From the data the growth of tumor spheroids determine doubling time using Mathematica and by hand

57 Views Asked by At

Task: From the data the growth of tumor spheroids determine doubling time Link to the data.

My idea was to use least squares method and Mathematica to helps me find result. But I think that this is nonlinear problem so I will have to use nonlinear least squares method but I don't know how to start. Also I have a question how does my result change if I add some data or throw away.

1

There are 1 best solutions below

0
On

I put (approximately) your data into Minitab statistical software:

T:    4.6      5.7      6.7      7.9      8.8      9.8     11.0     12.0     12.9     15.2
V: 0.0016   0.0032   0.0056   0.0119   0.0202   0.0275   0.0345   0.0661   0.0783   0.1550

@Did's hint to take the log of Volume works fine. (It doesn't matter whether natural logs or base 10.) Here is a plot of the data with Volume on a log scale (original numbers shown, scale distorted appropriately). The header shows the regression equation. The dotted band shows 95% prediction intervals for V at each T. Roughly, R-sq = 97.2% means that 97% of the variability in $\log_{10}V$ can be explained by regression on $T.$

enter image description here

However, notice that the residuals towards the center tend to be positive, while those at the extremes of T tend to be negative. That suggests that a slight improvement might be made by using $T^2$ as a second predictor variable. Results below. By eye, this looks a little better, and R-Sq(adj) is a little larger. You could use one of the usual criteria for a formal test whether the quadratic regression is a statistically significant improvement (for your exact data). If your only purpose is to find doubling time, then the quadratic regression makes that a little more complicated.

enter image description here

Notes: (1) Based on what I see here, I can't imagine why you would want to 'throw away' any of your data, except possibly the observation at $T = 11.$ Throwing away points might result in a loss of precision. (2) A quadratic fit to $V$ (instead of $\log V$) is pretty good.