Homework 12

Question 7.12

Consider the potting experiment in Problem 6.21. Analyze the data considering each replicate as a block.

Its a \(2^4\) design with 7 replicates. Each replicate has 16 runs, therefore each block will also have 16 runs

##              Df Sum Sq Mean Sq F value Pr(>F)   
## block         6    376    62.7   0.728 0.6284   
## A             1    917   917.1  10.647 0.0015 **
## B             1    388   388.1   4.506 0.0362 * 
## Residuals   103   8873    86.1                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Transformed data

##              Df Sum Sq Mean Sq F value  Pr(>F)   
## block         6   5.95   0.991   1.023 0.41438   
## A             1   7.37   7.373   7.613 0.00686 **
## B             1   6.98   6.976   7.202 0.00849 **
## Residuals   103  99.76   0.969                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

From the result, values of “Prob > F” less than 0.0500 indicate model terms are significant. In this case, factors A and B are the only significant model terms in the actual and transformed data analysis. The blocking has no effect on the significant factors, is a very different set of experimental condition at which the experiment is run under. Hence it cannot influence the response because its not one of the factors we are studying. What the blocking does is that it reduces the sums of squares and also reduces the degrees of freedom in the error.

Redsidual plots of Transformed data

The plots don’t appear to be normal as there are presence of an outliers on both normal probability plot and the plot of residuals vs fitted. The extreme values of the normal probability plot don’t appear to be on a straight line either.

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Question 7.20

Design an experiment for confounding a \(2^6\) factorial in four blocks. Suggest an appropriate confounding scheme, different from the one shown in Table 7.8.

Assignment of Four Blocks for Two Confounding Effects ABCE and ABDF using -1/+1 Coding System. This also confounds CDEF.

ABCE * ABDF = CDEF

Blocks assigned should not have same signs as the main effects A, B, C, D, E, F

##      t  A  B  C  D  E  F ABCE ABDF
## 1  (1) -1 -1 -1 -1 -1 -1    1    1
## 2    a  1 -1 -1 -1 -1 -1   -1   -1
## 3    b -1  1 -1 -1 -1 -1   -1   -1
## 4    c  1  1 -1 -1 -1 -1    1    1
## 5    d -1 -1  1 -1 -1 -1   -1    1
## 6    e  1 -1  1 -1 -1 -1    1   -1
## 7    f -1  1  1 -1 -1 -1    1   -1
## 8   ab  1  1  1 -1 -1 -1   -1    1
## 9   ac -1 -1 -1  1 -1 -1    1   -1
## 10  bc  1 -1 -1  1 -1 -1   -1    1
## 11  ad -1  1 -1  1 -1 -1   -1    1
## 12  bd  1  1 -1  1 -1 -1    1   -1
## 13  cd -1 -1  1  1 -1 -1   -1   -1
## 14  ae  1 -1  1  1 -1 -1    1    1
## 15  be -1  1  1  1 -1 -1    1    1
## 16  ce  1  1  1  1 -1 -1   -1   -1
## 17  de -1 -1 -1 -1  1 -1   -1    1
## 18  af  1 -1 -1 -1  1 -1    1   -1
## 19  bf -1  1 -1 -1  1 -1    1   -1
## 20  cf  1  1 -1 -1  1 -1   -1    1

##         t  A  B  C D E F ABCE ABDF
## 60  abcef  1  1 -1 1 1 1   -1    1
## 61  abdef -1 -1  1 1 1 1    1    1
## 62  acdef  1 -1  1 1 1 1   -1   -1
## 63  bcdef -1  1  1 1 1 1   -1   -1
## 64 abcdef  1  1  1 1 1 1    1    1

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