本人统计小白,请教各位一个在实际分析中遇到的问题。我在做y1对应一系列的x的GLM分析时,得到如下结果
glm(formula = y1 ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 +
x10 + x11 + x12 + x13 + x14 + x15 + x16 + x17 + x18 + x19 +
x20 + x21 + x22 + x23 + x24 + x25)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.6638 -0.8945 -0.0614 0.5987 7.2177
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.296e-01 1.614e-01 1.423 0.15760
x1 1.883e-01 1.510e-01 1.247 0.21489
x2 4.566e-01 1.653e-01 2.762 0.00672 **
x3 -1.047e-01 1.994e-01 -0.525 0.60047
x4 -1.761e-01 2.052e-01 -0.858 0.39290
x5 2.656e-02 1.746e-01 0.152 0.87938
x6 1.042e+02 4.220e+01 2.468 0.01510 *
x7 2.949e+02 1.185e+02 2.488 0.01434 *
x8 1.382e+01 5.782e+00 2.391 0.01848 *
x9 2.581e+01 1.045e+01 2.469 0.01507 *
x10 -4.465e+04 3.538e+04 -1.262 0.20962
x11 3.815e+01 1.520e+01 2.510 0.01350 *
x12 -4.725e+02 1.905e+02 -2.480 0.01464 *
x13 -3.960e-01 2.433e-01 -1.628 0.10639
x14 4.472e+04 3.538e+04 1.264 0.20889
x15 -6.051e-02 1.634e-01 -0.370 0.71192
x16 -2.534e-01 4.379e-01 -0.579 0.56395
x17 -1.180e+00 8.193e-01 -1.440 0.15256
x18 3.772e-01 8.851e-01 0.426 0.67085
x19 5.722e-01 2.661e-01 2.151 0.03368 *
x20 4.547e-02 9.679e-02 0.470 0.63941
x21 4.147e-02 1.849e-01 0.224 0.82289
x22 1.479e-01 1.275e-01 1.160 0.24855
x23 6.920e-01 8.658e-01 0.799 0.42586
x24 5.747e-01 8.482e-01 0.678 0.49947
x25 -6.797e-01 1.134e+00 -0.599 0.55009
请问,有没有必要把打星号的挑出来再重新做一遍GLM分析?谢谢!


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