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Mathmatic relationship (Linear,Quadratic,Exponential)

  1. Jun 29, 2014 #1
    1. The problem statement, all variables and given/known data

    State the mathematical relationship (Linear,Quadratic,Exponential) and determine the regression equation for this relationship.

    2. Relevant equations
    xX3KXds.png
    X| 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
    Y| 50 | 14 | 2 | 14 | 50| 50 |194|302|

    3. The attempt at a solution
    ?.? i input the information into my TI-83 Plus and i get an error when trying to make a regression equation and i cannot figure out the type of relationship it is(linear,quadratic, or exponential)

    anyone know how to do a table? i did the difference thing.
     
  2. jcsd
  3. Jun 30, 2014 #2

    adjacent

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    Look carefully at the graph.
    How does a quadratic graph normally look like?
    How does an exponential graph normally look like?
    And how does a linear graph normally look like?
    If you know all these, then you can answer your question. :wink:
     
    Last edited: Jun 30, 2014
  4. Jun 30, 2014 #3
    It would appear to be a quadratic but when i try to make the formula on my TI-83 it doesn't make the equation it gives and error.
     
  5. Jun 30, 2014 #4

    adjacent

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    The ##y## value for ##x=11## seems to be giving the error. Try removing that.
    Obviously this is not a linear graph -Linear graphs should be a straight line.
    This is also not a exponential graph. Do you know why?
     
    Last edited: Jun 30, 2014
  6. Jun 30, 2014 #5
    It is a Quadratic because it forms a parabola and not an exponential graph because an exponential graph would not lower at any point it would continune to go on a sharp incline. Also i still got an error when i removed x=11 / y=50

    Got another calculator my seems to malfunction when doing the calc. y=ax^2+bx+c A=3.267857 B= -35.357214 c=89.589285
     
    Last edited: Jun 30, 2014
  7. Jun 30, 2014 #6

    adjacent

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    Then your calculator is broken(Or reached some limit?)
    My calculator does that without malfunctioning. See:
    attachment.php?attachmentid=71004&stc=1&d=1404139720.png
     

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  8. Jun 30, 2014 #7

    SammyS

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    That agrees with adjacent's recent posted value.

    If you discard the point with x = 11, the rest of the points lie exactly on a parabola. For this parabola, A, B, and C are all integers.
     
  9. Jul 1, 2014 #8
    If there is a context to the problem, you may want to consider that. Sometimes output variables are non-negative. Examples would be height, age, test grades. If that is a constraint on this data set, it may be better to drop the point (11, 50) before running the regression.
     
  10. Jul 1, 2014 #9

    LCKurtz

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    Just for confirmation, using all the given points, Maple gives exact fractions:$$
    a=\frac {183}{56},~b=-\frac{495}{14},~c=\frac{5013}{56}$$which agrees with the other solutions given as decimals.
     

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    Last edited: Jul 1, 2014
  11. Jul 2, 2014 #10

    adjacent

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    Isn't that the parabola of best fit including that ##x=11## point?
    None of the points in the table lies in that(Your) parabola
     
  12. Jul 2, 2014 #11

    LCKurtz

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    Isn't that what I said above?

    You wouldn't expect them to be exactly on the best least squares fit.
     
  13. Jul 4, 2014 #12

    Ray Vickson

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    When I do it in Maple I get ##a = 183/56, b = -495/15, c = 5017/56##, so my ##c## is a bit different from what you wrote.

    If, instead of a least-squares solution we minimize the absolute error solution
    [tex] \sum_{i=1}^8 |Y_i - a - b X_i - c X_i^2| [/tex]
    we obtain ##a = 77, b = -30,\, c = 3##. The errors are all zero except at the point x = 11, where the absolute error is 60. Basically, the mean absolute-deviation method automatically de-emphasizes the point x = 11 and gives an exact fit to the rest of the data.

    For the benefit of the OP: the least-absolute solution can be obtained by solving a so-called linear programming problem:
    [tex] \text{minimize } \sum_{i=1}^8 z_i\\
    \text{subject to the constraints}\\
    z_i \geq Y_i - a - b X_i - c X_i^2, \: i=1, \ldots, 8\\
    z_i \geq a + b X_i + cX_i^2 - Y_i, \: i = 1, \ldots, 8[/tex]
    Here, the variables are ##a,b,c, z_1, z_2, \ldots, z_8##. Such problems can be solved using the EXCEL Solver tool, or using the (free) trial version of LINGO (from LINDO Systems, Inc.) plus numerous free "linear programming solution" tools available from the web.
     
  14. Jul 4, 2014 #13

    LCKurtz

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    I just ran it again. Apparently I mistyped the numerator for ##c## because I do get ##5017/56## as you do. But for ##b## I still get ##-495/14##.
     
  15. Jul 4, 2014 #14

    Ray Vickson

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    I mis-typed ##b##; it should have been as you state.
     
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