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原题如附件所示:
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题目要求判别病人的方法,以及确定主要的因素。
]8 ?1 i9 {' 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。 ( y3 O% K' Z6 `% n6 K
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: 1 C6 m" o; |) ~* c( _
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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) + q1 i( G) l Y% n% \ t
回归的代码如附件里的m文件所示: + X% H# Z$ v# E. s1 O) l; }
7 i! ~/ K4 m& r 运行的结果:
, d2 D$ Z' b' n- s. V+ V( tb = # W$ T1 r& X) N- i
0.85943269933448 -0.00026521067844 0.00045376919071 ) J0 u* k- p9 c- l- \
bint =
( t9 ^# K4 F' U& O8 | 0.68868335685722 1.03018204181175 -0.00033716969449 -0.00019325166239 -0.00002536250203 0.00093290088345 8 j9 y" o3 j: V% x1 c
r =
3 \- W! y; \7 a4 n7 K8 M 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 5 c7 s7 ^, ~/ v9 A
rint = # B- v; j, E. I' p6 G1 Z! L! m
-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 8 o2 N! [/ O/ I
s =
1 ^. O1 s& }% d: z 0.53107910778697 32.27784221875193 0.00000000042300 0.12340023479290 ! h. Y" o1 s3 N P( W9 u
得到回归方程: ; t, A- v. b/ S G
sorh=0.85943269933448-0.00026521067844.*ca+0.00045376919071.*k
# |3 C6 `! [4 F3 W这就是我们需要的模型。 0 {, r# O# s2 C3 |% c7 ^) h
然后判断表二中的30个病例。
( }; s9 z& ^0 {" g+ @5 L& ?0 r' gmatlab代码如下: - A, Z& t% L6 `0 d: O1 v$ }
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 $ i+ X" q3 t7 M$ ?
运行结果如下: - q- w2 D; s. A1 a8 T/ {
sorh = & P; O% ^: f/ K# z# o
Columns 1 through 5 + s I' Q- c. B9 H: ]) _
0.85499433533545 0.79918204271064 0.56425453206328 0.83402879546814 0.74658007076821 * o, j4 u( t2 y! ?3 N! B2 n; _" a
Columns 6 through 10 : e6 R) f) l7 o3 U; ^
0.67069889230140 0.77130142136514 0.67967325412561 0.82020036114713 0.38397110012925
3 \$ G5 O5 [' _, w4 s2 d Columns 11 through 15 ' e. D. n* {& k; Z7 f
1.13531091637190 0.89223866315360 0.82227702046971 0.29879868302170 0.45165511738824
* L; U& p* A& d! L t+ O, a Columns 16 through 20 ( p. e5 S, s9 [2 S
0.70253827538617 0.47247370213094 0.49482518478434 0.59924224155178 0.27184534930907
% j+ Z' }+ X* u Columns 21 through 25 ! Y) R) d% s2 X$ U) _& Q4 P
0.09537902075506 0.33448840420565 0.66927020826128 0.52980303692384 0.60023746186819 ' H/ W o3 D+ c1 N" ^- ^
Columns 26 through 30 7 G8 |3 S4 {2 U# ~8 h' m% O8 |
0.21725173887874 0.59906763929878 0.32095218605194 -0.10204363195679 0.41172691131176 ; ?9 H% I- f5 E. x) ?
定义: ) ]+ A- T2 X; ]( w" D
凡是sorh值大于0.55的为患有肾病,否则为健康 $ Q7 M# i+ ]* f: w2 ]
可以从结果中得到30个病例中的患病者。
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[此贴子已经被作者于2008-8-12 13:57:50编辑过] |