I'm trying to write some code to do a regression on data weight (x) and time (y). As best as I can tell, the model should be y = b1 + b2ln(x), but I don't know how you can do this by hand (I know how to in R...). I know how to do a simple linear regression by hand. Appreciate it.
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$\begingroup$ calculate logs: x1 = ln(x). Regress y on x1. Done. BTW your model is wrong unless your errors are all 0 (in which case you can just compute it from any two distinct points). That is, you either need to write $E(y)=$ or you need to add $+e$ on the end $\endgroup$– Glen_bCommented Dec 2, 2013 at 3:16
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$\begingroup$ can it be done using the least squares approach? $\endgroup$– CollinCommented Dec 2, 2013 at 3:29
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$\begingroup$ That's what I'm saying up there, yes. How are you doing it in R if not by least squares? $\endgroup$– Glen_bCommented Dec 2, 2013 at 3:47
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$\begingroup$ I guess I'm not exactly sure lol. So normally you would calculate SXX = sum(x-xbar)^2; SXY = sum((x-xbar)(y-ybar)); SYY = sum(y-ybar)^2. Then B1 = SXY/SXX. How would this differ given the log term? $\endgroup$– CollinCommented Dec 2, 2013 at 3:58
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$\begingroup$ The entire thing is in my first comment. Try reading it again, please. It has two steps. Tell me what you don't understand about step 1 (Calculate the logs, calling the result $x_1$). If you understand what that means, we can move to step 2. $\endgroup$– Glen_bCommented Dec 2, 2013 at 4:20
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1 Answer
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You say that you'd calculate the slope as follows:
So normally you would calculate
$$S_{XX} = \sum_i (x_i-\bar x)^2\\ S_{XY} = \sum_i ((x_i-\bar x)(y_i-\bar y))\\ S_{YY} = \sum(y_i-\bar y)^2$$
Then $b_2 = S_{XY}/S_{XX}$.
So imagine you have a set of x-values and y-values:
y x
1 2.3 0.36772
2 5.3 1.64873
3 6.5 7.38910
Step 1:
calculate a new $x$, $x_1 = \ln(x)$
y x1
1 2.3 -1.0
2 5.3 0.5
3 6.5 2.0
Now regress $y$ on this new $x_1$ as usual
SXX = 4.50; SXY= 6.30; SYY = 9.36
b2 = SXY/SXX = 1.4
b1 = mean(y) - b2 . mean(x) = 4.7 - 1.4 . 0.5 = 4
