# Thread: Finding line of best fit

1. ## 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

2. So no one knows how to do #'s 5 or 6 either ?

3. 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?

4. 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

5. Oh sorry, the y-value units are millions of metric tons

6. #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.

7. 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?

8. I'd stick with the linear model if I were you.

keep it simple.

9. 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 ?

10. 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.

11. Great help ! Cant thank you enough !