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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 workdaypay 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 nonadditivity 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 tinlead 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'. oilquenched. 0Q; airblown. 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 sirenSteward. 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
Soartr ti! squatu soul: 1" Value Ft. 5 1"
Model 15 157.9535DDUU 10.53o16667 5lﬂ 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 burnin test times chosen for this study were at 1hour 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 burnedin 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 WilliIN, 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 ‘ 6.51 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 ‘, unevendune Varietal3‘ “tum:  . 
. . ‘ 5km of P “Q!” ﬁnurea‘ , I' a , ‘ Dr I aquarun. _ square 7 Value: are : ti
Hedi1 ‘ * ,26 1636120“? 1.1550795 101.31 ' unseat
Error  5% 3.4565313 0.054D055
_ Cull”:th metJ. ' . 81: "Locations
llSﬁumtn ‘ _l:.1i‘. ' neat. Mes.  , rAtLuan stone
0315“: , 4.105995 otassooa ‘ , seasons:
. I . , ‘s
aﬂourun ‘ " or Move :3 .Nunn Square I" Vainoi 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 blocktreatment interaction? In increasingly severe oilwell environments. oil producers are interested in highstrength nicltel
alloys that are corrosionresistant. 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.] napindent: Vatinn a: II ELI: sun o! Hun antigen pl Squarii Equalso I vanilla: 1': a r
nodal 11 1i]1.73l5ll 11551123! 25173 00001
error 1.: 1.1.4!!!“ menJae
correctn! Total 21 lull'79.“! nSquuo ELY. Root HR: ream Nun b.955I01 1.8519‘1 Otﬂlﬂﬂil 45.720I31!
Souscl til' nnovn 59 item square 1' VII1H ‘l't‘.’ 1 1’
ALLOY 3 552.Dbﬂl157 Eilsﬁ50‘151 “1325 n.Dun1
KATCDHD 1 ESE3137569 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 41735ﬂﬂﬂ _ EUIGISUQ 3.57 610315
ALLDY‘HATCOHD'TIHE I ‘DIIITEQD 9.33 0.730! DMJ'i'SiWﬁ 3') Carry all relevant ANDVA Ftests 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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 Winter '10
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