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I need some help with labeling the trend of this graph, and also deciding what the relationship is between the two variables (X,Y). Here is a picture of it: enter image description here

enter image description here

The Excel document can be downloaded here: http://tempsend.com/5079DC65F4

Using Excels trend-line option it seems to fit either a exponential or moving average trend-line, and also to me seems like it has a horizontal asymptote. Can anyone help me with this? I am also looking at figuring out the relationship between the two variables, seeing i'm only familiar with linear relationship i have no clue.

Hatmix5
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  • If you need any pictures of the trend-lines excel gives me or the graph equation, just tell me if it helps with answering the question. Thanks. – Hatmix5 Sep 09 '14 at 08:00
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    This looks exponential, the vertical axis quantity roughly halves every increase of two units along the horizontal axis. – Travis Willse Sep 09 '14 at 08:04
  • Note that a "moving average" isn't a type of function exactly, rather it just refers to a function $\hat{f(t)}$ given by averaging another function $f(t)$ over some interval in $t$, sometimes with some kind of weighting. – Travis Willse Sep 09 '14 at 08:05
  • Could you post the points ? – Claude Leibovici Sep 09 '14 at 08:05
  • Points have been added. – Hatmix5 Sep 09 '14 at 08:08
  • @Travis, could you tell if it has a vertical asymptote or not? Thanks for confirming exponential trend though! – Hatmix5 Sep 09 '14 at 08:09
  • You could try to interpolate the points and find the polynomial (if you are interested to). here is the link to do that, you "only" need to copy your interpolation coordinates. – Bman72 Sep 09 '14 at 09:14
  • In the data you posted, the first column is $y$ but the second is ??. If you want me to play with your numbers, could you put them in a format such that I could edit them or send them to me by e-mail. Cheers :-) – Claude Leibovici Sep 09 '14 at 09:30
  • @Ale, can i do this in Excel? And it is {{x,y},{x,y},{x,y}} correct? So it would be: {{1.40746,2142.59},{1.175917,1495.618}.....}? – Hatmix5 Sep 09 '14 at 09:30
  • I don't think excel provides a function to interpolate data, but you can find the polynomial with wolframalpha (or with calculation, but i don't suggest it to you :P). It will be a polynomial with a very high degree. The order of input you wrote is correct. $x$ is meant the value on the $x$-axis, and $y$ the value of $f(x)$. – Bman72 Sep 09 '14 at 09:34
  • @Claude, do you have an email? I will send them in text form :) – Hatmix5 Sep 09 '14 at 09:42
  • EDIT: It seems i placed the wrong graph on the question, the real on is now of there. Still exponential, but would like to find out the relationship now. – Hatmix5 Sep 09 '14 at 09:46
  • Download link has been added for anyone to download which has the graph and the v and y values. Thanks! – Hatmix5 Sep 09 '14 at 09:59
  • It seems to be perfectly matched using $y=a x^2$ what you can do with Excel (ask for no intercept) $a \simeq 1080$ – Claude Leibovici Sep 09 '14 at 10:29
  • And a is a constant i'm guessing? What relationship would the two variables have then? – Hatmix5 Sep 09 '14 at 10:33
  • If the function is y=ax^2, does that mean it is a power trend? – Hatmix5 Sep 09 '14 at 10:41
  • Also i believe a = 1081.6, correct me if i'm wrong. – Hatmix5 Sep 09 '14 at 10:50
  • Yes and the value is correct ! You made it ! Cheers :-) – Claude Leibovici Sep 09 '14 at 10:55
  • If you want to see the power trend, plot $\log(y)$ as a function of $\log(x)$ or use the logarithmic scale for the axes. – Claude Leibovici Sep 09 '14 at 10:59
  • Yeah, thanks for your help! So i now know it is a power trend, with function: y=ax^2

    I just have to find out the relationship between the x and y axis, to see what kind of proportionality they are. Would the function suggest that it has a direct proportionality squared?

     – Hatmix5 Sep 09 '14 at 11:01
  • The moving average is not a model. And your curve has absolutely no horizontal asymptote. –  Nov 29 '18 at 21:03
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    You can indeed compare in Excel the various models supported, and try the parameters randomly, until you the the "best fit".

    But it is better to imagine what the model could be, given the origin of the data. Physical experiment ? Mathematical simulation ? Economical phenomenon... ? Avoid empirism and justify your choices.

     –  Nov 29 '18 at 21:09

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Desmos confirms the result that you and Claude reached; your set of data is almost perfectly modeled by the function $$y=1081.6x^2$$ I tried a few other kinds of fits (exponential, higher-order polynomial, etc.), and none worked as well as the above equation, so I think that one should work pretty well for you.

Robert Howard
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