This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 13.5 The results can be summarized as y versus x 3: versus 3;
Best ﬁt equation y = 4.351535 + 0.352473: )5 = —9.96T63 + 2.374101 Standard error 1.0650] 2. 754025
Correlation coefﬁcient 0.914%? 0.91476? We can also plot both lines on the same graph y12 Thus, the “best” ﬁt lines and the standard errors differ. This makes sense because different errors are being minimized depending on our choice of the dependent (ordinate) and independent
(abscissa) variables. In contrast, the correlation coefficients are identical since the same amount of uncertainty is explained regardless of how the points are plotted. 13.7 The function can be linearized by dividing it by J: and taking the natural logarithm to yield
ln(y." x) = lnrz4 + ,84x Therefore, if the model holds, a plot of ln(y.'x) versus I should yield a straight line with an
intercept of 111.124 and a slope of '34. X y InUn'x)
0.1 05 2.014003
0.2 1.25 1832581
0.4 1.45 1.2870354
0.8 1.25 0133069
0.9 0.85 —0.05?18
1.3 0.55 08802
1.5 0.35 —1 .45529
1.? 0.28 —1 80359
1.8 0.18 230259 y = 24233:: + 2.2532
R2= 0.9974 disdainthaw Therefore, ,21 = 4.4733 and m = 322532 = 9.661135. and the ﬁt is
y = assuage—“7331 This equation can be plotted together with the data: 13.11 The power ﬁt can be determined as WEKQ} 30 . 1.345033 0.322213
75 2.12 1.375031 0.323333
7? 2.15 1.333431 0.332433
30 2.2 1.30303 0.342423
321.313314 0.343353
341324233 0.343305
3? 1.333513 0.354103
301.354243 0.331323 0.32 I] 35 logA = 3323313913 0.3321 ' R2=0.9T11 0.35 0.34 0.33 0.32 ' 0.31 1.3 Therefore, the power is b = 0.3299 and the lead coefﬁcient is a = 10—03321 = 0.4149. and the ﬁt is A = 0.4143W"3"'99 Here is a plot of the ﬁt along with the original data: 2.35
2.3 2.25
2.2
2.15 2.1
2.05 . 1 .34 1.58 1 .92 1 .96 T0 The value of the surface area for a 95kg person can be estimated as T5 30 A = 0.4143(35)“3m = 2.34 m2 35 90 13.18 A logleg plot of pversua Tauggeata a linear relatienahip. 1]
1 .11 1 “:11 11mm 11.01 I
11.11111 .
11mm We regress. Ingmar versus 13ng to give legw y=4.5814Tl—3.6133810g13 T {r2 =naasma) 4.5 81421 Therefore, m = 16 = 38,142.94 and ,6; = —3~.61338= and the power model is ,u = 33.14194T‘3“1335 The model and the data can be plotted on lmtranafermed scales a5 2.5
2 y = 3314511341“
“5 R2= 0.915?
1
11.5
n 1'} 51:1 1 DO 1 51'} 2131131 251'} SDI] 35131 14.3 The data can be tabulated and the sums computed as :' x y ? F XI 5 1:5 xy ?y P}:
1.6 9 2T 81 243 1’29 4.8 14.4 43.2
3.6 16 64 256 11124 4096 14.4 57.6 230.4 4.4 25 125 625 3125 15625 22 111] 551]
3.4 49 343 2461 16867 117649 23.8 166.6 1166.2
2.2 64 512 4696 32768 262144 17.6 140.8 1126.4
2.8 81 7'29 6561 59649 531441 25.2 226.8 2641.2 UDU'IDLOM—i
moo401.500 2 11 3.3 121 1331 14541 151051
8 2 Q m 1723 20735 243332
2. 59 25.4 509 4959 49397 522999 Normal equations: 8 59 509 4859 “0 25.4
59 509 4859 4939? a] 204.8 509 4859 49397 522899 a; 1838.4
4859 4939'? 522899 5689229 03 18164 which can be solved for the coeﬂicients yielding the following bestﬁt polynomial y = —11.488'}' + 2.1438122: —1.04121x2 + 9.0465'1'5x3 Here is the resultiJJg ﬁt: The predicted values can be used to determined the sum of the squares. Note that the mean of the y values is 3.3. Tyred (JG—y)? I" x y 1 3 1.6 1.83213 2.8990
2 4 3.6 3.41452 9.0900
3 5 4.4 4.034“ 1.2190
4 T 3.4 3.508?5 9.0100
5 8 2.2 2.922?1 1.2160I
6 9 2.8 2.494? 9.2590
T 11 3.8 3.23302 9.2500
8 12 4.6 4.95946 1.6990
2 T8990 1771561
2985984 (2 —J1:r41)2
0.0539
0.0344
0.1934
0.0119
0.5223
0.0932 0.3215
0.1292
1.299? 41.8
55.2
5689229 204.8 459.8 505?.8
662.4 F9488
1 81 64 1 838.4 The coefﬁcient of determination can be computed as 2 _ 7.15 —1.2997
7.5 14.5 Because the data is curved. a linear regression will undoubtedly have too much error.
Therefore. as a ﬁrst try. ﬁt a parabola, r = 0.829 }} format long >> T = [0 5 10 15 20 25 30]: >> C = [14.6 12.8 11.3 10.1 9.09 8.26 7.56];
>> p = polyfit{T,c,2) p: 0.00439523809524 —0.363357l42857l4 14.55190476190477 Thus, the bestﬁt parabola would be c =14.55190476 — 0.36335714T + 0100439523813.1!"2 16 12 I] 5 1] 15 20 25 30 We can use this equation to generate predictions corresponding to the data. When these values are
rounded to the same munber of signiﬁcant digits the results are T c~data cpred rounded 0 14.6 14.55190 14.6
5 12.8 12.84500 12.8
10 11.3 11.35786 11.4
15 10.1 10.09048 10.1
20 9.09 9.04286 9.04
25 8.26 8.21500 8.22 30 7.56 7.60690 7.61 Thus, although the plot looks good= discrepancies occur in the third signiﬁcant digit. We can, therefore, ﬁt a thirdorder polynomial
}> 1:: = polyfit {T,::, 3} p :
D.00006949444444 0.007295231309524 —D.39557936507936 14.600238619523910 Thus, the bestﬁt cubic would be c=14.600238095 —0.395579365T +0.11072952381"2 —0'.000'(Milli4441"?r We can use this equation to generate predictions corresponding to the data. When these values are
rounded to the same munbcr of signiﬁcant digits the results are T cdata cpred rounded
0 14.6 14.60020 14.6
5 12.8 12.79663 12.8
10 11.3 11.30949 11.3
15 10.1 10.09044 10.1
20 9.09 9.09116 9.09
25 8.26 8.26331 8.26
30 7.56 7.55855 7.56 Thus, the predictions and data agree to three signiﬁcant digits. 14.6 The multiple linear regression model to evaluate is
0:130 +011" +a2c The [Z] and y matrices can be set up using LIA'I'LAB commands in a fashion similar to Example
14.4, >> format long >> 1: = [0 5 10 15 20 25 30]; >> T = [t t t]"': >> C = [zeros{size{t]] 10*ones{size{t)} 20*onesisizettJJJ'; >> 2 = [ones{5ize{T)] T c]; >> y = [14.6 12.8 11.3 10.1 9.09 8.26 7.56 12.9 11.3 10.1 9.03 8.17 7.46 6.35 11.4 10.3 6.96 8.06 7.35 6.73 6.2]'; The coefﬁcients can be evaluated as
>> a = ZXy
a : 13.52214285Tl4236 —0.20123809523810
—D.1049285T14285T Thus. the bestﬁt multiple regression model is a = 115221423 5T14286 — (3.2{31238'09 5238 [UT — 0.1049285? 14285?c We can evaluate the prediction at T= 12 and c = 15 and evaluate the percent relative error as
>> Cp 2 a{l}+a{2)*12+a{3)*15 Cp =
9.53335714285714 >> ea = abs[[9.09—cp]ﬂ9.09]*100 ea =
4.8774163130598? Thus, the error is considerable. This can be seen even better by generating predictions for all the data and then generating a plot of the predictions versus the data. A onetoone line is included to
show how the predictions diverge from a perfect ﬁt. 16 12 4 3 12 1E The cause for the discrepancy is because the dependence of oxygen concentration on the unknowns is signiﬁcantly nonlinear. It should be noted that this is particularly the case for the
dependency on temperature. 1=I.".l The multiple linear regression model to evaluate is
y =a. +5113!" +aﬁ’1"2 +£5131"3 + (:40 The [2'] matrix can be set up as in >> T = 0:5:30; >} T = [T T T]'; >} o = [0 0 0 0 0 0 0 10 10 10 10 10 10 10 20 20 20 20 20 20 20]';
>> 0 = [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]'; >} y = [14.6 12.8 11.3 10.1 9.09 8.26 7.56 12.9 11.3 10.1 9.03 8.17
T.46 6.05 11.4 10.3 0.96 8.06 3.35 6.?3 6.2]'; >15 Z = [D T T."2 T."3 G]: Then, the coefﬁcients can be generated by solving Eq.(l4.10) >} format long
>> a = {Z'*Z}\[Z'*Yl a:
14.02714285T1428T
—0.33642328042328 0.005T4444444444 —0.DDDD&3TDBTD3TD
—0.1049235T14285T Thus, the leastsquares ﬁt is
y==l¢027143——0336423$3+0ﬂ0574444T2—ﬂﬂ00043704T3—0J04928576 The model can then be used to predict values of oxygen at the same values as the data. These
predictions can be plotted against the data to depict the goodness of ﬁt. }} yp = Z*a
b} plotty,yp,'o'} 1B
14 12 1D 3 3 10 12 14 13
Finally, theprediction can he made at T= 12 and c =15= >> a{1]+a{2]*12+a{3]*12“2+a[4}*12‘3+a{5]*15
ans =
9.16T31492063485 which compares favorably with the true value of Qﬂg mg"L. 14.12 First, an Mﬁle function must he created to compute the sum of the squares= function f = fSSRia,xm,ym}
yp = a{1]*xm.*exp{a{2}*xml;
f = sum {rmrm 3‘2}; The data can then be entered as >} x
b} y [.1 .2 .4 .6 .9 1.3 1.5 1.? 1.5];
[0.25 1.25 1.45 1.25 0.35 0.55 0.35 0.28 0.13]; The minimization of the function is then implemented by >} a = fminsearchtﬂfSSR, [1Ir 1]Jr , x, y} 9.8545 —2.521T The beatﬁt mode] is therefore y = 9.3545m‘m‘“ The ﬁt along with the data can be displayed graphically. >} yp = a{1}*x.*exp{a121*xli
hr} P191313: Yr '0' Ix: YP}
15 Q6 14.14 (a) We regress y versus :1: to give y = 29.5 + 9.4945451: The model and the data can be plotted as 511
411 . I
:15
y = 5.45451: + 211.5
211 I, R2
111  5.5355
11 I} 11] 2!] ﬁll] 41] El] El} (II) We regress logmy versus logmx to give logm y = l9.99795 + 9.335977 logm 1: Therefore, a} = mamas = 9.952915 and ﬂ; = 9.385977. and the power model is y = vsssslsﬂmm The model and the data can be plotted as 5} 49 I
39
20 y = 9.952936]31151
10 a2 = 9.9553
9 I} 1!] 20 3D 4D 50 BI] (e) We regress l.'y versus 1.:"x to give 1 = U.ﬂ19963 + {31.1'9'3'4I54l y I Therefore, a; =1."0.01996322 = SﬂﬂQElE and ,3; = U.19T45357(50.09212) = 9.89131 and the saturationgrowthrate model is y = 50.059212—
98913? + x The model and the data can be plotted as 59
49
X
3“ y = 59.992—
20 x + 9.991999
19 a2 = 9.99919
9 D 1!] 20 30 4} 5!] El} ((1) We employ,F poljmomial regression to ﬁt a parabola y = —U.G16ﬂﬁx2 +1.3TT8TQI +11.T6667 The model and the data can be plotted as 5] 40
an
20 y =ﬂ.ﬂ161x2+1.3}'?9x+11.?6?
1a a2 = 0.93 n {I 10 21] 3] 4!] 5] 51] Comparison of ﬁts: The linear ﬁt is obviously inadequate. Although the power ﬁt follows the
general trend of the data, it is also inadequate because (1) the residuals do not appear to be
randomly distributed around the best ﬁt line and [2] it has a lower r2 than the saturation and
parabolic models. The best ﬁts are for the satLu'ationgrou‘thrate and the parabolic models. The}r both have
randomly distributed residuals and the},' have similar high coefﬁcients of determination. The
saturation model has a slifjwhtl}.r higher r2. Although the diEerence is probably not statistically
signiﬁcant, in the absence of additional information, we can conclude that the saturation model
represents the best ﬁt. ...
View
Full Document
 Spring '09
 RAPHAELHAFTKA

Click to edit the document details