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原题如附件所示：
6 h/ e; H2 A$ }; c2 g( P $ F7 K) g& _0 a
题目要求判别病人的方法，以及确定主要的因素。
- o- X" ^3 d4 ~# T4 m4 h+ f0 W
首先对题目给出的数据进行处理，将表一前面30个数据每种元素求平均值得到病人体内各种元素的平均含量。 ; {/ Z! Y( M) k7 F' h, b( t, w, \- }
再对表一后面30个元素求平均值，得到健康人体内个元素的平均含量，结果如下：
8 w5 X; _/ ~. P' m; K+ E+ N
0 ]1 v/ s, D8 d9 [2 J( O# L+ l
( m; M- Q) `$ J9 j, D9 L将ca k作为自变量，sorh作为因变量，进行回归。（得到的模型为线性概率模型，见《经计量学精要》，古亚拉提）
3 B& g6 {: `8 ^4 J/ e4 S0 k回归的代码如附件里的m文件所示： ) C; }- @8 P" t+ O) `
3 W b" h; h$ ~) M, L& s% e
运行的结果：
% }" q0 k v/ { I. q7 Qb =
, k3 E G# a7 P9 l 0.85943269933448
-0.00026521067844
0.00045376919071 ) @* u6 W% V+ w9 l
bint =
+ Z7 _* N4 x, u' [2 A
0.68868335685722 1.03018204181175
-0.00033716969449 -0.00019325166239
-0.00002536250203 0.00093290088345 9 ]: ^2 p( b$ o! Z' {
r =
/ K- s. f* P& Q, o 0.24499009043804
0.24298645516293
0.22596382129192
0.26974523197013
0.26938499769464
0.20134048077000
0.26428218646615
0.24546068238966
0.24204749612136
0.26329869754702
-0.30239364378778
0.22418882468333
0.36151397969142
0.27800877719166
0.22562971091175
0.25817997058727
0.34088284102996
0.47584435732540
0.39924147789994
0.10818014268404
0.03268403910683
0.23266006146459
0.42213189599121
0.15295625201459
0.32936275116498
0.23035596133112
0.27508838782562
0.19186969392530
-0.04387393419007
0.37180169649244
-0.28881576082096
-0.54037252177113
-0.47247370212743
-0.49482518478078
-0.27629228737278
-0.08212896037942
-0.34347303417696
-0.60782985983678
-0.50802211576599
0.82648437263390
-0.60023746186573
-0.21743324654913
-0.59323300437026
-0.32095218604696
0.10204363196572
-0.41168153438852
-0.22705964801276
-0.40337580412737
0.50019536866747
0.05178812516079
-0.08624980592198
-0.28795869072225
-0.34221336479873
-0.47546878418730
0.13032287365765
-0.06744638245026
0.12456342765303
-0.33184299743823
-0.32238525982281
-0.46743958519949 0 t! I' H! @. L7 x& z u
rint =
$ t3 x; x3 K d9 |3 j- P
-0.45268049797649 0.94266067885257
-0.45471901948541 0.94069192981128
-0.47070402952424 0.92263167210809
-0.42749772755413 0.96698819149440
-0.42530394206593 0.96407393745521
-0.49119576109825 0.89387672263825
-0.42929291658125 0.95785728951356
-0.45189766754183 0.94281903232114
-0.45372849228423 0.93782348452696
-0.43273992792445 0.95933732301849
-0.72468928783798 0.11990200026242
-0.47284729549112 0.92122494485777
-0.33268274520904 1.05571070459188
-0.41631507827795 0.97233263266126
-0.47200984810697 0.92326926993047
-0.43768034176985 0.95404028294440
-0.35542375701992 1.03718943907983
-0.21577903322429 1.16746774787510
-0.29583373430114 1.09431669010102
-0.57933084093047 0.79569112629854
-0.65245595836408 0.71782403657773
-0.46476517082366 0.93008529375283
-0.26956573150114 1.11382952348357
-0.53878440595437 0.84469690998355
-0.36446419214936 1.02318969447933
-0.46717750756131 0.92788943022355
-0.41904949253371 0.96922626818495
-0.50456761920479 0.88830700705539
-0.68119689782050 0.59344902944035
-0.32210495077523 1.06570834376012
-0.98555305629226 0.40792153465034
-1.22681878775805 0.14607374421580
-1.16263827940142 0.21769087514656
-1.18559446484527 0.19594409528372
-0.97334318994612 0.42075861520057
-0.77754513466812 0.61328721390927
-1.03962716392182 0.35268109556790
-1.28897250636556 0.07331278669200
-1.19830502970058 0.18226079816860
0.27837161793377 1.37459712733402
-1.28240289420555 0.08192797047410
-0.91514787090337 0.48028137780510
-1.27825077913241 0.09178477039189
-1.01775836842798 0.37585399633405
-0.58172439829518 0.78581166222663
-1.10573438258273 0.28237131380568
-0.92471005188033 0.47059075585482
-1.09775532080452 0.29100371254977
-0.13489900871102 1.13528974604595
-0.63713063617393 0.74070688649551
-0.78029430110123 0.60779468925727
-0.98542624488175 0.40950886343724
-1.03853703868559 0.35411030908814
-1.16731658432257 0.21637901594796
-0.55440318902529 0.81504893634058
-0.76249963681796 0.62760687191744
-0.55996646809892 0.80909332340498
-1.02839058696323 0.36470459208677
-1.01901501263124 0.37424449298562
-1.15831993205002 0.22344076165104
' \" t5 e# P! z+ R& q1 n
s =
: i0 Q& t0 k8 G
0.53107910778697 32.27784221875193 0.00000000042300 0.12340023479290
9 H" B, F) a2 Q. L得到回归方程：
1 i. N8 g# f _sorh=0.85943269933448-0.00026521067844.*ca+0.00045376919071.*k % m: Y1 a2 E8 x, g. L; L' r
这就是我们需要的模型。
8 Y: p n t E& b* b l! x9 `然后判断表二中的30个病例。 7 Y3 ?% d7 B4 N& x
matlab代码如下： , G1 @# y! Y( X1 ]+ C
ca=[323 542 1332 503 547 790 417 943 318 1969 1208 328 265 2220 1606 672 1521 1544 1062 2278 2993 2056 1025 1633 1068 2554 1211 2157 3870 1806];
k=[179 184 128 238 71 45.8 49.5 155 99.4 103 1314 264 73 62 40 47 36.2 98.9 47.3 36.5 65.5 44.8 180 228 53 77.5 134 74 143 68.9];
sorh=0.85943269933448-0.00026521067844.*ca+0.00045376919071.*k ) z- e# I- O! p9 {/ p7 t
运行结果如下：
' ^9 P3 p/ K& s7 D8 m5 |* ~- P6 bsorh =
) V0 O& [, t" V Columns 1 through 5 4 B* N! ]5 a/ E" W
0.85499433533545 0.79918204271064 0.56425453206328 0.83402879546814 0.74658007076821
* Q6 R7 l; l M" c Columns 6 through 10
, \- U; Q z6 x 0.67069889230140 0.77130142136514 0.67967325412561 0.82020036114713 0.38397110012925 7 E! D) [ J5 {7 O( j" _ ~: Q
Columns 11 through 15
l( O8 f1 A. O7 [ 1.13531091637190 0.89223866315360 0.82227702046971 0.29879868302170 0.45165511738824 $ ^" P" ^6 f+ l* [. x+ a
Columns 16 through 20 & B# F O# H# W$ @) ?
0.70253827538617 0.47247370213094 0.49482518478434 0.59924224155178 0.27184534930907
- }& D5 f" F* N; X4 @0 m8 h6 | Columns 21 through 25 5 `% _5 E% S$ M6 O- j) u
0.09537902075506 0.33448840420565 0.66927020826128 0.52980303692384 0.60023746186819
. ~5 \+ T. H; [ Columns 26 through 30
* ]# P9 c7 T+ Z 0.21725173887874 0.59906763929878 0.32095218605194 -0.10204363195679 0.41172691131176
2 |- y& F' e- H4 H9 V定义：
, ] m: C4 V5 O$ @8 i" f. Y凡是sorh值大于0.55的为患有肾病，否则为健康 8 l, ]% I( O! U1 U3 y( {
可以从结果中得到30个病例中的患病者。 ( Q6 j( V4 d) T+ o4 u& L
% H1 r# |0 v$ g% z5 r , j' |3 }9 ^0 q, u) A
[此贴子已经被作者于2008-8-12 13:57:50编辑过] | 
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发表于 2008-8-12 11:53
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