Experimental Design Lab Exercise III ASAAD, Al-Ahmadgaid B. July 6, 2012 al in the cloud: website: www.alstat.weebly.com blog: www.alstatr.blogspot.com email: [email protected]Questions 4-1. A chemist wishes to test the effect of four chemical agents on the strength of a particular type of cloth. Because there might be vari- ability from one bolt to another, the chemist decides to use a ran- domized block design, with the bolts of cloth considered as blocks. She selects five bolts and applies all four chemicals in random order to each bolt. The resulting tensile strength follow. Analyze the data from this experiment (use α = 0.05) and draw appropriate conclu- sions. Bolt Chemical 1 2 3 4 5 1 73 68 74 71 67 2 73 67 75 72 70 3 75 68 78 73 68 4 73 71 75 75 69 4-3. Plot the mean tensile strengths observed for each chemical type in Problem 4-1 and compare them to an appropriatety scaled t dis- tribution. What conclusions would you draw from this display? 4-9. Assuming that chemical types and bolts are fixed, estimate the model parameters τ i and β i in Problem 4-1. 4-11. Suppose that the obersvation for chemical type 2 and bolt 3 is missing in Problem 4-1. Analyze the problem by estimating the missing value. Perform the exact analysis and compare the results.
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Experimental Design Lab Exercise IIIASAAD, Al-Ahmadgaid B.July 6, 2012 al in the cloud:
Total SS or TSS = (262 + 162 + · · ·+ 172 + 142) - 7396 = 436
Acid SS or ASS =(892 + 882 + 922 + 832 + 782)
5- 7396 = 24.4
Batch SS or BSS =(902 + 892 + 862 + 832 + 822)
5- 7396 = 10
Times SS or TrSS =(1182 + 782 + 942 + 752 + 652)
5- 7396 = 342
experimental design lab exercise iii 20
Catalyst SS or CSS =(832 + 852 + 912 + 822 + 892)
5- 7396 = 12
Error SS or SSE = TSS - BSS - ASS - CSS - TrSS
= 436− 10− 24.4− 342− 12 = 47.6
ANOVA table for 5×5 Graeco-Latin Square (p=5)
SV DF SS MS F
Times p-1=4 342 85.5 14.37
Batch p-1=4 10 2.5 0.42
Acid p-1=4 24.4 6.1 1.025
Catalyst p-1=4 12 3 0.504
Error (p-1)(p-3)=8 47.6 5.95
Total 24 436
Decision: All FComputed of each Source Variation is less than the crit-
ical value, Fα,4,8 = 3.8379, except for the treatments which is 14.37.
And thus, the following decision is obtain,
a. The five standing times are significantly different.
b. The five batches of raw materials have no significant difference.
c. The five acid concentrations have no significant difference.
d. The five catalyst concentrations have no significant difference.
xiii. Multiple comparison for five standing times,
Using Tukey Honestly Significant Difference, the critical value is ob-
tain,
Tα = q(a, f )
√MSE
n= q(5, 8)
√5.95
5= 4.89
√5.95
5= 5.33
Treatment Means:
yA = 23.6, yC = 18.8, yB = 15.6, yD = 15, yE = 13
Thus, any pair of treatment averages differ by more than 5.33 would
imply that the corresponding pair of population means are signifi-
experimental design lab exercise iii 21
cantly different. The differences in averages are,
yA − yE = 23.6− 13 = 10.6 ∗
yA − yD = 23.6− 15 = 8.6 ∗
yA − yB = 23.6− 15.6 = 8 ∗
yA − yC = 23.6− 18.8 = 4.8
yC − yE = 18.8− 13 = 5.8 ∗
yC − yD = 18.8− 15 = 3.8
yC − yB = 18.8− 15.6 = 3.2
yB − yE = 15.6− 13 = 2.6
yB − yD = 15.6− 15 = 0.6
yD − yE = 15− 13 = 2
The starred values indicates pairs of mean that are significantly dif-
ferent.
experimental design lab exercise iii 22
Computation using SPSS Software
3-1. The tensile strength of portland cement is being studied. Four dif-
ferent mixing techniques can be used economically. The following
data have been collected:
Mixing Technique Tensile Strength (lb/in2)
1 3129 3000 2865 2890
2 3200 3300 2975 3150
3 2800 2900 2985 3050
4 2600 2700 2600 2765
Steps:
Step 1
Figure 2: The above table is entered inSPSS in this manner.
Step 2
Figure 3: The second step after inputtingyour data, go to analyze⇒compare
means⇒one-way anova.
experimental design lab exercise iii 23
Step 3
Figure 4: Next, enter the vari-able yield to the dependent
list (yield⇒dependent list)and treatment to factor
(treatment⇒factor). After thatyou can click the post hoc.. for choos-ing the test for multiple comparison.
Table 1: The output of the performedsteps. In multiple comparison table, thetest performed was Scheffé. You cancheck it in the post hoc.. section of theStep 3
4-1. A chemist wishes to test the effect of four chemical agents on the
strength of a particular type of cloth. Because there might be vari-
ability from one bolt to another, the chemist decides to use a ran-
domized block design, with the bolts of cloth considered as blocks.
She selects five bolts and applies all four chemicals in random order
experimental design lab exercise iii 24
to each bolt. The resulting tensile strength follow. Analyze the data
from this experiment (use α = 0.05) and draw appropriate conclu-
sions.
Bolt
Chemical 1 2 3 4 5
1 73 68 74 71 67
2 73 67 75 72 70
3 75 68 78 73 68
4 73 71 75 75 69
Solution
Step 1
Figure 5: The above table is entered inSPSS in this manner.
Step 2
Figure 6: The second step after inputtingyour data, go to analyze⇒general
linear model⇒univariate
experimental design lab exercise iii 25
Step 3
Figure 7: Next, enter the vari-able yield to the dependent list
(yield⇒dependent list) and treat-ment and block to fixed factor(s)(treatment and block⇒fixed fac-tor(s)). After that you can click thepost hoc.. for choosing the test formultiple comparison.
Step 4
Figure 8: Before clicking the ok but-ton, go first to the model (seen on Step3). In the univariate: model win-dow, click (custom) then put the treat-ment and block to the model box, asshown in the figure. Then, change thetype to main effects and uncheckedthe include intercept in model, be-fore the clicking the continue button.
output of the performed test. Refer to Manual Computation and
Graphical Illustration Section item 4-1 for the interpretation.
4-11. Suppose that the obersvation for chemical type 2 and bolt 3 is
missing in Problem 4-1. Analyze the problem by estimating the
experimental design lab exercise iii 26
missing value. Perform the exact analysis and compare the results.
Solution:
Bolt
Chemical 1 2 3 4 5
1 73 68 74 71 67
2 73 67 x 72 70
3 75 68 78 73 68
4 73 71 75 75 69
Solution: For missing value, just replace x to 75.25 as computed in
the Manual Computation and Graphical Illustration Section. After
that, perform the above steps in 4-1 of this section.
Table 2: This is the output of the testperformed.
Refer to Manual Computation and Graphical Illustration Section
item 4-11 for the interpretation.
1. Shown below the yield (ton per 1/4-ha.plots) of sugar cane in a
Latin square experiments comparing five (5) fertilizer levels. Where:
A=no fertilizer
C=10 tons manure/ha
E=30 tons manure/ha
B=complete inorganic fertilizer
D=20 tons manure/ha
Row Columns
1 2 3 4 5
1 14(A) 22(E) 20(B) 18(C) 25(D)
2 19(B) 21(D) 16(A) 23(E) 18(C)
3 23(D) 15(A) 20(C) 18(B) 23(E)
4 21(C) 25(B) 24(E) 21(D) 18(A)
5 23(E) 16(C) 23(D) 17(A) 19(B)
a. Analyze the data completely and interpret your results.
experimental design lab exercise iii 27
Solution:
Step 1
Figure 9: Enter the data to SPSS in thismanner.
Step 2
Figure 10: The second step af-ter inputting your data, go toanalyze⇒general linear
model⇒univariate
experimental design lab exercise iii 28
Step 3
Figure 11: Next, enter the vari-able yield to the dependent list
(yield⇒dependent list) and treat-ment, row and column to fixed fac-tor(s) (row and column⇒fixed fac-tor(s)). After that you can click thepost hoc.. for choosing the test formultiple comparison.
Step 4
Figure 12: Before clicking the ok button,go first to the model (seen on Step 3). Inthe univariate: model window, clickcustom then put the treatment, col-umn, and row to the model box, asshown in the figure. Then, change thetype to main effects and uncheckedthe include intercept in model, be-fore the clicking the continue button.
Figure 13: The output of the performedsteps
experimental design lab exercise iii 29
Figure 14: The output generated usingpost hoc..-lsd Method
Refer to the Manual Computation and Graphical Illustration Section for