Programming the statistical procedures from SAS

Linear Regression Log(y)vsLog(x) - equation parameter

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Linear Regression Log(y)vsLog(x) - equation parameter

Good afternoon, 

 

A rather dummy question, I transformed both y and x var to do a linear regression of Log(y) vs Log(x) - i got the SAS output on SAS studio v3.4 and want to know what to do with the Intercept value and slope value in order to build the true equation. How should I calculated the intercept and slop exactly please? And is SAS LOG(var) default to base 10, e or LN?  I wish to know, from this SAS output, what is the equation to be use to predict data from non transformed data... THANKS TO ALL !!! 

 

Parameter Estimates Variable DF Parameter
Estimate Standard
Error t Value Pr > |t| Standardized
Estimate Intercept 1 log_TT_bps_ 1

0.165250.500530.330.74190
1.048080.139557.51<.00010.56709

Screen Shot 2016-03-22 at 3.42.22 PM.jpg

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‎03-24-2016 12:12 AM
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Posts: 18,549

Re: Linear Regression Log(y)vsLog(x) - equation parameter

I would suggest you try and we'll be happy to help. @lvm has laid out the calculation - replace the a with your intercept value and b with your slope. X is your independent variable.  

The following is the formula in SAS (**=exponent)

 

Y = a*(x**B);

 

I would also recommend doing it via a data step so you can trace it out if required. 

 

 

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Super User
Posts: 10,857

Re: Linear Regression Log(y)vsLog(x) - equation parameter

The SAS function LOG is natural log (base e). Use Log10 for base 10 or Log2 for base 2.

 

If you share the code you used to generate the parameters there might be some better options than creating the model equation by hand.

Valued Guide
Valued Guide
Posts: 684

Re: Linear Regression Log(y)vsLog(x) - equation parameter

Using your intercept (a) and slope (b), you have log(y) = a + b*log(x).

So, use:

y = [exp(a)]*(x**b) = A*(x^b),

where A=exp(a).

Contributor
Posts: 20

Re: Linear Regression Log(y)vsLog(x) - equation parameter

Thanks to both ... I changed everything into Log10(y) vs Log10(x) - May lvm explain to me a bit more how to calculate the value then perhaps on a hand writing picture as its confusing with too many characters ;-))
Contributor
Posts: 20

Re: Linear Regression Log(y)vsLog(x) - equation parameter

from the sas Regression Log10(y) vs Log10(x)
My new Intercept is 0.00036 (sas output)
My new slope is 1.0075 (sas output)
...Im dreaming of an example calculation on these value PLEASE...will never thank enough ;-0)
Solution
‎03-24-2016 12:12 AM
Super User
Posts: 18,549

Re: Linear Regression Log(y)vsLog(x) - equation parameter

I would suggest you try and we'll be happy to help. @lvm has laid out the calculation - replace the a with your intercept value and b with your slope. X is your independent variable.  

The following is the formula in SAS (**=exponent)

 

Y = a*(x**B);

 

I would also recommend doing it via a data step so you can trace it out if required. 

 

 

Valued Guide
Valued Guide
Posts: 684

Re: Linear Regression Log(y)vsLog(x) - equation parameter

Here is a clarification to Reeza's post. If a is the intercept from the log(y):log(x) regression, then y is given by

y = (10**a)*(x**b)

if one is using base 10. If one is using base e (natural log), then it is:

y = (e**a)*(x**b)

 

Contributor
Posts: 20

Re: Linear Regression Log(y)vsLog(x) - equation parameter

Thank you kindly
Super User
Posts: 18,549

Re: Linear Regression Log(y)vsLog(x) - equation parameter

@jackice @lvm has the correct answer, not me...if you can change it. 

Contributor
Posts: 20

Re: Linear Regression Log(y)vsLog(x) - equation parameter

Don’t worry - Im the one reaching out - case closed Tks
☑ This topic is solved.

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