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Math 558- Design of Experiments Final Exam : W 2022
Due Date: April 22, 2022 Time:14:00 First name:
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Question 1.
[20 points]
Steel is normalized by heating above the critical temperature, soaking, and then air cooling. This process increases the strength of the steel, refines the grain, and homog- enizes the structure. An experiment is performed to determine the effect of tempera- ture and heat treatment time on the strength of normalized steel. Two temperatures and three times are selected. The experiment is performed by heating the oven to a randomly selected temperature and inserting three specimens. After 10 minutes one specimen is removed, after 20 minutes the second specimen is removed, and after 30 minutes the final specimen is removed. Then the temperature is changed to the other level and the process is repeated. Four shifts are required to collect the data, which are shown below. Analyze the data and draw conclusions, assuming both factors are fixed. Attach R-output and R-codes.
Temperature(0F )
Shift Time(min) 1500 1600
1 20 54 91 30 61 62 10 50 80
2 20 52 72 30 59 69 10 48 73
3 20 74 81 30 71 69 10 54 88
4 20 48 92 30 59 64
Solution: Q1.
Question 2.
[20 Points] An engineer is interested in the effects of cutting speed (A), tool geometry (B), and cutting angle (C) on the life (in hours) of a machine tool. Two levels of each factor are chosen, and three replicates of a 23 factorial design are run. The results are given in table below.
Treatment Replicates A B C Combinations I II
– – – (1) 22 31
+– a 32 43 -+- b 35 34 ++- ab 55 47 –+ c 44 45 +-+ ac 40 37 -++ bc 60 50 +++ abc 39 41
a. Write down a regression model for predicting tool parameters
III 25 29 50 46 38 36 54 47
life and estimate the model
b. Estimate the factor effects. Which effects appear to be large?
c. Based on the results from parts (a) and (b) write down the reduced model and estimate its parameters.
Note: Attach R-output and R-codes.
Solution Q2.
Question 3.
a. [12 Points] Three different washing solutions are being compared to study their effectiveness in retarding bacteria growth in five-gallon milk containers. The analysis is done in a laboratory, and only three trials can be run on any day. Because days could represent a potential source of variability, the experimenter decides to use a randomized block design. Observations are taken for four days, and the data are shown here. Analyze the data from this experiment (use
α = 0.05) and draw conclusions. Attach R-output and R-codes.
Days Solution 1 2 3 4
1 13 22 18 39
2 16 24 17 44
3 5 4 1 22
b. [8 Points] Prove that k ai=1 Q2i is the adjusted sum of squares for treatments in a λa
Balanced Incomplete Block Design(BIBD). Here a is the number of treatments, k is the block size and λ is the concurrence of a pair of treatments (number of times a pair of treatments occur in the blocks).
Solution Q3.
Question 4.
a. [14 Points] Seven different hardwood concentrations are being studied to determine their effect on the strength of the paper produced. However, the pilot plant can only produce three runs each day. As days may differ, the analyst uses the balanced incomplete block design that follows. Analyze this experiment (use α = 0.05) and draw conclusions. Attach R-output and R-codes.
Hardwood Days Concentration% 1 2 3 4 5 6 7
2 4 6 8 10 12 14
136 130 127
b. [3 Points] Give the incidence matrix for the design given in part (a).
c. [3 Points] Give the concurrence matrix for the design given in part (a).
Solution Q4.
Question 5. Show that
a. [7 Points] SSTr = ||P Y||2 = t r τˆ2
VT i=1 i i b. [7 Points] E(||PW Y||2) = (N − t)σ2
c. [3 Points] You want to compare four treatments in blocks of two runs. Find a Balanced Incomplete Block Design for this experiment with six blocks.(You just have give all the parameters of this BIBD)
d. [3 Points] Verify that a BIBD with the parameters a = 8, r = 8, k = 4, and b = 16 does not exist.
Solution Q5.
Extra page
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