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Math Help - Finding line of best fit

  1. #1
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    HELP PLEAASSEE

    I have this set of numbers

    Year Gass
    1982 | 5464
    1984 | 5710
    1986 | 5718
    1988 | 6113
    1990 | 6115
    1992 | 6200
    1994 | 6392
    1996 | 6607
    1998 | 6690
    2000 | 6933
    2002 | 6891

    I plugged this in to a table on my TI-84 to find the regression and do the scatter plot


    There are 6 questions associated with this data. I THINK I solved the first 5 right, here is what i have.

    1) Do scatter plot and find linear regression with only the first 10
    -Did the graph fine and went to stat/calc/ LinReg (ax+b) and used this 73.67272727x - 140498.5273

    2) what does your linear model predict will be the emissions for 2002, how does this compare with actual emissions.
    - I pluged 2002 in for x so I had: 73.67272727(2002) - 140498.5273
    The linear model predicted 7042 compared to the actual of 6891 which was a difference 151.

    3)
    Apply ALL data to find the best linear model.
    - For this I am not sure if I did this right. I went to stat/calc/ and tried the LinReg, QuadRed, and CubReg to find which would give me the closest value to the actual when plugging a yearin for x. I found that the CubReg was the closest.

    4) which model do you have the most confidence to predict emissions levels.
    - Again I dont think I did this right, I found the LinReg, QuadReg, and CubicReg equations and plugged a year into each of the x for each equation and found that cubic came the closest to the actual so I said that was

    5) Interpret practical meaning of the slope of the regression line
    - I have no idea
    6) In what year did the best model predict the emissions will reach 10 billion tons
    -Need help on this one too


    PLEASE, someone help me with this I will be very greatful
    Last edited by aforce20; September 18th 2008 at 03:48 PM.
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  2. #2
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    So no one knows how to do #'s 5 or 6 either ?
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  3. #3
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    5. Slope of the regression line is the rate of change of the y-value units per year. I can't be more specific because you never have stated what the y-value units are in the first place.

    6. same problem here ... units?
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  4. #4
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    the x values are the years and the y values are the emissions levels

    x ______y
    1982 | 5464
    1984 | 5710
    1986 | 5718
    1988 | 6113
    1990 | 6115
    1992 | 6200
    1994 | 6392
    1996 | 6607
    1998 | 6690
    2000 | 6933
    2002 | 6891
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  5. #5
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    Oh sorry, the y-value units are millions of metric tons
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  6. #6
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    #5. slope equals the rate of change of emission levels in millions of metric tons per year.

    #6. 10 billion = 10000 million

    use your regression equation to solve for the year when y reaches 10000 million metric tons.
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  7. #7
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    Yeah I tried that , I set the regression equation ( I found that the Cubic Regression was the most accurate) equal to 10,000

    So,
    -.027972028x^3 + 166.54662x^2 - 330461.6783x + 218519934 = 10,000

    When I tried to solve for X, the number I had was off the wall and looked no where near right, so I decided to plug it into my TI-89 and the solution said "false". Im thinking because if you graph that regression line the highest it goes is in the 8,000's so it never reaches 10,000 which is why I am getting the "false" message. So am I using the wrong regression equation?
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  8. #8
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    I'd stick with the linear model if I were you.

    keep it simple.
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  9. #9
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    ok so the linear model is

    73.67272727x - 140498.5273

    So I set this equal to 10,000

    10,000 = 73.67272727x - 140498.5273
    x = -140363


    This cant be right, what am I doing wrong ?
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  10. #10
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    try starting your x-values at 1982 = 0, adjust the other years so that your time list is 0, 2, 4, 6, 8, ....

    my linear regression (rounded) using all the data is

    y = 73.7x + 5520.8

    I graphed y = 10000 and intersected with my regression line.

    when y = 10000, x = 60.8 years from 1982 ... year 2043 approximately.
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  11. #11
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    Great help ! Cant thank you enough !
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