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We know, In statistics, simple linear regression is the least squares estimator of a linear regression model with a single explanatory variable. In other words, simple linear regression fits a straight line through the set of n points in such a way that makes the sum of squared residuals of the model (that is, vertical distances between the points of the data set and the fitted line) as small as possible.

suppose in simple linear regression model

$y= \alpha + \beta x + \epsilon $

, from a random instance we gather this information:

$ \bar{x}=\bar{y}, \Sigma x_iy_i = \Sigma x_{i}^{2} $

I want to calculate estimation $(\alpha, \beta)$ with least square method. how I can solve it?

Infact I'm a Biology Background, and for experiment need this value. thanks.

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closed as off-topic by Glen_b, gung, Nick Cox, kjetil b halvorsen, Scortchi Mar 16 '15 at 15:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Self-study questions (including textbook exercises, old exam papers, and homework) that seek to understand the concepts are welcome, but those that demand a solution need to indicate clearly at what step help or advice are needed. For help writing a good self-study question, please visit the meta pages." – Glen_b, gung, Nick Cox, kjetil b halvorsen, Scortchi
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ There is no reason in general why $\sum x_i y_i = \sum x_i^2$: in fact that minimally requires the same units of measurement. You mean something else, I imagine, but it is not clear what. $\endgroup$ – Nick Cox Mar 16 '15 at 14:41
  • $\begingroup$ @NickCox, we gather this from our information. $\endgroup$ – Mina Akram Mar 16 '15 at 14:42
  • $\begingroup$ Your question sounds increasingly like a self-study question and should be flagged as such. $\endgroup$ – Nick Cox Mar 16 '15 at 14:46
  • $\begingroup$ @NickCox, what do you mean by "self-study"? $\endgroup$ – Mina Akram Mar 16 '15 at 14:47
  • $\begingroup$ stats.stackexchange.com/help/on-topic Note that @statchrist made the same comment in their answer. $\endgroup$ – Nick Cox Mar 16 '15 at 14:49
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This question appears to be a self-study question.

One way to solve the problem ist to look at the mathematical expression for the estimator in matrix notation:

$\hat{\beta}_{OLS}=(X'X)^{-1}X'y$.

Insert the matrices for your model in matrix notation: $y = X\beta_{OLS} + \epsilon$.

For instance with $\beta_{OLS} = \begin{pmatrix} a \\ \beta \end{pmatrix}$ and $X=\begin{pmatrix} 1 & X_1 \\ \vdots & \vdots \\ 1 & X_n \end{pmatrix}$. Next step would be to manipulate this equation in order to get expressions in terms of the mean and insert the additional information you have.

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  • $\begingroup$ i couldent get it, you means final value is (0,1) ?, is it possible learn me ? $\endgroup$ – Mina Akram Mar 16 '15 at 13:58
  • $\begingroup$ oooh, i got it thanks. but last step is how we can get numerical value for our experiment ? $\endgroup$ – Mina Akram Mar 16 '15 at 14:11
  • $\begingroup$ it dosent satisfy me, is it possible to make it clear?‌it's (0,1) ? $\endgroup$ – Mina Akram Mar 16 '15 at 14:26
  • $\begingroup$ Yes, have you tried and computed the inverse of X'X/N? $\endgroup$ – statchrist Mar 16 '15 at 14:36
  • $\begingroup$ Is it possible to add it to your answer, no, this is difficult for me. if i didnt waste your time, please add... $\endgroup$ – Mina Akram Mar 16 '15 at 14:38

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