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Stat 402 A - Spring 2009 - Exam 2 Answers
Additional explanations are included as Notes.
- Swamp tree growth. My SAS code for parts d-g:
data nyssa; infile ’nyssa.txt’; input block pool light species $ water growth; run; proc mixed method=type3; class block light species water; model growth = block light|species|water / ddfm=satterth; random block*light; estimate ’light 100-35’ light -1 0 1; estimate ’water 0 - 10’ water 1 0 -1; run;
(a) 2 pts. Two e.u.’s: Pool and pot. Note: Light levels are assigned to pools; water level and species are assigned to pots (b) 3 pts. Trt design: 3 way factorial (or 3x3x2 complete factorial) Expt design: split plot with RCBD at the main plot level and CRD at the split plot level (c) 10 pts. Source d.f. F/R Block 4 F or R Light 2 F main plot trt BL 8 R Main plot error Species 1 F split plot trts Water 2 F “ SW 2 F “ LS 2 F mainsplit interactions LW 4 F “ LSW 4 F error 60 R split plot error c.total 89 Note: There are 15 pools (main plot eu), so main plot d.f. add up to 14. There are 90 observations so all df add up to 89. (d) 5 pts. Strong evidence that species respond differently to water: F for waterspecies = 46.28, p < 0. No evidence that species respond differently to light: F for lightspecies = 1.09, p = 0. No evidence that species have different 2way interations between light and water: F for specieslight*water = 0.70, p = 0. Note: These questions all ask whether species have the same pattern of response, i.e. whether the curves (e.g. height vs water level) for each species are parallel. These are questions about the interaction between species and another factor.
(e) 5 pts. F for light = 26.0, p = 0.0003. Strong evidence that light affects height. (f) 5 pts. The difference in mean height between 100% and 35% light, averaged over species and water levels, is estimated to be 15.5 with a s.e. of 2.3. Note: I deducted a point if the difference was taken the other way around. That really changes the conclusion! Note: A few people reported means, not the mean difference. Please answer the question that I ask. (g) 5 pts. The difference in mean height between 10cm and 0cm water depth, averaged over species and water levels, is estimated to be 31.0 with a s.e. of 1.5. (h) 5 pts. No they are not. The difference between water levels is more precisely estimated because water is a split plot factor. “Miniblocking” increases the precision. Note: A lot of folks answered that the split plot was more precise because the split plot error had more d.f. D.f. have nothing to do with the magnitude of the s.e. D.f. tell you much information was used to calculate that variance. So, they tell you how precisely you have estimated a variance. That has nothing to do with the size of the s.e. When you test a hypothesis or calculate a confidence interval, the d.f. influence the power of the test and width of the ci because d.f. determine the t or F quantiles. However, s.e. and d.f. are unrelated. Note: The other common explanation of unequal s.e.’s, different sample sizes, doesn’t apply here. Water means are based on 6 cells (3 light x 2 spp); light means are based on 6 cells (3 water x 2 spp). Each cell has 5 observations.
- Potato scab. My approach combined a 1 way and 2 way ANOVA. SAS code for all parts using this approach :
data scab; infile ’c:/philip/stat 402/scab.txt’; input block trtcode1 trtcode2 amount season $ disease; run; proc glm; class trtcode2 block; model disease = block trtcode2; estimate ’fall - spring, when applied’ trtcode2 0 1 -1 1 -1 1 -1 / divisor=3; contrast ’interaction when applied’ trtcode2 0 1 -1 -1 1 0 0, trtcode2 0 1 -1 0 0 -1 1; contrast ’amount in fall incl 0’ trtcode2 1 -1 0 0 0 0 0, trtcode2 1 0 0 -1 0 0 0, trtcode2 1 0 0 0 0 -1 0; contrast ’amount in spring incl 0’ trtcode2 1 0 -1 0 0 0 0, trtcode2 1 0 0 0 -1 0 0 , trtcode2 1 0 0 0 0 0 -1 ; contrast ’amount, ave over season’ trtcode2 2 -1 -1 0 0 0 0, trtcode2 2 0 0 -1 -1 0 0, trtcode2 2 0 0 0 0 -1 -1; estimate ’control - fungcide’ trtcode2 6 -1 -1 -1 -1 -1 -1 / divisor=6; title ’Analysis using trtcode2 ’; run;
(d) 5 pts. This could be reasonably answered either using the 2 way factorial and slicing, or by constructing contrasts among 7 treatments. The answers differ slightly because the 0 group is defined differently in the two approaches. If you used contrasts with trtcode (8 treatments), you get the same answer as the 2 way factorial. Using 2 way factorial: fall, p = 0.0008; spring, p = 0. Using contrasts: fall, p = 0.0027; spring p = 0. There is strong evidence that amount of fungicide applied in the fall affects the amount of disease. There is no evidence that amount applied in the spring affects the amount of disease. (e) 5 pts. Yes, it does. The estimated difference between control (0 fungicide) and the average of all fungicide treatments is 9.29 units with a s.e. of 2.83.
- Barley-fungus microarray studies All studies have the same treatment design. They differ in the experimental design.
(a) 8 pts. This is a CRD. There are 18 chambers, each with 1 flat. The e.u. is the combina- tion of chamber and flat. Each combination of barley and fungus is randomly assigned to an e.u. Source d.f. error term Barley 2 error Fungus 1 error BF 2 error Error 12 c.total 17 Note: Some folks thought this was a split plot because fungus went to chamber and barley to flat. It’s not a split plot because 1 chamber is 1 flat. If you did this please bring your exam and see me. (b) 8 pts. This is a RCBD with 3 blocks, one for each repetition of the study. It is identical to the previous problem, except the 18 chambers are grouped into 3 repetitions. Source d.f. error term Block 2 Barley 2 error Fungus 1 error BF 2 error Error 10 c.total 17 (c) 8 pts. This is a split plot. The two eu’s are chamber (fungus, in blocks) and flat (barley). There are three flats in this chamber. Chamber is the main plot to which fungus is assigned. Flat is the split plot to which barley is assigned. Source d.f. error term Block 2 Fungus 1 BlockF BlockF 2 (= main error) Barley 2 split error B*F 2 split error split error 8 c.total 17
(d) 8 pts. This is also a split plot, with same same eu’s as before, except only one repetition of the main plot trt. There are only 2 e.u.’s, so only one d.f., at the main plot level. There is no main plot error. Source d.f. error term Fungus 1 none main error 0 Barley 2 split error B*F 2 split error split error 12 c.total 17 Note: I talked about pseudoreplication after the cutoff for what was covered on this exam. Everything above can be figured out very easily from the same principles used in the previous part. I gave full credit if you claimed that the split error could be used to test fungus AND explicitly said you were committing pseudoreplication.
