Module 4 Solved Practice Problems

Exam Practice - Work through each problem by hand first using scratch paper and a scientific calculator. Click "View Step-by-Step Solution" to verify your arithmetic, ANOVA partition tables, and final statistical decisions.

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Problem 1: Completely Randomized Design (CRD)

Problem Statement

Three fertilizer compounds ($A, B, C$) are applied to 12 homogeneous experimental plots ($t = 3, r = 4, N = 12$). The measured crop yields (in kg) are:

• Fertilizer A: $18, 20, 19, 22$
• Fertilizer B: $24, 22, 21, 23$
• Fertilizer C: $26, 25, 27, 28$

Test at the $\alpha = 0.05$ significance level whether there is a statistically significant difference in mean yield among the three fertilizers. (Given: Critical $F_{0.05}(2, 9) \approx 4.26$).

View Step-by-Step Solution

1. Hypothesis Formulation

• Null Hypothesis ($H_0$): $\mu_A = \mu_B = \mu_C$ (The mean yields of all three fertilizers are equal).
• Alternative Hypothesis ($H_1$): At least one fertilizer mean yield is different.

2. Totals & Correction Factor ($CF$)

• $T_A = 18 + 20 + 19 + 22 = 79$
• $T_B = 24 + 22 + 21 + 23 = 90$
• $T_C = 26 + 25 + 27 + 28 = 106$
• $\text{Grand Total } G = 79 + 90 + 106 = 275$
• $CF = \frac{G^2}{N} = \frac{275^2}{12} = \frac{75625}{12} \approx 6302.0833$

3. Sum of Squares Computations

• $\sum y_{ij}^2 = 18^2 + 20^2 + 19^2 + 22^2 + 24^2 + 22^2 + 21^2 + 23^2 + 26^2 + 25^2 + 27^2 + 28^2 = 6413$
• $\text{SST} = \sum y_{ij}^2 - CF = 6413 - 6302.0833 = 110.9167$
• $\text{SSTreat} = \frac{79^2 + 90^2 + 106^2}{4} - CF = \frac{6241 + 8100 + 11236}{4} - 6302.0833 = 6394.25 - 6302.0833 = 92.1667$
• $\text{SSE} = \text{SST} - \text{SSTreat} = 110.9167 - 92.1667 = 18.7500$

4. CRD ANOVA Table

Source	$\text{df}$	$\text{SS}$	$\text{MS}$	Calculated $F$	Tabulated $F_{0.05}$
Fertilizers (Treatments)	$3 - 1 = 2$	$92.1667$	$\frac{92.1667}{2} = 46.0833$	$\frac{46.0833}{2.0833} \approx 22.12$	$4.26$
Error (Residual)	$12 - 3 = 9$	$18.7500$	$\frac{18.7500}{9} = 2.0833$	-	-
Total	$12 - 1 = 11$	$110.9167$	-	-	-

5. Decision & Academic Conclusion

• Comparison: $F_{\text{cal}} = 22.12 > F_{\text{tab}} = 4.26$.
• Decision: Reject $H_0$ at the $5\%$ level of significance.
• Conclusion: "Since the calculated $F$-value ($22.12$) is greater than the tabulated $F$-value ($4.26$) at the $5\%$ level of significance, the null hypothesis ($H_0$) is rejected, indicating that there is a statistically significant difference in mean yield among the three fertilizers."

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Problem 2: Randomized Block Design (RBD)

Problem Statement

An experiment measures cholesterol content (%) across 4 different diet foods ($D_1, D_2, D_3, D_4$) tested across 4 distinct laboratories ($A, B, C, D$) to control for inter-laboratory testing bias ($t = 4, r = 4, N = 16$).

Laboratory (Block)	$D_1$	$D_2$	$D_3$	$D_4$
A	2	5	4	3
B	3	7	4	2
C	5	5	6	4
D	5	5	8	5

Analyze the data using RBD and state your conclusion at $\alpha = 0.05$. (Given: Critical $F_{0.05}(3, 9) = 3.86$).

View Step-by-Step Solution

1. Totals Calculation

• Treatment Totals ($T_i$):
  • $T_{D1} = 2 + 3 + 5 + 5 = 15$
  • $T_{D2} = 5 + 7 + 5 + 5 = 22$
  • $T_{D3} = 4 + 4 + 6 + 8 = 22$
  • $T_{D4} = 3 + 2 + 4 + 5 = 14$
• Block Totals ($B_j$):
  • $B_A = 2 + 5 + 4 + 3 = 14$
  • $B_B = 3 + 7 + 4 + 2 = 16$
  • $B_C = 5 + 5 + 6 + 4 = 20$
  • $B_D = 5 + 5 + 8 + 5 = 23$
• $\text{Grand Total } G = 15 + 22 + 22 + 14 = 73$
• $CF = \frac{G^2}{N} = \frac{73^2}{16} = \frac{5329}{16} = 333.0625$

2. Sum of Squares Partitioning

• $\sum y_{ij}^2 = 4 + 25 + 16 + 9 + 9 + 49 + 16 + 4 + 25 + 25 + 36 + 16 + 25 + 25 + 64 + 25 = 373$
• $\text{SST} = 373 - 333.0625 = 39.9375$
• $\text{SSTreat} = \frac{15^2 + 22^2 + 22^2 + 14^2}{4} - CF = \frac{1389}{4} - 333.0625 = 347.25 - 333.0625 = 14.1875$
• $\text{SSBlock} = \frac{14^2 + 16^2 + 20^2 + 23^2}{4} - CF = \frac{1381}{4} - 333.0625 = 345.25 - 333.0625 = 12.1875$
• $\text{SSE} = \text{SST} - \text{SSTreat} - \text{SSBlock} = 39.9375 - 14.1875 - 12.1875 = 13.5625$

3. RBD ANOVA Table

Source	$\text{df}$	$\text{SS}$	$\text{MS}$	Calculated $F$	Critical $F_{0.05}$
Diets (Treatments)	$4 - 1 = 3$	$14.1875$	$\frac{14.1875}{3} \approx 4.7292$	$\frac{4.7292}{1.5069} \approx 3.14$	$3.86$
Labs (Blocks)	$4 - 1 = 3$	$12.1875$	$\frac{12.1875}{3} \approx 4.0625$	-	-
Error	$(4-1)(4-1)=9$	$13.5625$	$\frac{13.5625}{9} \approx 1.5069$	-	-
Total	$16 - 1 = 15$	$39.9375$	-	-	-

4. Conclusion

• $F_{\text{cal}} = 3.14 < F_{\text{tab}} = 3.86$.
• Fail to reject $H_0$. At the $5\%$ level of significance, there is no statistically significant difference in cholesterol content among the four diet foods after controlling for laboratory variation.

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Problem 3: Latin Square Design (LSD)

Problem Statement

Four treatments ($A, B, C, D$) are allocated in a $4 \times 4$ Latin square to control for two independent nuisance sources (e.g., Row = Machine, Column = Operator). The experimental layout and observations are:

• Row 1: $A(18), B(22), C(25), D(28)$
• Row 2: $B(21), C(24), D(26), A(20)$
• Row 3: $C(19), D(23), A(22), B(25)$
• Row 4: $D(20), A(21), B(23), C(27)$

Test at $\alpha = 0.05$ whether treatment differences are significant. (Given: Critical $F_{0.05}(3, 6) = 4.76$, $\sum y^2 = 8408$).

View Step-by-Step Solution

1. Totals Calculation ($p = 4, N = 16$)

• Treatment Totals:
  • $T_A = 18 + 20 + 22 + 21 = 81$
  • $T_B = 22 + 21 + 25 + 23 = 91$
  • $T_C = 25 + 24 + 19 + 27 = 95$
  • $T_D = 28 + 26 + 23 + 20 = 97$
• Row Totals:
  • $R_1 = 18 + 22 + 25 + 28 = 93$
  • $R_2 = 21 + 24 + 26 + 20 = 91$
  • $R_3 = 19 + 23 + 22 + 25 = 89$
  • $R_4 = 20 + 21 + 23 + 27 = 91$
• Column Totals:
  • $C_1 = 18 + 21 + 19 + 20 = 78$
  • $C_2 = 22 + 24 + 23 + 21 = 90$
  • $C_3 = 25 + 26 + 22 + 23 = 96$
  • $C_4 = 28 + 20 + 25 + 27 = 100$
• $G = 364$
• $CF = \frac{364^2}{16} = 8281.0$

2. Sum of Squares Partitioning

• $\text{SST} = 8408.0 - 8281.0 = 127.0$
• $\text{SSTreat} = \frac{81^2 + 91^2 + 95^2 + 97^2}{4} - 8281.0 = \frac{33276}{4} - 8281.0 = 8319.0 - 8281.0 = 38.0$
• $\text{SSRow} = \frac{93^2 + 91^2 + 89^2 + 91^2}{4} - 8281.0 = \frac{33132}{4} - 8281.0 = 8283.0 - 8281.0 = 2.0$
• $\text{SSCol} = \frac{78^2 + 90^2 + 96^2 + 100^2}{4} - 8281.0 = \frac{33400}{4} - 8281.0 = 8350.0 - 8281.0 = 69.0$
• $\text{SSE} = 127.0 - 38.0 - 2.0 - 69.0 = 18.0$

3. LSD ANOVA Table

Source	$\text{df}$	$\text{SS}$	$\text{MS}$	Calculated $F$	Critical $F_{0.05}$
Rows	$4 - 1 = 3$	$2.0$	$0.6667$	-	-
Columns	$4 - 1 = 3$	$69.0$	$23.0000$	-	-
Treatments	$4 - 1 = 3$	$38.0$	$12.6667$	$\frac{12.6667}{3.0000} \approx 4.222$	$4.76$
Error	$(4-1)(4-2)=6$	$18.0$	$3.0000$	-	-
Total	$16 - 1 = 15$	$127.0$	-	-	-

4. Conclusion

• $F_{\text{cal}} = 4.222 < F_{\text{tab}} = 4.76$.
• Fail to reject $H_0$. After adjusting for row and column effects, there is no statistically significant treatment effect at the $5\%$ level.

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Problem 4: $2^2$ Full Factorial Design

Problem Statement

A chemical engineer investigates the effect of Temperature ($A$) and Pressure ($B$) on product yield (%). Both factors are set at two coded levels: Low ($-1$) and High ($+1$) with $r=2$ replicates ($N = 2^2 \times 2 = 8$).

Run	Factor A (Temp)	Factor B (Pressure)	Replicate 1	Replicate 2
1	$-1$	$-1$	42	43
2	$+1$	$-1$	50	49
3	$-1$	$+1$	46	45
4	$+1$	$+1$	62	61

Perform a complete $2^2$ factorial analysis: compute contrasts, main effects, interaction effect, construct the ANOVA table, and test for significance at $\alpha = 0.05$. (Given: Critical $F_{0.05}(1, 4) = 7.708$).

View Step-by-Step Solution

1. Treatment Totals

• Run 1 ($-1, -1$): $y_1 = 42 + 43 = 85$
• Run 2 ($+1, -1$): $y_2 = 50 + 49 = 99$
• Run 3 ($-1, +1$): $y_3 = 46 + 45 = 91$
• Run 4 ($+1, +1$): $y_4 = 62 + 61 = 123$
• $\text{Grand Total } G = 85 + 99 + 91 + 123 = 398$

2. Contrasts & Effects ($2^{k-1} \cdot r = 2^{2-1} \cdot 2 = 4$)

• $C_A = -85 + 99 - 91 + 123 = 46 \implies \hat{A} = \frac{46}{4} = 11.5$
• $C_B = -85 - 99 + 91 + 123 = 30 \implies \hat{B} = \frac{30}{4} = 7.5$
• $C_{AB} = +85 - 99 - 91 + 123 = 18 \implies \widehat{AB} = \frac{18}{4} = 4.5$

3. Sum of Squares ($2^k \cdot r = 8$)

• $\text{SS}_A = \frac{46^2}{8} = 264.5$
• $\text{SS}_B = \frac{30^2}{8} = 112.5$
• $\text{SS}_{AB} = \frac{18^2}{8} = 40.5$
• $\sum y^2 = 42^2 + 43^2 + \dots + 61^2 = 20220$
• $CF = \frac{398^2}{8} = 19800.5 \implies \text{SST} = 20220 - 19800.5 = 419.5$
• $\text{SSE} = \text{SST} - (\text{SS}_A + \text{SS}_B + \text{SS}_{AB}) = 419.5 - (264.5 + 112.5 + 40.5) = 2.0$

4. Factorial ANOVA Table

Source	$\text{df}$	$\text{SS}$	$\text{MS}$	Calculated $F$	Critical $F_{0.05}$	Decision
A (Temperature)	1	$264.5$	$264.5$	$\frac{264.5}{0.5} = 529.0$	$7.708$	Reject $H_0$
B (Pressure)	1	$112.5$	$112.5$	$\frac{112.5}{0.5} = 225.0$	$7.708$	Reject $H_0$
AB (Interaction)	1	$40.5$	$40.5$	$\frac{40.5}{0.5} = 81.0$	$7.708$	Reject $H_0$
Error (Residual)	4	$2.0$	$\text{MSE} = 0.5$	-	-	-
Total	7	$419.5$	-	-	-	-

5. Conclusion

All factors ($A, B,$ and interaction $AB$) are statistically significant at $p < 0.001$. Temperature increases yield by $11.5\%$, Pressure increases yield by $7.5\%$, and positive interaction indicates the benefits of temperature are greatest when paired with high pressure.

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Problem 5: Taguchi $L_9$ Orthogonal Array Optimization

Problem Statement

A manufacturing engineer seeks to optimize three 3-level parameters to maximize tensile strength ($y$ in $\text{MPa}$):

• Factor A (Temperature): Level 1 = $200^\circ\text{C}$, Level 2 = $220^\circ\text{C}$, Level 3 = $240^\circ\text{C}$
• Factor B (Holding Time): Level 1 = $10\,\text{min}$, Level 2 = $20\,\text{min}$, Level 3 = $30\,\text{min}$
• Factor C (Pressure): Level 1 = $5\,\text{bar}$, Level 2 = $7\,\text{bar}$, Level 3 = $9\,\text{bar}$

The measured single-run responses in an $L_9$ array are:

• Run 1: (1, 1, 1) $\implies y = 42\,\text{MPa}$
• Run 2: (1, 2, 2) $\implies y = 48\,\text{MPa}$
• Run 3: (1, 3, 3) $\implies y = 50\,\text{MPa}$
• Run 4: (2, 1, 2) $\implies y = 55\,\text{MPa}$
• Run 5: (2, 2, 3) $\implies y = 60\,\text{MPa}$
• Run 6: (2, 3, 1) $\implies y = 58\,\text{MPa}$
• Run 7: (3, 1, 3) $\implies y = 65\,\text{MPa}$
• Run 8: (3, 2, 1) $\implies y = 62\,\text{MPa}$
• Run 9: (3, 3, 2) $\implies y = 59\,\text{MPa}$

Using the Larger-the-better $S/N$ ratio formulation:

1. Compute the $S/N$ ratio for each trial.
2. Build the factor-level response table and rank the factors.
3. Determine the optimal parameter combination.
4. Predict the optimum tensile strength performance.

View Step-by-Step Solution

1. S/N Computation ($S/N = 20 \log_{10}(y)$)

• $\text{Run 1}: 20 \log_{10}(42) = 32.4650\,\text{dB}$
• $\text{Run 2}: 20 \log_{10}(48) = 33.6248\,\text{dB}$
• $\text{Run 3}: 20 \log_{10}(50) = 33.9794\,\text{dB}$
• $\text{Run 4}: 20 \log_{10}(55) = 34.8073\,\text{dB}$
• $\text{Run 5}: 20 \log_{10}(60) = 35.5630\,\text{dB}$
• $\text{Run 6}: 20 \log_{10}(58) = 35.2686\,\text{dB}$
• $\text{Run 7}: 20 \log_{10}(65) = 36.2583\,\text{dB}$
• $\text{Run 8}: 20 \log_{10}(62) = 35.8478\,\text{dB}$
• $\text{Run 9}: 20 \log_{10}(59) = 35.4170\,\text{dB}$
• Overall Average: $\overline{S/N} = 34.8035\,\text{dB}$

2. Factor-Level S/N Response Table

• Factor A (Temp):
  • Level 1: $\frac{32.4650 + 33.6248 + 33.9794}{3} = 33.3564\,\text{dB}$
  • Level 2: $\frac{34.8073 + 35.5630 + 35.2686}{3} = 35.2130\,\text{dB}$
  • Level 3: $\frac{36.2583 + 35.8478 + 35.4170}{3} = 35.8410\,\text{dB}$
  • $\Delta_A = 35.8410 - 33.3564 = 2.4846\,\text{dB}$ (Rank 1)
• Factor B (Time):
  • Level 1: $\frac{32.4650 + 34.8073 + 36.2583}{3} = 34.5102\,\text{dB}$
  • Level 2: $\frac{33.6248 + 35.5630 + 35.8478}{3} = 35.0119\,\text{dB}$
  • Level 3: $\frac{33.9794 + 35.2686 + 35.4170}{3} = 34.8883\,\text{dB}$
  • $\Delta_B = 35.0119 - 34.5102 = 0.5017\,\text{dB}$ (Rank 3)
• Factor C (Pressure):
  • Level 1: $\frac{32.4650 + 35.2686 + 35.8478}{3} = 34.5271\,\text{dB}$
  • Level 2: $\frac{33.6248 + 34.8073 + 35.4170}{3} = 34.6164\,\text{dB}$
  • Level 3: $\frac{33.9794 + 35.5630 + 36.2583}{3} = 35.2669\,\text{dB}$
  • $\Delta_C = 35.2669 - 34.5271 = 0.7398\,\text{dB}$ (Rank 2)

3. Optimal Setting Selection

Always pick the level with the highest mean $S/N$:

• Factor A: Level 3 ($240^\circ\text{C}$)
• Factor B: Level 2 ($20\,\text{min}$)
• Factor C: Level 3 ($9\,\text{bar}$)
• Optimal Setting: $A_3 B_2 C_3$

4. Predicted Optimum Performance

$\overline{S/N}_{\text{pred}} = 34.8035 + (35.8410 - 34.8035) + (35.0119 - 34.8035) + (35.2669 - 34.8035) = 36.5128\,\text{dB}$ $\text{Predicted Tensile Strength: } \hat{y} = 10^{\frac{36.5128}{20}} \approx 66.93\,\text{MPa}$

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Problem 6: Genetic Algorithm Optimization

Problem Statement

A Genetic Algorithm is used to maximize the function $f(x) = x^2$ over the integer domain $0 \le x \le 31$ encoded as 5-bit binary strings. The initial random population consists of:

• $C_1 = 10110_2$
• $C_2 = 00101_2$
• $C_3 = 11100_2$
• $C_4 = 01011_2$

Perform one full generation of the algorithm:

1. Decode each chromosome to decimal and evaluate its fitness.
2. Select the top two chromosomes as parents.
3. Perform single-point crossover after the 2nd bit to produce two children.
4. Mutate the last bit of Child 2.
5. Form the next generation using the children and the top two elite parents from the original generation, and compute the new average population fitness.

View Step-by-Step Solution

1. Decoding & Fitness ($f(x) = x^2$)

• $C_1 = 10110_2 = 16 + 4 + 2 = 22 \implies f(22) = 22^2 = 484$
• $C_2 = 00101_2 = 4 + 1 = 5 \implies f(5) = 5^2 = 25$
• $C_3 = 11100_2 = 16 + 8 + 4 = 28 \implies f(28) = 28^2 = 784$
• $C_4 = 01011_2 = 8 + 2 + 1 = 11 \implies f(11) = 11^2 = 121$
• Gen 0 Average Fitness: $\bar{f}_0 = \frac{484 + 25 + 784 + 121}{4} = 353.5$

2. Parent Selection

The two highest fitness individuals are:

• Parent 1: $C_3 = 11100$ ($x = 28, f = 784$)
• Parent 2: $C_1 = 10110$ ($x = 22, f = 484$)

3. Crossover (Cut after Bit 2)

• $\text{Parent 1: } [11 \mid 100]$
• $\text{Parent 2: } [10 \mid 110]$
• $\text{Child 1: } [11 \mid 110] = 11110_2 = 30 \implies f(30) = 30^2 = 900$
• $\text{Child 2: } [10 \mid 100] = 10100_2 = 20 \implies f(20) = 20^2 = 400$

4. Mutation

Flip the 5th bit of Child 2 ($10100 \to 10101$):

• $\text{Mutated Child 2} = 10101_2 = 21 \implies f(21) = 21^2 = 441$

5. New Generation Formation (Elitism)

1. Child 1: $11110_2 \implies x = 30, f = 900$
2. Mutated Child 2: $10101_2 \implies x = 21, f = 441$
3. Elite Parent 1: $11100_2 \implies x = 28, f = 784$
4. Elite Parent 2: $10110_2 \implies x = 22, f = 484$

• Gen 1 Average Fitness: $\bar{f}_1 = \frac{900 + 441 + 784 + 484}{4} = \frac{2609}{4} = 652.25$
• Improvement: Average fitness increased by $\frac{652.25 - 353.5}{353.5} \times 100\% \approx 84.5\%$.

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Problem 7: Response Surface Methodology (RSM) Optimization

Problem Statement

A chemical company wants to maximize product yield ($Y$, in $\%$). Two continuous factors are investigated:

• Temperature ($X_1$): Low level ($-1$) = $100^\circ\text{C}$, High level ($+1$) = $140^\circ\text{C}$ (Center point $0 = 120^\circ\text{C}$).
• Pressure ($X_2$): Low level ($-1$) = $10\,\text{psi}$, High level ($+1$) = $20\,\text{psi}$ (Center point $0 = 15\,\text{psi}$).

The experimental results from an initial $2^2$ factorial design matrix are:

Run	$X_1$ (Temperature)	$X_2$ (Pressure)	Yield $Y$ (%)
1	$-1$	$-1$	70
2	$-1$	$+1$	76
3	$+1$	$-1$	80
4	$+1$	$+1$	86

Perform the following:

1. Assume a first-order model $Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2$ and calculate the coefficients $\beta_0, \beta_1, \beta_2$.
2. Write the fitted mathematical model.
3. Verify the model at run $(X_1 = +1, X_2 = +1)$.
4. Predict the yield at the center point $(X_1 = 0, X_2 = 0)$.
5. Determine the direction of improvement and identify which factor has the stronger effect.
6. Execute the Method of Steepest Ascent: determine the direction vector and list the coded and physical factor coordinates for the first 3 steps using step increments $\Delta X_1 = 0.5$ and $\Delta X_2 = 0.3$.
7. Suppose subsequent trials exhibit curvature and a second-order model is fitted: $Y = 90 + 4X_1 + 3X_2 - 2X_1 X_2 - 5X_1^2 - 4X_2^2$ If mathematical optimization predicts an optimum at $(X_1 = 0.4, X_2 = 0.3)$ yielding $92\%$, validate the model if an actual confirmation trial produces $91.5\%$.

View Step-by-Step Solution

1. Calculate First-Order Regression Coefficients

Using the orthogonal properties of the $2^2$ coded matrix ($N = 4$):

• Calculate $\beta_0$ (Intercept / Average Response): $\beta_0 = \frac{\sum Y_i}{N} = \frac{70 + 76 + 80 + 86}{4} = \frac{312}{4} = 78$ $\boxed{\beta_0 = 78}$

• Calculate $\beta_1$ (Main Effect of Temperature $X_1$): $\beta_1 = \frac{\sum X_{1i} Y_i}{N} = \frac{(-1)(70) + (-1)(76) + (+1)(80) + (+1)(86)}{4} = \frac{-70 - 76 + 80 + 86}{4} = \frac{20}{4} = 5$ $\boxed{\beta_1 = 5}$

• Calculate $\beta_2$ (Main Effect of Pressure $X_2$): $\beta_2 = \frac{\sum X_{2i} Y_i}{N} = \frac{(-1)(70) + (+1)(76) + (-1)(80) + (+1)(86)}{4} = \frac{-70 + 76 - 80 + 86}{4} = \frac{12}{4} = 3$ $\boxed{\beta_2 = 3}$

2. Write the Fitted Mathematical Model

$\boxed{Y = 78 + 5X_1 + 3X_2}$ where:

• $78$ = Intercept (baseline expected yield at the center point $X_1 = 0, X_2 = 0$).
• $5$ = Main effect coefficient of Temperature ($X_1$).
• $3$ = Main effect coefficient of Pressure ($X_2$).

3. Verify the Model

Substitute $X_1 = 1, X_2 = 1$ (Run 4): $\hat{Y} = 78 + 5(1) + 3(1) = 78 + 5 + 3 = 86$

• Observed value: $86\%$
• Predicted value: $86\%$
• Residual: $e = 86 - 86 = 0$.
• Conclusion: The model fits the orthogonal factorial data points perfectly.

4. Predict Response at Center Point

Substitute $X_1 = 0, X_2 = 0$ (Temperature $= 120^\circ\text{C}$, Pressure $= 15\,\text{psi}$): $\hat{Y} = 78 + 5(0) + 3(0) = 78$

• Predicted yield at center point: $\mathbf{78\%}$.

5. Direction of Improvement & Factor Significance

• Coefficients: Both $\beta_1 = +5$ and $\beta_2 = +3$ are positive.
• Direction: Increase both Temperature and Pressure to increase yield.
• Relative Strength: Since $5 > 3$, Temperature has a stronger positive effect on yield than Pressure.

6. Method of Steepest Ascent

Move in the direction of the gradient vector: $\nabla Y = (5, 3)$ Taking step increments proportional to the $(5 : 3)$ ratio with $\Delta X_1 = 0.5$ and $\Delta X_2 = 0.3$:

Step	Coded $X_1$	Coded $X_2$	Physical Temp ($^\circ\text{C}$)	Physical Pressure ($\text{psi}$)
Start	0.0	0.0	$120.0^\circ\text{C}$	$15.0\,\text{psi}$
1	0.5	0.3	$130.0^\circ\text{C}$	$16.5\,\text{psi}$
2	1.0	0.6	$140.0^\circ\text{C}$	$18.0\,\text{psi}$
3	1.5	0.9	$150.0^\circ\text{C}$	$19.5\,\text{psi}$

Continue running experiments along this path until observed yield ceases to increase.

7. Second-Order Model & Confirmation Experiment

• Second-Order Model: When curvature exists, the quadratic model accounts for diminishing returns ($-5X_1^2, -4X_2^2$) and factor interaction ($-2X_1 X_2$): $Y = 90 + 4X_1 + 3X_2 - 2X_1 X_2 - 5X_1^2 - 4X_2^2$
• Confirmation Trial:
  • Predicted Yield: $\hat{Y} = 92\%$ at $(X_1 = 0.4, X_2 = 0.3)$.
  • Actual Experimental Yield: $Y_{\text{actual}} = 91.5\%$.
• Validation Decision: Since the actual experimental yield ($91.5\%$) closely matches the model prediction ($92\%$) within experimental uncertainty, the fitted model is validated.