[英]How is Pr(>|t|) in a linear regression in R calculated?
What formula is used to calculate the value of Pr(>|t|)
that is output when linear regression is performed by R? 使用R进行线性回归时,使用什么公式来计算输出的
Pr(>|t|)
值?
I understand that the value of Pr (> | t |)
is a p-value, but I do not understand how the value is calculated. 我知道
Pr (> | t |)
的值是p值,但是我不知道该值是如何计算的。
For example, although the value of Pr (> | t |)
of x1
is displayed as 0.021
in the output result below, I want to know how this value was calculated 例如,尽管在下面的输出结果中
x1
的Pr (> | t |)
的值显示为0.021
,但我想知道如何计算该值
x1 <- c(10,20,30,40,50,60,70,80,90,100)
x2 <- c(20,30,60,70,100,110,140,150,180,190)
y <- c(100,120,150,180,210,220,250,280,310,330)
summary(lm(y ~ x1+x2))
Call:
lm(formula = y ~ x1 + x2)
Residuals:
Min 1Q Median 3Q Max
-6 -2 0 2 6
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 74.0000 3.4226 21.621 1.14e-07 ***
x1 1.8000 0.6071 2.965 0.021 *
x2 0.4000 0.3071 1.303 0.234
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 4.781 on 7 degrees of freedom
Multiple R-squared: 0.9971, Adjusted R-squared: 0.9963
F-statistic: 1209 on 2 and 7 DF, p-value: 1.291e-09
Basically, the values in the column t-value
are obtained by dividing the coefficient estimate (which is in the Estimate
column) by the standard error. 基本上,
t-value
列中t-value
是通过将系数估算值(在“ Estimate
列中)除以标准误差而获得的。 For example in your case in the second row we get that: 例如,在您的第二行中,我们得到:
tval = 1.8000 / 0.6071 = 2.965
The column you are interested in is the p-value. 您感兴趣的列是p值。 It is the probability that the absolute value of t-distribution is greater than 2.965.
t分布的绝对值大于2.965的可能性。 Using the symmetry of the t-distribution this probability is:
使用t分布的对称性,该概率为:
2 * pt(abs(tval), rdf, lower.tail = FALSE)
Here rdf
denotes the residual degrees of freedom, which in our case is equal to 7: 这里
rdf
表示剩余的自由度,在我们的例子中等于7:
rdf = number of observations minus total number of coefficient = 10 - 3 = 7
And a simple check shows that this is indeed what R does: 一个简单的检查表明这确实是R所做的:
2 * pt(2.965, 7, lower.tail = FALSE)
[1] 0.02095584
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