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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 workdaypay 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 nonadditivity 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=#B2/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 tnlcad solder joint (journal of Materials Science, May 1986).
Tin—lead solder specimens were prepared using one of four possible cooling tncthOds (water
qucnched, WQ; oilquenched. OQ'; airblown, AB;’and furnacecooled, 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 SHEARSTRENGTH. % 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: sat113110211 _ 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
coxmend roe1 01 111349915501
0 RSqulxn 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
mommoo 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 BURN1N. 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 Ilan 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 rot1 '0 17245606000 RSqut ¢:.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 blocktreatment 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
lOyear olds, 3 8year olds, 3 6year olds, and 3 4year 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 Rsq 0.9462
6.30559
Parameter Estimates
Parameter Standard T for H0:
Estimate Error Parameter=0 Prob > IT!
115.722931 1.71991214 67.284 ,‘00001
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 Rsquare 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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 Spring '10
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