5-19. The yield of a chemical process is being studied. The two factors of interest are temperature and pressure. Three levels of each factor are selected; however, only nine runs can be made in one day. The experimenter runs a complete replicate of the design on each day. The data are shown in the following table. Analyze the data, assuming that the days are blocks.

Temperature Low Medium High

250 86.3 88.5 89.1

Day 1 Pressure 260 84.0 87.3 90.2

270 85.8 89.0 91.3

250 86.1 89.4 91.7

Day 1 Pressure 260 85.2 89.9 93.2

270 87.3 90.3 93.7

Analyze using Minitab 14.

MTB > Name C8 "RESI1" MTB > GLM 'Responses' = C3 C1 C2 C1*C2; SUBC> GFourpack C6; SUBC> Brief 2; SUBC> Means C1 C2 C1*C2; SUBC> Residuals 'RESI1'.

General Linear Model: Responses versus Block, Temperature, Pressure Factor Block Temperature Pressure

Type fixed fixed fixed

Levels 2 3 3

Values 1, 2 High, Low, Medium 250, 260, 270

Analysis of Variance for Responses, using Adjusted SS for Tests Source DF Block 1 Temperature 2 Pressure 2 Temperature*Pressure 4 Error 8 Total 17 S = 0.728869

R-Sq = 96.66%

Seq SS 13.005 99.854 5.508 4.452 4.250 127.069

Adj SS 13.005 99.854 5.508 4.452 4.250

Adj MS 13.005 49.927 2.754 1.113 0.531

F 24.48 93.98 5.18 2.10

P 0.001 0.000 0.036 0.173

R-Sq (adj) = 92.89%

According to the output above; block, temperature and pressure are significant

influence the yield of the chemical process. That were shown by the P-value was smaller than α = 0.05 or F calc < F table, that is, F (1, 8; 0.05) = 5.32 and F (2, 8; 0.05) = 4.46. The other way, interaction between temperature and pressure is not significant

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with P-value 0.173 that is bigger than α = 0.05, or we can conclude from the F calc = 2.10 < F (4, 8; 0.05) = 3.84 which mean fail to reject H 0.

Then we were going to analyze with removing the insignificant, the interaction between temperatures and pressures from the model, and we got, MTB > Name C9 "RESI2" MTB > GLM 'Responses' = C3 C1 C2; SUBC> GFourpack C6; SUBC> Brief 2; SUBC> Means C1 C2; SUBC> Residuals 'RESI2'.

General Linear Model: Responses versus Block, Temperature, Pressure Factor Block Temperature Pressure

Type fixed fixed fixed

Levels 2 3 3

Values 1, 2 High, Low, Medium 250, 260, 270

Analysis of Variance for Responses, using Adjusted SS for Tests Source DF Block 1 Temperature 2 Pressure 2 Error 12 Total 17

Seq SS 13.005 99.854 5.508 8.702 127.069

S = 0.851578 R-Sq = 93.15%

Adj SS 13.005 99.854 5.508 8.702

Adj MS 13.005 49.927 2.754 0.725

F 17.93 68.85 3.80

P 0.001 0.000 0.053

R-Sq(adj) = 90.30%

With α = 0.05, block and temperature are significant influence the yield of the

chemical process, and pressure is significant at α = 0.10. R-sq = 93.15% shown that model are presented the 93.15% of the data.

Then we were going to analyze using Tukey’s Method, to see the difference of the factor/treatment effect where the sample size is the same. Here we got the result,

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Tukey 95.0% Simultaneous Confidence Intervals Response Variable Responses All Pairwise Comparisons among Levels of Temperature Temperature = High subtracted from: Temperature Low Medium

Lower -7.061 -3.777

Temperature = Low Temperature Medium

Lower 1.973

Center -5.750 -2.467

Upper -4.439 -1.156

+---------+---------+---------+-----(---*--) (---*---) +---------+---------+---------+------7.0 -3.5 0.0 3.5

subtracted from: Center 3.283

Upper 4.594

+---------+---------+---------+-----(--*---) +---------+---------+---------+------7.0 -3.5 0.0 3.5

Tukey Simultaneous Tests Response Variable Responses All Pairwise Comparisons among Levels of Temperature Temperature = High subtracted from:

Temperature Low Medium

Difference of Means -5.750 -2.467

Temperature = Low

Temperature Medium

SE of Difference 0.4917 0.4917

T-Value -11.70 -5.02

Adjusted P-Value 0.0000 0.0008

T-Value 6.678

Adjusted P-Value 0.0001

subtracted from:

Difference of Means 3.283

SE of Difference 0.4917

From the above output, we got that treatment on temperatures are significant with

three levels; high, medium and low, according to the different of lower, center, and upper value that is not exceed the null. It was also shown through the hypothesis

testing with P-value < α = 0.05, that was, reject H0 or there was significant effect

from the three temperatures given. Those different might because of the levels that were just said to be high, medium, and low without any further explanation of what high, medium, and low are.

From the output bellow, the three pressures given are not significantly different

yielding the chemical process. It was shown that among lower, center, and upper value are exceed the null or testing hypothesis that gave the P-value bigger than α = 0.05, that is, fail to reject H0. It might because of the distance of 250, 260, and 270 were just 10 that were particularly not different enough.

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Tukey 95.0% Simultaneous Confidence Intervals Response Variable Responses All Pairwise Comparisons among Levels of Pressure Pressure = 250 subtracted from: Pressure 260 270

Lower -1.527 -0.261

Pressure = 260 Pressure 270

Center -0.2167 1.0500

Upper 1.094 2.361

---+---------+---------+---------+--(----------*----------) (----------*----------) ---+---------+---------+---------+---1.2 0.0 1.2 2.4

subtracted from:

Lower -0.04399

Center 1.267

Upper 2.577

---+---------+---------+---------+--(----------*---------) ---+---------+---------+---------+---1.2 0.0 1.2 2.4

Tukey Simultaneous Tests Response Variable Responses All Pairwise Comparisons among Levels of Pressure Pressure = 250 subtracted from:

Pressure 260 270

Difference of Means -0.2167 1.0500

Pressure = 260

Pressure 270

SE of Difference 0.4917 0.4917

T-Value -0.4407 2.1356

Adjusted P-Value 0.8994 0.1241

T-Value 2.576

Adjusted P-Value 0.0587

subtracted from:

Difference of Means 1.267

SE of Difference 0.4917

Then we have to check whether the residual was normal, and from the graph bellow we can conclude that with P-value = 0.138, that is, bigger than α = 0.05 then we fail to reject H0 or the residual was normal.

Residual Plots for Responses Probability Plot of RESI3 Normal

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Mean StDev N AD P-Value

95 90

7.105427E-15 0.7155 18 0.546 0.138

Percent

80 70 60 50 40 30 20 10 5

1

-2

-1

0 RESI3

1

2

4