matlab實(shí)驗(yàn)11培訓(xùn)講學(xué)

1、Good is good, but better carries it.精益求精，善益求善。matlab實(shí)驗(yàn)11-成績遼寧工程技術(shù)大學(xué)上機(jī)實(shí)驗(yàn)報(bào)告實(shí)驗(yàn)名稱實(shí)驗(yàn)十一回歸分析院系理學(xué)院專業(yè)信科班級(jí)姓名學(xué)號(hào)日期20131121實(shí)驗(yàn)?zāi)康暮喪霰敬螌?shí)驗(yàn)?zāi)康模?、了解回歸分析基本內(nèi)容2、掌握用matlab軟件求解回歸分析問題。實(shí)驗(yàn)準(zhǔn)備你為本次實(shí)驗(yàn)做了哪些準(zhǔn)備：經(jīng)過看書和看ppt完成了此次實(shí)驗(yàn)。實(shí)驗(yàn)進(jìn)度本次共有6個(gè)練習(xí)，完成6個(gè)。實(shí)驗(yàn)總結(jié)日本次實(shí)驗(yàn)的收獲、體會(huì)、經(jīng)驗(yàn)、問題和教訓(xùn)：通過本次實(shí)驗(yàn)學(xué)會(huì)了有關(guān)回歸分析的有關(guān)知識(shí)，怎么求線性與非線性回歸模型，知道了怎么找系數(shù)和置信區(qū)間，知道了有關(guān)回歸分析有關(guān)的matla

2、b命令，在實(shí)驗(yàn)的過程中遇到了很多的難題，比如怎么觀察模型是不是線性的，比如怎么找到最合適的模型，這時(shí)通過看書和認(rèn)真思考學(xué)會(huì)了怎么做。本次實(shí)驗(yàn)收獲了很多，知道了自己的不足，以后還需要勤加訓(xùn)練！教師評(píng)語1、考察溫度x對(duì)產(chǎn)量y的影響，測(cè)得下列10組數(shù)據(jù)：溫度（）20253035404550556065產(chǎn)量（kg）13.215.116.417.117.918.719.621.222.524.3求y關(guān)于x的線性回歸方程，檢驗(yàn)回歸效果是否顯著，并預(yù)測(cè)x=42時(shí)產(chǎn)量的估值及預(yù)測(cè)區(qū)間（置信度95%）.x=20253035404550556065;X=ones(10,1)x;y=13.215.116.417.1

3、17.918.719.621.222.524.3;r=-0.38180.40300.58790.1727-0.1424-0.4576-0.6727-0.1879-0.00300.6818rint=-1.28580.5221-0.56751.3736-0.36391.5397-0.92931.2748-1.26320.9783-1.51230.5972-1.61790.2725-1.25630.8806-1.03521.0291-0.07631.4399corrcoef(x,y)ans=1.00000.99100.99101.0000b,bint,r,rint,stats=regress(y,X

4、)b=9.12120.2230bint=8.021110.22140.19850.2476stats=0.9821439.83110.00000.2333置信區(qū)間8.0211，10.2214置信區(qū)間0.1985，0.2476r2=0.9821F=439.8311p=0.0000px=02468101214161820;y=0.62.04.47.511.817.123.331.239.649.761.7;X=100124141616361864110100112144114196116256118324120400p,S=polyfit(x,y,2)p=0.14030.19711.0105S=R

5、:3x3doubledf:8normr:1.1097Y=polyconf(x,y,S)回歸模型：y=0.1403x2+0.1971x+1.0105X=ones(11,1)x(x.2)b,bint,r,rint,stats=regress(y,X);Y=polyconf(p,x,S)plot(x,y,k+,x,Y,r)Y=1.01051.96604.04417.244911.568317.014223.582831.274040.087850.024261.08323、在研究化學(xué)動(dòng)力學(xué)反應(yīng)過程中，建立了一個(gè)反應(yīng)速度和反應(yīng)物含量的數(shù)學(xué)模型，形式為其中是未知參數(shù)，是三種反應(yīng)物（氫，n戊烷，異構(gòu)戊烷）

6、的含量，y是反應(yīng)速度.今測(cè)得一組數(shù)據(jù)如下表，試由此確定參數(shù)，并給出置信區(qū)間.的參考值為（1，0.05,0.02,0.1,2）.序號(hào)反應(yīng)速度y氫x1n戊烷x2異構(gòu)戊烷x318.554703001023.79285801034.8247030012040.024708012052.754708010614.391001901072.54100806584.3547019065913.0010030054108.50100300120110.05100801201211.3228530010133：x1=47028547047047010010047010010010028

7、5285;x2=3008030080801908019030030080300190;x3=1010120120101065655412012010120;x=x1x2x3;y=8.553.794.820.022.7514.392.544.3513.008.500.0511.323.13;f=(beta,x)(beta(1).*x(:,2)-(1/beta(5).*x(:,3).*(1+beta(2).*x(:,1)+beta(3).*x(:,2)+beta(4).*x(:,3).(-1);bint=0.1254-0.1508-0.08230.03990.12020.07020.00080.3

8、200-0.02820.12700.0891-0.1619-0.2862beta0=10.050.020.12;opt=optimset(TolFun,1e-3,TolX,1e-3);beta,bint=nlinfit(x,y,f,beta0,opt)beta=1.12920.05660.03570.10181.3160得到beta的擬合值及95%的置信區(qū)間4、混凝土的抗壓強(qiáng)度隨養(yǎng)護(hù)時(shí)間的延長而增加，現(xiàn)將一批混凝土作成12個(gè)試塊，記錄了養(yǎng)護(hù)日期x（日）及抗壓強(qiáng)度y（kg/cm2）的數(shù)據(jù)：養(yǎng)護(hù)時(shí)間x234579121417212856抗壓強(qiáng)度y354247535965687376828699試求

9、型回歸方程.對(duì)將要擬合的非線性模型，建立M文件volum.m如下x=234579121417212856;y=354247535965687376828699;beta0=51;beta,r,J=nlinfit(x,y,volum,beta0);beta=21.005319.5287functionyhat=volum(beta,x)yhat=beta(1)+beta(2)*log(x);即得回規(guī)模型為5、下表給出了某工廠產(chǎn)品的生產(chǎn)批量與單位成本（元）的數(shù)據(jù)，從散點(diǎn)圖，可以明顯的發(fā)現(xiàn)，生產(chǎn)批量在500以內(nèi)時(shí)，單位成本對(duì)生產(chǎn)批量服從一種線性關(guān)系，生產(chǎn)批量超過500時(shí)服從另一種線性關(guān)系，此時(shí)單位成

10、本明顯下降。希望你構(gòu)造一個(gè)合適的回歸模型全面地描述生產(chǎn)批量與單位成本的關(guān)系。生產(chǎn)批量650340400800300600720480440540750單位成本2.484.454.521.384.652.962.184.044.203.101.50記生產(chǎn)批量x1500時(shí)，單位成本為y2。為了大致地分析y與x的關(guān)系，首先利用表中表中數(shù)據(jù)分別作出y1對(duì)x1和y2對(duì)x2的散點(diǎn)圖。由圖像可知兩段程線性關(guān)系，所以做以下程序：兩段直線，x小于500時(shí)：r=-0.08310.1728-0.0070-0.0594-0.0233rint=-0.40610.23990.05530.2902-0.29510.2811

11、-0.32850.2097-0.39810.3514x1=340,400,300,480,440;y1=4.45,4.52,4.65,4.04,4.20;X=ones(size(x1)x1;b,bint,r,rint,stats=regress(y1,X)b=5.5863-0.0031bint=4.57436.5983-0.0056-0.0006stats=0.833214.98680.03050.0136stepwise(X,y1,1,2)回歸模型：y=5.5863-0.0031xrcoplot(r,rint)從結(jié)果可以看出，應(yīng)將第二個(gè)點(diǎn)去掉后再進(jìn)行擬合；兩段直線，x大于500時(shí)：r=0.0

12、222-0.00280.14390.2239-0.1460-0.2411rint=-0.53980.5843-0.44940.4437-0.32720.6151-0.19910.6469-0.48740.1953-0.60150.1192x2=650,800,600,720,540,750;y2=2.48,1.38,2.96,2.18,3.10,1.50;X=ones(size(x2)x2;b,bint,r,rint,stats=regress(y2,X)b=7.1158-0.0072bint=5.43168.8000-0.0096-0.0047stats=0.942065.01530.001

13、30.0377回歸模型：y=7.1158-0.0072xstepwise(X,y2,1,2)rcoplot(r,rint)由圖可知，數(shù)據(jù)無異常點(diǎn)。若直接考慮全組數(shù)據(jù)，對(duì)整個(gè)11組數(shù)據(jù)直接擬合。整組數(shù)據(jù)：x1=650,340,400,800,300,600,720,480,440,540,750;y1=2.48,4.45,4.52,1.38,4.65,2.96,2.18,4.04,4.20,3.10,1.50;r=-0.0202-0.23340.2592-0.0638-0.31510.10770.17280.34260.2209-0.1749-0.2959rint=-0.57460.5343-0

14、.71650.2498-0.24200.7603-0.55390.4264-0.74250.1123-0.44880.6642-0.34210.6877-0.14720.8324-0.30380.7456-0.72340.3737-0.75910.1673X=ones(size(x1)x1;b,bint,r,rint,stats=regress(y1,X)b=7.0779-0.0070bint=6.48457.6713-0.0081-0.0060stats=0.9631234.89360.00000.0612回歸模型：y=7.10779-0.0070 xstepwise(X,y1,1,2)rc

15、oplot(r,rint)我們已經(jīng)可以發(fā)現(xiàn)整組數(shù)據(jù)本身就服從置信度較高的線性關(guān)系。但是題目卻仍然告訴我們：生產(chǎn)批量在500以內(nèi)時(shí)，單位成本對(duì)生產(chǎn)批量服從一種線性關(guān)系，生產(chǎn)批量超過500時(shí)服從另一種線性關(guān)系。于是我們開始考慮再引入一個(gè)虛擬變量A。,并加入一項(xiàng)再次進(jìn)行擬合。y=2.48004.45004.52001.38004.65002.96002.18004.04004.20003.10001.5000;x=111111111116503404008003006007204804405407501500030001002200040250;r=-0.0632-0.10360.25020.090

16、4-0.0929-0.00100.22180.14870.1194-0.3624-0.2074rint=-0.54190.4156-0.52030.3130-0.17020.6706-0.29250.4733-0.43030.2446-0.47620.4742-0.19790.6416-0.26630.5637-0.33770.5766-0.6734-0.0515-0.61120.1964b,bint,r,rint,stats=regress(y,x)b=6.1621-0.0047-0.0036bint=5.03687.2874-0.0074-0.0020-0.00760.0003stats=

17、0.9763164.71430.00000.0443stepwise(x,y,1,2,3)能高達(dá)97.63%。是所有模型中準(zhǔn)確度最高的。6、一礦脈有13個(gè)相鄰樣本點(diǎn),人為設(shè)定一個(gè)原點(diǎn),現(xiàn)測(cè)得各樣本點(diǎn)對(duì)原點(diǎn)的距離x,與該樣本點(diǎn)某種金屬含量y的一組數(shù)據(jù)如下,畫出散點(diǎn)圖觀察二者的關(guān)系,試建立合適的回歸模型,如二次曲線,雙曲線,對(duì)數(shù)曲線等.x23457810y106.42109.20109.58109.50110.00109.93110.49x111415151819y110.59110.60110.90110.76111.00111.20 x=23457810111415151819;y=106.4

18、2109.20109.58109.50110.00109.93110.49110.59110.60110.90110.76111.00111.20;rint=-2.5001-1.8727-1.14571.9846-0.94012.1906-1.26042.0026-1.12832.1739-1.41311.9704-1.18822.1688-1.27002.1023-1.77171.5790-1.62591.6849-1.76421.5433-1.92891.1429-1.86191.1276plot(x,y)排除第一個(gè)點(diǎn)，很明顯成線性關(guān)系，采用一元線性回歸分析：x=23457810111415151819;y=106.42109.2109.58109.5110109.93110.49110.59110.6110.9110.76111111.2;X=ones(size(x),x;r=-2.18640.41940.62530.37110.52280.27860.49030.4162-0.09630.0295-0.1105-0.3930-0.3671c,cint,r,rint,stats=regress(y,X,0.05);c=108.25810.1742