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KJ4UTS

  • one year ago

The following table shows total forest and timberland in the United States in millions of acres in the indicated year. Calculate the SSE for the quadratic regression function (with coefficients rounded to three decimal places) from the previous question. Round your answer to one decimal place.

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  1. KJ4UTS
    • one year ago
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  2. KJ4UTS
    • one year ago
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    I am not sure how to calculate the SSE (sum of the squares error)? Is it the R^2 0.8823... or 0.9 rounded to on decimal place?

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  3. KJ4UTS
    • one year ago
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    This is the quadratic equation y=0.036x^2-1.976x+761.454

  4. anonymous
    • one year ago
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    Honestly I've never done this before. A google search gave this formula \[\sum_{i=1}^{n}(x_i-\bar x)^2\] The best I can figure is that you're supposed to use the number in the table for \(x_i\) and the corresponding point from your regression function for \(\bar x\).

  5. KJ4UTS
    • one year ago
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    This is an example I found in my textbook. I guess instead of y=x^2 I would use y=0.036x^2-1.976x+761.454 but its still a little confusing.

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  6. anonymous
    • one year ago
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    ok. the numbers in the timber land row of 1st table are the actual values. You have to use your function to get the other value. What year did your function start at?

  7. KJ4UTS
    • one year ago
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    1962 but I think you represent that with 0 1970 = 8 1977 = 15 1987 = 25 1992= 30

  8. anonymous
    • one year ago
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    ok great now find the corresponding y values from your regression function.

  9. anonymous
    • one year ago
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    It really is easier to set this up in a table like they have it

  10. KJ4UTS
    • one year ago
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    759 754 737 731 737

  11. anonymous
    • one year ago
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    ok. those are your actual values. You have to plug in the year after 1962 into your regression formula to get the 3rd column|dw:1442969120793:dw|

  12. anonymous
    • one year ago
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    y=0.036x^2-1.976x+761.454 So for x = 0 you get y = 761.454 For x = 8 you get y = 747.95 and so on

  13. anonymous
    • one year ago
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    do you follow?

  14. KJ4UTS
    • one year ago
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    yeah I figured out how to plug it in on my calculator

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  15. anonymous
    • one year ago
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    something's not right with the third column. They shouldn't be that far off from the actuals.

  16. KJ4UTS
    • one year ago
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    im not sure what happened?

  17. anonymous
    • one year ago
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    this is what I got|dw:1442970171548:dw|

  18. KJ4UTS
    • one year ago
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    oh ok I must have made a mistake somewhere

  19. anonymous
    • one year ago
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    oh ok. once you get the right values, sum the L5 column and that's the SSE.

  20. KJ4UTS
    • one year ago
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    so add everything up on L5. I wonder why my calculator got the first two right and not the rest on the same column it was the same equation.

  21. KJ4UTS
    • one year ago
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    because with what I got now it would be 1557.127

  22. anonymous
    • one year ago
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    Yeah, that's huge. I think the smaller the number the better the regression curve is

  23. KJ4UTS
    • one year ago
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    oh ok thank you for your time and help :)

  24. anonymous
    • one year ago
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    you're welcome

  25. KJ4UTS
    • one year ago
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    @jim_thompson5910 I was wondering if you might know where I went wrong?

  26. jim_thompson5910
    • one year ago
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    let me check

  27. KJ4UTS
    • one year ago
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    oh thank you I have just been confused with this SSE

  28. jim_thompson5910
    • one year ago
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    what data did you type in to get y=0.036x^2-1.976x+761.454

  29. KJ4UTS
    • one year ago
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    the x an y's on the chart I checked and I got this formula y=0.036x^2-1.976x+761.454 as the right answer but now I just have to figure out the SSE

  30. jim_thompson5910
    • one year ago
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    did you use 1962? or just 62?

  31. jim_thompson5910
    • one year ago
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    if possible, show me a screenshot of the L1 & L2 lists you typed in

  32. KJ4UTS
    • one year ago
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    0 for 1962

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  33. jim_thompson5910
    • one year ago
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    ok thanks

  34. jim_thompson5910
    • one year ago
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    if you plugged x = 0 into your regression equation y=0.036x^2-1.976x+761.454, what is y?

  35. KJ4UTS
    • one year ago
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    @jim_thompson5910 761.454

  36. jim_thompson5910
    • one year ago
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    yes, and I see how that is at the top of L3. Good

  37. jim_thompson5910
    • one year ago
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    L4 looks like y-yhat and L5 looks like (y-yhat)^2

  38. jim_thompson5910
    • one year ago
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    the last thing you do is add up the values in L5

  39. KJ4UTS
    • one year ago
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    I did add them and got 1557.127 but I though that looked to big I though it was supposed to be close to 1

  40. jim_thompson5910
    • one year ago
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    you're thinking of the correlation coefficient r maybe? there is no restriction on the SSE. The larger the SSE, the more error we have so to speak

  41. KJ4UTS
    • one year ago
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    oh so the answer would be 1557.1 rounded to one decimal place then. Thank you for taking the time to check everything over :)

  42. jim_thompson5910
    • one year ago
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    on page 4 of this pdf, they computed the SSE to be approx 96 thousand. So it's definitely possible to get a large SSE http://www.public.iastate.edu/~alicia/stat328/Regression%20inference-part3.pdf

  43. jim_thompson5910
    • one year ago
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    `oh so the answer would be 1557.1 rounded to one decimal place then` I agree

  44. KJ4UTS
    • one year ago
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    Thank you :)

  45. jim_thompson5910
    • one year ago
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    np

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