See the Excel function LOGEST, e.g.

https://www.automateexcel.com/functions/logest-formula-excel/

You must enter LOGEST as matrix formula, including the results for the model statistics. This will give you the F-value and the degrees of freedom. From this you can calculate the p-value using the F.DIST.RT function. The first degrees of freedon is 1, the second degrees of freedom is given in the output of LOGEST.

So it is possible in Excel. But you can use R as well. For this, organize your data in an Excel sheet as "variables in columns", that is, one column contains the x-values and the other column contains the corresponding y-values. Use column names in the first row (e.g. "x" and "y" or whatever the names of your variables are). Use short names without spaces and special characters. Select the two columns and copy them into the clipboard. The go to R:

read the data into an R object with: df

This inexpensive and simple program will give you a good estimation of the reliability of your parameters: www.lerenisplezant.be/fitting.htm.

If you can send me your data I can show it for you.

If the model you want can be made linear, the solution proposed by Jochen Wilhelm, with R, is pretty straightforward. I might suggest performing this procedure with something like Jamovi, which is R-based, GUI, free software. It might be a little more friendly if you are not used to using R. (Importing data as .csv is easy; transforming a variable took me a bit, but it's not difficult once you get it; fitting the linear model and producing p-value, r-squared, other fit statistics, diagnostic plots, all real easy).

R can also fit nonlinear or curvilinear functions of many types with the nls() function. If you aren't used to using R, this can be a little tricky. The advantage is that you can fit a model in it's original form (without transforming the variables or parameters), and you can fit all kinds of models. The disadvantages are that you have to define the function you want to fit the data to; you may need to come up with initial estimates of the parameters; there are additional steps to get the p-value, some kind of pseudo r-squared value, or confidence intervals for the parameters.

Sal Mangiafico Transformations are not a good idea: Method Non-linear regression -Why you shouldn't take the logarithms...

Koen Van de Moortel , thanks for the reference. ... I agree. I think it's better to fit the model in it's original form. It's also undesirable that there are limited models available when using e.g. Excel. It's better to be able to have the flexibility to fit models suggested by theory. For example, I've used linear-plateau and quadratic plateau models in my work. An Excel user might be tempted to fit these with an exponential or polynomial model, but these would might make less sense theoretically. There's an example that I've use in classes of a plot of life expectancy vs. health care spending that the Organisation for Economic Co-operation and Development produced. (I.e. compare the follow two: https://rcompanion.org/handbook/images/image040.png ; https://rcompanion.org/handbook/images/image041.png ).

Joseph Whittaker , P.S., I'll make the offer that if you are interested in using the nls() function in R to fit your data to an e.g. exponential model without using the log transformations, I'll help with the code. You'd probably have to be willing to share the data with me privately, tho.

Sal Mangiafico Whatever function you use, it has to have a horizontal asymptote! Could you give us the raw data file? I would like to put it in my software.

I wish to thank everyone for their helpful answers.

It will take me some time to work through them, as I am a novice in statistics, but will report back on how things go.

Follow-up note 1: For those interested, my discussion on the plot from the Organisation for Economic Co-operation and Development, can be found here: https://rcompanion.org/handbook/C_04.html . Search for the word organisation.

Follow-up note 2: I was working with Joseph Whittaker 's data, fitting an exponential model, either by linearizing the function as I believe Excel does, or using the nls() function in R to fit the model by iterative least squares without linearizing the function. A) If you look very carefully at the results, you could see that the fitted curves were slightly different for the different approaches, but practically there was little difference. R-squared values were very close. (Although they increases a smidgen if I used a model with an additional parameter). B) Interestingly, I couldn't replicate the Excel results precisely in R. I would suspect that because the solution should be a closed mathematical one (OLS), I should get precisely the same results. My only guess is that there are some small internal rounding differences with the software, and that these make a noticeable difference when you have a function containing e(c * x) .

Update: Sal Mangiafico was able to help me do the significance testing and fit slightly superior models with R, as described in his post above.

What dou you think about these models (for example Japan), both with horizontal asymptote:

Model 1: Logistic MDLS

limit: 84.51 +- 0.91 years

0$ --> 66.68 years

Model 2: Plateau minus exponential MDLS

limit: 84.28 +- 0.40 years

0$ --> 64.54 years

Data from: https://ourworldindata.org/the-link-between-life-expectancy-and-health-spending-us-focus

Remark with my previous comment: I didn't know the precisions of the data, so I assumed them to be the same.

Joseph Whittaker , Sal Mangiafico One more remark: the logistic model seems to fit more stable (see the dotted lines).

Koen Van de Moortel , those models look good, and it does make sense for this kind of data to have a horizontal asymptote. For this specific case, there's a question as to how much faith we put in the value of the asymptote. I guess it's about 84 or 85 years of age for your models. Do we suspect that life expectancy continues to climb to this level with increasing health expenditures ? I don't know. The model suggests that there a (diminishing) marginal increase in life expectancy with more expenditure, even at relatively high rates of expenditure. On the other hand, the way I looked at the OECD data across nations, it looks like there's little benefit to increasing expenditures above about $3500, and that the plateau is about 82 years. (Of course, Japan may be different than the range of nations in my OECD plot. And the dollar units on these plots may not be quite the same.) ... My point here is simply that the model we choose sometimes has implications for what conclusions we draw from the data.

As part of general discussion ---- Sometimes we are able to start with a theoretical model. Like the Mitscherlich–Bray equation is sometimes used to model crop yield in response to fertilization. Or Michaelis–Menten equation for kinetics in chemistry. Interestingly, with the Mitscherlich–Bray model, we often look at a value of say 85% of asymptotic yield, since it's not practical to try to achieve 100% of theoretical maximum crop yield. It might be politically fraught, but a similar approach may make sense for the Japan models of life expectancy above. If there is a small increase in life expectancy above, say, expenditures of $3500 or $4000, is this worth the expense ?

In other cases, there's no obvious theoretical model to consider. In the case of Joseph Whittaker 's data, the goal was to simply fit a model that shows a decreasing trend, has some curvature, and allows reporting a p-value and r-squared value. Probably no one would try to draw traditional conclusions from the chosen model: it's just a curved line.

Sal Mangiafico Of course, the choice of the model determines that limit value! That's why we have to choose it wisely. And that's why any linear or polynomic model is completely nonsensical because you can not extrapolate them. I'm doing some more comparisons now.... I'll keep you informed.

Sal Mangiafico In https://rcompanion.org/handbook/C_04.html you use a "quadratic plateau" and "linear plateau" model but I don't see any formula of those... Do you have them?

Koen Van de Moortel ,

Yes, those models are used relatively commonly in agriculture.

One advantage of them is that they predict a "critical x" value above which you don't expect y to increase. (As opposed to an exponential curve with a asymptote.)

Critical x values can vary depending on whether you fit a linear plateau, quadratic plateau, or, say, a Cate-Nelson model.

LINEAR PLATEAU

linplat = function(x, a, b, clx)           {ifelse(x < clx, a + b * x,                            a + b * clx)}

QUADRATIC PLATEAU

quadplat = function(x, a, b, clx) {            ifelse(x  < clx, a + b * x   + (-0.5*b/clx) * x   * x,                             a + b * clx + (-0.5*b/clx) * clx * clx)}

Thanks. Oops, I don't like those models with a discontinuous derivative, since they are not natural. Why not something like y=a - b/(x-c)? This works also fine with the life expectancy data, although the most logical and the best fitting is the logistic function (starting from 0).

Koen Van de Moortel ,

The quadratic plateau has a continuous derivative, since it forces the start of the plateau line at the point where the quadratic line reaches its maximum. (Both have a slope of 0).

Further discussion on linear plateau, quadratic plateau, Cate-Nelson, and Mitscherlich–Bray (exponential) models for interested readers:

In my experience, the linear plateau models and quadratic plateau models are useful because they predict explicitly an x-value above which the y-values are essentially plateaued.

Consider the following data, which plots the relative cotton yield vs. soil potassium concentration. We want to know from these data what a sufficient concentration of soil potassium is to reasonably maximize yield.

The linear plateau model really captures the idea that yield doesn't increase meaningfully above a soil test of 75 mg/kg (plateau=95%)

https://gradcylinder.org/post/how-to-linear-plateau-R_files/figure-html/lp-4-1.png

The same data could be fit with a quadratic plateau model. Again, this emphasizes the plateau and identifies a critical x value (here, 86 mg/kg) (plateau=95%).

https://gradcylinder.org/post/how-to-quad-plateau-R_files/figure-html/qp-4-1.png

Having these kinds of results is simply practically useful in these applications.

For those who are interested, a Cate-Nelson analysis, where critical y is set at 95%, a critical x of 75 mg/kg predicts 79% of data points correctly. But a critical x as low as 49 mg/kg works just as well.

If I fit a Mitscherlich–Bray model, the asymptote comes out at 96%, which is reasonable. The question then is, How do you choose a critical x value above which you don't expect much yield response. As I've mentioned above, I think, sometimes people will use an arbitrary point on the y-axis, like 85% of the asymptote. Here, that would correspond to about 50 mg/kg on the x-axis.

The upshot is that for these data, if that any of the models yields similar results, though the differences are likely to have practical implications. The "plateau" yield is about 95% - 96% of maximum yield. The critical x value ranges from, say, 49 to 86. This is probably a large enough range to have practical importance. But I could imagine someone breaking the soil test results into three categories: Low, < 50; Medium = 50 - 86 or 100. Sufficient > 100.

I would probably come to about the same conclusion by eye-balling the data.

( Note to self: agridat::cate.potassium, linplat, quadplat, cateNelsonFixedY, a + b * exp(-1 * c * (x)), a=96.38182, b=-143.53670, c=0.04584. )

Sal Mangiafico Well, maybe the first derivatives may be the same, but still there is an artificial jump. I bet that more natural models like logistic functions will fit nicely as well. And of course this "cut-off" point depends a bit on the glasses you look through (the model); there is no "objectively certain point". Do you have the raw data from the graphs? If possible including the measurement uncertainties. I would like to have a look at them.

Koen Van de Moortel , the cotton / potassium data are taken from:

https://www.rdocumentation.org/packages/agridat/versions/1.20/topics/cate.potassium

The values have no uncertainty reported. The data are:

potassium = (26, 28, 30, 31, 34, 35, 40, 44, 49, 56, 68, 75, 77, 78, 78, 102, 118, 118, 131, 133, 133, 152, 193, 211)

yield = (53.5, 64.8, 63.0, 40.8, 79.5, 70.3, 63.0, 64.0, 94.0, 99.0, 66.5 103.0, 97.3, 85.3 101.3, 97.0, 96.8, 98.0, 85.8, 92.3, 96.8, 88.3 106.8, 97.5)

Sal Mangiafico Thanks. I tried 3 different models. Of course they give a different outcome. The only way to judge the validity of the models is, in my humble opinion: how do they reflect reality if there is no or very little potassium? So you need more data to decide.

Sal Mangiafico This model works fine too, IF it is true that zero potassium means zero yield.

Koen Van de Moortel , with this kind of soil test experiment, the zero soil test potassium won't necessarily correspond to a zero yield. One reason for this is that the soil test value is somewhat arbitrary. It's supposed to mimic "plant-extractable" nutrients, but that's not absolute. ... There's probably no desirable way to get data at the lower end of the x-axis. In theory, you could swap out extractable potassium in soil with some other ion, but no one would want to do this. ... Ultimately, with this kind of experiment, the goal is to determine the range on the x-axis where additional fertilization would increase yield and that range where it wouldn't. ... I think there's an interesting discussion about whether fitting a model helps with this goal, or what kind of model would be best. Using eyeballs and judgement probably works as well as any other kind procedure.

Sal Mangiafico That is exactly why we DO need data with no added potassium! To see which model fits the best!

I found some similar data and there it is clear that zero ADDED K doesn't produce zero yield, I guess because of the K already in the soil. But this measurement influences the curvature of the graph!!! And hence the predictions!

Koen Van de Moortel , yeah, these are studies tend to be very empirical and situation-specific. That is, the results vary based on the climate, soil, extractant used in the soil test, weather during the experiment, and so on. ... It's possible that in that most recent image, their rate of 400 kg / ha of added K over 3 years wasn't enough to produce a plateau in the yield. (Although with few data points, it's difficult to have much confidence.)

Sal Mangiafico With an exponential model the plateau comes quicker.