matlab多变量二次多项式拟合

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1、X1X2X3X4X5X6X7X8Y1Y2Y3Y总0.600.50.76121.60.1619.48303685.480.60.30.91.1081.20.164.6914.792746.480.60.61.30.36140.80.124.551.443641.990.60.90.10.70100.80.127.772.413343.18100.51.56140.40.0826.322.743766.0610.31.30.30121.60.083.4910.812842.310.60.10.7681.60.0

2、43.923.842835.7610.90.51.50121.20.0416.6793661.671.400.90.39100.80.163.663.523138.181.40.31.30.73140.40.165.7711.693552.461.40.60.11.59100.40.1220.554.343761.891.40.90.90.3381.60.123.853.353239.21.801.31.19121.20.08615.723758.721.80.30.11.5381.20.08301.435

3、66.41.80.60.50.79140.80.0414.0214.383866.41.80.90.91.13100.40.043.3315.323856.65一、对Y总做线性多项式拟合：设置显著性水平为0.05，拟合得到：B=[,,……….,]=[-60.034912.58092.2002-12.986320.41450.02665.143017.2416151.6779]对应的置信区间为：-161.405841.3359-7.587032.7488-25.570629.9709-33.50897.536

4、2-0.309641.1386-2.59892.65200.98309.3030-3.281037.7642-64.0209367.3767r2=0.7454(越接近于1，回归效果越显著)，F=2.5616，p=0.1163，（p>0.05,可知回归模型不成立）。残差图如下：从残差图可以看出，除第一个数据和最后一个数据的残差离零点均较远，说明这两个数据可视为异常点，去掉这两个数据之后再做拟合得到：B=[,,……….,]=[-478.815.71.8-85.3432.824.7135.31131.9]对应的置

5、信区间为：-1048.791.17.523.9-811.6-183.512.810.575.5-1.16.7-251.4-25.8296.4-206.72470.4r2=0.9690(越接近于1，回归效果越显著)，F=19.5530，p=0.0023，（p<0.05,可知回归模型成立）。残差图如下：从残差图可以看出，数据的残差离零点均较近，且残差的置信区间均包含零点，这说明回归模型能较好的符合原始数据。预测值与实测值的比较：Y总预测值Y总实测值相对误差42.745846.48-8.03%40.500841.

6、99-3.55%44.935843.184.07%66.135866.060.11%42.910842.31.44%35.7635.760.00%64.450861.674.51%40.960838.187.28%52.4652.460.00%62.500861.890.99%39.275839.20.19%60.475858.722.99%64.910866.4-2.24%62.665866.4-5.62%从上表可以看出，预测值和实测值的误差都在10%以内，说明该拟合模型能很好的预测实验值。最优解：使Y总

7、取最大值的X为：X=[,]=[1.80.90.11.59141.60.16]此时Y总的预测值为:375.7516二、对Y总做纯二次多项式拟合：一共有17个系数，B=[,,……….,,,……….,]=[-0.2-102.32.98.1-12.60.315.6-48.3-109.647.5-5.9-16.216-0.1-0.5311144.8]置信区间为：-5.15-34.914.4-15.416-14.416-21.619.1-1.51.6-3.86.9-24.514.9-228.7206.8-5.515-1

8、7.115.9-11.98.7-9.112.3-0.20.2-0.30.2-6.612.8-938.61167.6r2=0.9980残差图如下：预测值与实测值的比较：Y总预测值Y总实测值相对误差84.19185.48-1.51%47.121746.481.38%41.756241.99-0.56%44.017443.181.94%67.601966.062.33%42.894742.31.41%36.731635