 467.2837
- df_A = B
- MS_A = 467.2837 (given)
- **Factor B**:
- SS_B = 1547.8312
- df_B = 3
- MS_B = SS_B / df_B
- **Residuals**:
- SS_Residuals = 503.26
- df_Residuals = C
- MS_Residuals = SS_Residuals / df_Residuals
- **Total**:
- SS_Total = SS_A + SS_B + SS_Residuals
- df_Total = A = df_A + df_B + df_Residuals = B + 3 + C
### Step 2: Calculate Degrees of Freedom
1. **Factor A (df_A)**: The degrees of freedom for Factor A is given by:
\[
df_A = k_A - 1
\]
where \( k_A \) is the number of levels for Factor A.
2. **Factor B (df_B)** = 3 (given).
3. **Residuals (df_Residuals)**:
Let’s denote it as C.
4. **Total (df_Total)**: The total degrees of freedom is given by:
\[
df_Total = N - 1
\]
where \( N \) is the total number of observations.
Since we know \( df_Total = df_A + df_B + df_Residuals \):
\[
df_Total = B + 3 + C
\]
5. The value of \( A \) is given as 7.
Since \( A \) is actually \( df_{Total} + 1 \):
\[
df_{Total} = 7 - 1 = 6
\]
This means:
\[
6 = B + 3 + C
\]
From this, we can solve for \( B + C = 3 \).
### Step 3: Calculating the Mean Squares
#### For Factor B:
\[
MS_B = \frac{SS_B}{df_B} = \frac{1547.8312}{3} = 515.9437
\]
#### For Residuals:
Now we know:
\[
MS_{Residuals} = \frac{SS_{Residuals}}{df_{Residuals}} = \frac{503.26}{C}
\]
### Solving for Unknowns
From the previous equations:
\[
B + C = 3
\]
Taking two plausible values that add to 3, we can set \( B = 1 \) and \( C = 2 \). Thus:
\[
df_A = B = 1 \Rightarrow k_A = 2 \text{ (2 levels for Factor A)}
\]
\[
df_{Residuals} = C = 2
\]
### Step 4: Completing the ANOVA Table
#### Total Sum of Squares
\[
SS_{Total} = SS_A + SS_B + SS_{Residuals} = 467.2837 + 1547.8312 + 503.26 = 2518.375
\]
#### Mean Squares for Residuals:
\[
MS_{Residuals} = \frac{503.26}{2} = 251.63
\]
### Step 5: F-values
#### F for Factor A:
\[
F_A = \frac{MS_A}{MS_{Residuals}} = \frac{467.2837}{251.63} \approx 1.858
\]
#### F for Factor B:
\[
F_B = \frac{MS_B}{MS_{Residuals}} = \frac{515.9437}{251.63} \approx 2.046
\]
### Step 6: Ftab value
Assuming an alpha level of 0.10, degrees of freedom for Factor A (1, 2) and Factor B (3, 2), we will consult an F-distribution table. The critical value \( F_{tab} \) for \( df_1 = 1, df_2 = 2 \) and \( \alpha = 0.10 \) is approximately \( 5.55 \) and for \( df_1 = 3, df_2 = 2 \) is approximately \( 4.78 \).
### Step 7: Decision
- Compare calculated F-values with Ftab.
- For Factor A:
- \( F_A \approx 1.858 < F_{tab} \approx 5.55 \) (fail to reject null).
- For Factor B:
- \( F_B \approx 2.046 < F_{tab} \approx 4.78 \) (fail to reject null).
### Conclusion
At the 10% significance level, there is no evidence to support a difference among the treatment means for both Factor A and Factor B based on this ANOVA analysis.
### Completed ANOVA Table
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Source} & SS & df & MS & F & F_{tab} \\
\hline
\text{Factor A} & 467.2837 & 1 & 467.2837 & 1.858 & 5.55 \\
\text{Factor B} & 1547.8312 & 3 & 515.9437 & 2.046 & 4.78 \\
\text{Residuals} & 503.26 & 2 & 251.63 & & \\
\hline
\text{Total} & 2518.375 & 6 & & & \\
\hline
\end{array}
\]
### Step 1: Define the given information
- **Factor A**:
- SS_A = 467.2837
- df_A = B
- MS_A = 467.2837 (given)
- **Factor B**:
- SS_B = 1547.8312
- df_B = 3
- MS_B = SS_B / df_B
- **Residuals**:
- SS_Residuals = 503.26
- df_Residuals = C
- MS_Residuals = SS_Residuals / df_Residuals
- **Total**:
- SS_Total = SS_A + SS_B + SS_Residuals
- df_Total = A = df_A + df_B + df_Residuals = B + 3 + C
### Step 2: Calculate Degrees of Freedom
1. **Factor A (df_A)**: The degrees of freedom for Factor A is given by:
\[
df_A = k_A - 1
\]
where \( k_A \) is the number of levels for Factor A.
2. **Factor B (df_B)** = 3 (given).
3. **Residuals (df_Residuals)**:
Let’s denote it as C.
4. **Total (df_Total)**: The total degrees of freedom is given by:
\[
df_Total = N - 1
\]
where \( N \) is the total number of observations.
Since we know \( df_Total = df_A + df_B + df_Residuals \):
\[
df_Total = B + 3 + C
\]
5. The value of \( A \) is given as 7.
Since \( A \) is actually \( df_{Total} + 1 \):
\[
df_{Total} = 7 - 1 = 6
\]
This means:
\[
6 = B + 3 + C
\]
From this, we can solve for \( B + C = 3 \).
### Step 3: Calculating the Mean Squares
#### For Factor B:
\[
MS_B = \frac{SS_B}{df_B} = \frac{1547.8312}{3} = 515.9437
\]
#### For Residuals:
Now we know:
\[
MS_{Residuals} = \frac{SS_{Residuals}}{df_{Residuals}} = \frac{503.26}{C}
\]
### Solving for Unknowns
From the previous equations:
\[
B + C = 3
\]
Taking two plausible values that add to 3, we can set \( B = 1 \) and \( C = 2 \). Thus:
\[
df_A = B = 1 \Rightarrow k_A = 2 \text{ (2 levels for Factor A)}
\]
\[
df_{Residuals} = C = 2
\]
### Step 4: Completing the ANOVA Table
#### Total Sum of Squares
\[
SS_{Total} = SS_A + SS_B + SS_{Residuals} = 467.2837 + 1547.8312 + 503.26 = 2518.375
\]
#### Mean Squares for Residuals:
\[
MS_{Residuals} = \frac{503.26}{2} = 251.63
\]
### Step 5: F-values
#### F for Factor A:
\[
F_A = \frac{MS_A}{MS_{Residuals}} = \frac{467.2837}{251.63} \approx 1.858
\]
#### F for Factor B:
\[
F_B = \frac{MS_B}{MS_{Residuals}} = \frac{515.9437}{251.63} \approx 2.046
\]
### Step 6: Ftab value
Assuming an alpha level of 0.10, degrees of freedom for Factor A (1, 2) and Factor B (3, 2), we will consult an F-distribution table. The critical value \( F_{tab} \) for \( df_1 = 1, df_2 = 2 \) and \( \alpha = 0.10 \) is approximately \( 5.55 \) and for \( df_1 = 3, df_2 = 2 \) is approximately \( 4.78 \).
### Step 7: Decision
- Compare calculated F-values with Ftab.
- For Factor A:
- \( F_A \approx 1.858 < F_{tab} \approx 5.55 \) (fail to reject null).
- For Factor B:
- \( F_B \approx 2.046 < F_{tab} \approx 4.78 \) (fail to reject null).
### Conclusion
At the 10% significance level, there is no evidence to support a difference among the treatment means for both Factor A and Factor B based on this ANOVA analysis.
### Completed ANOVA Table
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Source} & SS & df & MS & F & F_{tab} \\
\hline
\text{Factor A} & 467.2837 & 1 & 467.2837 & 1.858 & 5.55 \\
\text{Factor B} & 1547.8312 & 3 & 515.9437 & 2.046 & 4.78 \\
\text{Residuals} & 503.26 & 2 & 251.63 & & \\
\hline
\text{Total} & 2518.375 & 6 & & & \\
\hline
\end{array}
\]