Intercept least squares

Note that m and b do not have subscripts because there is only one slope and one intercept they are the minimization parameters for the least squares function. Now there are two minimization conditions... 

Expand the three detemiinants D, Dt, and for the least squares fit to a linear function not passing through the origin so as to obtain explicit algebraic expressions for b and m, the y-intercept and the slope of the best straight line representing the experimental data. 

A straight line may be fitted to the (X, Y) or (X, Y) pairs of data when plotted on log-log graph paper from which the slope N and the intercept log K with X = 1 may be read. Alternatively, the method of least squares may be used to estimate the values of K and N, giving the best fit to the available data. 

The first of these reactions was carried out in 1,4-cyclohexadiene over a temperature range of 39 to 100 °C. It is fairly slow the half-times were 20 h and 3.4 min at the extremes. Reaction (7-11) is quite fast the second-order rate constant, kn, was evaluated over the range 6.4 to 47.5 °C. Values of feio and fen are presented in Table 7-1. The temperature profiles are depicted in Fig. 7-1 from their intercepts and slopes the activation parameters can be obtained. A nonlinear least-squares fit to Eq. (7-1) or... 

Multiple Linear Least-Squares Fits with a Common Intercept Determination of the Intrinsic Viscosity of Macromolecules in Solution. Journal of Chemical Education 80(9), 1036-1038. 

The data are also represented in Fig. 39.5a and have been replotted semi-logarithmically in Fig. 39.5b. Least squares linear regression of log Cp with respect to time t has been performed on the first nine data points. The last three points have been discarded as the corresponding concentration values are assumed to be close to the quantitation limit of the detection system and, hence, are endowed with a large relative error. We obtained the values of 1.701 and 0.005117 for the intercept log B and slope Sp, respectively. From these we derive the following pharmacokinetic quantities ... 

A least squares linear regression has been applied to the data pertaining to the p-phase, yielding the values of 1.745 and 0.005166 for the intercept log B and slope Sp, respectively. Using these results, we can compute the extrapolated plasma concentrations between 0 and 20 minutes. From the latter, we subtract the observed concentrations C which yields the concentrations of the a-phase C ... 

These data are also shown in the semilogarithmic plot in Fig. 39.13a, which clearly shows two distinct phases. A straight line has been fitted by least-squares regression through the data starting from the observation at 90 minutes down to the last one. This yields the values of 1.086 and -0.001380 for the intercept log and slope ip, respectively. From these results we have computed the extrapolated p-phase values between 2 and 60 minutes. These have been subtracted from the experimental Cp values in order to yield the a-phase concentrations C ... 

The residual a-phase concentrations C are shown in the semilogarithmic plot of Fig. 39.13b. Least-squares linear regression of log C upon time produced 1.524 and -0.02408 for the intercept log and the slope respectively. 

Ordinary least squares regression of MV upon MX produces a slope of 9.32 and an intercept of 2.36. From these we derive the parameters of the simple Michaelis-Menten reaction (eq. (39.116)) ... 

The amount of tebuconazole residue is calculated by using a least-squares fitting algorithm to generate the best line which can be used to calculate the corresponding concentration for a given peak area or peak height. Calculate the slope and the intercept of the standard calibration curve. 

Next, by trial and error an estimate value of K is sought such that the [IH + ] values obtained yield the best fit with the straight line for eqn. 3.77 (e.g., by means of the least-squares method) finally, K and Kh2o are calculated from the intercept and the slope of this line. 

Once kon is known, k+l can be estimated in at least three different ways. First, an independent estimate of k, can be obtained from dissociation studies as described above, where, from Eq. (5.15), k+l = (kon — k, )/ E. Second, koa can be measured at several different concentrations of L and a plot of kon against [L] constructed in which, according to Eq. (5.15), k+l is given directly by the slope. This plot will also provide an estimate of k, (intercept). Third, it is possible to perform a simultaneous nonlinear least-squares lit of a family of onset curves (obtained by using different concentrations of L), the fitting routine providing estimates of k+l, k, and /imax (Problem 5.2 provides an opportunity to calculate binding rate constants). 

Linear least squares analysis of points 3-8 yields the following slope = 0.0073 intercept = 7.36... 

Also determine the slope and intercept of the least-squares line for this set of data. Determine the concentration and standard ... 

The method of least squares is a procedure by which the best straight line through a series of data points is mathematically determined. More details are given in Section 6.4.4. It is useful because it eliminates guesswork as to the exact placement of the line and provides the slope and y-intercept of the line. 

Linearity is evaluated by appropriate statistical methods such as the calculation of a regression line by the method of least squares. The linearity results should include the correlation coefficient, y-intercept, slope of the regression line, and residual sum of squares as well as a plot of the data. Also, it is helpful to include an analysis of the deviation of the actual data points for the regression line to evaluate the degree of linearity. 

This suggests that a plot of r0 vs pA (Fig. 4) or r0/pA vs pA should be linear. It is very difficult to reject this model on the basis of data curvature, even though it is evident that some curvature could exist in Fig. 4. However, Eq. (16) demands that Fig. 4 also exhibit a zero intercept. In fact, the 99.99% confidence interval on the intercept of a least-square line through the data does not contain zero. Hence the model could be rejected with 99.99% certainty. 

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