How to find the equation of a line of best fit?

Question

I was given some points on a graph. The points were in year and weight of garbage format. points:- (1991, 523), (1992, 528), (1993,534), (1994, 542), (1995, 551), (1996,562), (1997, 574), (1998, 583), (1999, 594). I plotted these points and made a line of best fit. THe questions are:-
a) Determine an algebric expression that models the amount of waste received each year? (the x axis is years, and the y - axis is weight)
b) Use your model to make a prediction of the last year that the landfill site handled less then 500t of waste.
c)Using the predictions of model, write a short newspaper article on the effectiveness of recycling campaign.

I was given some points on a graph. The points were in year and weight of garbage format. points:- (1991, 523), (1992, 528), (1993,534), (1994, 542), (1995, 551), (1996,562), (1997, 574), (1998, 583), (1999, 594). I plotted these points and made a line of best fit. THe questions are:-
a) Determine an algebric expression that models the amount of waste received each year? (the x axis is years, and the y - axis is weight)
b) Use your model to make a prediction of the last year that the landfill site handled less then 500t of waste.
c)Using the predictions of model, write a short newspaper article on the effectiveness of recycling campaign.
THE PROBLEM IS THAT I DONT HAVE A GRAPHING CALCULATOR>> AND THE PART C IS STUPID>> so i know thats not math but i have to write something..

Answer 1

Here's a casual rendition of several things you might try.

LINEAR LEAST SQUARES FIT TO DATA.
(N data points.)

P=0 : Q=0 : R=0 : S=0 : T=0
For i=1 to N
P = P + x[i]
Q = Q + x[i]^2
R = R + y[i]
S = S + x[i] y[i]
T = T + y[i]^2
next i
A = (NS - PR) / (NQ - P^2)
B = R/N - AP/N
y = Ax + B

In your case N=9, and after you put in all your data points and run the algorithm, you get...
y = 9.073529 x - 17547.14
...where x is the year (four digits) and y is the weight of the garbage. The slope of the line (9.073529) is the rate, in additional tons of garbage per year, that your annual garbage weight is increasing.

The linear model says you had 491.035 tons of garbage in 1988 and 500.1092 tons of garbage in 1989.

The correlation coefficient of the data is...
cc = (NS - PR) / { (NQ - P^2) (NT - R^2) }^(1/2)
...and in your case,
cc = 0.9885332

QUADRATIC LEAST SQUARES FIT TO DATA.
(N data points.)

P=0 : Q=0 : R=0 : S=0 : T=0 : U=0 : V=0
For i=1 to N
P = P + x[i]
Q = Q + x[i]^2
R = R + x[i]^3
S = S + x[i]^4
T = T + y[i]
U = U + x[i] y[i]
V = V + x[i]^2 y[i]
next i
M = QR - PS
L = Q^2 - PR
K = QU - PV
G = PQ - NR
F = P^2 - NQ
E = PT - NU
A = (KF - LE) / (FM - LG)
B = (E - AG) / F
C = (T - AQ - BP) / N
y = A x^2 + B x + C

According to the quadratic model, your data would give this equation...
y = 0.000686646 x^2 + 6.34375 x - 14834.1
...where x is the year (four digits) and y is the weight of the garbage.

The quadratic model says you had 490.999 tons of garbage in 1988 and 500.0742 tons of garbage in 1989.

LAGRANGE INTERPOLATING POLYNOMIAL OF DEGREE 2.

Just because I'm a curious fellow, I'll generate the Lagrange interpolating polynomial of degree two (i.e., a quadratic equation) based on the data for 1991, 1995, and 1999. This is sort of like what you do when you draw a line from the first point and the last point and try to estimate with that line what's going on elsewhere, except using a parabola instead of a line.

Here's the algorithm:

y = y1 { (x-x2) (x-x3) } / { (x1-x2) (x1-x3) }
+ y2 { (x-x1) (x-x3) } / { (x2-x1) (x2-x3) }
+ y3 { (x-x1) (x-x2) } / { (x3-x1) (x3-x2) }

Where...
x1 = 1991, y1 = 523
x2 = 1995, y2 = 551
x3 = 1999, y3 = 594

y = (523/32) { (x-1995) (x-1999) }
- (551/16) { (x-1991) (x-1999) }
+ (594/32) { (x-1991) (x-1995) }

y = (523/32) {x^2 - 3994 x + 3988005}
- (551/16) {x^2 - 3990 x + 3980009}
+ (594/32) {x^2 - 3986 x + 3972045}

y = 0.46875 x^2 - 1861.4375 x + 1848482.09375

You're really not supposed to use Lagrange interpolating polynomials for extrapolation. They are for estimating points within a spread of the domain for which you already have some data.

But, if you were to break the rules and use it for extrapolation anyway, this model predicts...

514.625 tons of garbage for 1989
511.844 tons of garbage for 1988

Answer 2

What? first of all, x isn't the prediction, its the predictor, or autonomous variable you plug in. The y is the prediction, the based variable. the value b is the slope and the value a is the y-intercept of the line of terrific extra wholesome. this could nicely be a ordinary line equation which you ought to in all probability know at this ingredient. you could't know the values of a or b without needing an information set to charm to on. So do you like us to grant you particular values, or what?

Answer 3

If you've made a line that is the best fit then you just use the equation of the line for part b.

Btw, if you have a TI-83 calculator, you can enter the set of (x,y) points and it will give you the best fit.

Part c illustrates what's wrong with US public education. There's no mathematics in that, it's indoctrination into politically correct thinking.