KJ4UTS
  • KJ4UTS
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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KJ4UTS
  • KJ4UTS
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KJ4UTS
  • KJ4UTS
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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KJ4UTS
  • KJ4UTS
This is the quadratic equation y=0.036x^2-1.976x+761.454

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anonymous
  • anonymous
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 \(\bar x\).
KJ4UTS
  • KJ4UTS
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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anonymous
  • anonymous
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?
KJ4UTS
  • KJ4UTS
1962 but I think you represent that with 0 1970 = 8 1977 = 15 1987 = 25 1992= 30
anonymous
  • anonymous
ok great now find the corresponding y values from your regression function.
anonymous
  • anonymous
It really is easier to set this up in a table like they have it
KJ4UTS
  • KJ4UTS
759 754 737 731 737
anonymous
  • anonymous
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|
anonymous
  • anonymous
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
anonymous
  • anonymous
do you follow?
KJ4UTS
  • KJ4UTS
yeah I figured out how to plug it in on my calculator
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anonymous
  • anonymous
something's not right with the third column. They shouldn't be that far off from the actuals.
KJ4UTS
  • KJ4UTS
im not sure what happened?
anonymous
  • anonymous
this is what I got|dw:1442970171548:dw|
KJ4UTS
  • KJ4UTS
oh ok I must have made a mistake somewhere
anonymous
  • anonymous
oh ok. once you get the right values, sum the L5 column and that's the SSE.
KJ4UTS
  • KJ4UTS
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.
KJ4UTS
  • KJ4UTS
because with what I got now it would be 1557.127
anonymous
  • anonymous
Yeah, that's huge. I think the smaller the number the better the regression curve is
KJ4UTS
  • KJ4UTS
oh ok thank you for your time and help :)
anonymous
  • anonymous
you're welcome
KJ4UTS
  • KJ4UTS
@jim_thompson5910 I was wondering if you might know where I went wrong?
jim_thompson5910
  • jim_thompson5910
let me check
KJ4UTS
  • KJ4UTS
oh thank you I have just been confused with this SSE
jim_thompson5910
  • jim_thompson5910
what data did you type in to get y=0.036x^2-1.976x+761.454
KJ4UTS
  • KJ4UTS
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
jim_thompson5910
  • jim_thompson5910
did you use 1962? or just 62?
jim_thompson5910
  • jim_thompson5910
if possible, show me a screenshot of the L1 & L2 lists you typed in
KJ4UTS
  • KJ4UTS
0 for 1962
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jim_thompson5910
  • jim_thompson5910
ok thanks
jim_thompson5910
  • jim_thompson5910
if you plugged x = 0 into your regression equation y=0.036x^2-1.976x+761.454, what is y?
KJ4UTS
  • KJ4UTS
@jim_thompson5910 761.454
jim_thompson5910
  • jim_thompson5910
yes, and I see how that is at the top of L3. Good
jim_thompson5910
  • jim_thompson5910
L4 looks like y-yhat and L5 looks like (y-yhat)^2
jim_thompson5910
  • jim_thompson5910
the last thing you do is add up the values in L5
KJ4UTS
  • KJ4UTS
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
jim_thompson5910
  • jim_thompson5910
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
KJ4UTS
  • KJ4UTS
oh so the answer would be 1557.1 rounded to one decimal place then. Thank you for taking the time to check everything over :)
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
`oh so the answer would be 1557.1 rounded to one decimal place then` I agree
KJ4UTS
  • KJ4UTS
Thank you :)
jim_thompson5910
  • jim_thompson5910
np

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