Find a line of regression with the following points

Question

Find a line of regression in the form \begin{align} y &= \frac{c}{1+a e^{-b(x)}} \\[4mm] \end{align}

with the points $(69.99, 44), (84.71, 54), (93.66, 60), (116.36, 74)$

I can easily find the answer using a calculator but I need to know if it's possible by hand, and if so, what the steps are. Thanks.

Answer 1

$$y(x)=\frac{c}{a e^{-b(x)}}$$ $$\begin{cases} \ln(44) \simeq \ln(\frac{c}{a})+69.99\:b \\ \ln(60) \simeq \ln(\frac{c}{a})+93.66\:b\\ \ln(74) \simeq \ln(\frac{c}{a})+116.36\:b \end{cases} \qquad\text{Let}\quad \frac{c}{a}=e^A$$ Without specification about the expected kind of regression and about criteria of fiting, an example of solution (among many):

For example. Linear regression in order to compute $A$ and $b$ : $$y=A+b\,x\qquad \text{Three points}\quad\begin{cases} y_1=\ln(44)\quad,\quad x_1=69.99 \\ y_2=\ln(60)\quad,\quad x_2=93.66\\ y_3=\ln(74)\quad,\quad x_3=116.36 \end{cases}\quad\;\quad$$

IN ADDITION after you modify the question :

There is one point more. Thus an equation more $y_4\simeq A+b\,x_4$ : $$y_4=\ln(54)\quad,\quad x_4=84.71$$ This changes nothing in the example of linear regression. Of course the numerical results for $A$ and $b$ will be slightly different.

Answer 2

The equation in the quest was changed, consequently my answer becomes almost obsolete. The hint to LS remains valid.

Answer 3

The question has been changed so that the answer becomes very different. A distinc answer is given below in order to avoid confusion with the previous answer.

Your first equation was : $\quad y = \frac{c}{a e^{-b(x)}}\quad$. Your new equation is : $$y = \frac{c}{1+a e^{-b(x)}} $$ For the first equation a simple linear regression $Y=A+b\,x$ was sufficient in considering $Y=\ln(y)$.

For the second equation the problem is very different because one cannot linearise with $Y=\ln(y)$ as before. The regression is non longer linear but is non-linear.

The usual way for non-linear regression is to use iterative method of calculus starting from "guessed" values of the parameters. Many softwares are available to do it.

When no particular criteria of fitting is required (This is the case in the present wording of the question) a more direct method (not iterative) can be proposed. Eventually the numerical calculus could be done by hand but this would be rather boring because the matix inversion.

METHOD DESCRIPTION :

For explanations about the general principle see https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales

NUMERICAL CALCULUS :

RESULT :

If a particular criteria of fitting is required nonlinear regression is necessary. Use any software for nonlinear regression. If difficulies appear with the iteratve process take the values found above as good initial values.

For comparison : With a required criteria of fitting LMSE (for example) the result from a nonlinear regression software is : $a=8.55143534$ ; $b=0.0263881773$ ; $c=103.35740847$ . With another criteria of fitting the result would be slightly different. All these small differences are indistinguishable on the drawing of the curve.

NOTE : The logistic curve is badly defined with only four points. As a consequence the result can be more or less different depending on the method of fitting and/or the method of calculus.