# Intercept parameter, $\beta_0$

I have a hypothetical data given below that consists of 11 pairs of points (xi, yi ), to which the simple linear regression mean function $\mathbb E(y|x) = β_0 + β_1x$ is fit.:

 X     Y
10    8.04
8    6.95
13    7.58
9    8.81
11    8.33
14    9.96
6    7.24
4    4.26
12    10.84
7    4.82
5    5.68


I have got intercept parameter,$\beta_0=3.001$

But the plot of the data is not showing the y-intercept is $3.001$. Rather the y-intercept is more than $3.001$. WHY?

I have used R software to calculate the parameters, $\beta_0$,$\beta_1$ and also to produce the plot.

 x1 <- c(10,8,13,9,11,14,6,4,12,7,5)
y1 <- c(8.04,6.95,7.58,8.81,8.33,9.96,7.24,4.26,10.84,4.82,5.68)

lm(y1~x1)

plot(y1~x1)
abline(lm(y1~x1))


EDIT

  ht <- c(169.6,166.8,157.1,181.1,158.4,165.6,166.7,156.5,168.1,165.3)
wt <- c(71.2,58.2,56.0,64.5,53.0,52.4,56.8,49.2,55.6,77.8)

lm(wt~ht)

windows(9,6)
par(mfrow=c(1,2))

plot(wt~ht)
abline(lm(wt~ht))

plot(wt~ht,xlim=c(0,180),ylim=c(0,75))
abline(lm(wt~ht))


How can i get the y-intercept? By expanding the straight line(population regression line) to negative axis of Y ?

• why do you say so? we do not see the $\{x=0\}$ part of the graph. – mookid Mar 13 '14 at 2:57
• Exactly, if you extend the line for $x1<4$, which is not shown in the graph, it will probably be $y1=3.001$ for $x1=0$. – Nameless Mar 13 '14 at 11:46
• I will migrate to Cross Validated as suggested. As you have already registered an account there, you can edit in the suitable tags yourself on the website. – Willie Wong Mar 13 '14 at 11:55
• This question seems to be predicated on a different understanding of "intercept": it is not where the graph meets the left hand of its box, but rather its value at $0$, whether or not $0$ appears within the extent of the plot. – whuber Mar 13 '14 at 13:39
• In your plot, x=0 is not included.. try to use a larger x range. – TYZ Mar 13 '14 at 13:41

The intercept is in the right place. The problem is that plot() doesn't usually show the origin unless there are data points there. Note how the plot(y1~x1) starts at roughly $(4,4)$, not $(0,0)$. This isn't an error, just a result of us confusing plot borders with plane axes. Anyway, in order to force R to show the intercept where we'd expect it to be, you have to use the parameters xlim and ylim of the plot function. Try running this:

plot(y1 ~ x1, xlim = c(0, 14), ylim = c(0, 11))
abline(lm(y1 ~ x1))


And it give you what you want, i.e.:

Try

plot(y1~x1, xlim=c(0,14))

• I get the true picture while using the command plot(y1~x1,xlim=c(0,14),ylim=c(0,14)). But how does the y-intercept change for changing the origin $(x,y)$$=(0,0)$ – user 31466 Mar 13 '14 at 5:38
• How can i get the appropriate picture of y-intercept parameter ,$\beta_0$ , if the data are ht <- c(169.6,166.8,157.1,181.1,158.4,165.6,166.7,156.5,168.1,165.3) wt <- c(71.2,58.2,56.0,64.5,53.0,52.4,56.8,49.2,55.6,77.8) and lm(wt~ht) ? I have edited the graph of this data. – user 31466 Mar 13 '14 at 6:12

You can try this

x1 <- c(169.6,166.8,157.1,181.1,158.4,165.6,166.7,156.5,168.1,165.3)
y1 <- c(71.2,58.2,56.0,64.5,53.0,52.4,56.8,49.2,55.6,77.8)

lm1 <- lm(y1 ~ x1)\$coef
plot(y1 ~ x1, xlim = c(0, max(x1)), ylim = c(min(0,lm1[1]), max(y1)))
abline(lm1)


Of course it's practically meaningless to look at the intercept in the first place (since y can't be negative), but if you want to get this point across...