用最小二乘法求解線性模型及對(duì)模型的分析

1、用最小二乘法求解線性模型及對(duì)模型的分析作者 :鄧春亮1、研究 30 名兒童體重為因變量與身高為自變量的關(guān)系，兒童體重與身高的記錄如下表：編號(hào)體重 Y (kg)身高 X(cm)122.60119.80221.50121.70319.10121.40421.80124.40521.50120.00620.10117.00718.80118.00822.00118.80921.30124.201024.00124.801123.30124.701222.50123.101322.90125.301419.50124.201522.90127.401622.30128.201722.70126.101

2、823.50128.601921.50129.402025.50126.902125.00126.502226.10128.202327.90131.402426.80130.802527.20133.902624.40130.402724.40131.302823.00130.202926.30136.003028.80138.00試用計(jì)算機(jī)完成下面統(tǒng)計(jì)分析：（ 1）應(yīng)用最小二乘法求經(jīng)驗(yàn)回歸方程；（ 2）以擬合值 ? 為橫坐標(biāo)，殘差? 為縱坐標(biāo)，作殘差圖，分析Gauss-Markou假設(shè)對(duì)本例的適用性；yiei1（ 3）考慮因變量的變換 UY 2 ,再對(duì)新變量 U 和 X 重復(fù)（ 1）和（

3、2）的統(tǒng)計(jì)分析 ;（ 4）將 Box-Cox 變換應(yīng)用到本例，計(jì)算變換參數(shù)的值，并做討論。說明：第一題的數(shù)據(jù)和結(jié)果文件見附件1，下面第二題的數(shù)據(jù)文件和結(jié)果文件見附件2，必要時(shí)可參看。解：（ 1）在 SPSS 窗口中錄入數(shù)據(jù)，首先進(jìn)行異常值檢測(cè)，探查對(duì)回歸估計(jì)有異常大影響的數(shù)據(jù)。先利用 SPSS畫出體重與身高的散點(diǎn)圖130.0028.0026.00體重 24.00Y22.0020.0018.00115.00120.00125.00130.00135.00140.00身高 X圖 1從圖 1 可以看出沒有明顯不一致的點(diǎn)。也可以通過 SPSS 軟件計(jì)算 COOK 統(tǒng)計(jì)量，看下表表 1編號(hào)殘差 ?學(xué)生化

4、殘差 ricenter hiiCOOK 統(tǒng)計(jì)量 Diei11.883781.27241.05491.078352.03312.02204.02770.000023-2.24835-1.49944.03138.077784-.73361-.48247.00489.004635.70477.47518.05161.010486.49003.34183.11182.009927-1.20506-.82971.08920.0480781.678871.14538.07294.078009-1.15460-.75974.00594.01180101.30835.85964.00310.0139711.6

5、4786.42576.00351.00347212.48000.31710.01355.0024713.01081.00710.00143.0000014-2.95460-1.94417.00594.0772615-.81887-.53756.00139.0052016-1.73494-1.14067.00434.0254717-.50526-.33146.00008.0019018-.69298-.45611.00643.0043119-3.00905-1.98610.01183.09329201.978671.29824.00038.02940211.636701.07368.00003.

6、01989222.065061.35771.00434.03608232.600781.73550.03249.10611241.737831.15517.02522.0415025.91306.62283.07268.0230026-.50413-.33434.02088.0032027-.85971-.57329.03121.0113428-1.82512-1.20912.01887.0402629-.81662-.57199.11878.0293530.89321.64671.17316.05442.從上面數(shù)據(jù)看殘差值和中心化的杠桿率center hii 的值沒有異常大的，數(shù)據(jù),這里 (

7、 hii = center hii +1/n) ， COOK 統(tǒng)計(jì)量 Di 值也沒有異常大的數(shù)據(jù)，一般來說，殘差值和杠桿率越大，COOK 統(tǒng)計(jì)量就越大，殘差值和杠桿率越小，COOK 統(tǒng)計(jì)量就越小?？梢娺@些數(shù)據(jù)是比較一致的。接下來對(duì)這些數(shù)據(jù)求解經(jīng)驗(yàn)回歸方程。然后利用最小二乘法，在 SPSS 中 Analyze 菜單下依次選擇 Regress:2-Stage Least Square,選擇因變量和自變量執(zhí)行可輸出結(jié)果如下表：表 2MODEL: MOD_3.Equation number:1Dependent variable.體重 YListwise Deletion of Missing Dat

8、aMultiple R.80301R Square.64483Adjusted R Square.63215Standard Error1.55047Analysis of Variance:DFSum of SquaresMean SquareRegression1122.20765122.20765Residuals2867.311022.40396F = 50.83587Signif F = .0000-Variables in the Equation-VariableBSE BBetaT Sig T身高 X.395087 .055412.803014 7.130 .00003(Con

9、stant) -26.615154 7.007449-3.798 .0007Correlation Matrix of Parameter Estimates身高 X身高 X1.0000000這里可以看出所求經(jīng)驗(yàn)回歸方程的常數(shù)項(xiàng)(Constant)為 -26.615154，身高 X 的系數(shù)為 0.395087 。故經(jīng)驗(yàn)回歸方程為：yi =-26.615154+0.395087 xi（ 2）通過 SPSS,可得擬合值與殘差如下表表 3：擬合值與殘差表體重 Y身高 X擬合值 ?殘差 ?yiei22.60119.8020.716221.8837821.50121.7021.46688.0331219

10、.10121.4021.34835-2.2483521.80124.4022.53361-.7336121.50120.0020.79523.7047720.10117.0019.60997.4900318.80118.0020.00506-1.2050622.00118.8020.321131.6788721.30124.2022.45460-1.1546024.00124.8022.691651.3083523.30124.7022.65214.6478622.50123.1022.02000.4800022.90125.3022.88919.0108119.50124.2022.4546

11、0-2.9546022.90127.4023.71887-.8188722.30128.2024.03494-1.7349422.70126.1023.20526-.5052623.50128.6024.19298-.6929821.50129.4024.50905-3.0090525.50126.9023.521331.9786725.00126.5023.363301.6367026.10128.2024.034942.0650627.90131.4025.299222.6007826.80130.8025.062171.7378327.20133.9026.28694.9130624.4

12、0130.4024.90413-.5041324.40131.3025.25971-.8597123.00130.2024.82512-1.8251226.30136.0027.11662-.8166228.80138.0027.90679.89321yiei以擬合值 ? 為橫坐標(biāo)，殘差? 為縱標(biāo)，得殘差圖4殘差圖laudiseRdezidradnatsnU3.000002.000001.000000.00000-1.00000-2.00000-3.00000-4.0000018.0000020.0000022.0000024.0000026.0000028.00000Un sta n d a

13、 rd iz ed P re dic te d Va lu e圖 2從圖中可以看出，殘差圖沒有明顯的不一致的征兆，則可以認(rèn)為Gauss-Markou 假設(shè) e : N 0, 2I對(duì)本例基本上是合理的。1（3）作變換 U Y 2,這時(shí)用同樣的方法可求得經(jīng)驗(yàn)回歸方程為：ui =-0.314471+0.040641 xi其預(yù)測(cè)值與殘差如下表U擬合值殘差4.754.554264.554264.644.631484.631484.374.619294.619294.674.741214.741214.644.562394.562394.484.440474.440474.344.481114.48111

14、4.694.513624.513624.624.733084.733084.904.757474.757474.834.753404.7534054.744.688384.688384.794.777794.777794.424.733084.733084.794.863134.863134.724.895654.895654.764.810304.810304.854.911904.911904.644.944414.944415.054.842814.842815.004.826564.826565.114.895654.895655.285.025705.025705.185.00131

15、5.001315.225.127305.127304.944.985054.985054.945.021635.021634.804.976934.976935.135.212645.212645.375.293925.29392.以擬合值y?i 為橫坐標(biāo)，殘差e?i 為縱坐標(biāo)，作殘差圖得6U=Y0.5 的殘差圖laudiseRdezirdadntasnU0.300000.200000.100000.00000-0.10000-0.20000-0.30000-0.400004.400004.600004.800005.000005.200005.40000Un sta n da rd iz e

16、d P re d ic te d Va lu e圖 3從圖 3 看，此時(shí)的殘差圖也沒有明顯的不一致的趨勢(shì)，認(rèn)為Gauss-Markou 假設(shè) e : N 0,2 I 對(duì)本例基本上是合理的。（ 4）將因變量 Y 進(jìn)行 Box-Cox 變換，YY1 ,ln Y,變換后原來的因變量 Y y1, y2 ,L , yn變?yōu)?Yy1, y2 ,L , yn計(jì)算不同值對(duì)應(yīng)的殘差平方和 RSS, z ，7yi1 ,nnziyii 1n1nln yiyi,i 1這里分別取i ,i 1,2,L 7 值為 -1.5， -1，-0.5， 0， 0.5，1， 1.5，計(jì)算分別計(jì)算zi ，然后計(jì)算對(duì)應(yīng)的殘差平方和RSS

17、, z，這里 n=30,計(jì)算得到 z如表所示，這里Zi 表示 zi1 。i表 5編號(hào)UZ1Z2Z3Z4Z5Z6Z714.751707.19513.21176.1772.2536.1421.6014.7424.641705.94511.99174.9871.0935.0120.5013.6734.371702.58508.85172.0568.3532.4518.1011.4244.671706.30512.34175.3171.4235.3220.8013.9654.641705.94511.99174.9871.0935.0120.5013.6764.481704.11510.25173.3

18、369.5333.5419.1012.3474.341702.09508.40171.6467.9832.1217.8011.1584.691706.53512.56175.5371.6335.5321.0014.1594.621705.70511.76174.7670.8834.8120.3013.48104.901708.57514.59177.5673.6437.5423.0016.14114.831707.91513.92176.8872.9636.8422.3015.44124.741707.08513.10176.0672.1536.0421.5014.64134.791707.5

19、0513.52176.4872.5636.4421.9015.04144.421703.22509.43172.5768.8332.8918.5011.79154.791707.50513.52176.4872.5636.4421.9015.04164.721706.86512.89175.8571.9435.8421.3014.45174.761707.30513.31176.2772.3536.2421.7014.84184.851708.10514.12177.0773.1637.0422.5015.64194.641705.94511.99174.9871.0935.0120.5013

20、.67205.051709.85515.91178.9275.0538.9924.5017.69215.001709.44515.49178.4874.5938.5124.0017.17225.111710.30516.39179.4375.5939.5625.1018.33235.281711.54517.72180.8677.1341.2326.9020.27245.181710.81516.93180.0076.2040.2125.8019.08255.221711.08517.23180.3276.5440.5826.2019.51264.941708.93514.96177.9374

21、.0337.9323.4016.55274.941708.93514.96177.9374.0337.9323.4016.55284.801707.61513.62176.5872.6636.5422.0015.14295.131710.45516.55179.5975.7639.7525.3018.54305.371712.08518.32181.5277.8742.0427.8021.27.通過 SPSS 軟件運(yùn)行得到的方差分析表，可知道相應(yīng)的殘差平方和，具體數(shù)據(jù)如下表所示：8表 6-1.5-1-0.500.511.5RSS73.14370.51468.63867.49267.05167.

22、31168.277通過表 6 的簡(jiǎn)單比較可以看出當(dāng)0.5 時(shí)，殘差平方和 RSS , z達(dá)到最小， 因此我們可以近似地認(rèn)為0.5 就是變換參數(shù)的最優(yōu)選擇 .2、研究?jī)和捏w重 Y 與身高 X1 和胸圍 X 2 之間的關(guān)系是具有一定現(xiàn)實(shí)意義的，因?yàn)檫@種關(guān)系使我們能夠用簡(jiǎn)單的方法從 X1 和 X 2 的值去估計(jì)一個(gè)兒童的體重，下表是一組觀測(cè)數(shù)據(jù):表編號(hào)體重 Y身高 X1胸圍 X2122.60119.8060.50221.50121.7055.50319.10121.4056.50421.80124.4060.50521.50120.0057.70620.10117.0057.00718.80118

23、.0057.10822.00118.8061.70921.30124.2058.401024.00124.8060.801123.30124.7060.001222.50123.1060.001322.90125.3065.201419.50124.2053.701522.90127.4059.501622.30128.2060.101722.70126.1057.401823.50128.6060.401921.50129.4052.002025.50126.9061.502125.00126.5063.902226.10128.2063.002327.90131.4063.102426.8

24、0130.8061.502527.20133.9065.802624.40130.4062.602724.40131.3059.502823.00130.2062.502926.30136.0060.003028.80138.0063.70試用計(jì)算機(jī)完成下面的統(tǒng)計(jì)分析：（ 1）先假設(shè) Y 與 X1和 X2有如下線性關(guān)系： Y1X12 X 2 e ，做最小二乘分析， 并做相應(yīng)的殘差圖。9試計(jì)算 Box-Cox 變換參數(shù)的值 .（ 2）對(duì)（ 1）中計(jì)算出的變換參數(shù)值，做相應(yīng)的 Box-Cox 變換，并對(duì)變換后的因變量做對(duì)X1和 X2的最小二乘回歸，并做殘差圖。解：（ 1）先計(jì)算中心化杠桿率cent

25、er hii 和 COOK 統(tǒng)計(jì)量 Di 的值表 2-1編號(hào)擬合值?殘差 ?學(xué)生化殘差ricenter hiiCOOK統(tǒng)計(jì)量Diy1ei121.517031.08297.99976.07509.04052220.275301.224701.12888.07238.05021320.54753-1.44753-1.31902.05156.05380422.89020-1.09020-.97103.00889.01386520.56375.93625.85394.05330.02305619.41496.68504.64630.11302.02387719.74965-.94965-.88474.

26、09125.03713821.65266.34734.33076.12873.00705922.07076-.77076-.68711.01058.007231023.11814.88186.78544.00882.009051122.79886.50114.44527.00418.002581222.32124.17876.15985.01640.000451324.85923-1.95923-1.85996.12356.214591420.37038-.87038-.83580.14264.049731523.42396-.52396-.46553.00413.002811623.8798

27、4-1.57984-1.40437.00510.026271722.27615.42385.38119.02725.003121824.10778-.60778-.54071.00666.004061921.30763.19237.21088.33436.008622023.998271.501731.33663.00754.025382124.74713.25287.23152.06029.001852224.929011.170991.05440.02950.024842325.920431.979571.79703.04463.091022425.162471.637531.47137.

28、02553.045142527.64352-.44352-.42282.13059.011682625.44102-1.04102-.93760.02995.019802724.58817-.18817-.17089.04540.000832825.34514-2.34514-2.10925.02738.095852926.17207.12793.12316.14686.001113028.10769.69231.67800.17443.04018從表中 2-1 的計(jì)算結(jié)果可以看出， 第 19 個(gè)觀測(cè)的杠桿率最高為 0.33436.。因此， 第 19 個(gè)樣本點(diǎn)最有可能對(duì)模型擬合造成較大的影響。

29、然后求解經(jīng)驗(yàn)回歸方程， 從運(yùn)行結(jié)果的方差分析表 2-2（ ANOVA(b) ）可以看出 F 統(tǒng)計(jì)量的 P-值（ Sig.）為 0.000，這表明模型在總體中是顯著的。表 2-2ANOVA(b)10AN OV AbSum ofModelSquaresdfMean SquareFSig.1Regression153.984276.99258.501.000 aResidual35.534271.316Total189.51929a.Pre dict ors : (Cons tant), 胸 圍 X2,身高X1b.Dependent Va r iable: 體 重 Y表2-3Coefficients

30、aUnstandardizedStandardizedCoeff icientsCoef f icientsModelBStd. ErrorBetatSig.1(Constant)-36.1335.535-6.528.000身高X1.299.045.6076.565.000胸圍X2.362.074.4544.914.000a.Depe ndent Varia ble : 體 重 Y從回歸系數(shù)計(jì)算分析表 2-3（ Coefficients(a) ），可知道回歸方程的常數(shù)項(xiàng)為 -36.133，自變量身高和胸圍相對(duì)應(yīng)的未標(biāo)準(zhǔn)化的回歸系數(shù)（ Unstandardized Coefficients ）分

31、別為 0.299、0.362，因而回歸方程為yi36.1330.299 x1i0.362 x2i且從表中可知3個(gè)系數(shù)的 t統(tǒng)計(jì)量的 P值均為 0.000，這表明模型在總體中是顯著的。以擬合值y?i 為橫坐標(biāo)，殘差e?i 為縱坐標(biāo)，作殘差圖：圖 2-1 殘差圖11laudiseRdezidradnatsnU2.000001.000000.00000-1.00000-2.00000-3.0000015.0000018.0000021.0000024.0000027.0000030.00000Un s ta n d a rd iz e d P re d ic te d Va lu e從圖 2-1 可

32、以看出，殘差圖從左至右逐漸散開呈漏斗狀，這是誤差方差不相等的征兆?？紤]將因變量 Y 進(jìn)行 Box-Cox變換，跟第一題的（4）問同樣。這里同樣分別取i , i1,2,L 7 值為 -1.5， -1，-0.5，0，0.5， 1，1.5，計(jì)算分別計(jì)算 zi，然后計(jì)算對(duì)應(yīng)的殘差平方和RSS , z，這里 n=30, 計(jì)算得到 z如表 1-5所示，然后計(jì)算對(duì)應(yīng)自變量 X1i和 X 2的殘差平方和 RSS , z。得 Z1, Z2 L Z7 方差分析表如下ANOVA bSum ofModelSquaresdfMean SquareFSig.1Regression151.509275.75453.967.

33、000 aResidual37.900271.404Total189.40929a. Predictors: (Constant), 胸圍 X2,身高X1b. Dependent Variable: Z112ANOVA bSum ofModelSquaresdfMean SquareFSig.1Regression150.580275.29056.081.000 aResidual36.248271.343Total186.82829a. Predictors: (Constant), 胸圍 X2,身高X1b. Dependent Variable: Z2ANOVA bSum ofModelS

34、quaresdfMean SquareFSig.1Regression150.361275.18057.657.000 aResidual35.206271.304Total185.56729a. Predictors: (Constant), 胸圍 X2,身高X1b. Dependent Variable: Z3ANOVA bSum ofModelSquaresdfMean SquareFSig.1Regression150.852275.42658.610.000 aResidual34.747271.287Total185.59829a. Predictors: (Constant),

35、胸圍 X2,身高X1b. Dependent Variable: Z4ANOVA bSum ofModelSquaresdfMean SquareFSig.1Regression152.051276.02658.889.000 aResidual34.857271.291Total186.90929a. Predictors: (Constant), 胸圍 X2,身高X1b. Dependent Variable: Z5ANOVA bSum ofModelSquaresdfMean SquareFSig.1Regression153.984276.99258.501.000 aResidual

36、35.534271.316Total189.51929a. Predictors: (Constant), 胸圍 X2,身高X1b. Dependent Variable: Z613ANOVA bSum ofModelSquaresdfMean SquareFSig.1Regression156.674278.33757.496.000 aResidual36.787271.362Total193.46129a. Predictors: (Constant),胸圍 X2, 身高 X1b. Dependent Variable: Z7從上面的方差分析表中可以得到i , i1,2,L,7 對(duì)應(yīng)的殘

37、差平方和RSSi , zi表2-4-1.5-1-0.500.511.5RSS37.90036.24835.20634.74734.85735.53436.787從這個(gè)表中可的簡(jiǎn)單比較可以看出當(dāng)0 時(shí)，殘差平方和RSS, z34.747 達(dá)到最小，而0.5 對(duì)應(yīng)的殘差平方和次之為 34.857，且從的方差分析表可知它們對(duì)應(yīng)的P值都為 0.000，都具有顯著性?，F(xiàn)在再看0 和0.5 時(shí) ,對(duì)應(yīng)因變量 Z4 和 Z5 對(duì)應(yīng)的回歸系數(shù)分析表。Coe fficients aUnstandardizedStandardizedCoefficientsCoefficientsModelBStd. ErrorBetat