
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 TI84 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 (Crying)

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

5. Slope of the regression line is the rate of change of the yvalue units per year. I can't be more specific because you never have stated what the yvalue units are in the first place.
6. same problem here ... units?

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

Oh sorry, the yvalue units are millions of metric tons

#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.

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 TI89 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?

I'd stick with the linear model if I were you.
keep it simple.

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 ?(Headbang)

try starting your xvalues 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.

Great help ! Cant thank you enough ! (Clapping)