""Eachof us is a statistical impossibility around which hover a millionother lives that were never destined to be born." -- Loren Eiseley

4.1 MODELING NON-LINEAR DATA (Pages 176 -197)

OVERVIEW: If a scatterplot shows a curved pattern, itcan perhaps be conveniently modeled by an exponential growthor decay function of the form

y = abx

or a power function of the form

y = axb

In these situations, we can linearize the data by making use oflogarithms. Among the advantages of using logarithms is the fact thatuse of logarithms produces smaller numbers, making graphical displaysmore convenient to construct.

Definition: logbx = y if and only ifby = x. [x> 0, b > 0 and b is not equal to 1]

Rules for logarithms

1. log(AB) = logA+ logB

2. log(A/B) = logA - logB

3. logAp = plogA

Note that y = abx (exponential function)

==> logy = loga + logbx

==> logy = loga + xlogb.

This is a linear relationship between the variables x andlogy since loga and log b are constants.

Also, y = axb (power function)

==> logy = loga + logxb

==> logy = loga + blogx.

This is a linear relationship between the variables logxand logy.

Here is a simple example using four points. (Use TI-83 to verifycalculations.)

 x y logy logx 2 3 0.47712 0.30103 7 41 1.6128 0.8451 10 168 2.2253 1.0000 16 625 2.7959 1.2041

The points {(x,y)} form a curved pattern.

Fitting a least squares regression line to {(x,y)} yields

y (hat) = -186.6715 + 45.2482x

r2 = .8573, r = .9259

For x = 12, the predicted y value is y(hat) =-186.6715+45.2482(12) = 356.31.

Fitting an exponential function to {(x,y)} yields

y (hat) = (2.1347)(1.4640)x

r2 = .9526, r = .9760

For x = 12, the predicted value is y(hat) =(2.1347)(1.4640)12 = 206.93

Fitting a least squares regression line to {(x,logy)}, weget

log[y(hat)] = 0.3293 + 0.1655x

r2 = .9526, r = .9760

For x = 12, we have logy = 2.3153, and y(hat) =102.3153 = 206.68

Fitting a power function to {(x,y)} yields

y(hat) = (0.4416)x2.5464

r2 = .9843, r = .9921

For x = 12, y(hat) = (0.4416)(12)2.5464 = 247.20

Fitting a least squares regression line to {(logx,logy)}yields

log[y(hat)] = -0.3550 + (2.5464)logx

r2 = .9843, r = .9921

For x = 12, we have log[y(hat)] = -0.3550 + (2.5464)(log12) =2.3930, and y(hat) = 102.3930 = 247.19

Don't forget residuals. These are useful in determining thebest model to fit to a data set.