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原题如附件所示:
1 j- T( d& O3 K# d 5 f6 c S; Y$ d5 u# r4 Y
题目要求判别病人的方法,以及确定主要的因素。
, E# w- c3 H* h; C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。 9 o6 Z, c8 t: E' Z$ S
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 f9 |3 h3 A2 M/ L1 O
( s, O7 V6 a0 s
* @1 y9 U! d) ?3 k ^3 S; S
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) / [( f3 |3 R' v9 l) x
回归的代码如附件里的m文件所示: : l/ V" H8 P2 ?% e- z- I
9 j/ x! g: _3 {% G
运行的结果: 9 S+ n0 S& y) x
b =
3 Q! Q% S7 H i# d! X 0.85943269933448 -0.00026521067844 0.00045376919071
: I( S* F8 f: {) z: M; z+ P/ @ bint = # w- K2 \1 J% L
0.68868335685722 1.03018204181175 -0.00033716969449 -0.00019325166239 -0.00002536250203 0.00093290088345
8 Z" e$ C. I$ ^/ l r = ; G: ]& \ ^8 s" V$ Z9 s: C
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 ]) u7 V# p6 l, Z
rint =
( J' t$ W, F( e" I1 l0 v -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
6 t7 o4 m* A3 @8 w5 q2 e0 Y4 ~ s = & e, F: }* D. n; A! _# {
0.53107910778697 32.27784221875193 0.00000000042300 0.12340023479290 ! Q" b( Z6 X* m1 a& H" Q* b9 l' B
得到回归方程:
7 q4 \% p7 L+ V/ k' t. Lsorh=0.85943269933448-0.00026521067844.*ca+0.00045376919071.*k
# y& t3 B }+ c( V这就是我们需要的模型。 7 M6 V2 p, l9 r! O c/ {- l/ q, g8 c: ]
然后判断表二中的30个病例。
* z3 y" s) o9 N6 Xmatlab代码如下: 8 S! R- l C$ G
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 & h% n {: L! t/ F9 M
运行结果如下:
. q, p# M/ M3 h, i% I, ksorh =
' P8 j# z; P; Q4 I& I$ b Columns 1 through 5 ) h* n4 \8 y9 a& k/ b
0.85499433533545 0.79918204271064 0.56425453206328 0.83402879546814 0.74658007076821 1 _& p9 O- n3 c$ }% }! j. I6 v
Columns 6 through 10 + O/ w& T" ? v7 e8 b
0.67069889230140 0.77130142136514 0.67967325412561 0.82020036114713 0.38397110012925 2 j$ b3 N' b5 P8 G7 I
Columns 11 through 15
. Z1 U6 W; L2 D3 c4 H Q6 E 1.13531091637190 0.89223866315360 0.82227702046971 0.29879868302170 0.45165511738824 - g. k- D/ f9 T: G! I. S
Columns 16 through 20 + w; N$ F! R9 f. d7 q1 U: z+ _
0.70253827538617 0.47247370213094 0.49482518478434 0.59924224155178 0.27184534930907
D# l# N. H/ u6 t1 g( ?7 z Columns 21 through 25 $ g4 J2 `; |1 @$ r/ f4 p8 m
0.09537902075506 0.33448840420565 0.66927020826128 0.52980303692384 0.60023746186819
: h/ t4 O. U% |; w Columns 26 through 30 4 C9 }, E0 }5 V! l. ?
0.21725173887874 0.59906763929878 0.32095218605194 -0.10204363195679 0.41172691131176 & D3 a6 r7 i7 O8 S
定义:
; J8 ~7 F8 Z8 P凡是sorh值大于0.55的为患有肾病,否则为健康
, v6 H2 h8 s8 q! q9 |: L# P可以从结果中得到30个病例中的患病者。 ) d0 {* y3 g7 r7 x. v
( V2 u- F, O( N5 H 8 \8 J W3 V9 E: H0 {( \
[此贴子已经被作者于2008-8-12 13:57:50编辑过] |