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I am not a mathematician, but I have a basic question about comparing numbers with exponents near $1$. I am evaluating how well a "model" works and I want to make a compelling case. For example, my model suggests

$$\frac{x_1}{x_2} = \sqrt{\frac{L_1}{L_2}}$$

Given $L_1$ and $L_2$, my model predicts the ratio $\frac{x_1}{x_2}$, that I will call $X_{ideal}$. This is what I will hopefully find when I take the quotient of the values I find for $x_1$ and $x_2$. So $X_{ideal}$ is the ideal, expected situation given those two values, $L_1$ and $L_2$.

But I measured $x_1$ and $x_2$, and calculated the actual ratio, which I'll call $X_{actual}$. $X_{actual}$ does not quite equal $X_{ideal}$, but the two numbers are close. I thought perhaps I could convince someone how close they were using exponents. I took the original model and just called $\frac{x_1}{x_2} = X_{ratio}$.

$$X_{ratio}^n = \sqrt{\frac{L_1}{L_2}}$$

Obviously $n$ is $1$ when $X_{ratio} = X_{ideal}$, because the model predicts $\sqrt{\frac{L_1}{L_2}}= X_{ideal}$ when $L_1$ and $L_2$ are known. Then I put in my $X_{actual}$ value though and wanted to see how close its exponent $m$ was to $1$: $$X_{ideal} = \sqrt{\frac{L_1}{L_2}} = X_{actual}^m$$

Turns out, it is really close! $m=0.9937$. Is that convincing? I am not sure why I did that, so is this even a good way to convince someone that my model predicted close to the actual values, or is it just unnecessarily complicated?

daOnlyBG
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britta
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1 Answers1

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In fact, the question you address here is relevant from regression analysis. Basically, your assumed model is $$Y=X^m$$ which is nonlinear with respect to parameter $m$. The question is : is a value of $m$ close to $1$ significant ? There are statistical tests to check for that.

In any manner, remember that $100^{0.99}=95.5$, $1000^{0.99}=933.3$, $10000^{0.99}=9120.1$ and that shows significant deviations compared to the values obtained for $m=1$.

Among the classical tests, you have the standard deviation on parameters.

In any manner, for illustration, report on the same graph the experimental data as well as the line $Y=X$ and the curve $Y=X^m$. This could easily show the validity of your model.

For illustration purposes, I generated data using $X=100, 200,\cdots, 1000$, $m=0.99$ and the exact $Y$ values where changed using a random relative error between $-5$% and $+5$%. The nonlinear regression leads to $m=0.992815$ (estimated parameter error is $0.001321$) and the corresponding confidence interval is $\left( \begin{array}{cc} 0.989826 & 0.995804 \end{array} \right)$ which reveals that the model is very significant.

Claude Leibovici
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