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 (Crying)
Sep 18th 2008, 03:03 PM
So no one knows how to do #'s 5 or 6 either ?
Sep 18th 2008, 03:12 PM
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?
Sep 18th 2008, 03:13 PM
the x values are the years and the y values are the emissions levels
Oh sorry, the y-value units are millions of metric tons
Sep 18th 2008, 03:25 PM
#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.
Sep 18th 2008, 03:30 PM
Yeah I tried that , I set the regression equation ( I found that the Cubic Regression was the most accurate) equal to 10,000
-.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?
Sep 18th 2008, 03:37 PM
I'd stick with the linear model if I were you.
keep it simple.
Sep 18th 2008, 03:42 PM
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)
Sep 18th 2008, 04:11 PM
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.