实验五因子分析

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1、实验五因子分析Ui练习卜•表资料为25名健康人的7项生化检验结果，7项生化检验指标依次命名为XI至X7,请对该资料进行因子分析。XIX2X3X4X5X6X73.763.660.545.289.7713.744.78&594.991.3410.027.5010.162.136.226.144.529.842.172.731.097.577.287.0712.661.792.100.829.037.082.5911.764.546.221.285.513.981.306.925.337.302.403.270.620.443.367.638.848.39&747.003.

2、3111.683.534.761.129.649.491.0313.5713.1318.522.359.731.331.009.879.8711.063.708.592.981.179.177.859.912.627.125.493.689.722.643.431.194.693.012.175.982.763.552.015.511.341.275.814.575.383.431.661.611.572.801.782.093.725.905.761.558.845.407.501.979.849.271.5113.609.0212.671.758.394.922.

3、5410.053.965.241.434.944.381.036.686.499.062.817.232.301.777.794.395.372.279.467.311.0412.0011.5816.182.429.555.354.2511.742.773.511.054.944.524.50&071.792.101.29&213.082.429.103.754.661.729.416.445.1112.502.453.100.91输出结果如图所示:KMOandBartletfsTestKaiser-Meyer-OIkinMeasureofSamplingAdequa

4、cy..321Bartlett'sTestofApprox.Chi-Square326.285Sphericitydf21Sig..000巴特利特球度检验统计量的观测值326.285,相应的概率P接近于0,拒绝零假设，认为相关系数矩阵与单位矩阵有显著性差异。同时，KMO值为0.882,根据Kaiser给出了KMO度量标准可知原有变量适合因了分析。TotalVarianceExplainedComoonontInitialEiaervaluesExtractionSumsofSquaredLoadinqsRotationSumsofSquaredLoadinqsTota

5、l%ofVarianeeCumulative%Total%ofVarianceCumulative%Total%ofVarianceCumulative%13.39548.50348.5033.39548.50348.5033.30647.23147.23122.80640.09088.5932.80640.09088.5932.89541.36188.5933.4366.23694.8284.2763.946987755.0811.16099.9356.004.05999.9947.000.006100.000ExtractionMethod:PrincipalCo

6、mporientAnalysis.第1,2因子的累积方差贡献率达88.593%,对解释原有变量的贡献最大,因此提取这两个因子是合适的。ComponentMatrix3Component12x4.911.389x7-.886.219x2.796.372X1.746.489x3.709-.597x6-.177.972x5-.234.963Xl=0.746fl+0.489f2X2=0.796fl+0.372f2X3=0.709fl-0.597f2X4=0.911fl+0.389f2X5=-0.234fl+0.963f2X6=-0.177fl+0.972f2X7=-0.886

7、fl+0219f2XI、X2、X3、X4、X7在第一个因子上的载荷都很高，意味着他们与第一个因子的相关程度高，fl主要解释这几个变量。而X5、X6在f2上的载荷很高,主要解释这两个变量。ComponentScoreCoefficientMatrixComporient12x1.270.075x2.268.031x3.110-.277x4.301.023x5.070.343x6.087.339x7-.210.173F1=0.270X1+0.268X2+0.110X3+0.301X4+0.074X5+0.087X6+0.21X7F2=0.075X1+0.