Consider the following partially completed analysis of variance table:

Question

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.

Answer 1

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

\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Source} & SS & df & MS & F & F_{tab} \ \hline \text{Factor A} & 467.2837 & 1 & 467.2837 & 1.858 & 5.55 \ \text{Factor B} & 1547.8312 & 3 & 515.9437 & 2.046 & 4.78 \ \text{Residuals} & 503.26 & 2 & 251.63 & & \ \hline \text{Total} & 2518.375 & 6 & & & \ \hline \end{array} \]