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Topic: **Worst case scenario excel****Question:**
Hi there,
I'm working on some research and I'm having a little trouble understanding my statistics. How practical is an R^2 value for a nonlinear regression model? Does it depend on the model? What factors in a model determine how useful it is? Is 1.0 the ideal value as with a linear regression, and how can I determine how well my curves fit if the values deviate greatly from 1.0 (e.g. < 0)?
So confused and appreciate any help
@Bob: You're amazing, thank you. I honestly didn't expect any responses at 1am. That helped clarify things for me quite a bit. Looking at the graphs turns out to be pretty revealing.

May 24, 2019 / By Dee

Let's see. An R^2 value basically tells you how close the data points are to the curve- the further away they are from the curve, the lower the R^2 value would be. You're correct in saying that the usefulness of R^2 does depend on the model. In all cases, R^2=1 is the ideal case, where all the points lie dead on the line. In some cases however, the model can be completely wrong and still get a very high R^2 value. A common example for this is very steep curves- these are often approximately linear in some regions, so fitting a straight line to a steep curve often gives you quite a good R^2 value even though the model is clearly not a straight line. To determine how useful the R^2 value is, you basically need to use a bit of common sense. This involves using other tests like the Chi-squared test, but simply looking at the data and seeing how well it fits is also very important. Different scenarios will have different R^2 cutoffs for a "good" fit. In a lot of my work, I'll consider a model to be useless if R^2 is less than 0.95, but in some experiments that would be a really good R^2 value. Again, it takes a bit of intelligent thought to work out just what information this statistic gives you (which is also true of any other statistic). In just about any case, however value of R^2 < 0 simply means the data fit is really, really bad, and if you get a value like this, chances are the model you're using is wrong. One thing you should also know is that microsoft excel has a bit of a bug when it comes to calculating R^2 values, so it can give negative values when the fit is good, or values greater than 1. This usually happens if some of the cells that are supposed to contain the data are blank, so be careful when using it for data fitting (when this bug occurs, it also usually results in seriously weird fits which are clearly wrong when viewed on a graph). So if you use excel, make sure all the cells you're using are filled in with data. Another thing you can do is use Qtiplot, which is a free open-source program with far more robust data-fitting algorithms than excel. It's also far more accurate and produces better quality graphs. I use it for most of my work. You can get it here: http://www.cells.es/Members/cpascual/doc... EDIT- I'm in a different time zone :)

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The function f(x) = 4x / (2x + 3) has a horizontal asymptote at y = 2 and a vertical asymptote at x = -3/2 On the left of the VA, the graph is above the HA On the right of the VA, the graph is below the HA So everything from -∞ to -3/2 is above y=2 Edit Graph: http://www.jari.sk/math/png12.png

Steps: a million. positioned all expressions in one section and exhibit as a single fraction. element thoroughly. 2. observe the zeros of the numerator and denominator, they are referred to as intense factors. 3. Create a variety line and place all intense factors there. 4. confirm the sign of the rightmost era by watching the indicators of the coefficients of all factors. ideally commerce the indicators after passing the intense factors. 5. decide for the wanted sign. occasion: a million. x/(x+2) - 6x > 0 [x - 6x(x+2)] / [x+2] > 0 [-6x² - 11x] / [x+2] > 0 -x (6x+11) / (x+2) > 0 ... crit nos: -2,-11/6,0 ... arranged lowest to optimal. observe that extra beneficial than 0, the expression is detrimental, as a effect the rightmost sign is detrimental. commerce between the intense numbers. ..... (-2) ..... (-11/6) ..... (0) ...... ..+ ........ - ............... + ........... - .. decide for advantageous. as a effect x < -2 or -11/6 < x < 0. attempt the stairs for the final 2 numbers.©

I want to expand on one point that Bob made. The "meaningfulness" of R2 really does depend on context. He stated that for his research, an R2 of .95 is useful to him. For those of us who work in the behavioral/social sciences, and R2 of .5 is absolutely tremendous. And as my second statistics instructor (out of the 15 or so that I've had) once told me, an R2 of .02 would undoubtedly be meaningful if, for example, it provided unbelievable strong evidence (e.g., good experimental design, proper analysis of data, lack of alternative explanations, etc.) for the presence of ESP in a group of individuals. This is why I've always despised the tendency to describe effect sizes as large, medium, small, etc. It all really depends on what you are looking for.

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The answer is A.It's because the regression is linear. B is not true, because non-linear regression might improve the correlation coefficient.C and D are not true,because the coefficient is zero.

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Since you are talking about word documents (I assume Word documents) you can do this easily in Visual Basic. Too much to go into here, but grab a computer savvy friend to do this. Read your starting words into a string array, then create and place random text boxes on a page. This could be built in maybe 2 hours for someone familiar with VBA in Word.

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