1.3 Using Linear Equations
So far we’ve covered how to take data and generate an equation by using the
slope intercept form, finding the slope, and using an ordered pair to find the y
intercept. Now, we need to start making use of the equations we find.
Different ways to look at slope –
From your text:
In the book’s example about blood pressure (which I’m sure
you read in detail), the initial
A millimeter of mercury is a
value for slope that was
manometric unit of pressure,
determined from the data
formerly defined as the extra
pressure generated by a column of
was
mercury one millimetre high and now
Slope = 0.69.
defined as precisely 133.322387415
Rewriting the slope as a fraction and attaching
pascals.[1] It is denoted by the symbol
units, it looks like:
0.69mmHg
1lb
"mmHg".[2]
from: Wikipedia
.69 𝑚𝑚𝐻𝑔
Again the slope (
) can be very useful, meaning the blood
1 𝑙𝑏
pressure will go up about .7 mmHg for every pound gained. The slope
−.69 𝑚𝑚𝐻𝑔
could also be written
, meaning the blood pressure will go down
−1 𝑙𝑏
about .7 mmHg for every pound lost.
Another option can be found dividing both numerator and denominator
−1 𝑚𝑚𝐻𝑔
by -.69,
, meaning you can lower your blood pressure by one
−1.45 𝑙𝑏𝑠
“point” for every pound and a half you lose.
1 𝑚𝑚𝐻𝑔
−2 𝑚𝑚𝐻𝑔
Probably the most useful and memorable version is
≈
.
1.45 𝑙𝑏𝑠
−3 𝑙𝑏𝑠
Meaning you can lower your blood pressure two “points” for every three
pounds you lose.
The point is, just because the slope turns out to be some weird number, doesn’t
necessarily me you need to leave it that way.
Least Squares Regression
Meet your new best friend: “least squares regression”
This is a little tool, actually is a GREAT TERRIFIC tool you
will be using on your calculator. This application is going
to save you a lot of work and time. It’s also going to give
us equations that are generally much more accurate than
what we can do by hand. Just remember, garbage in garbage out.
Least squares regression on your calculator, gives us the best possible fit for a
trend line, given a set of data. This takes the guessing out of where you would
place the trend line if you were doing it by hand.
Front your text:
The concept is simple; consider the 5 points in the figure and the vertical distance
each is away from the line. The distances above the line are positive and below
the line are negative. The easiest method to find the best trend line is to square
the distances to remove the
negatives, and then add
them to find the least sum
possible, hence the name
least squares regression.
See this excellent YouTube video for an animated
demonstration of least squares regression
http://www.youtube.com/watch?v=jEEJNz0RK4Q
How to use your calculator. This is
one of the videos in your list
(5. Regression and Correlation).
See this excellent YouTube video for an animated
demonstration of regression on the Ti 83/84 calculator
http://www.youtube.com/watch?v=nw6GOUtC2jY
Example Problem #2 from text –
A1
Homework: 1, 3, 5, 7, 8 – Your work should be as demonstrated in class.
A2
Sample Problem 1.3.4
A tree’s diameter and volume are naturally related.
Round slopes and y-intercepts to 3 decimal places
a) Find the equation for the trend line using the regression
feature of your graphing calculator.
b) Use your regression equation to predict the volume of a
5.5 cm diameter tree, accurate to 1 decimal place.
c) Use your regression equation to predict the diameter a
tree would need to have a 500 m3 volume, accurate to 2
decimal places.
Solutions:
a)
After entering the data in two lists, use the linear regression function to generate the equation.
y  54.9520 x  242.9704
y  54.952 x  242.970
b)
Using the equation from a),
substitute 5.5 in for x, solve for y.
y  54.952 • 5.5  242.970
y  59.266
y  59.3 m3
c)
Using the equation from a),
substitute 500 in for y, solve for y.
500  54.952 x  242.970
x  13.5203
x  13.52 cm