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Unformatted text preview: 'I' 0 (a) Response: Incidence rates of headaches. Explanatory: Usage of aspartame.
(b) Control: Subjects taking placebos. Treatments: Subjects given aspartame. (c) It reduces the variability that occurs when diﬂ'erent subjects take the treatment and
control. If some people are more likely to get headaches than others (due to stress,
health, etc...) a crossover design helps to equalize this factor for both treatments. 3.21 Response: Mileage. Treatments: Gasoline additive, Concentrations. Blocks: Make of car. i 3.22. Response: Occurrence of stomach irritation. Treatment factors: Dosing schedule, form of
administration. Confounding variable: Severity of infection, weight, age. 23.23 (a) Let A=method l, B=method 2, and C=method 3; Then one CR design is
{A,B,B,G,A,A,B,O,B} Ib) One randomized block design is { A,C,B , B,A,C , B,C,A ‘—v—’ W “v”
Person 1 Person 2 Person 3 I' 3.24 (a) Let C=Column sum, R=Row sum. Then one CR design assigns the problems to the 20
students as follows: _{qaqonqaanqaqqaonndam (b) Classify the students into fast and slow adders. Then within each group of students,
randomize 5 C and 5 R questions to the students. Then one randomized block design is {unaqqadaaonnnpnncnnn}
WWW—d
aw as 3.25' (a) Let Azaerobic, S=strengthening, and C=control. Then one CR design is
' {A,C,'C,S,A,A,C,A,A,C,S,S,S,C,A,C,S,S,C,S,A,S,A,C } (b) One possible randomized block design is {AacssysaC,C.A:5,C.A=S:A151A,C,G,A,A:S.0144.055}
—.._v............._.r \———_—V__....a
' 5579 80 and older 3.26 (a) Let S=Standard method, N=New method. Then one GR design is
{SINQSISTN1N15’NIS’STN1N18’SisiNisiN3N38INYN!SIN!S!N?NIS!S;N} (b) One randomized block design is {SrNisisiNaNiS1NisiN } NISISlNINISINiNISISI SiNiszNisasiNiSiNaN}v
‘—/———J ‘—.—v———/ —W——’
Tech 1 7 Tech 2 ‘ Tech 3 39 Chapter 4 Solutions iE  _  Solutions for Section 4.1 I 4.1 (a), Categorical (Ordinal).
(b) Categorical (Nominal)
(c) Numerical (Practically
(d) Categorical (Nominal) Continuous) . 4.2 (a) Engine Size: Numerical (Continuous) ,
Number of Cylinders: Numerical (Discrete)
Size of Car: Categorical (Ordinal) '
Type of 'Itansmission: Categorical (Nominal)
Gas Guzzler Tax: Categorical (Nominal) 
Dealer Cost: Numerical (Practically Continuous)
Theft Rate Index: Numerical (Practically Continuous) (b) Engine Size: Ratio scale
Number of Cylinders: Ratio scale Dealer Cost: Ratio scale 4.3 (a) Enrollment: Numerical (Practically Continuous)
Required Entrance Tests: Categorical (Nominal)
Annual Tuition: Numerical (Practically Continuous) Fields of Study: Categorical (Nominal) Selectivity: Categorical (Ordinal)
Percent of Applicants Selected: Numerical (Continuous) r (b) Enrollment: Ratio scale Annual Tuition: Ratio scale . Percentof Applicants Selected: Ratio scale 4.4 (a) Sex: Categorical (Nominal)
Age at Graduation: Numerical (Practically Continuous)
Time to Complete Degree: Numerical (Practically Continuous) Future Plans: Categorical (Nominal)
Citizenship: Categorical (Nominal)
Class Rank: Categorical. (Crdinal) Grade Point Average: Numerical (Continuous) (‘0) Age at Graduation: Ratio scale
lete Degree: Ratio scale Time to Comp
e: Interval scale Solutions for Section 4.2 4.5 40 Pareto Chart ior No. of Complaints Pareto Chart for Total Demerit Score mntd COMP LTS
I'J EMEHITB E(45%), 095%), and 305%) agcount for more than 80% of all cominlaints.. 4.6 Number of Degrees I Relative Frequency
Physical Sciences ‘ 6276 0.168 ‘ Engineering  5212 0.139
Life Sciences 6928 0.185
Social Sciences . 6127 0.164
Humanities 4094 0.109
Education 6397 , 0.171
Other 2417 0.065 Total 37451  1.000 R ' I  Pareto Chart 1 _ ‘ Total Number of Degrees Awarded
3 ‘ 40000 luemad 30000 2m 00 10000 41, The‘degrees awarded are pretty errenly distributed among the professional ﬁelds. 4,7 (:1) The number of doctorate degrees awarded to women were: 916 in physical sciences, 453
in engineering, 2674 in life sciences, 3027 in social sciences, 1904 in humanities, 3717 in
education, and 836 in other professionalﬁelds. The number of doctorate degrees awarded to men were: 5360 in physical sciences, 4759 in engineering, 4254 in life sciences, 3100
in social sciences, 2190 in humanities, 2680 in education, and 1581 in other professional ﬁelds . Pareto Chart. _ '   Pareto Chart Number of Degrees Awarded to Men Number of Degrees Awarded to Women 13W 3mm 3, 1
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[:11de mg Bland and em tunrun and human cum (1)) The degrees awarded most to men and 'women are different. Men are more likely to earn
degrees in physical sciences, engineering, and life sciences, 'while women are more likely
to earn degrees in life sciences, social sciences, humanities, and education. 4.8 ' (a) Pareto Chart 1M 1200 '3‘
meme; 10m E 8
a Number cl Acquisitions
“e' s The US. and Britain have the largest numbers of acquisitions, accounting for almost
50% of all acquisitions, ' 42 (b) Pareto Chart 70 mauled 3:
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40
SCI 5!!
20 10 Value at Acqulsillons Country The U.S. has the largest acquisition values, followed by ﬁance. Together they account
for about 40% of all acquisition values. ' Solutions for Section 4.3 4.9 (a) The median will be larger because the long left tail will shift the mean left.
(is) The sample skewness will be negative. (c) One would expect Q1 to be farther from the median because the long left tail shjﬁ,
the ﬁrst quartile farther to the left. ' r ‘ 4.10. (a) The distribution is positively skeWed. (b) ‘One would expect Q3 to be farther from the median because the distribution is positively
skewed.
(C)  Q1 = {Bu2:75) = mug) + 0.75 (was)  mug) = 13 + (U.75)(14 — 13) = 13.75. Median: Q2 & 11795.5) = $95) + ($95) —' 113125)) = 20 + (0.5)(20 — 20) = 20. Q3 = $9.25) _= 50355) + 0.23 (egg) — 3:93)) = 33 + (0.25)(34 — 33) = 33.23. As expected, Q3 is farther from the median. (d) The meaniwill be larger than the median because the positive skew will shift the mean
111). Check: :7; = 25.3 > 92 = 20.0. . (e) The IQR is 33.25 — 13.75 = 19.5. The lower fence is 13.75 — 1.5 x 19.5 = ~15.5. The 2'
upper fence is 33.25 + 1.5 x 19.5 = 62.5. Therefore, 64 and 69 should be outliers. Mean: 5 = 97.0 + 97.23: + 99.8 = 98_563__ Median: Q2 = {5055} = 3‘15) + 0.5 ($05) — 3.15)) = 98.6 +_(9.5)(9s.9  98.6) = 98.6. . (97.9  98.563)2 + (97.2 — 99.593)2 + + (99.3 m 99.563)2
30 — 1 Q1 .—. mm = mm + 0.75 (5(3) gem) = 97.9 + (0.75)(98.2 — 97.9) = 99.125. SD: 3 = = 0.751. Q3 2 $03.25} = $(23) + 0.25 (${24) — $93)) +‘(D.25) (99.2 *' = 99.125. (b) There are no outliers, since no observations fall ontside the fences (LF = 98.125 — 1.5 x
1 = 96.625 and UF = 99.125 + 1.5 x l = 100.625).  ' ('3) Histogram of Temperature 97.00 97.50 93.00 53.50 99.90 95.50
5735 97.75 63.25 53.75 99.25 99.75 TEMP The shape is unimodal with .110 outliers or skew.
4.12 (a)‘ 1 .
Min: 452. Q1 = mm) = 850.
Median: Q2 = 3(24) = 1331.
‘Q3 = W35) = 1737.
Max: 3830. This summary suggests a symmetric distribution, since thequartiles are equidistant 3
from the median. Note, hcnaureverI that the maximum is much farther from Q3 than the
minimum is ﬁom Q1, possibly because it is an outlier. ' 44 (b)
N IQR _ 1737 — 850 Approx; 3 ~ 1.34 — 134 = 661.94
14 — 2 ._ 2 __ 2
Actual: S ___ 68 1369) +(909 47:16—62) + +(850 1369) #:69367 The actual SD is roughly the same as the approx. SD. (:1) A 10% trimmed mean trims off the outside no: = 47 x 0.1 = 4.7 = 4 observations when
computing'the mean. So 5am) =‘ 96(5) + 3(6) +3" + mes) I ' 47 _ 2 x 4 =— 1292.64. This is substantially lower than the sample mean of 1369, which suggests that there are
some outliers on the high end of the data. 4.13 (a) RAINFALL StemandLeaf Plot Frequency . Stem E Leaf 2.00 0 . 44
17.00 0 . 555677788885995
11.00 1 . 11123333334
11.00 1 . 55577778888
1.00 2 . 0
4.00 2 . 5667
1.00 Extremes (>=3830) Stem width: 1000.00 Each leaf: 1 case(s) The shape is rightskewed and unimodal, with one outlier. (b) ...
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This note was uploaded on 04/09/2008 for the course MATH 3200 taught by Professor Sawyer during the Spring '08 term at Washington University in St. Louis.
 Spring '08
 SAWYER

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