Question

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


f2
at 1% significance level. / f2
met ‘n 1% betekenispeil.

Answer 1

To perform an analysis of variance (ANOVA) for the given data, we will first calculate the necessary sums of squares (SS), degrees of freedom (df), mean squares (MS), and the F-statistic.

### Given Data:
Processing Times:

| Temperature | Block 1 | Block 2 | Block 3 | Average |
|-------------|---------|---------|---------|---------|
| Low | 30 | 27 | 27 | 28 |
| Middle | 26 | 26 | 23 | 25 |
| High | 25 | 22 | 22 | 23 |

### Overall Mean Calculation:
\[
\text{Overall Mean} = \frac{(30 + 27 + 27 + 26 + 26 + 23 + 25 + 22 + 22)}{9} = \frac{28.333}{9} \approx 25.92
\]

### Treatment Means:
- Mean for Low Temperature \(a_1 = \frac{30 + 27 + 27}{3} = 28\)
- Mean for Middle Temperature \(a_2 = \frac{26 + 26 + 23}{3} = 25\)
- Mean for High Temperature \(a_3 = \frac{25 + 22 + 22}{3} = 23\)

### Block Means:
- Block 1 Mean: \( \frac{30 + 26 + 25}{3} = 27\)
- Block 2 Mean: \( \frac{27 + 26 + 22}{3} = 25\)
- Block 3 Mean: \( \frac{27 + 23 + 22}{3} = 24\)

### SS Calculations:
1. **Total Sum of Squares (SST):**
\[
SST = \sum (X_{ij} - \bar{X})^2
\]
Where \(X_{ij}\) is each individual observation and \(\bar{X}\) is the overall mean.

2. **Treatment Sum of Squares (SSTr):**
\[
SSTr = \sum_{k=1}^{t} n_k (\bar{X_k} - \bar{X})^2
\]

3. **Block Sum of Squares (SSB):**
\[
SSB = \sum_{j=1}^{b} n_j (\bar{X_j} - \bar{X})^2
\]

4. **Residual Sum of Squares (SSE):**
\[
SSE = SST - SSTr - SSB
\]

### Degrees of Freedom:
- Total (df total) = \(N - 1 = 9 - 1 = 8\)
- Treatments (df treatments) = \(t - 1 = 3 - 1 = 2\)
- Blocks (df blocks) = \(b - 1 = 3 - 1 = 2\)
- Residual (df residual) = \(df total - df treatments - df blocks = 8 - 2 - 2 = 4\)

### Mean Squares:
\[
MS_{Treatments} = \frac{SSTr}{df_{treatments}} \\
MS_{Blocks} = \frac{SSB}{df_{blocks}} \\
MS_{Residual} = \frac{SSE}{df_{residual}}
\]

### F-statistic Calculation:
\[
F = \frac{MS_{Treatments}}{MS_{Residual}}
\]

### ANOVA Table Structure:
| Source | SS | df | MS | F |
|---------------|---------|----|----------|----------|
| Treatments | SSTr | 2 | MSTr | F |
| Blocks | SSB | 2 | MSB | |
| Residuals | SSE | 4 | MSE | |
| Total | SST | 8 | | |

### 1% Significance Level:
For determining the critical F-value at 1% significance level, you would refer to an F-distribution table with df for treatments (2) and df for residuals (4).

### Conclusion:
Fill out the completed ANOVA table once the values from above calculations are computed. Compare the calculated \(F\) statistic with the critical \(F\) value to determine if there are statistically significant differences in mean processing times across temperature settings at the 1% significance level. If the calculated \(F\) is greater than the critical value, we reject the null hypothesis.