286 PART 5 Looking for Relationships with Correlation and Regression

Applying both of these substitutions, you get the equivalent model:

C

2

/

Dose

vd

Time

, which produces exactly the same fitted curve as the original

model. But it has the tremendous advantage of giving you exactly the PK param-

eters you want, which are Vd and λ, rather than C 0 and ke which require post-

processing with additional calculations.

From the original description of this example, you already know that Dose = 10,000

μg, so you can substitute this value for Dose in the formula to be fitted. You’ve

already estimated λ (variable tHalf) as 4 hours. Also, you estimated C 0 as about 50

μg/dL from looking at Figure 19-6, as we describe earlier. This means you can

estimate Vd (variable Vd) as 10 000 50

,

/

, which is 200 dL. With these estimates, the

final R statement is

summary(nls(Conc (10000 Vd) 2 ( Time tHalf)

start

list(Vd

~

/

*

^

/

,

200 tHalf

4)))

,

which produces the output shown in Figure 19-9.

From Figure 19-9, you can see the direct results for Vd and tHalf. Using the output,

you can estimate that the Vd is 168 2

6 5

.

.

dL (or 16 8

0 66

.

.

liters), and λ is

4 24

0 43

.

.

hours.

Smoothing Nonparametric

Data with LOWESS

Sometimes you want to fit a smooth curve to a set of points that don’t seem to

conform to a common, recognizable distribution, such as normal, exponential,

logistic, and so forth. If you can’t write an equation for the curve you want to fit,

you can’t use linear or nonlinear regression techniques. What you need is essen-

tially a nonparametric regression approach, which would not try to fit any formula/

model to the relationship, but would instead just try to draw a smooth line through

the data points.

FIGURE 19-9:

Nonlinear

regression that

estimates the

PK parameters

you want.