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从关于数据的简单线性回归 model 的参数可以得出什么结论

[英]What to conclude from parameters of the simple linear regression model about data

原文 2017-11-04 15:52:59 4 2 r/ math/ statistics

I had a dataset for which I needed to provide a linear regression model that represents diameter as a function of length.Data which has length in first column and diameter in second looked like: 0.455,0.365 0.44,0.365我有一个数据集，我需要为它提供一个线性回归 model，它表示直径为 function 的长度。第一列长度和第二列直径的数据看起来像：0.455,0.365 0.44,0.365

I carried out the required operations on the given dataset in R,and plotted the regression line for the data我对 R 中的给定数据集执行了所需的操作，并绘制了数据的回归线

I am just confused about what to conclude from the parameters(slope=0.8154, y intercept:-0.019413, correlation coefficient:0.98 ).我只是对从参数得出的结论感到困惑（斜率=0.8154，y 截距：-0.019413，相关系数：0.98）。 Can I conclude anything other than line is a good fit.我可以得出结论，除了线以外的任何东西都很合适。 I am new to statistics.我是统计新手。 Any help would be appreciated.任何帮助，将不胜感激。

2 个解决方案

From the parameters slope and intercept, you cannot conclude if the line is a good fit.从参数斜率和截距，您无法得出该线是否合适的结论。 The correlation coefficient says that they depend highly and that a straight line could fit your model.相关系数表明它们高度依赖并且直线可以适合您的 model。 However, from the p-values for the slope and intercept, you can conclude if your fit is good.但是，根据斜率和截距的 p 值，您可以得出拟合是否良好的结论。 If they are small (say below 0.05) you can conclude that the fit is pretty good.如果它们很小（比如低于 0.05），您可以得出结论，拟合非常好。

Slope 0.8154 informs you that each unit increase for lenght causes increase of diamater in 0.8154*unit.斜率 0.8154 告诉您，长度的每个单位增加都会导致直径增加 0.8154*unit。 Intercept -0.019413 is probably statistically insignificant in this case.在这种情况下，截距 -0.019413 可能在统计上不显着。 To verify that you have to look at t -statistics for example.例如，要验证您是否必须查看t -statistics。

On this page you can find nice course with visualizations about simple linear regression and other statistical methods answering your questions.在此页面上，您可以找到关于简单线性回归和其他统计方法的可视化课程来回答您的问题。