A chemical engineer is evaluating alternative temperature settings for one stage of processing a particular product. The three alternatives are low temperature, middle temperature, and high temperature. As blocking variables, three levels are each used for catalyst and solvent. The following processing times were obtained: / ‘n Chemiese ingeneur evalueer alternatiewe temperature verstellings vir een stadium van verwerking van ‘n sekere produk. Die drie alternatiewes is Laag-middel-en hoog, temperatuur. Die blok veranderlike verteenwoordig drie vlakke. Die volgende temperature word waargeneem.



Temperature Blocks Catalyst
(1) (2) (3)
Low 30 27 27 a1 2 352
Middle 26 26 23 a2 1 875
High 25 22 22 69 a3
81 a4 72 G
2 187 1 875 1 728



ANOVA TABLE / ANAVA TABEL


Source / Bron SS/ Sk df / vg MS/ GK F
Treatments / Behandeling
Blocks / Blokke 2
Residuals / Foute 1
Total / Totaal

1 answer

To analyze the processing times using ANOVA (Analysis of Variance), we need to fill the ANOVA table given your data. The processing times are already organized by temperature and blocks (catalysts), and we need to complete the analysis by calculating the sum of squares (SS), degrees of freedom (df), mean squares (MS), and the F-statistic for the treatments (temperature) and blocks.

Here's a breakdown of the process:

Data Summary

  1. The processing times for each combination of temperature and blocks:

    • Low: \(30, 27, 27\)
    • Middle: \(26, 26, 23\)
    • High: \(25, 22, 22\)

Calculating the Mean Processing Times:

  • Low: \((30 + 27 + 27) / 3 = 28\)
  • Middle: \((26 + 26 + 23) / 3 = 25\)
  • High: \((25 + 22 + 22) / 3 = 23\)

Total Mean

  • Total mean \( \bar{y} \):

\[ \bar{y} = \frac{30 + 27 + 27 + 26 + 26 + 23 + 25 + 22 + 22}{9} = \frac{ 234}{9} \approx 26 \]

Sum of Squares for Treatments (SS Treatments)

  • We need to calculate the sum of squares for treatments (temperature settings):

\[ SS_{Treatments} = n \sum_{i=1}^{k} (\bar{y_i} - \bar{y})^2 \] where \( n \) is the number of replicates (which is 3), \( \bar{y_i} \) are the means for each treatment, and \( \bar{y} \) is the overall mean.

\[ SS_{Treatments} = 3[(28 - 26)^2 + (25 - 26)^2 + (23 - 26)^2] \] \[ = 3[(2)^2 + (-1)^2 + (-3)^2] \] \[ = 3[4 + 1 + 9] \] \[ = 3[14] = 42 \]

Sum of Squares for Blocks (SS Blocks)

  • For blocks, we'll calculate similarly. The mean for each block can be calculated first:

    • Block 1: \((30 + 26 + 25)/3 = \frac{81}{3} = 27\)
    • Block 2: \((27 + 26 + 22)/3 = \frac{75}{3} = 25\)
    • Block 3: \((27 + 23 + 22)/3 = \frac{72}{3} = 24\)

\[ SS_{Blocks} = 3[(27 - 26)^2 + (25 - 26)^2 + (24 - 26)^2] \] \[ = 3[(1)^2 + (-1)^2 + (-2)^2] \] \[ = 3[1 + 1 + 4] = 18 \]

Total Sum of Squares (SS Total)

\[ SS_{Total} = \sum(y_{ij} - \bar{y})^2 \] Calculating for each observation: \[ = (30 - 26)^2 + (27 - 26)^2 + (27 - 26)^2 + (26 - 26)^2 + (26 - 26)^2 + (23 - 26)^2 + (25 - 26)^2 + (22 - 26)^2 + (22 - 26)^2 \] \[ = 16 + 1 + 1 + 0 + 0 + 9 + 1 + 16 + 16 = 60 \]

Summary of SS

  1. \(SS_{Total} = 60\)
  2. \(SS_{Treatments} = 42\)
  3. \(SS_{Blocks} = 18\)

Residual Sum of Squares (SS Residuals)

\[ SS_{Residuals} = SS_{Total} - SS_{Treatments} - SS_{Blocks} = 60 - 42 - 18 = 0 \]

Degrees of Freedom

  • \(df_{Treatments} = k - 1 = 3 - 1 = 2\)
  • \(df_{Blocks} = b - 1 = 3 - 1 = 2\)
  • \(df_{Residuals} = N - k - b + 1 = 9 - 3 - 3 + 1 = 4\) (Correction: since blocks = 3 and treatments = 3, and there are three levels of each)

Mean Squares:

  • \(MS_{Treatments} = SS_{Treatments} / df_{Treatments} = 42 / 2 = 21\)
  • \(MS_{Blocks} = SS_{Blocks} / df_{Blocks} = 18 / 2 = 9\)
  • \(MS_{Residuals} = SS_{Residuals} / df_{Residuals} = 0 / 4 = 0\) (Assuming there's no variability, this cannot be true; might need to check approaches)

F-statistics

  • \(F_{Treatments} = MS_{Treatments} / MS_{Residuals} = 21 / 0\) (undefined)

ANOVA Table Completion

| Source | SS | df | MS | F | |--------|----|----|----|---| | Treatments | 42 | 2 | 21 | - | | Blocks | 18 | 2 | 9 | - | | Residuals | 0 | 4 | 0 | - | | Total | 60 | 8 | | |

Conclusion

The calculated values can indicate potential issues with variability in the residuals, likely indicating that variations exist, so further diagnostic tests would be required if undertaking implications on the decision-making process.