# How to create a model in which one predictor variable creates an x% increase in response?

I have two different predictors "A" and "B" that can be used (not at the same time) to affect a response variable, "C".

Previous data has shown a very strong linear relationship between A and C, and B and C. When SLR models were fit to these relationships, both had R-squared values above 95%. Scatterplots confirm the highly linear relationships.

A claim has been made that when predictor "B" is applied, there is a 5% increase in the response "C", compared to when "A" is applied.

The two linear models fit to (A~C) and (B~C) have different slopes and intercepts. So for lower levels of A (or B), one predictor is associated with higher "C" measurements, but this is slightly reversed at higher levels.

Does anyone have an idea on how to test the 5% claim mentioned above? I was thinking it might be appropriate to create a model like this one:

$C = (U + V*X)*(1 + Y*B)$, where $U, V$ and $Y$ are constants.

X is the predictor level (either A or B). "B" in this model is an indicator variable such that B = 1 if B is being used, or B = 0 if A is used. If the 5% claim were true, we would expect Y to be .05.

The problem is that this model is not linear in the parameters, and as far as I know cannot be calculated using the common least-squares matrix operations.

Can anyone think of how to validate this model, or offer a better alternative to test the desired claim?

Thanks a lot for your help.

• Are your predictors of the type Applied/Not applied, or can they be applied with different intensity level. – AlainD Aug 22 '18 at 8:10
• I am sorry but I don't understand your question. – Greg Aug 22 '18 at 14:08

Adding 5% is tantamount to multiply by 1.05 [$x+5\%x = x + 0.05\cdot x = 1.05 \cdot x$].
So to test if there is a % increase on $C$, you should test the model $C \approx b^U \cdot c$, where $U = 1$ when $B$ is applied and $U = 0$ when $A$ is applied. You can linearize by taking the log on both sides $\log(C) \approx log(b) \cdot U + log(c)$.
• The linear regression of the logs will give you $\log(b)$ and its p-sign, so that you will know of $B$ is significant or not.
• The measure the percent increase is given by $p = (\exp(log(b))-1) \times 100$.
• To test if it is equal to 5%, compute its Student by dividing the deviation by its Std-error (given by the regression) $t= \frac{ \log(b) - \log(1.05)} {std_{log(b)}}$.