I have been trying to linearise this set of data for a while now. It seems to fit a curve of $y=-x^2$ but I am not sure how to transform it. I have tried $x^2$, $y^2$, $1/x^2$, $\log x$ etc but have had no success. Here is an image of the data set plotted on a graph:
Any help would be much appreciated, thanks!
Using the data you provided here, I assumed the model to be something like $$y=a+b\, e^{cx}$$ which is highly nonlinear withe respect to the parameters as long as $c$ needs to be adjusted. If this is not the case, fix $c$ ata guessed value, define $z_i=e^{cx_i}$ and you just face a linear regression problem. So, for each value of $c$, you have the sum of suares of residuals and, plotting the results, you could identify around which value you get a minimum. This can be easily done using Excel or any linear regression tool.
Since the data do not vary much, we can expect $c$ to be small and run cases. The results, given in the next table, $$\left( \begin{array}{cc} c & SSQ \\ 0.01 & 39.642 \\ 0.02 & 25.243 \\ 0.03 & 14.306 \\ 0.04 & 7.036 \\ 0.05 & 3.033 \\ 0.06 & 1.632 \\ 0.07 & 2.152 \\ 0.08 & 4.015 \\ 0.09 & 6.770 \end{array} \right)$$ reaveal a minimum close to $c=0.06$; for this value, we have, from the linear regression $a=22.44$ and $b=-0.048$.
So, we have all the required guesses for the parameters and we can start confidently a nonlinear regression. As a result, what was obtained is $$y=22.3916 -0.0411834\, e^{0.0617983 x}$$ Concerning the parameters, $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 22.3916 & 0.128716 & \{22.1172,22.6659\} \\ b & -0.0411834 & 0.0131941 & \{-0.0693059,-0.0130608\} \\ c & 0.0617983 & 0.00362314 & \{0.0540758,0.0695208\} \\ \end{array}$$ tehy are quite significant and we also obtain $R^2=0.9998$ which is very good.
The last table reports your original data as well as the values computed from the correlation. $$\left( \begin{array}{ccc} x & y & y_{calc} \\ 0 & 22.2213 & 22.3504 \\ 5 & 22.1259 & 22.3355 \\ 10 & 22.1829 & 22.3152 \\ 15 & 22.2114 & 22.2875 \\ 20 & 22.439 & 22.2499 \\ 25 & 22.1171 & 22.1985 \\ 30 & 22.1175 & 22.1286 \\ 35 & 22.0893 & 22.0334 \\ 40 & 21.8352 & 21.9038 \\ 45 & 21.8176 & 21.7271 \\ 50 & 21.6959 & 21.4866 \\ 55 & 21.1582 & 21.1589 \\ 60 & 20.8114 & 20.7126 \\ 65 & 20.2859 & 20.1048 \\ 70 & 19.7753 & 19.2768 \\ 75 & 18.0794 & 18.1491 \\ 80 & 16.3294 & 16.6131 \\ 85 & 13.6814 & 14.5210 \\ 90 & 12.2500 & 11.6714 \end{array} \right)$$ It is possible that other models give as good or even better results.