How to manually compute the p-value of t-statistic in linear regression
Asked Answered
Z

2

8

I did a linear regression for a two tailed t-test with 178 degrees of freedom. The summary function gives me two p-values for my two t-values.

t value Pr(>|t|)
5.06    1.04e-06 ***
10.09    < 2e-16 ***
...
...
F-statistic: 101.8 on 1 and 178 DF, p-value: < 2.2e-16

I want to calculate manually the p-value of the t-values with this formula:

p = 1 - 2*F(|t|)

p_value_1 <- 1 - 2 * pt(abs(t_1), 178)
p_value_2 <- 1 - 2 * pt(abs(t_2), 178)

I don't get the same p-values as in the model summary. Therefore, I want to know how the summary function Pr(>|t|) is different from my formula, as I can't find the definition of Pr(>|t|).

Can you help me? Thanks a lot!

Zamia answered 6/1, 2017 at 11:57 Comment(1)
Yeah! Both answers helped me! I both voted them up :-) But your answer convinced me a bit more because it did exactly what I wanted and was so short. Thanks again!Zamia
E
4

It is

2 * pt(-abs(t_value), df)

For example:

2 * pt(-5.06, 178)
#[1] 1.038543e-06

2 * pt(-10.09, 178)
#[1] 3.223683e-19

Alternatively, use

2 * pt(abs(t_value), df, lower.tail = FALSE)
Englishism answered 6/1, 2017 at 12:1 Comment(0)
T
4

We can compute the p value Pr(>|t|) in the following different ways:

tval <- 5.06
df <- 178

# compute area under the t-pdf 
integrate(function(x) dt(x, df), -Inf, -tval)$value + integrate(function(x) dt(x, df), tval, Inf)$value # sum of two areas
# [1] 1.038543e-06
1-integrate(function(x) dt(x, df), -tval, tval)$value
# [1] 1.038543e-06
# 2-sided t-test: Pr_T(|t|>|tval|) = 2*(1 - F_T(|tval|)) = 2*F_T(-|tval|), where T~t(df=178)
2*(1 - pt(tval, df)) 
# [1] 1.038543e-06
2*pt(tval, df, lower.tail = FALSE)
# [1] 1.038543e-06
1 - (pt(tval, df) - pt(-tval, df))
# [1] 1.038543e-06
2*pt(-tval, df)
# [1] 1.038543e-06

The following illustrates the same geometrically with a different (less extreme) value of the t-statistic, as we can see, there are two (symmetric) blue regions that together represent the corresponding probability, under the 2-sided t-test.

df <- 178
x <- seq(-6, 6,0.01)
y <- dt(x, df)
tval <- 1.25
plot(x, y, type='l', main='t-distribution and p-value (5% significance level, 2-sided t-test)')
abline(h=0)
abline(v = c(tval, -tval), col='red')
index1 <- which(x >= -tval)[1]
index2 <- which(x >= tval)[1]
polygon(x = c(x[1:index1], x[index1], x[1]), 
        y = c(y[1:index1], 0, 0),
        col = "blue")
polygon(x = c(x[index2], x[index2], x[index2:length(x)]), 
        y = c(0, y[index2], y[index2:length(y)]),
        col = "blue")

enter image description here

Tumular answered 6/1, 2017 at 13:15 Comment(0)

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