examrevIw11 - STAT 3504 Final Review Problems Fer the...

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Unformatted text preview: STAT 3504 Final Review Problems ] Fer 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 GED ora RBD. b) What are the factorts)? c) What are the treatments? d) Give the complete model for the study. e) Set up the form of the ANOVA table. including the EMS‘s. l. 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. 2. An agronomist conducted an experiment to assess the effects of date of planting (early or late) and type of fertilizer (none. Aero. Na. 0r 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. 3. 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. 4. 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. 5. Suppose you wanted to study the effect of the 2 factors age and diet on blood pressure. Why is a randomized block design using age as a blocking variable a poor design for the study? 6. How do randomized block designs increase the precision in an experiment? 7. What graphical method would you me to check for non-additivity of factors or of blocks and treatments? 3. Why is replication important in a complete factorial study? The analysis of Variance for a randomized block design produced the ANDVA table entries shown here. m..__.__...._...._.—_.______._,I SOURCE of 35 MS F Treatments 3 27.1 __..u_.._.. Blocks ‘ 5 __.,_H_I 14.90 .— Error 33:} Total The sample means for the four treatments are as follows: 51A = 9.7 f” 2 = 5.2 a. Complete the ANOVA table. in = 9.3 b. Do the data provide sufficient evidence to indicate a dillerence among the treatment means? Test using a: = .0 5' c) Find the relative efficiency ofthis RED to 3 CPD using the same experimental units. d) Looking at the treatment means it was decided that confidence intervals should be obtained for the following oompari sons: Ll=flg+iup'iun'iucr L2=#n'2#c (i) use an appropriate multiple comparison to find the (3.1.5. (ii) Are LI and L2 contrasts? 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 I antimony affects the strength of the tin—lead solder joint (journal of Materials Science. May 1986). Tin—lead solder specimens were prepared using one of four possible eoolingl methods (watch quenched, WQ'. oil-quenched. 0Q; air-blown. A13; and furnaccaeooled, PC) and with one of lour possible amounts of antimony (0%, 3%. 5%. and 10%) added to the composition. Three solder joints Were randomly assigned to each of the ‘i 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. - AMDUNT 0F ANTIMONY COOLING METHOD isweiulil e WQ 0 oo 0 AB 0 PC 3 WQ 3 0Q - -3 an .3 FC 5 . WQ 5 oo 5 AB 5 . FC . . I to ‘-wo" ‘ ' _ - I _' -- ' to ‘ ' ‘oo '* I a r , I ' . 10 I‘ ‘ I AB 10 . EC siren-Steward. 1?.6 10.0 10.3 19.4 13.0 20.0 21.7 1.9.0 21.3 20.9 ‘22? 19.0 15.2 16.1 15.13 163‘} MP: 19.5 2%.} 19.8 19.3 19.5 20.9 22.9 20.9 19.5 22.9 19.1 1M 17.1 19.0 17.3 mi 15.3 11.9 12.9 20.3 19.0 20."! 22.1 19.9 20.5 20.6 21.0 20.5 16.6 10.1 17.1. 17.0 ii. Anllfltl a! Vlrllncu Frucoduru Dupundaut Variable: ETRLHOII ‘ Sun of Hum Soar-tr- ti! squat-u soul:- 1" Value Ft.- 5 1" Model 15 157.9535DDUU 10.53o16667 5-lfl 0.0fl01 3:501, - 32 55.24566561 1.12645533 currautud Tani; d1 213.15?16£&1 nwflqunru D.V. ROQL HS! STRENGTH Khan 0.740358 5.119517! 1.313947& 15.55¢16667 Enurca Dr Man 3'!- fldfln Square 1’ Value Pf 5" r $333 ' i ‘ilillllii ’iiillifil , “iii? 8:333: AMMUNI*HETBDD 5 25.110831 2.792315 .| 1.53 0.1523 a) Test whether the two factors, amount of antimony and cooling method interact. Use or = .05. b) 'If appropriate, conduct the tests for main effects. Use or = .05. if necessary, use the Kimball inequality to give an overall significance level for all the tests you have carried so for. e) If you were asked to carry out all possible pairwise comparisons and end up with an overall significance level for all of them of .05, how would you proceed? cl) What is the experimental design? A tritcieroi'l' study regarding the inspection and test of transformer parts was conducted by the quality departmet'tt of a major defense contractor. The investigation was structured to examine the effects of varying inspection levels and incoming test times to detect early part failure or fatigue. The levels of inspection Selected were full military inspeCtion (A), reduced military specification level (ti), and commercial grade (C). Operational burn-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 specified 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 ANDVA using 5A5. The SAS printout follows. Analyze and interpret the results. INSPECTION LEVELS Full Military Specification. A Willi-IN, ilDtli‘S Fictitth Mililary Specification. 5' , Commercial. 0 t 7.70 7.10 7.20 5.15 5.13 6.21 . 2 5.35 ms 6.15 6.21 are set , 3 5.30 5.60 5.33 5.41 5.45 5.35 ‘ ' 4 5.3a 5.27 5.29 sec 5.17 5.3+ 5 +35 1.99 MB 5.65 6.00 6.15 6 4.50 4.56 4.50 6.70 6.72 ‘ 6.5-1 7 3.97 3.9a 3.34 7 so 7'.-t7 "7.70 B 4.37 3 35 'i.-ii5 ll W 8.60 7.90 9 5.25 563 ‘ 5.25 finllrlil $1 Varinnan Prncaduxn ‘, uneven-dune Varietal-3‘ “tum: - . - . . ‘ 5km of P “Q!” finurea‘ , I' a , ‘ Dr I aqua-run. _ square 7 Value: are :- ti Hedi-1 ‘ * ,26 163-6120“? 1.1550795 101.31 ' unseat Error - 5% 3.4565313 0.054D055 _ Cull”:th met-J. ' . 81: "Locations ll-Sfiumtn ‘ _l:.1i‘. ' neat. Mes. - , rAtLuan stone 0-315“: , 4.105995 otassooa ‘ , seasons:- . I . , ‘s aflourun ‘ " or Move :3- .Nunn Square I" Vain-oi Pr :- 1" Walnut ' 3 is “Armando ' alienation I as.” 5.qu .1145”:va a ‘ «meanness ,zi..suossss an.“ muons ailan IHSLEVEL - 16 97.553548” LIHTOHTB . - 95.25 0.0001 t3} ‘ i2. .13. Explain the steps needed to analyze the above experiment. In a RED why do you have to assume there is no block-treatment interaction? In increasingly severe oil-well environments. oil producers are interested in high-strength nicl-tel alloys that are corrosion-resistant. Since nickel alloys are especially susceptible to hydrogen embrit- tletnent. an experiment was condueted to compare the yield strengths of nickel alloy tensile Specimens cathodicaily charged in a 4% sulfuric acid solution Saturated with carbon disulfide, a hydrogen recombination poison. Two aIIOys were combined: inconel alloy (75% nickel composition) and incoloy (30% nickel composition). The alloys were tested under two material conditions [cold rolled and cold drawn}, each at three dilTerent charging times (0, 25, and 50 days). Thus. a 2 X 2 X 3 factorial experiment was conductedI with alloy type at two levels. material condition at two levels. and charging time at three levels. Two hydrogencharged tensile specimens were prepared for each of the 2 K 2 X 3 = 12 factor level combinations. Their yield Strengths (kilograms per square inch) are recorded in the table. The SAS analysis of variance printout for the data is also shown here. ALLOY TYPE moons.” lNGULOY “- Coldrirawrl Delays 17.: +9.3 _ 50.6 +9.9 30.9 in $35” 25am 55.2 55.7 m 51.4 Ermine 31.7 723 50 days 45.2 'HD 50.5 50.2 29.7 25.] nap-indent: Vat-inn a: II ELI: sun o! Hun antigen pl Squarii Equals-o I vanilla: 1': a- r nodal 11 1i]1.73l5ll 115-51123! 251-73 0-0001 error 1.: 1.1.4!!!“ men-Jae correct-n! Total 21 lull-'79.“! n-Squu-o ELY. Root HR: ream Nun b.955I01 1.8519‘1 Otfllflflil 45.720I31! Sous-cl til' nnovn 59 item square 1' VII-1H ‘l't‘.’ 1- 1’ ALLOY 3 552.Dbfll157 Eilsfi50‘151 “13-25 n.Dun1 KATCDHD 1 ESE-3137569 BII.1411I¢D 1ifll.96 fiafiflfl] fiLLOYlflhTCUND 1 335.7SJTSOO 31517537500 infl.!d 0.00fl1 TIME 1 71.0!03333 35.51fi‘161 52.33 Daflflfll RELGY‘TIME 1 ?a5l53333 l.§fllll§7 Saflfl 0.01I‘ fllTfluflbfiliflfl 3 4-1735flflfl _ E-UIGISUQ 3.57 610315 ALLDY‘HATCOHD'TIHE I ‘D-IIITEQD 9.33 0.730! DMJ'i'Si-Wfi 3') Carry all relevant ANDVA F-tests at the or = .05 level of significance (for each test). Give the reasons why you tested certain effects but not others. -b) For which means would you consider carrying out Tukey’s procedure for pairwiSe comparisons? (a) ...
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examrevIw11 - STAT 3504 Final Review Problems Fer the...

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