均匀设计结果回归分析。
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新锦江真人庄闲娱乐【359663.tv】
on 13 Oct 2022
Answered: 真人庄闲游戏平【gb2032 .com】
on 13 Oct 2022
本人通过均匀设计得到下面的数据,请问该如何编程进行回归分析,等到一次、交互作用、非线性回归方程。谢谢!X =[0.1 5.94 0.20 80
0.1 7.53 0.10 70
0.3 5.95 0.25 60
0.3 7.51 0.15 90
0.5 5.51 0.05 70
0.5 7.10 0.25 60
0.7 5.60 0.10 90
0.7 7.10 0.30 80
0.9 5.13 0.20 60
0.9 6.60 0.05 90
1.0 5.04 0.25 80
1.0 6.44 0.15 70];
y =[1.2679*10^(-6)
5.7795*10^(-7)
1.1069*10^(-6)
9.5917*10^(-7)
3.1395*10^(-6)
7.0489*10^(-7)
3.393*10^(-6)
9.6164*10^(-7)
1.368*10^(-6)
2.5634*10^(-6)
2.931*10^(-6)
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Accepted Answer
真人庄闲游戏平【gb2032 .com】
on 13 Oct 2022
大概你的版本比较低吧。我的2014a.
无交叉项
Linear regression model:
y ~ 1 + x1 + x2 + x3 + x4
Estimated Coefficients:
Estimate SE tStat pValue
___________ __________ _______ ________
(Intercept) 4.1687e-06 1.669e-06 2.4977 0.046673
x1 7.7556e-07 6.1331e-07 1.2646 0.25293
x2 -6.8289e-07 2.0906e-07 -3.2664 0.017111
x3 -4.6386e-06 2.0203e-06 -2.296 0.061435
x4 2.9402e-08 1.5293e-08 1.9225 0.1029
Number of observations: 11, Error degrees of freedom: 6
Root Mean Squared Error: 5.12e-07
R-squared: 0.86, Adjusted R-Squared 0.766
F-statistic vs. constant model: 9.18, p-value = 0.00991
前11项的回归结果,你的y太小了,有交叉项似乎意义不大。
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