Problem 8.4 Implantable heart pacemakers contain small circuit boards called sub
ID: 3332977 • Letter: P
Question
Problem 8.4 Implantable heart pacemakers contain small circuit boards called sub- strates. These substrates are assembled, cut to shape, and fired. Some of the substrates will separate, or delaminate, making them useless. The pur- pose of this experiment was to study the effects of three factors on the rate of delamination. The factors were A: firing profile time, 8 versus 13 hours with the theory suggesting 13 hours is better; B: furnace airflow, low versus high, with theory suggesting high is better; and C: laser, old versus new, with theory suggesting new cutting lasers are better A large number of raw, assembled substrates are divided into sixteen groups. These sixteen groups are assigned at random to the eight factor- level combinations of the three factors, two groups per combination. The substrates are then processed, and the response is the fraction of substrates that delaminate, Data from Todd Kerkow, 8 hrs 13 hrs Low High Low High Old.83 .68 .78 90 18 ·25 16 ·20 New.86 .72 .67 81 30 10 23 14 Analyze these data to determine how the treatments affect delaminationExplanation / Answer
Solution:
Before we even proceed to do any analysis, it is clear that low firing profile times, 8 hours, lead to higher (worse) delamination, confirming the theoretical suggestion that 13 hours would be better. The effects of furnace airflow, B, and laser age, C, are unclear from the raw data.
The two response values from each experiment can be averaged, and then you proceed with
the regular calculations for the experiment. In standard order, the responses would be: y = [0.805, 0.17, 0.79, 0.225, 0.765, 0.265, 0.765, 0.12].
The estimated factors by hand are: I =1/8(+0.805 + 0.17 + 0.79 + 0.225 + 0.765 + 0.265 + 0.765 + 0.12) = 0.488
A = 1/8(0.805 + 0.17 0.79 + 0.225 0.765 + 0.265 0.765 + 0.12) = 0.293
B = 1/8(0.805 0.17 + 0.79 + 0.225 0.765 0.265 + 0.765 + 0.12) = 0.013
C = 1/8(0.805 0.17 0.79 0.225 + 0.765 + 0.265 + 0.765 + 0.12) = 0.009
AB = 1/8(+0.805 0.17 0.79 + 0.225 + 0.765 0.265 0.765 + 0.12) = 0.009
AC = 1/8(+0.805 0.17 + 0.79 0.225 0.765 + 0.265 0.765 + 0.12) = 0.006
BC = 1/8(+0.805 + 0.17 0.79 0.225 0.765 0.265 + 0.765 + 0.12) = 0.023
ABC = 1/8(0.805 + 0.17 + 0.79 0.225 + 0.765 0.265 0.765 + 0.12) = 0.027
Here you can also use the calculation method where you calculate the half-differences for the interactions: you will get the exact same answers once you halve the values (to make the results compatible with the least squares model).
The above results show that increasing firing profile time, factor A, will decrease the delamination fraction; none of the other factors, nor any of the interaction terms have a significant effect. In particular, one coded unit of factor A represents 138/2 = 2.5hours: so every 2.5 hours that we increase the firing profile time by is expected to give a decrease, on average, in delamination fraction of 0.29, or 29%.