测试简单线性回归中的斜率是否等于R中的给定常数 [英] Test if the slope in simple linear regression equals to a given constant in R

问题描述

我想测试简单线性回归中的斜率是否等于给定的常数，而不是零.

I want to test if the slope in a simple linear regression is equal to a given constant other than zero.

> x <- c(1,2,3,4)
> y <- c(2,5,8,13)
> fit <- lm(y ~ x)
> summary(fit)

Call:
lm(formula = y ~ x)

Residuals:
   1    2    3    4 
 0.4 -0.2 -0.8  0.6 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)  -2.0000     0.9487  -2.108  0.16955   
x             3.6000     0.3464  10.392  0.00913 **
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7746 on 2 degrees of freedom
Multiple R-squared:  0.9818,    Adjusted R-squared:  0.9727 
F-statistic:   108 on 1 and 2 DF,  p-value: 0.009133
> confint(fit)
                2.5 %   97.5 %
(Intercept) -6.081855 2.081855
x            2.109517 5.090483

在此示例中，我想测试斜率是否等于5.我知道我不会拒绝它，因为5在95％CI中.但是有没有可以直接给我p值的函数?

In this example, I want to test if the slope is equal to 5. I know I won't reject it since 5 is in the 95% CI. But is there a function which can give me the p-value directly?

推荐答案

测试拟合是否与特定系数显着不同的一种方法是构建偏移"，其中将该系数用作应用于系数的因数. x值.您应该将其视为至少将零斜率"重置为零".估计，拦截仍然是免费"的，可以四处走动".

One approach to testing whether a fit is significantly different than a particular coefficient is to construct an "offset" in which that coefficient is used as a factor applied to the x-value. You should think of this as re-setting the "zero" at least the zero-for-the-slope. The Intercept is still "free" to "move around", er, to be estimated.

 fit2 <- lm( y~x +offset(5*x) )
#----------------
 summary(fit2)
#--------
Call:
lm(formula = y ~ x + offset(5 * x))

Residuals:
   1    2    3    4 
 0.4 -0.2 -0.8  0.6 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  -2.0000     0.9487  -2.108   0.1695  
x            -1.4000     0.3464  -4.041   0.0561 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7746 on 2 degrees of freedom
Multiple R-squared:  0.9818,    Adjusted R-squared:  0.9727 
F-statistic:   108 on 1 and 2 DF,  p-value: 0.009133

现在将其与您的fit对象的结果进行比较. x的系数恰好相差5.模型拟合统计量是相同的，但是您怀疑x变量的p值要低得多，而要高得多，即不那么重要.

Now compare to the results from your fit-object. The coefficients for x differ by exactly 5. The model fit statistics are the same, but as you suspected the p-value for the x-variable is much lower ... er, higher, i.e. less significant.

> summary(fit)

Call:
lm(formula = y ~ x)

Residuals:
   1    2    3    4 
 0.4 -0.2 -0.8  0.6 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)  -2.0000     0.9487  -2.108  0.16955   
x             3.6000     0.3464  10.392  0.00913 **
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7746 on 2 degrees of freedom
Multiple R-squared:  0.9818,    Adjusted R-squared:  0.9727 
F-statistic:   108 on 1 and 2 DF,  p-value: 0.009133

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