多元统计分析方法 covariance 协方差分析.doc

协方差分析一个协变量的协方差分析例：为研究三种饲料(A1(g=1)，A2(g=2)，A3(g=3)对猪催肥效果，用每种饲料喂养8头猪，实验用猪的初始体重(x)未控制。喂养一段时间后，观察小猪的增重(y)。所得资料如表21，试分析三种饲料对猪催肥效果是否相同。资料结构：(文件名covariance1.dta)xyg1585113831116511276112801169111484117901179721690218100218952211032221062199921894222893249132083323953251003271023301053321103对于不考虑初始体重影响而评价三种饲料的统计分析为单因素方差分析(One-way ANOVA)，由于小猪的增重与初始体重有关，因此在分析三种饲料对增重的关系时，应该考虑校正初始体重对增重的影响。并假定初始体重与增重呈线性统计关系以及要求初始体重与饲料不构成交互作用。称校正变量(初始体重)为协变量，分组变量为因子变量。因此可用协方差分析上述统计问题，相应的角模型如下：A1(g=1)A2(g=2)A3(g=3)不校正初始体重校正初始体重用STATA命令为： anova y g x g*x,class(g) Number of obs = 24 R-squared = 0.9297 Root MSE = 3.15855 Adj R-squared = 0.9102 Source | Partial SS df MS F Prob F Model | 2376.3819 5 475.27638 47.64 0.0000 | g | 24.4661579 2 12.233079 1.23 0.3168 x | 830.415407 1 830.415407 83.24 0.0000 g*x | 48.0381359 2 24.019068 2.41 0.1184 | Residual | 179.576433 18 9.97646848 Total | 2555.95833 23 111.128623 由g*x项的P值0.11840.05，说明初始体重与饲料不构成交互作用。anova y g x,class(g) Number of obs = 24 R-squared = 0.9109 Root MSE = 3.37353 Adj R-squared = 0.8976 Source | Partial SS df MS F Prob F Model | 2328.34376 3 776.114588 68.20 0.0000 g | 707.218765 2 353.609382 31.07 0.0000 x | 1010.76043 1 1010.76043 88.81 0.0000Residual | 227.614568 20 11.3807284 Total | 2555.95833 23 111.128623 regress Source | SS df MS Number of obs = 24-+- F( 3, 20) = 68.20 Model | 2328.34376 3 776.114588 Prob F = 0.0000 Residual | 227.614568 20 11.3807284 R-squared = 0.9109-+- Adj R-squared = 0.8976 Total | 2555.95833 23 111.128623 Root MSE = 3.3735 y Coef. Std. Err. t P|t| 95% Conf. Interval_cons 35.93518 6.575471 5.47 0.000 22.21899 49.65137g 1 12.79324 3.408989 3.75 0.001 5.682214 19.90427 2 17.33559 2.409151 7.20 0.000 12.31019 22.36099 3 (dropped)x 2.401569 .2548332 9.42 0.000 1.869996 2.933142，A1 vs A3：，H0：a1=0 vs H1:a10对应的P值为0.0010.05，因此认为两组总体均数不同，由a1的95可信区间可认为A1的均数大于A3的均数，差别有统计意义。A2 vs A3：，H0：a2=0 vs H1:a20对应的P值为0.001 F = 0.0424对应的P值为0.04240.05，因此认为两组总体均数不同，由于点估计为：，P值小于0.05，因此可认为A2的均数大于A1的均数，差别有统计意义。结论：1)A2饲料喂养的小猪增重最高，A1饲料喂养的小猪增重也高于A3饲料喂养的小猪的增重，差别均有统计意义，P值均小于0.05。2)小猪的增重与初始的呈正相关，P F -+- Model | 7.93878728 2 3.96939364 14.39 0.0001 | group | 5.74578681 1 5.74578681 20.83 0.0001 x1 | 1.40547531 1 1.40547531 5.09 0.0323 | Residual | 7.44916457 27 .275894984 -+- Total | 15.3879518 29 .530619029 说明：疗效与基线情况有关，并且两组干预的疗效有差异。. regress Source | SS df MS Number of obs = 30-+- F( 2, 27) = 14.39 Model | 7.93878728 2 3.96939364 Prob F = 0.0001 Residual | 7.44916457 27 .275894984 R-squared = 0.5159-+- Adj R-squared = 0.4801 Total | 15.3879518 29 .530619029 Root MSE = .52526- d Coef. Std. Err. t P|t| 95% Conf. Interval-_cons 1.876583 2.471592 0.76 0.454 -3.194705 6.947871group 1 -.8815209 .1931656 -4.56 0.000 -1.277864 -.4851778 2 (dropped)x1 .0402676 .0178409 2.26 0.032 .0036612 .076874-校正了基线以后，两组疗效的差异为0.8815209，并且有统计学意义。(注：未校正前的两组疗效的差异为0.9333318)一个协变量、二个因子的协方差分析例22 某园艺家研究鲜花的种类(因子A：花种LP(a=1)和花种WB(a=2)和湿度（因子B：湿度低(b=1)和湿度高(b=2)对出售鲜花量(y)的影响。因为试验田的大小不等，故把试验田的大小(x)作为协变量，每个试验田重复6次，资料如书(p28)所述,试分析出售鲜花量与这2个因子的关系。数据结构：yxab9815116041177711809119514116451171101280121286141282131246212553125542160521758216572187132178112176112268102243222473226272270922由于试验田大小(x)同样可以影响出售鲜花量(y)，所以可用协方差分析：花种湿度低(b=1)湿度高(b=2)无x(ANOVA)LP(a=1)WB(a=2)含x(Co-ANOVA)LP(a=1)WB(a=2)anova y a b a*b x a*b*x,class(a b) Number of obs = 24 R-squared = 0.9786 Root MSE = 2.60706 Adj R-squared = 0.9693 Source | Partial SS df MS F Prob F Model | 4977.25219 7 711.036027 104.61 0.0000 | a | 38.8083118 1 38.8083118 5.71 0.0295 b | 44.0347332 1 44.0347332 6.48 0.0216 a*b | .027656949 1 .027656949 0.00 0.9499 x | 3703.3091 1 3703.3091 544.87 0.0000 a*b*x | 10.7333748 3 3.5777916 0.53 0.6704 |Residual | 108.747808 16 6.796738 Total | 5086.00 23 221.130435 由于协变量x与因子a和b的交互项a*b*x的检验的P值0.67040.05，所以可以认为因子a和b与协变量x无交互作用。 Number of obs = 24 R-squared = 0.9765 Root MSE = 2.50768 Adj R-squared = 0.9716 Source | Partial SS df MS F Prob F Model | 4966.51882 4 1241.6297 197.45 0.0000 | a | 96.6018263 1 96.6018263 15.36 0.0009 b | 323.849473 1 323.849473 51.50 0.0000 x | 3994.51882 1 3994.51882 635.21 0.0000 a*b | 16.0422442 1 16.0422442 2.55 0.1267 |Residual | 119.481183 19 6.28848331 Total | 5086.00 23 221.130435 anova y a b x ,class(a b) Number of obs = 24 R-squared = 0.9734 Root MSE = 2.60311 Adj R-squared = 0.9694 Source | Partial SS df MS F Prob F Model | 4950.47657 3 1650.15886 243.52 0.0000 | a | 97.5515084 1 97.5515084 14.40 0.0011 b | 324.433906 1 324.433906 47.88 0.0000 x | 3978.47657 1 3978.47657 587.13 0.0000 | Residual | 135.523427 20 6.77617135 Total | 5086.00 23 221.130435 regressyCoef.Std.Err.tP|t|95% Conf.Interval_cons37.338021.34187527.830.00034.5389240.13712a14.1044181.081753.790.0011.8479286.3609082 (dropped)b17.3681391.0648466