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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
3-1 The tensile strength o portlan! "ement is #eing stu!ie!. Four !i erent mi$ing te"hni%ues "an #euse! e"onomi"all&. The ollowing !ata ha'e #een "olle"te!:
(i$ingTe"hni%ue Tensile Strength )l#*in 2+
1 312 3000 2- 5 2- 0
2 3200 3300 2 /5 31503 2-00 2 00 2 -5 30502 00 2/00 2 00 2/ 5
(a) Test the h&pothesis that mi$ing te"hni%ues a e"t the strength o the "ement. se α 0.05.
Design $pert 4utputResponse: Tensile Strengthin lb/in 2̂
ANOVA for Selected F ctori l !odelAn l"sis of # ri nce t ble $% rti l s&' of s(& res)
S&' of !e n FSo&rce S(& res *F S(& re V l&e %rob + F(o!el .- / 005 3 1. 32 005 12./3 0.0005 signi i"ant
A 4.897E+005 3 1.632E+005 12.73 0.00056esi!ual 1.53 005 12 12-25.
Lack of Fit 0.000 0 Pure Error 1.539E+005 12 12825.697or Total . 3 005 15
The (o!el F8'alue o 12./3 implies the mo!el is signi i"ant. There is onl&a 0.059 "han"e that a (o!el F8;alue this large "oul! o""ur !ue to noise.
Tre t'ent !e ns ,Ad &sted. f Necess r"0sti' ted St nd rd!e n rror
181 2 /1.00 5 . 3 2-2 3156.25 56.63 3-3 2933.75 56.63 4-4 2666.25 56.63
!e n St nd rd t for Tre t'ent *ifference *F rror 4oeff5 %rob + 6t6 1 's 2 81-5.25 1 -0.0- 82.31 0.03 2 1 's 3 3/.25 1 -0.0- 0. / 0. 501 1 's 30 ./5 1 -0.0- 3.-1 0.0025 2 's 3 222.50 1 -0.0- 2./- 0.01 / 2 's 0.00 1 -0.0- .12 < 0.0001 3 's 2 /.50 1 -0.0- 3.3 0.005
The F 8'alue is 12./3 with a "orrespon!ing P 8'alue o .0005. (i$ing te"hni%ue has an e e"t.
(b) 7onstru"t a graphi"al !ispla& as !es"ri#e! in Se"tion 385.3 to "ompare the mean tensile strengths or the our mi$ing te"hni%ues. =hat are &our "on"lusions>
.255.
/12-25
..
n
MS S
E .i
===
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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
2700 2800 2900 3000 3100
Tensile Strength
Scaled t Distribution
(3) (2)(1)(4)
?ase! on e$amination o the plot, we woul! "on"lu!e that 1 µ an! 3 µ are the same@ that µ !i ers rom
1 µ an! 3 µ , that 2 µ !i ers rom 1 µ an! 3 µ , an! that 2 µ an! µ are !i erent.
(c) se the Fisher SD metho! with α 0.05 to make "omparisons #etween pairs o means.
,51/-5. 121/,2
/12-252
2
1.0250
2
... LS!
".# t LS!
n MS
t LS!
$.
a % $
E
==
=
=
−
−α
Treatment 2 's. Treatment 315 .250 8 2 .250 0.000 B 1/ . 5
Treatment 2 's. Treatment 3 315 .250 8 2 33./50 222.500 B 1/ . 5Treatment 2 's. Treatment 1 315 .250 8 2 /1.000 1-5.250 B 1/ . 5Treatment 1 's. Treatment 2 /1.000 8 2 .250 30 ./50 B 1/ . 5Treatment 1 's. Treatment 3 2 /1.000 8 2 33./50 3/.250 < 1/ . 5Treatment 3 's. Treatment 2 33./50 8 2 .250 2 /.500 B 1/ . 5
The Fisher SD metho! is also presente! in the Design8 $pert "omputer output a#o'e. The results agreewith the graphi"al metho! or this e$periment.
(d) 7onstru"t a normal pro#a#ilit& plot o the resi!uals. =hat "on"lusion woul! &ou !raw a#out the'ali!it& o the normalit& assumption>
There is nothing unusual a#out the normal pro#a#ilit& plot o resi!uals.
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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
%e#&'i(ue
e ' s
i l e ) t r e ' *
t &
+'e ,a#tor "lot
1 2 3 4
25 9.!1
2 59.26
2939.51
3119. 5
33!!
22
372 )a+ 6ework part )#+ o Cro#lem 381 using Tuke& s test with α 0.05. Do &ou get the same"on"lusions rom Tuke& s test that &ou !i! rom the graphi"al pro"e!ure an!*or the Fisher SDmetho!>
(inita# 4utputTuke&Es pairwise "omparisons
Famil& error rate 0.0500n!i'i!ual error rate 0.011/
7riti"al 'alue .20
nter'als or )"olumn le'el mean+ 8 )row le'el mean+
1 2 3
2 8 23 53
3 8201 815 2/5 0
/ 252 30 5 3 /2- 505
Go, the "on"lusions are not the same. The mean o Treatment is !i erent than the means o Treatments 1,2, an! 3. Howe'er, the mean o Treatment 2 is not !i erent rom the means o Treatments 1 an! 3a""or!ing to the Tuke& metho!, the& were oun! to #e !i erent using the graphi"al metho! an! the Fisher
SD metho!.
(b) $plain the !i eren"e #etween the Tuke& an! Fisher pro"e!ures.
?oth Tuke& an! Fisher utili e a single "riti"al 'alue@ howe'er, Tuke& s is #ase! on the stu!enti e! rangestatisti" while Fisher s is #ase! on t !istri#ution.
3-8 A pro!u"t !e'eloper is in'estigating the tensile strength o a new s&ntheti" i#er that will #e use! tomake "loth or men s shirts. Strength is usuall& a e"te! #& the per"entage o "otton use! in the #len! o
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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
materials or the i#er. The engineer "on!u"ts an e$periment with i'e le'els o "otton "ontent an!repli"ate! the e$periment i'e times. The !ata are shown in the ollowing ta#le.
7otton=eight
Cer"entage 4#ser'ations
15 / / 15 1120 12 1/ 12 1- 1-25 1 1 1 1- 1-30 1 25 22 1 2335 / 10 11 15 11
)a+ s there e'i!en"e to support the "laim that "otton "ontent a e"ts the mean tensile strength> se α
0.05.Minitab OutputOne-way A O!A" Tensile Strength #ersus $otton %ercentage
Analysis of Variance for TensileSource DF SS MS F PCotton P 4 475.76 118.94 14.76 .!rror " 161." 8. 6Total "4 6#6.96
Ies, the F 8'alue is 1 ./ with a "orrespon!ing P 8'alue o 0.000. The per"entage o "otton in the i#er appears to ha'e an a e"t on the tensile strength.
)#+ se the Fisher SD metho! to make "omparisons #etween the pairs o means. =hat "on"lusions "an&ou !raw>
Minitab OutputFis$er%s &air'ise co(&arisons
Fa(ily error rate ) ."64*n+i,i+ual error rate ) . 5
Critical ,alue ) ". 86
*nter,als for -colu(n le,el (ean / -ro' le,el (ean
15 " "5 #
" /9.#46 /1.854
"5 /11.546 /5.946 /4. 54 1.546
# /15.546 /9.946 /7.746 /8. 54 /".454 / ."54
#5 /4.746 .854 #. 54 7. 54 ".746 8.#46 1 .546 14.546
n the (inita# output the pairs o treatments that !o not "ontain ero in the pair o num#ers in!i"ates thatthere is a !i eren"e in the pairs o the treatments. 159 "otton is !i erent than 209, 259 an! 309. 209"otton is !i erent than 309 an! 359 "otton. 259 "otton is !i erent than 309 an! 359 "otton. 309"otton is !i erent than 359.
)"+ Anal& e the resi!uals rom this e$periment an! "omment on mo!el a!e%ua"&.
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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
The resi!ual plots #elow show nothing unusual.
Residual
P e r c e n t
5.02.50.0-2.5-5.0
99
95
90
80
70
60504030
20
10
5
1
Normal Probability Plot of the Residuals(response is ensi!e "tren#t$)
Fitted Value
R e s i d
u a
l
22201816141210
5.0
2.5
0.0
-2.5
-5.0
Residuals Versus the Fitted Values(response is ensi!e "tren#t$)
3-1 An e$periment was run to !etermine whether our spe"i i" iring temperatures a e"t the !ensit& o a
"ertain t&pe o #ri"k. The e$periment le! to the ollowing !ata:
Temperature Densit&100 21.- 21. 21./ 21. 21./125 21./ 21. 21.5 21.150 21. 21.- 21.- 21. 21.51/5 21. 21./ 21.- 21.
(a) Does the iring temperature a e"t the !ensit& o the #ri"ks> se α 0.05.
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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
Go, iring temperature !oes not a e"t the !ensit& o the #ri"ks. 6e er to the Design8 $pert output #elow.
Design $pert 4utputResponse: *ensit" ANOVA for Selected F ctori l !odelAn l"sis of # ri nce t ble $% rti l s&' of s(& res)
S&' of !e n FSo&rce S(& res *F S(& re V l&e %rob + F(o!el 0.1 3 0.052 2.02 0.15 not signi i"ant
A 0.16 3 0.052 2.02 0.15696esi!ual 0.3 1 0.02
Lack of Fit 0.000 0 Pure Error 0.36 14 0.026 7or Total 0.52 1/
The (o!el F8'alue o 2.02 implies the mo!el is not signi i"ant relati'e to the noise. There is a15. 9 "han"e that a (o!el F8'alue this large "oul! o""ur !ue to noise.
Tre t'ent !e ns ,Ad &sted. f Necess r"0sti' ted St nd rd!e n rror
18100 21./ 0.0/2
2-125 21.50 0.080 3-150 21.72 0.072 4-175 21.70 0.080
!e n St nd rd t for Tre t'ent *ifference *F rror 4oeff5 %rob + 6t6 1 's 2 0.2 1 0.11 2.23 0.0 25 1 's 3 0.020 1 0.10 0.20 0.- 5 1 's 0.0 0 1 0.11 0.3/ 0./15 2 's 3 80.22 1 0.11 82.05 0.0 01 2 's 80.20 1 0.11 81./ 0.0 3 's 0.020 1 0.11 0.1 0.-552
(b) s it appropriate to "ompare the means using the Fisher SD metho! in this e$periment>
The anal&sis o 'arian"e tells us that there is no !i eren"e in the treatments. There is no nee! to pro"ee!with Fisher s SD metho! to !e"i!e whi"h mean is !i eren"e.
)"+ Anal& e the resi!uals rom this e$periment. Are the anal&sis o 'arian"e assumptions satis ie!> Thereis nothing unusual a#out the resi!ual plots.
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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
Residual
N o r m a
l % p
r o b a b i l i t y
Normal plot of residuals
-!.3 -!.1 5 -!.!5 !.! 5 !.2
1
5
1!
2!
3!
5!
!
8!
9!
95
99
22
22
22
"redi#ted
R e s
i d u a
l s
Residuals $s. "redi#ted
-!.3
-!.1 5
-!.!5
!.! 5
!.2
21.5! 21.56 21.62 21.68 21. 4
(d)%onstruct a #rap$ica! disp!a& of t$e treat'ents as described in "ection 3-5.3.
oes t$is #rap$ ade uate!& su''ari*e t$e resu!ts of t$e ana!&sis of +ariance inpart (b). ,es.
21&2 21&3 21&' 21&( 21&) 21&7 21&8
*ean Density
Scaled t Distribution
(125) (1 5-15!-1!!)
)e+ (& pro"e!ures or anal& ing the !ata with a regression mo!el are liste! #elow:
1. Fit the 4ne8wa& AG4;A mo!el using the original !ata. The result is not signi i"ant@2. Fit the 6egression mo!el with Temp an! TempJJ2 as pre!i"tors. The result is not signi i"ant@3. Fit the 6egression mo!el with Temp, TempJJ2 an! TempJJ3 as pre!i"tors. The result is not
signi i"ant@. (inita# output marke! o#s 1- as an unusual o#ser'ation. t ma& #e a potential outlier, so,
remo'e! this point, an! re!o the a#o'e pro"e!ures.5. Fit the 4ne8wa& AG4;A mo!el using the new !ata. The result is signi i"ant )p 0.0 +@
. Fit the 6egression mo!el with Temp an! TempJJ2 as pre!i"tors. C8'alues or Temp an! TempJJ2are aroun! 0.05@
/. Fit the 6egression mo!el with Temp, TempJJ2 an! TempJJ3 as pre!i"tors. C8'alues or the threeterms are all aroun! 0.05.
-. The o'erall mo!el )a ter remo'ing the outlier+ or the "u#i" pol&nomial regression is a #etter itthan the %ua!rati", #ase! on #oth the p8'alue or the mo!el an! the a!Kuste! 68s%uare! 'alue.
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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
(inita# output:
MT0 2et c7 ) %Te(&% 3 %Te(&%MT0 2et c8 ) %Te(&% 3 %Te(&% 3 %Te(&%MT0 ne'ay %Density% %Te(&%.
One-way A O!A" Density #ersus Te+,
Analysis of Variance for DensitySource DF SS MS F PTe(& # .1561 . 5" ". " .157!rror 14 .#6 . "57Total 17 .5161 *n+i,i+ual 95 C*s For Mean 0ase+ on Poole+ StDe,2e,el Mean StDe, /////// ///////// ///////// /////////1 5 "1.74 .114 -/////////3////////1"5 4 "1.5 .141 -//////////3/////////15 5 "1.7" .164 -////////3/////////175 4 "1.7 ."16 -/////////3//////////
/////// ///////// ///////// /////////Poole+ StDe, ) .16 "1.44 "1.6 "1.76MT0 e ress %Density% " %Te(&% C7:S;0C Constant:S;0C 0rief ".
egression Analysis" Density #ersus Te+,. $7
T$e re ression e<uation isDensity ) "#.1 / . ""1 Te(& . 8" C7
Pre+ictor Coef S! Coef T PConstant "#.1 1."1" 19. 6 .Te(& / . "" 9 . 18#4 /1." ."47
C7 . 8178 . 67 1 1."" ."41
S ) .1768 /S< ) 9.1 /S<-a+= ) .
Analysis of Variance
Source DF SS MS F Pe ression " . 47"" . "#61 .76 .487esi+ual !rror 15 .46889 . #1"6
Total 17 .51611
Source DF Se< SSTe(& 1 . 67C7 1 . 4656
;nusual >ser,ations>s Te(& Density Fit S! Fit esi+ual St esi+
18 175 "1.4 "1.7#89 . 859 / .##89 /".19
+enotes an o>ser,ation 'it$ a lar e stan+ar+i?e+ resi+ual
MT0 e ress %Density% # %Te(&% C7 C8:S;0C Constant:S;0C 0rief ".
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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
egression Analysis" Density #ersus Te+,. $7. $8
T$e re ression e<uation isDensity ) 41.# / .4#8 Te(& . #17 C7 / . 7 C8
Pre+ictor Coef S! Coef T PConstant 41.# 8.91" 4.6# .Te(& / .4#77 ." "7 /".16 . 49C7 . #168 . 15 1 ".11 . 5#C8 / . 747 . #6# /". 6 . 59
S ) .16 4 /S< ) # ." /S<-a+= ) 15.#
Analysis of Variance
Source DF SS MS F Pe ression # .15611 . 5" 4 ". " .157esi+ual !rror 14 .#6 . "571
Total 17 .51611
Source DF Se< SSTe(& 1 . 67C7 1 . 4656C8 1 .1 889
;nusual >ser,ations>s Te(& Density Fit S! Fit esi+ual St esi+
18 175 "1.4 "1.7 . 8 " / .# /".16
+enotes an o>ser,ation 'it$ a lar e stan+ar+i?e+ resi+ual
MT0 ne'ay %Density% %Te(&%.
One-way A O!A" Density #ersus Te+,
Analysis of Variance for DensitySource DF SS MS F PTe(& # .1976 . 659 #.57 . 44!rror 1# ."4 . 185Total 16 .4#76 *n+i,i+ual 95 C*s For Mean 0ase+ on Poole+ StDe,2e,el Mean StDe, /// ///////// ///////// ///////// ///1 5 "1.74 .114 -//////3//////1"5 4 "1.5 .141 -//////3//////15 5 "1.7" .164 -//////3//////175 # "1.8 .1 -///////3///////
/// ///////// ///////// ///////// ///Poole+ StDe, ) .1#6 "1.4 "1.6 "1.8 "".MT0 e ress %Density% " %Te(&% %Te(&33"%:
S;0C Constant:S;0C 0rief ".
egression Analysis" Density #ersus Te+,. Te+,//2
T$e re ression e<uation isDensity ) "#.8 / . ##" Te(& . 1"7 Te(&33"
Pre+ictor Coef S! Coef T P
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STAT 503, Fall 2005 Homework Solution 2 Due: Sept. 15, 2005
Constant "#.765 1. 66 ""."9 .Te(& / . ##"# . 16"4 /". 5 . 6Te(&33" . 1"71# . 598# ".1" . 5"
S ) .15 8 /S< ) "7." /S<-a+= ) 16.8
Analysis of Variance
Source DF SS MS F Pe ression " .1191 . 5955 ".6" .1 8esi+ual !rror 14 .#1855 . ""75
Total 16 .4#765
Source DF Se< SSTe(& 1 . 16#7Te(&33" 1 .1 "7#
MT0 e ress %Density% # %Te(&% %Te(&33"% %Te(&33#%:S;0C Constant:S;0C 0rief ".
egression Analysis" Density #ersus Te+,. Te+,//2. Te+,//3
T$e re ression e<uation isDensity ) #9.# / .#88 Te(& . "77 Te(&33" / . 6 Te(&33#
Pre+ictor Coef S! Coef T PConstant #9.# 7.59" 5.18 .Te(& / .#884 .17"8 /"."5 . 4#Te(&33" . "768 . 1"81 ".16 . 5Te(&33# / . 64 . #1 /". 6 . 6
S ) .1#59 /S< ) 45." /S<-a+= ) #".5
Analysis of Variance
Source DF SS MS F Pe ression # .19765 . 6588 #.57 . 44esi+ual !rror 1# ."4 . 1846
Total 16 .4#765
Source DF Se< SSTe(& 1 . 16#7Te(&33" 1 .1 "7#Te(&33# 1 . 7855
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