Could you please check if what I've done is correct? and how could I improve some of them? Thank you in advance.

Suppose I have the following data (the original data its like 20 data with decimal numbers):

$$ \begin{array}{c|lcr} y &2&4&3&70&9 \\ \hline x & 0.24&21&33&11&10 \end{array} $$

  1. Draw a dispersion diagram of the data.

  2. Adjust a simple linear regression model.

  3. Find the SSE,SCModel, SCTotal.

  4. Find the model's errors.


  1. Using the instruction plot(x,y,xlab="These are x's",ylab="These are y's").

  2. I think we want to find $\beta_0$ and $\beta_1$ in our segment line, that is we need to calculate $\hat \beta_0=\frac{(\sum y_i)(\sum x_i^2)-(\sum x_iy_i)(\sum x_i)}{n\sum x_i^2-(\sum x_i)^2}$ and $\hat \beta_1=\frac{n(\sum x_iy_i)-(\sum x_i)(\sum y_i)}{n\sum x_i^2-(\sum x_i)^2}$ with $n=5$ and substitute them in the model $y=\beta_0+\beta_1x$

Am I correct?

  1. I think this is calculated with $SSE=y'(I-M)y,\ where \ y=(2,4,3,70,9)',M=X(X'X)^{-1}X'$ and X=$\left( \begin{matrix} 1 & .24 \\ 1 & 21 \\ 1 & 33 \\ 1 & 11 \\ 1 & 10 \end{matrix}\right) $

$SCTotal=SSE+SSModel$ and $SCTotal=y'y$ so $SSModel$ can be calculated with $SCTotal-SSE$

Am I correct?

  1. Here am I asked to find $error=y-\hat y$ right?

How can I do that?


1 Answer 1

  1. It is called scatter plot generally.
  2. I would replace "Adjust" by "Fit"

Get the $\hat\beta_1$ first, then $\hat\beta_0 = \bar y - \hat\beta_1 \bar x$. Your $\hat\beta_0$ is complicated and I did not verify it.

  1. Suppose you mixed up SC and SS. All of them should be SS. SSTotal = $(y-\bar y)'(y-\bar y)$

  2. In the simple linear regression model $$ y=\beta_0 + \beta_1 x + \epsilon $$ $\epsilon$ is called error or error term. Its estimate is $y-\hat y$ and is called residual.

The question "Find the model's error" maybe means to find something that violates the assumptions of simple linear regression. For example the linearity between $y$ and $x$.

  • $\begingroup$ Hi, I was observing and I wonder why $\hat\beta_0 = \bar y - \hat\beta_1 \bar x$ ? Why the equality holds? Why did you take $\overline y$ and $\overline x$? $\endgroup$ Commented Oct 30, 2018 at 1:13
  • $\begingroup$ By least square method, you get that $\hat \beta_0$. See en.wikipedia.org/wiki/Simple_linear_regression. Which equality holds? Taking $bar y$ and $\bar x$ are easy for calculation? $\endgroup$
    – user158565
    Commented Oct 30, 2018 at 1:20
  • $\begingroup$ Ah I see, I did not know the alternative way to calculate $\hat\beta_0: (\hat\beta_0 = \bar y - \hat\beta_1 \bar x)$. I was referring to that same equality $(\hat\beta_0 = \bar y - \hat\beta_1 \bar x)$. I think yes, $\overline x,\overline y$ are easy to calculate using R or other software. $\endgroup$ Commented Oct 30, 2018 at 1:28
  • $\begingroup$ Any other questions? My answers are understandable? $\endgroup$
    – user158565
    Commented Oct 30, 2018 at 1:33
  • 1
    $\begingroup$ supposing $\hat β_0=5,\hat β_1=10$, then $ϵ_1=2−5−10*0.24 = -5.4$, $ϵ_2=4−5−10*21 = -211$, ... $\endgroup$
    – user158565
    Commented Oct 30, 2018 at 2:43

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