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.

- KJ4UTS

- schrodinger

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- KJ4UTS

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- 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

This is the quadratic equation y=0.036x^2-1.976x+761.454

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- 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

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

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

1962 but I think you represent that with 0
1970 = 8
1977 = 15
1987 = 25
1992= 30

- anonymous

ok great now find the corresponding y values from your regression function.

- anonymous

It really is easier to set this up in a table like they have it

- KJ4UTS

759
754
737
731
737

- 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

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

do you follow?

- KJ4UTS

yeah I figured out how to plug it in on my calculator

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- anonymous

something's not right with the third column. They shouldn't be that far off from the actuals.

- KJ4UTS

im not sure what happened?

- anonymous

this is what I got|dw:1442970171548:dw|

- KJ4UTS

oh ok I must have made a mistake somewhere

- anonymous

oh ok. once you get the right values, sum the L5 column and that's the SSE.

- 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

because with what I got now it would be 1557.127

- anonymous

Yeah, that's huge. I think the smaller the number the better the regression curve is

- KJ4UTS

oh ok thank you for your time and help :)

- anonymous

you're welcome

- KJ4UTS

@jim_thompson5910 I was wondering if you might know where I went wrong?

- jim_thompson5910

let me check

- KJ4UTS

oh thank you I have just been confused with this SSE

- jim_thompson5910

what data did you type in to get y=0.036x^2-1.976x+761.454

- 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

did you use 1962? or just 62?

- jim_thompson5910

if possible, show me a screenshot of the L1 & L2 lists you typed in

- KJ4UTS

0 for 1962

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- jim_thompson5910

ok thanks

- jim_thompson5910

if you plugged x = 0 into your regression equation y=0.036x^2-1.976x+761.454, what is y?

- KJ4UTS

@jim_thompson5910 761.454

- jim_thompson5910

yes, and I see how that is at the top of L3. Good

- jim_thompson5910

L4 looks like y-yhat
and L5 looks like (y-yhat)^2

- jim_thompson5910

the last thing you do is add up the values in L5

- 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

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

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

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

`oh so the answer would be 1557.1 rounded to one decimal place then`
I agree

- KJ4UTS

Thank you :)

- jim_thompson5910

np

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