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Unformatted text preview: Stat 500 lab notes c Philip M. Dixon, 2009 Week 11: Model construction Case study: Constructing models For each of these four situations, construct an appropriate model, describe any additional X variables, and indicate how you would estimate (or test) the quantities of interest. Warranty claims on cars. A major car manufacturer is interested in modeling the number of warranty CLAIMS per month for two different car TYPEs, A and B, as a function of TIME, the number months since the type was introduced. 1. There are no claims in month 0, but then claims increase linearly with time, perhaps at a different rate for the two car types. The quantity of interest is the difference in claims (between car types) in month 12. 2. 1 There is an initial spike in claims in month 1. After that, claims increase linearly. The quantities of interest are: a) the difference in slopes from month 1 to 12, and b) the difference in claims in month 12. Corn yield. Farmers add nitrogen fertilizer, N, to corn to increase the YIELD. Corn produces grain without added N, but adding small or moderate amounts of N increases the yield. Large amounts of N provide no additional benefit. 3. Assume that the yield response is approximately linear from 0 lb N /acre to 100 lb N /acre. Above 100 lb N/acre, the yield is constant. You wish to estimate the yield at 100 lb N/acre and the N response (increase in yield for each additional lb N/acre) when less than 100 lb N/acre is added. 4. The linear curve described above is a simplification because most biolog ical processes don’t have sharp breaks. The common model for yield is a quadratic model with a maximum at N=100 for N < 100. You want to model yield using only N values < 100. You want to estimate the difference in yield between 50 lb N/acre and 100 lb N/acre. 1  Stat 500 lab notes c Philip M. Dixon, 2009 bacillus2.sas options ls=75 formdlim=’’ nonumber nodate; data bacillus; infile ’bacillus2.txt’ ; input trt $ pre post; /* reference group coding: Placebo is the ref. group */ if trt = ’Ab1’ then a1 = 1; else a1 = 0; /* a short cut way to the same thing */ a2 = (trt = ’Ab2’ ); a3 = (trt = ’Pl’ ); /* to show what happens */ /* effects coding */ b1 = (trt = ’Ab1’ ); b2 = (trt = ’Ab2’ ); if trt = ’Pl’ then do; b1 = 1; b2 = 1; end; run; proc print; var trt a1a3 b1b2; title ’Leprosy study: indicator variables’; run; 2  Stat 500 lab notes...
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 Fall '08
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 Regression Analysis, Philip M. Dixon

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