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A linear model:

$y_{i}=\alpha +\beta x_{i}+u_{i}$

was estimated using least squares and the following estimate was calculated:

$\widehat{\beta }=\frac{\sum (x_{i}-\overline{x})y_{i}}{\sum (x_{i}-\overline{x})x_{i}}$

Now we want to change the units of the independent variable from Celsius to Fahrenheit:

$z_{i}=32+1.8x_{i}$

The OLS is now:

$\widehat{\gamma }=\frac{\sum (z_{i}-\overline{z})y_{i}}{\sum (z_{i}-\overline{z})z_{i}}$

I need to show that:

$\widehat{\gamma }=\frac{5}{9}\widehat{\beta }$

And I can't do it, trying for a couple of hours. Can you please assist ? Once I prove this, I already know how to prove the equation for the new intercept.

Thank you !

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Insert $z_i$ into $\hat \gamma$. For the numerator you obtain $\sum(1.8(x_i-\bar x_i))$, as the 32 in each $z_i$ cancels with that in $\bar z$. The same happens in the denominator, leaving you with $\sum(1.8(x_i-\bar x_i)) (32+1.8x_i)$ But as $\sum(x_i-\bar x_i) $=0, this equals $\sum(1.8(x_i-\bar x_i)) 1.8x_i$. Thus $\hat \gamma=\frac{1}{1.8}\hat\beta$.

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  • $\begingroup$ How come you use the fact that the sum of xi-xbar is 0 if this term is multiplied by another term in the sum ? $\endgroup$ – user3275222 Mar 28 '19 at 23:23
  • $\begingroup$ I use this only with the 'other term' being 32 (which can be taken out of the sum). The term involving the $1.8x_i$ is kept as it is and leads to the 1/1.8. $\endgroup$ – user1587692 Mar 28 '19 at 23:34

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