i have the following equation for calculating the t statistics of a simple linear regression model.
t= beta1/SE(beta1)
SE(beta1)=sqrt((RSS/var(x1))*(1/n-2))
If i want to do this for an simple example wit R, i am not able to get the same results as the linear model in R.
x <- c(1,2,4,8,16)
y <- c(1,2,3,4,5)
mod <- lm(y~x)
summary(mod)
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5
-0.74194 0.01613 0.53226 0.56452 -0.37097
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.50000 0.44400 3.378 0.0431 *
x 0.24194 0.05376 4.500 0.0205 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.6558 on 3 degrees of freedom
Multiple R-squared: 0.871, Adjusted R-squared: 0.828
F-statistic: 20.25 on 1 and 3 DF, p-value: 0.02049
If i do this by hand i get a other value.
var(x)
37.2
sum(resid(mod)^2)
1.290323
beta1=0.24194
SE(beta1)=sqrt((1.290323/37.2)*(1/3))
SE(beta1)=0.1075269
So t= 0.24194/0.1075269=2.250042
So why is my calculation exact the half of the value from R? Has it something to do with one/two tailed tests? The value for t(0.05/2) is 3.18
Regards,
Jan
The different result was caused by a missing term in your formula for se(beta). It should be:
se(beta) = sqrt((1 / (n - 2)) * rss / (var(x) * (n - 1)))
The formula is usually written out as:
se(beta) = sqrt((1 / (n - 2)) * rss / sum((x - mean(x)) ^ 2))
rather than in terms of var(x).
For the sake of completeness, here's also the computational check:
reprex::reprex_info()
#> Created by the reprex package v0.1.1.9000 on 2017-10-30
x <- c(1, 2, 4, 8, 16)
y <- c(1, 2, 3, 4, 5)
n <- length(x)
mod <- lm(y ~ x)
summary(mod)
#>
#> Call:
#> lm(formula = y ~ x)
#>
#> Residuals:
#> 1 2 3 4 5
#> -0.74194 0.01613 0.53226 0.56452 -0.37097
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 1.50000 0.44400 3.378 0.0431 *
#> x 0.24194 0.05376 4.500 0.0205 *
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 0.6558 on 3 degrees of freedom
#> Multiple R-squared: 0.871, Adjusted R-squared: 0.828
#> F-statistic: 20.25 on 1 and 3 DF, p-value: 0.02049
mod_se_b <- summary(mod)$coefficients[2, 2]
rss <- sum(resid(mod) ^ 2)
se_b <- sqrt((1 / (n - 2)) * rss / (var(x) * (n - 1)))
all.equal(se_b, mod_se_b)
#> [1] TRUE
Related
I want to fit beta-binomial regression. I don't have counts but proportions that I want to fit. Here's example:
library(dplyr)
library(gamlss)
df <- tibble(
cluster = LETTERS[1:20]
) |>
mutate(
p = rbeta(n(), 1, 1),
n = as.integer(3 * runif(n()))
)
fit <- gamlss(
p ~ log(n),
weights = n,
data = df,
family = BB(mu.link='identity')
)
I get error:
Error in while (abs(olddv - dv) > cc && itn < cyc) { :
missing value where TRUE/FALSE needed
In addition: There were 50 or more warnings (use warnings() to see the first 50)
Warnings look like:
In dbinom(x, size = bd, prob = mu, log = log) : non-integer x = 0.834502
Note that I DON'T want to rounded number of successes such as mutate(y = round(p * n)).
The help file for the BB() family suggests that the dependent variable is expected to be a two-column matrix of the numbers of successes and failures. If you've got p (the probability of success) and you've got n (the number of trials), then you can make both the number of successes k=floor(p*n) and the number of failures notk = n-k. Then, you can do as I did below.
library(dplyr)
library(gamlss)
df <- tibble(
cluster = LETTERS[1:20]
) |>
mutate(
p = rbeta(n(), 1, 1),
n = as.integer(100 * runif(n()))
)
df <- df %>%
mutate(k = floor(p*n),
notk = n-k)
fit <- gamlss(
cbind(k, notk) ~ cluster,
data = df,
family = BB(mu.link='logit')
)
#> ******************************************************************
#> Family: c("BB", "Beta Binomial")
#>
#> Call:
#> gamlss(formula = cbind(k, notk) ~ cluster, family = BB(mu.link = "logit"),
#> data = df)
#>
#> Fitting method: RS()
#>
#> ------------------------------------------------------------------
#> Mu link function: logit
#> Mu Coefficients:
#> Estimate Std. Error z value Pr(>|z|)
#> (Intercept) 8.130e-01 3.836e-01 2.119 0.03406 *
#> clusterB -8.130e-01 2.036e+00 -0.399 0.68973
#> clusterC -3.686e+01 1.000e+05 0.000 0.99971
#> clusterD -2.970e+00 6.922e-01 -4.291 1.78e-05 ***
#> clusterE 3.618e-01 4.843e-01 0.747 0.45508
#> clusterF -3.381e-01 5.317e-01 -0.636 0.52479
#> clusterG -3.569e+00 6.506e-01 -5.485 4.13e-08 ***
#> clusterH -1.118e+00 4.356e-01 -2.566 0.01030 *
#> clusterI -1.712e+00 4.453e-01 -3.845 0.00012 ***
#> clusterJ 1.825e+00 6.315e-01 2.889 0.00386 **
#> clusterK -3.686e+01 1.000e+05 0.000 0.99971
#> clusterL -5.247e-01 4.602e-01 -1.140 0.25419
#> clusterM 1.439e+00 7.167e-01 2.008 0.04464 *
#> clusterN 9.161e-02 4.721e-01 0.194 0.84613
#> clusterO -2.405e+00 1.000e+05 0.000 0.99998
#> clusterP 3.034e-01 5.583e-01 0.543 0.58686
#> clusterQ -1.523e+00 5.389e-01 -2.826 0.00471 **
#> clusterR -2.498e+00 6.208e-01 -4.024 5.73e-05 ***
#> clusterS 1.006e+00 5.268e-01 1.910 0.05619 .
#> clusterT -6.228e-02 4.688e-01 -0.133 0.89433
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> ------------------------------------------------------------------
#> Sigma link function: log
#> Sigma Coefficients:
#> Estimate Std. Error z value Pr(>|z|)
#> (Intercept) -36.034363 0.005137 -7014 <2e-16 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> ------------------------------------------------------------------
#> No. of observations in the fit: 20
#> Degrees of Freedom for the fit: 21
#> Residual Deg. of Freedom: -1
#> at cycle: 9
#>
#> Global Deviance: 69.49748
#> AIC: 111.4975
#> SBC: 132.4079
#> ******************************************************************
Created on 2023-01-20 by the reprex package (v2.0.1)
I am trying to model the relation between a scar acquisition rate of a wild population of animals, and I have calculated yearly rates before.
If you see below the plot, it seems to me that rates rise through the middle of the period and than fall again. I have tried to fit a polynomial LM with the code
model1 <- lm(Rate~poly(year, 2, raw = TRUE),data=yearlyratesub)
summary(model1)
model1
I have plotted using:
g <-ggplot(yearlyratesub, aes(year, Rate)) + geom_point(shape=1) + geom_smooth(method = lm, formula = y ~ poly(x, 2, raw = TRUE))
g
The model output was:
Call:
lm(formula = Rate ~ poly(year, 2, raw = TRUE), data = yearlyratesub)
Residuals:
Min 1Q Median 3Q Max
-0.126332 -0.037683 -0.002602 0.053222 0.083503
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -8.796e+03 3.566e+03 -2.467 0.0297 *
poly(year, 2, raw = TRUE)1 8.747e+00 3.545e+00 2.467 0.0297 *
poly(year, 2, raw = TRUE)2 -2.174e-03 8.813e-04 -2.467 0.0297 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.0666 on 12 degrees of freedom
Multiple R-squared: 0.3369, Adjusted R-squared: 0.2264
F-statistic: 3.048 on 2 and 12 DF, p-value: 0.08503
How can I enterpret that now? The overall model p value is not significant but the intercept and single slopes are?
Should I rather try another fit than x² or even group the values and test between groups e.g. with an ANOVA? I know the LM has low fit but I guess it's because I have little values and maybe x² might be not it...?
Would be happy about input regarding model and outcome interpretation..
Grouping
Since the data was not provided (next time please provide a complete reproducible question including all inputs) we used the data in the Note at the end. We see that that the model is highly significant if we group the points using the indicated breakpoints.
g <- factor(findInterval(yearlyratesub$year, c(2007.5, 2014.5))+1); g
## [1] 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3
## Levels: 1 2 3
fm <- lm(rate ~ g, yearlyratesub)
summary(fm)
giving
Call:
lm(formula = rate ~ g, data = yearlyratesub)
Residuals:
Min 1Q Median 3Q Max
-0.064618 -0.018491 0.006091 0.029684 0.046831
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.110854 0.019694 5.629 0.000111 ***
g2 0.127783 0.024687 5.176 0.000231 ***
g3 -0.006714 0.027851 -0.241 0.813574
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.03939 on 12 degrees of freedom
Multiple R-squared: 0.7755, Adjusted R-squared: 0.738
F-statistic: 20.72 on 2 and 12 DF, p-value: 0.0001281
We could consider combining the outer two groups.
g2 <- factor(g == 2)
fm2 <- lm(rate ~ g2, yearlyratesub)
summary(fm2)
giving:
Call:
lm(formula = rate ~ g2, data = yearlyratesub)
Residuals:
Min 1Q Median 3Q Max
-0.064618 -0.016813 0.007096 0.031363 0.046831
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.10750 0.01341 8.015 2.19e-06 ***
g2TRUE 0.13114 0.01963 6.680 1.52e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.03793 on 13 degrees of freedom
Multiple R-squared: 0.7744, Adjusted R-squared: 0.757
F-statistic: 44.62 on 1 and 13 DF, p-value: 1.517e-05
Sinusoid
Looking at the graph it seems that the points are turning up at the left and right edges suggesting we use a sinusoidal fit. a + b * cos(c * year)
fm3 <- nls(rate ~ cbind(a = 1, b = cos(c * year)),
yearlyratesub, start = list(c = 0.5), algorithm = "plinear")
summary(fm3)
giving
Formula: rate ~ cbind(a = 1, b = cos(c * year))
Parameters:
Estimate Std. Error t value Pr(>|t|)
c 0.4999618 0.0001449 3449.654 < 2e-16 ***
.lin.a 0.1787200 0.0150659 11.863 5.5e-08 ***
.lin.b 0.0753754 0.0205818 3.662 0.00325 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.05688 on 12 degrees of freedom
Number of iterations to convergence: 2
Achieved convergence tolerance: 5.241e-08
Comparison
Plotting the fits and looking at their residual sum of squares and AIC we have
plot(yearlyratesub)
# fm0 from Note at end, fm and fm2 are grouping models, fm3 is sinusoidal
L <- list(fm0 = fm0, fm = fm, fm2 = fm2, fm3 = fm3)
for(i in seq_along(L)) {
lines(fitted(L[[i]]) ~ year, yearlyratesub, col = i, lwd = 2)
}
legend("topright", names(L), col = seq_along(L), lwd = 2)
giving the following where lower residual sum of squares and AIC (which takes into account the number of paramters) are better. We see that fm fits the most closely based on residual sum of squares but with fm2 fitting almost as well; however, when taking the number of parameters into account by using AIC fm2 has the lowest and so is most favored by that criterion.
cbind(RSS = sapply(L, deviance), AIC = sapply(L, AIC))
## RSS AIC
## fm0 0.05488031 -33.59161
## fm 0.01861659 -49.80813
## fm2 0.01870674 -51.73567
## fm3 0.04024237 -38.24512
Note
yearlyratesub <-
structure(list(year = c(2004, 2005, 2006, 2007, 2008, 2009, 2010,
2011, 2012, 2013, 2014, 2015, 2017, 2018, 2019), rate = c(0.14099813521287,
0.0949946651016247, 0.0904788394070601, 0.11694517831575, 0.26786193592875,
0.256346628540479, 0.222029818828298, 0.180116679856725, 0.285467976459104,
0.174019208113095, 0.28461698734932, 0.0574827955982996, 0.103378448084776,
0.114593695172686, 0.141105952837639)), row.names = c(NA, -15L
), class = "data.frame")
fm0 <- lm(rate ~ poly(year, 2, raw = TRUE), yearlyratesub)
summary(fm0)
giving
Call:
lm(formula = rate ~ poly(year, 2, raw = TRUE), data = yearlyratesub)
Residuals:
Min 1Q Median 3Q Max
-0.128335 -0.038289 -0.002715 0.054090 0.084792
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -8.930e+03 3.621e+03 -2.466 0.0297 *
poly(year, 2, raw = TRUE)1 8.880e+00 3.600e+00 2.467 0.0297 *
poly(year, 2, raw = TRUE)2 -2.207e-03 8.949e-04 -2.467 0.0297 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.06763 on 12 degrees of freedom
Multiple R-squared: 0.3381, Adjusted R-squared: 0.2278
F-statistic: 3.065 on 2 and 12 DF, p-value: 0.0841
Recently, I stumbled upon the fact that Stata and R handle regressions without intercept differently. I'm not a statistician, so please be kind if my vocabulary is not ideal.
I tried to make the example somewhat reproducible. This is my example in R:
> set.seed(20210211)
> df <- data.frame(y = runif(50), x = runif(50))
> df$d <- df$x > 0.5
>
> (tmp <- tempfile("data", fileext = ".csv"))
[1] "C:\\Users\\s1504gl\\AppData\\Local\\Temp\\1\\RtmpYtS6uk\\data1b2c1c4a96.csv"
> write.csv(df, tmp, row.names = FALSE)
>
> summary(lm(y ~ x + d, data = df))
Call:
lm(formula = y ~ x + d, data = df)
Residuals:
Min 1Q Median 3Q Max
-0.48651 -0.27449 0.03828 0.22119 0.53347
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.4375 0.1038 4.214 0.000113 ***
x -0.1026 0.3168 -0.324 0.747521
dTRUE 0.1513 0.1787 0.847 0.401353
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.2997 on 47 degrees of freedom
Multiple R-squared: 0.03103, Adjusted R-squared: -0.0102
F-statistic: 0.7526 on 2 and 47 DF, p-value: 0.4767
> summary(lm(y ~ x + d + 0, data = df))
Call:
lm(formula = y ~ x + d + 0, data = df)
Residuals:
Min 1Q Median 3Q Max
-0.48651 -0.27449 0.03828 0.22119 0.53347
Coefficients:
Estimate Std. Error t value Pr(>|t|)
x -0.1026 0.3168 -0.324 0.747521
dFALSE 0.4375 0.1038 4.214 0.000113 ***
dTRUE 0.5888 0.2482 2.372 0.021813 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.2997 on 47 degrees of freedom
Multiple R-squared: 0.7196, Adjusted R-squared: 0.7017
F-statistic: 40.21 on 3 and 47 DF, p-value: 4.996e-13
And here is what I have in Stata (please note that I have copied the filename from R to Stata):
. import delimited "C:\Users\s1504gl\AppData\Local\Temp\1\RtmpYtS6uk\data1b2c1c4a96.csv"
(3 vars, 50 obs)
. encode d, generate(d_enc)
.
. regress y x i.d_enc
Source | SS df MS Number of obs = 50
-------------+---------------------------------- F(2, 47) = 0.75
Model | .135181652 2 .067590826 Prob > F = 0.4767
Residual | 4.22088995 47 .089806169 R-squared = 0.0310
-------------+---------------------------------- Adj R-squared = -0.0102
Total | 4.3560716 49 .08889942 Root MSE = .29968
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x | -.1025954 .3168411 -0.32 0.748 -.7399975 .5348067
|
d_enc |
TRUE | .1512977 .1786527 0.85 0.401 -.2081052 .5107007
_cons | .4375371 .103837 4.21 0.000 .2286441 .6464301
------------------------------------------------------------------------------
. regress y x i.d_enc, noconstant
Source | SS df MS Number of obs = 50
-------------+---------------------------------- F(2, 48) = 38.13
Model | 9.23913703 2 4.61956852 Prob > F = 0.0000
Residual | 5.81541777 48 .121154537 R-squared = 0.6137
-------------+---------------------------------- Adj R-squared = 0.5976
Total | 15.0545548 50 .301091096 Root MSE = .34807
------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x | .976214 .2167973 4.50 0.000 .5403139 1.412114
|
d_enc |
TRUE | -.2322011 .1785587 -1.30 0.200 -.5912174 .1268151
------------------------------------------------------------------------------
As you can see, the results of the regression with intercept are identical. But if I omit the intercept (+ 0 in R, , noconstant in Stata), the results differ. In R, the intercept is now captured in dFALSE, which is reasonable from what I understand. I don't understand what Stata is doing here. Also the degrees of freedom differ.
My questions:
Can anyone explain to me how Stata is handling this?
How can I replicate Stata's behavior in R?
I believe bas pointed in the right direction, but I am still unsure why both results differ.
I am not attempting to answer the question, but provdide deeper understanding of what stata is doing (by digging into the source of R's lm() function. In the following lines I replicate what lm() does, but jumping over sanity checks and options such as weights, contrasts, etc...
(I cannot yet fully understand why in the second regression (with NO CONSTANT) the dFALSE coefficient captures the effect of the intercept in the default regression (with constant)
set.seed(20210211)
df <- data.frame(y = runif(50), x = runif(50))
df$d <- df$x > 0.5
lm() With Constant
form_default <- as.formula(y ~ x + d)
mod_frame_def <- model.frame(form_default, df)
mod_matrix_def <- model.matrix(object = attr(mod_frame_def, "terms"), mod_frame_def)
head(mod_matrix_def)
#> (Intercept) x dTRUE
#> 1 1 0.7861162 1
#> 2 1 0.2059603 0
#> 3 1 0.9793946 1
#> 4 1 0.8569093 1
#> 5 1 0.8124811 1
#> 6 1 0.7769280 1
stats:::lm.fit(
y = model.response(mod_frame_def),
x = mod_matrix_def
)$coefficients
#> (Intercept) x dTRUE
#> 0.4375371 -0.1025954 0.1512977
lm() No Constant
form_nocon <- as.formula(y ~ x + d + 0)
mod_frame_nocon <- model.frame(form_nocon, df)
mod_matrix_nocon <- model.matrix(object = attr(mod_frame_nocon, "terms"), mod_frame_nocon)
head(mod_matrix_nocon)
#> x dFALSE dTRUE
#> 1 0.7861162 0 1
#> 2 0.2059603 1 0
#> 3 0.9793946 0 1
#> 4 0.8569093 0 1
#> 5 0.8124811 0 1
#> 6 0.7769280 0 1
stats:::lm.fit(
y = model.response(mod_frame_nocon),
x = mod_matrix_nocon
)$coefficients
#> x dFALSE dTRUE
#> -0.1025954 0.4375371 0.5888348
lm() with as.numeric()
[as indicated in the comments by bas]
form_asnum <- as.formula(y ~ x + as.numeric(d) + 0)
mod_frame_asnum <- model.frame(form_asnum, df)
mod_matrix_asnum <- model.matrix(object = attr(mod_frame_asnum, "terms"), mod_frame_asnum)
head(mod_matrix_asnum)
#> x as.numeric(d)
#> 1 0.7861162 1
#> 2 0.2059603 0
#> 3 0.9793946 1
#> 4 0.8569093 1
#> 5 0.8124811 1
#> 6 0.7769280 1
stats:::lm.fit(
y = model.response(mod_frame_asnum),
x = mod_matrix_asnum
)$coefficients
#> x as.numeric(d)
#> 0.9762140 -0.2322012
Created on 2021-03-18 by the reprex package (v1.0.0)
Suppose I have this small data T
69 59 100 70 35 1
matplot(t(T[1,]), type="l",xaxt="n")
I want find a polynomial which is fit to data. (even over fit is ok)
is there any way that I can do it in R?
First the data.
y <- scan(text = '69 59 100 70 35 1')
x <- seq_along(y)
Now a 2nd degree polynomial fit. This is fit with lm.
fit <- lm(y ~ poly(x, 2))
summary(fit)
#
#Call:
#lm(formula = y ~ poly(x, 2))
#
#Residuals:
# 1 2 3 4 5 6
# 7.0000 -20.6571 17.8286 0.4571 -6.7714 2.1429
#
#Coefficients:
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 55.667 6.848 8.128 0.00389 **
#poly(x, 2)1 -52.829 16.775 -3.149 0.05130 .
#poly(x, 2)2 -46.262 16.775 -2.758 0.07028 .
#---
#Signif. codes:
#0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
#Residual standard error: 16.78 on 3 degrees of freedom
#Multiple R-squared: 0.8538, Adjusted R-squared: 0.7564
#F-statistic: 8.761 on 2 and 3 DF, p-value: 0.05589
Finally, the plot of both the original data and of the fitted values.
newy <- predict(fit, data.frame(x))
plot(y, type = "b")
lines(x, newy, col = "red")
What formula is used to calculate the value of Pr(>|t|) that is output when linear regression is performed by R?
I understand that the value of Pr (> | t |) is a p-value, but I do not understand how the value is calculated.
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
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.
For example in your case in the second row we get that:
tval = 1.8000 / 0.6071 = 2.965
The column you are interested in is the p-value. It is the probability that the absolute value of t-distribution is greater than 2.965. Using the symmetry of the t-distribution this probability is:
2 * pt(abs(tval), rdf, lower.tail = FALSE)
Here rdf denotes the residual degrees of freedom, which in our case is equal to 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:
2 * pt(2.965, 7, lower.tail = FALSE)
[1] 0.02095584