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Final rev 1 - STAT 3 5504 Final Review Problems I For the...

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Unformatted text preview: STAT 3 5504 Final Review Problems I For the studies in each of questions 1 - 6 below a) State whether it is an observational study or a designed experiment. If it is a designed experiment state whether the design is a CRD or a RED. b) What are the factor(s)? c) What are the treatments? d) Give the complete model for the study. e) Set up the form of the ANOV A table, including the EMS's. A pharmaceutical company is investigating the comparative effects of 3 proposed compounds by injecting rats with the compounds and recording the pertinent reaction. A litter of 11 rats is available. 4 of the 11 rats are chosen at random and injected with compound A, 4 more are chosen at random and injected with compound B and the remaining 3 rats are injected with compound C. A nutritionist wishes to assess the relative effects of 4 newly developed rations on the weight- gaining ability of rats. He has 20 rats available for experimentation consisting of 4 rats from each of 5 litters. He assigns each of the 4 types of rations at random to the 4 rats within each litter. An engineer wishes to assess the relative effects of 8 different treatments on the life of a particular type of battery. He has 8 batteries from each of 8 different production lots and assigns the treatments at random to the batteries within each production lot. An agronomist conducted an experiment to assess the effects of date of planting (early or late) and type of fertilizer (none, Aero, Na, or K) on the yield of soybeans. 32 experimental plots were available and the treatments were randomly assigned to the plots with each treatment being assigned to 4 plots. An agronomist has 28 experimental plots available for testing the relative effects of 4 different fertilizers (none, Aero, Na, K) on the yield of a particular variety of oats. 7 plots are randomly assigned to receive no fertilizer, 7 Aero, 7 Na, and 7 K. It is desired to investigate the effect of hourly pay rate (3 levels) and length of work week (3 different plans) on some measure of worker productivity. Worker productivity is measured for random sample of 6 workers from each of the possible workday-pay rate combinations. 10. Why is a randomized block design a poor design to study the effect of two factors of interest on a response? How do randomized block designs increase the precision in an experiment? Suppose you wish to investigate the effect of 3 factors on a response. Explain why a factorial selection of treatments is better than varying each factor, one at a time, while holding the remaining 2 factors constant. What graphical method would you use to check for non-additivity of factors or of blocks and treatments? (1) l 1. Why is replication important in a complete factorial study? 12- , The analysis of variance for a randomized block design produced the ANOVA table entries shown here. ' SOURCE til SS MS Treatments 3 27.1 __._. Blocks 5 14.911 Error 33.4 Total The sample means for the four treatments are as follows: y,‘ = 9.7 ya = 12.1 yc = 5.2 in = 9.3 a. Complete the ANOVA table. . 1). Do the data provide sufficient evidence to indicate a di using a = .0 5' * ilerence among the treatment means? Test 0) Find the relative efficiency of this RBD to a CRD using the same experimental units. d) Looking at the treatment means it was decided that confidence intervals should be obtained for the following comparisons: L1=fla+flo'#3‘#cr L2=#B-2/tc (i) use an appropriate multiple comparison to find the C.I.s. (ii) Are L1 and L2 contrasts? 13. The chemical element antimony is sometimes added to tin—lead solder to replace the more expensive tin and to reduce the cost of soldering. A factorial experiment was conducted to determine how , antimony affects the strength of the tn-lcad solder joint (journal of Materials Science, May 1986). Tin—lead solder specimens were prepared using one of four possible cooling tncthOds (water- qucnched, WQ; oil-quenched. OQ'; air-blown, AB;’and furnace-cooled, FC) and with one of four possible amounts of antimony (0%, 3%. 5%. and 10%) added to the composition. Three solder joints were randomly assigned to each of the 4 X 4 = 16 treatments and the shear strength of each measured. The experimental results, shown in the accompanying table, Were subjected to an ANOVA using SAS. The SAS printout is also shown. AMOUNT OF ANTIMONY COOLING METHOD SHEAR-STRENGTH. % weight MPa 0 WQ 17.6 19.5 18.3 0 0Q 20.0 24.3 21.9 0 AB 18.3 19.8 22.9 0 PC 19.4 19.8 20.3 3 WQ 18.6 19.5 19.0 3 0Q 20.0 20.9 20.4 3 All 21.7 22.9 22.1 3 PC 19.0 120.9 19.9 5 WQ 22.3 19.5 20.5 5 0Q 20.9 22.9 20.6 5 AB 22.9 19.7 21.6 5 PG 19.6 16.4 20.5 10 WQ‘ 15.2 17.1 16.6 10 0Q 16.1 19.0 18.1 10 . ' AB 15.8 17.3 17.1 10 .FC . IGA 17.6 17.6 14. Analysis ‘0! Variance Procedur- nupondnne Variable: sat-113110211 _ Sun of loan 190::- D! Squares Squat. r V1.10. Pr > r 1101101. ' 15 1151.95250000 10.53016667 5.10 0.0001 lrtor ' 32 55.215656“ 1.1:515033 cox-mend roe-1 01 111349915501 0 R-Squlxn C.V. Root. HS: STRENGTH Mann 0.1100511 5.1195215 1.3139015 19.55415551 source I)! Anov'n s0 "our Square at V1111. , Pr > P mom 3 100.194151 34.131309 20.12 0.0001 1151000 3 20.521500 9.502500 ‘ 5.53 0.0030 mom-moo 9 25 . 130033 2.192315 1 .62 0. 1523 a) Test whether the two factors, amount of antimony and cooling method Interact. Use 0! = .05. b) If appropriate, conduct the tests for main effects. ’Use 0! = .05. If necessary, use the Kimball inequality to give an overall significance level for all the tests you have carried so far 0) If you were asked to carry out all possible pairwise comparisons and end up with an overall significance level for all 1 :1f them of .,05 how would you proceed? A trade—off study regarding the insp: :ction‘and test of transformer parts was conducted by the quality department of a major defense contractor. The investigation was structured to examine the elfects of varying inspection levels and inc .1ming test times to detect early part failure or fatigue. The levels of inspection selected were full mili::11y inspection (A), reduced militmy specification level (13), and commercial grade (C) Operational liurn- in test times chosen for this study were at 1 hour increments from 1 hour to 9 hours. The response was failures per thousand pieces obtained from samples taken from lot sizes inspected to a spec: f1ed level and burned- ~in over a prescribed time length. Three replications were randomly sequenced under each condition, making this a complete 3 X 9 factorial experiment (a total of 81 observations) The data for the study (shown in the table) were subjected to an ANOVA using SAS. The SAS printout follows. Analyze and interpret the results lNSPECTlON LEVELS 3. 00 3.92 4.22 . . » 7.90 7.17 ‘ 7.70 3. 70 3.68 3.80 . . 11.40 8.60 7.90 3.40 3.55 3.45 . . 8.82 9.76 9.52 BURN-1N. hours ’Fuii Military Specification. A Reduced Military Specification. B Commercial. C l 7.60 7.50 7.67 . . . 6.16 6.13 5.21 2 6.54 7.46 ' 6.8+ . . . 6.21 5.50 5.64 3 6.53 5.85 6.38 . . . 5.41' 5.415 5.35 4 5.66- 5.98 5.37 . . . 5.68 5.47 5.84 5 5.00 . 5.27 5.39 . . . 5.65 6.00 5.15 6 4. 20 3.60 4.20 . . . . 6.70 6.72 6.51 7 B 9 Mu— Analylil of Variance Procedure anondlnt. Vatinhlo: urama . , - 31101 of Ila-n Souruo' . D! p . Square-- Square 1' Value Pr > 1' 1 Model T V :6 “8.6110561 6.0850195 101.31 0.0001 Error 54 3 .0565133 0.0640099 . corroctod rot-1 '0 17245606000 R-Squ-t- ¢:.V. ’ ' loot K83 . - _FAXLUR£ Moan 0.919912 4.002990 ' 0.2521002 5.7422222! “source 01' Anon. 5s Mean Squat. r Value . Pr > F 0mm ' 0' 21 .91110000 339550000 50.53 0.0001 INSLIVEL 2 43.030121“! ,31.5¢205926 336.56 0.0001 BUMXH'INSLML 16 97 .553511015 0 .09709676 95.25 0.000). w... (3} 15. 16. 17. 18. a) Explain the steps needed to analyze the above experiment. In a RBD why do you have to assume there is no block-treatment interaction? The following data were generated from a RBD. Treatments "A B c D 1 8.3 10.9 8.0 8.1 2 [1.5 13.6 11.6 10.8 3 7.4 9.3 10.1 9.1 4 8.8 9.5 8.7 8.8 5 110.5 13.1 9.8 9.5 a) At the .10 level of significance, can we conclude that there are differences among the 4 treatments? b) If appropriate, carry out all pairwise comparisons between the treatment means and summarize your results using the underlining method. In marketing children's products, it's important to produce television commercials that hold the attention of the children who View them. A psychologist hired by a marketing research firm wants to determine whether differences in attention span exist among advertisements for different types of products. Each type of commercial is randomly assigned to the 3 children within each group of: 3 lO-year olds, 3 8-year olds, 3 6-year olds, and 3 4-year olds. Their attention spans (in sec) are measured with the following results. Type of Product Toys/Games Food/Candy Clothes a) Identify the treatments. b) Identify the response variabi e. c) Identify the experimental design. d) Give the complete model for the experiment (assuming fixed levels). For text problem #219, a) what is the experimental design? b) What are the treatments? c) Set up the AN OVA table putting; in actual numbers only for the df column. EMS are not needed. (a) , [1% For the data shown below on 3 treatments (fixed) and covariate X a) Give the complete covariance model. State the reduced model for testing for adjusted treatment effects. ‘ b) State the model to be used :f'or testing whether or not the treatment regression lines have the same slope. c) Use the SAS output given to carry out the test of part (b). Use 0! = .01. d) Test for adjusted treatment effects with a significance level of .05 if appropriate, otherwise state why it is not appropriate. Model: MODELl Dependent Variable: Y Analysis of Variance Sum of Mean Source DF Squares Square F Value Prob>F Model 5 16396.23091 3279.24618 89.154 0.0001 Error 12 441.38020 36.78168 C Total 17 16837.61111 Root MSE 6.06479 R—square 0.9738 Dep Mean 115.72222 Adj R—sq 0.9629 C.V. 5.24082 - Parameter Estimates Parameter Standard T for H0: . Variable DF Estimate Error Parameter=0 Prob > |T| INTERCEP 1 114.487284 1.54231370 74.231 0.0001 X 1 3.545822 C.28535447 12.426 0.0001 11 1 1.156839 2.10954999 0.548 0.5935 12 1 19.974309 2.25812898 8.846 0.0001 INTERACl 1 0.696452 C.41516070 1.678 0.1193 INTERACZ 1 0.238049 C 0.532 0.6043 .44730053 (5‘) Model: MODELZ Dependent Variable: Source Model Error C Total Root MSE Dep Mean C.V. Variable DF INTERCEP l X 1 I1 1 I2 1 Model: MODEL3 Dependent Variable: Source Model Error C Total Root MSE Dep Mean C.V. Variable DF INTERCEP l X 1 Analysis of Variance Sum of Mean DF Squares Square F Value Prob>F 3 16092.17047 5364.05682 100.741 0.0001 14 745.44064 53.24576 17 16837.61111 7.29697 R—square 0.9557 115.72222 Adj R-sq 0.9462 6.30559 Parameter Estimates Parameter Standard T for H0: Estimate Error Parameter=0 Prob > IT! 115.722931 1.71991214 67.284 ,‘0-0001 3.187940 0.30003032 10.625 0.0001 0.506700 2.43802768 0.208 0.8384 20.162143 2.53572682 7.951 0.0001 Analysis of Variance Sum of Mean DF Squares Square F Value Prob>F 1 11721.23132 11721.23132 36.655 0.0001 16 5116.37979 319.77374 17 16837.61111 17.88222 R-square 0.6961 115.72222 Adj R—sq 0.6771 15.45271 Parameter Estimates Parameter Standard T for H0: Estimate Error Parameter=0 Prob > ITI 115.723145 4.21487931 27.456 0.0001 4.152947 0.68594795 6.054 0.0001 ...
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