 4. Process Modeling
4.6. Case Studies in Process Modeling
Calibration Obtaining a Numerical Calibration Value To solve for the numerical estimate of the load associated with the observed deflection, the observed value substituting in the regression function and the equation is solved for load. Typically this will be done using a root finding procedure in a statistical or mathematical package. That is one reason why rough bounds on the value of the load to be estimated are needed. $$\begin{eqnarray} 1.239722 & = & 0.673618\times10^{-3} \\ & + & (0.732059\times10^{-6})L \\ & - & (0.316081\times10^{-14})L^2 \\ & \Downarrow & \\ L & = & 1705106 \end{eqnarray}$$
+/- What? The final step in the calibration process, after determining the estimated load associated with the observed deflection, is to compute an uncertainty or confidence interval for the load. A single-use 95 % confidence interval for the load, is obtained by inverting the formulas for the upper and lower bounds of a 95 % prediction interval for a new deflection value. These inequalities, shown below, are usually solved numerically, just as the calibration equation was, to find the end points of the confidence interval. For some models, including this one, the solution could actually be obtained algebraically, but it is easier to let the computer do the work using a generic algorithm. $$\begin{eqnarray} 1.239722 & > & f(L;\hat{\vec{\beta}}) + t_{0.975,37} \cdot \hat{\sigma}_p \\ & \Downarrow & \\ L & > & 1704513 \end{eqnarray}$$ $$\begin{eqnarray} 1.239722 & < & f(L;\hat{\vec{\beta}}) - t_{0.975,37} \cdot \hat{\sigma}_p \\ & \Downarrow & \\ L & < & 1705697 \end{eqnarray}$$ The three terms on the right-hand side of each inequality are the regression function ($$f$$), a t-distribution multiplier, and the standard deviation of a new measurement from the process ($$\hat{\sigma}_p$$). Regression software often provides convenient methods for computing these quantities for arbitrary values of the predictor variables, which can make computation of the confidence interval end points easier. Although this interval is not symmetric mathematically, the asymmetry is very small, so for all practical purposes, the interval can be written as $$1705106 \pm 593$$ if desired. 