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Method 2
The weights are a smooth function of the dependent variable, such as:

σ i = ay i b
2
and
1
1
w i = ----- = --------
2 b
σ i ay i
In the instrument calibration problem, the relation between x and y is linear, or very nearly so, which
makes it possible to express equivalent weights in terms of x instead of y. That is:
σ i = a′x i b′
2
and

1
1
w i = ----- = -----------
2 b′
σ i a′x i
The value of a(or a′ ) is irrelevant because it appears in each term of the sum of squares function and
can be factored out. Therefore, the weights will be:
1
1
w i ∝ ----- ∝ ----
2 b
σ i y i
or
1
1
w i ∝ ----- ∝ ------
2 b′
σ i x i
The value of b is the slope of a plot of log(σ ) against log( y), or b′ is the slope of a plot of log(σ )
2
2

against log(x). For some instruments (e.g., ICP), we expect b = 1 (or b′ = 1) and the weights are
proportional to the inverse of the concentration. For others (e.g., HPLC), we expect b = 2.
The absolute values of the weights are not important. The numerical values of the weights can be
scaled in any way that is convenient, as long as they convey the relative precision of the different
2 2
measurements. The absolute values of weights w i = 1/ will be numerically much different than w i = 1/ ,x i s i
but in both cases the weights generally decrease with concentration and have the same relative values.

Case Study: Solution

Table 37.2 shows the variances and the weights calculated by Methods 1 and 2. The relative weights in the
two right-hand columns have been scaled so the least precisely measured concentration level has w = 1.
The calculated calibration curve will be the same whether we use the weights or the relative weights.
5
For this nitrate calibration data, the measurements at low concentrations are, roughly speaking, 10 times
more precise (in terms of variance) than those at high concentrations.
2
The calibration curve obtained using Minitab to do weighted least squares with weights w i = 1/ is:
s i
y ˆ = 4.402 + 6908.57x + 105.501x – 1.4428x 3
2
The diagnostic output is given in Table 37.3. The constant (intercept) has a small t value and can be
made zero. The quadratic and the cubic terms are justified in the equation. This can be seen by examining
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