MATH6027

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UNIVERSITY OF SOUTHAMPTON MATH6027W1 SEMESTER 2 EXAMINATION 2018/19 MATH6027 Design of Experiments Duration: 120 min (2 hours) This paper contains 4 questions. Question 1 is worth 30 marks, Question 2 is worth 40 marks, and Questions 3 and 4 are worth 15 marks each. Answer ALL questions. An outline marking scheme is shown in brackets to the right of each question. Formula Sheet FS/MATH6027/2018/19 will be available. Only University approved calculators may be used. A foreign language direct ‘Word to Word’ translation dictionary (paper version) ONLY is permitted provided it contains no notes, additions or annotations. Page 1 of 6 + Appendices (Formula Sheet) Copyright 2019 v01 ⃝c University of Southampton Page 1 of 6 Source Blocks Extra due to treatments Residual Total degrees of freedom Sum of squares Mean Square 2 MATH6027W1 1. (a) An incomplete block design was used to investigate four types of resistor (treatments) using four mountings (blocks), each of which could hold four resistors. The design and data (current noise ×100) are given below. Treatment (resistor) Block 1234 (mounting) 1 110 9580 2 170120 95 3 160 110 150 4 120 150 120 (i) Write down an appropriate and estimable linear model for this experiment. Write down the model matrices for treatments (X) and blocks (Z) such that treatment 4 and block 4 are used as baselines. [4 marks] (ii) Given that the residual sum of squares from fitting a null model is 8816.67, the residual sum of squares from fitting the model with just the intercept and block parameters is 5366.67 and the residual sum of squares from fitting the model with intercept, block and treatment parameters is 672.92, complete the below analysis of variance table, and test the null hypothesis of no difference between treatments at the 5% level. [9 marks] (iii) From these data, the estimates of the treatment parameters are βˆ = 50.63, 1 βˆ = 3.75 and βˆ = 33.13. Use Bonferroni’s method at an experimentwise 23 5% level to decide which pairs of treatments produce a significantly different mean response. [7 marks] Copyright 2019 v01 ⃝c University of Southampton Page 2 of 6 3 MATH6027W1 (b) (i) An engineer would like to compare 5 new materials, A, B, C, D, E, and a control material F. The machine in which he tests the materials can only hold 2 units at a time, so each run of the machine is considered as a block. He suggests the following design for the 6 materials in 5 blocks of 2 units: Materials 1AF 2BF Blocks 3 C F 4DF 5EF Find the degrees of freedom associated with each source of variation from the ANOVA table. Do you find any problem? Can the problem be solved by using a different allocation of the 6 treatments to 5 blocks of size 2? [4 marks] (ii) The engineer receives a research grant, from which he buys a new machine that can test 12 units simultaneously. He suggests to have 7 units of material F and one each of A, B, C, D and E (Design 1). His PhD student would prefer to test 2 units of each of the 6 materials (Design 2). Using material F as the baseline treatment, the covariance matrices for βˆ = (βˆ ,βˆ ,βˆ ,βˆ ,βˆ ,βˆ ) for Design 1 and Design 2 are, respectively, 0ABCDE  1−1−1−1−1−1 −1 8 1 1 1 1 1−1 1 8 1 1 1 Var1(βˆ)=  σ2, 7−1 1 1 8 1 1 −1 1 1 1 8 1 −1 1 1 1 1 8  1−1−1−1−1−1 −1 2 1 1 1 1 1−1 1 2 1 1 1 Var2(βˆ)=  σ2. 2−1 1 1 2 1 1 −1 1 1 1 2 1 −1 1 1 1 1 2 Explain, without performing explicit calculations, how you would find these covariance matrices. Which design would you prefer for estimating the treatment differences? Give your reasons. [6 marks] [Total: 30 marks] TURN OVER Copyright 2019 v01 ⃝c University of Southampton Page 3 of 6 4 MATH6027W1 2. (a) Two choices of a 27−2 factorial design are being considered for use in an experiment. One design, d1, aliases F1F2F3F4 and F3F4F5F6F7 with the mean; the other design, d2, aliases F1F2F3F4 and F2F3F4F5F6F7 with the mean. (i) Write down the full defining relation for each design. (ii) State the resolution of each design. (iii) Which design would you prefer? Give your reasons. [4 marks] [2 marks] [4 marks] (b) A design is required for five factors, each at two levels, that can estimate all main effects independently of each other, when higher-order interactions are negligibly small. (i) Write down the defining relation for a fractional factorial design with the smallest number of runs that allows all the required effects to be estimated. [4 marks] (ii) What is the resolution of your chosen design? Could there be a design with smaller or larger resolution, with the design still satisfying the requirements of part (i)? Explain why (or why not). [4 marks] (c) A 34 experiment is to be laid out in three blocks with 27 units per block. It has been decided that certain contrasts from the highest order interaction should be confounded with blocks. (i) If the component F1F21F31F41 is confounded with blocks, which other component is also confounded? [3 marks] (ii) Write down a congruence whose solution generates the initial block of the design. Determine five treatments from this initial block, and indicate how the other two blocks could be found. [8 marks] (iii) Give the partition of the degrees of freedom for your experiment. [6 marks] (iv) If the experiment is carried out as a 1/3 replicate using only one of these blocks, write down the defining relation of the design. Can all main effects be estimated if three-factor and four-factor interactions are negligible? Justify your answer. [5 marks] Copyright 2019 v01 ⃝c University of Southampton Page 4 of 6 [Total: 40 marks] 5 MATH6027W1 3. A chemist performs an experiment to study the yield of a reaction. There are two factors to investigate, the time of the reaction (A) and the temperature of the reaction (B). A full factorial design was used, with 5 additional centre points. The results are: A B Response -1 -1 58.9 +1 -1 60.5 -1 +1 60.1 +1 +1 61.2 0 0 60.2 0 0 60.4 0 0 60.6 0 0 60.5 0 0 60.3 (a) For these data, use an appropriate test (at the 5% level) to test the null hypothesis of no curvature in the response surface. [8 marks] (b) Suppose the chemist’s ultimate aim is to find the combination of time and temperature where the maximum yield of the reaction is achieved. She is asking you for advice on what she should do next, in the light of the result of the test in (a). Describe, without performing any explicit calculations, which strategy she should pursue. [4 marks] (c) Assume that, through your recommended strategy, the chemist has found a region in which the maximum is likely to lie. She now wishes to fit a second-order model to observations from this region. How could she augment the given first order design? [3 marks] [Total: 15 marks] TURN OVER Copyright 2019 v01 ⃝c University of Southampton Page 5 of 6 6 MATH6027W1 4. (a) Define D-, A-, V – and G-optimality and state their respective objective functions in terms of the information matrix X′X. State a situation in which you would use a V -optimal design rather than a D-optimal design. [5 marks] (b) Consider the following model: Y =β0+β1x+ε, where β0 and β1 are unknown constants to be estimated from the data and ε is a normally distributed random error term. The errors for different observations are assumed to be independent and have constant variance. Three different designs, d1, d2 and d3, each having 12 runs, are considered to collect data to estimate this model. The designs are summarized in the table below. x -1 0 1 d1 6 0 6 d2 4 4 4 d3 3 6 3 For example, design d1 takes 6 observations at x = −1, no observations at x = 0, and 6 observations at x = 1. (i) Write down the model matrix X and the information matrix X′X for an arbitrary design with 12 observations at points x1, x2, . . . , x12. For each design (d1, d2 and d3), calculate the information matrix X′X from this general formula. [5 marks] (ii) For each design, calculate the value of the D-optimality objective function. Which design is preferred under D-optimality? Can you think of a situation where you would recommend the design that comes second in terms of D-optimality? [5 marks] [Total: 15 marks] END OF PAPER Copyright 2019 v01 ⃝c University of Southampton Page 6 of 6

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