Linear least squares intercept

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. 

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

The derived enthalpies for isomerization of n-BuOH to f-BuOH are —31.9 kJmol (Iq) and —36.7 kJmol (g). From the enthalpies of formation of t-BuOOH, the calculated enthalpies of formation of n-BuOOH are —261.7 kJmol (Iq) and —209.3 kJmol (g). Using the more recently reported enthalpy of formation of teri-BuOOH in the gas phase, the derived enthalpy of formation of n-BuOOH is —198.2 kJmoU. The linear least-squares analysis of the derived liquid enthalpies of formation of n-PrOOH and n-BuOOH and the experimental enthalpy of formation of n-HexOOH gives an intercept of — 161.3 kJ mol and a slope of —23.4 kJmoU, the latter value very close to that for the C3-C6 n-alkanols of —25 kJmoU. From the regression constants, the liquid enthalpy of formation of n-HeptOOH is —325 kJmoU, a value that is less negative, and thus more plausible when compared to its secondary isomers. 

Table 3 summarizes the 1979 annual average particle extinction coefficient and the mass concentrations of the fine aerosol chemical species estimated by statistical analysis of the 61 filter samples. Organics and sulfates dominated the chemically determined fine aerosol mass at China Lake in 1979. A linear least squares fit between molar concentrations of NHt and SO gave a zero intercept, a slope of 1.87 and a correlation coefficient of 0.98. It is therefore assumed that the fine sulfate aerosol was in the form of ammonium sulfate. The mass concentration of carbonaceous and sulfate aerosols were, on the average, comparable in magnitude. 

Prove that the least squares intercept estimator in the classical regression model is the minimum variance linear unbiased estimator. 

Figure 6. Plot of fa vs. fixed carbon. The solid line is derived from the linear least-squares fit. The errors in fa are estimated at .10%. Intercept, 0.11 slope,...

Calculate solution using Eq. (17) for pycnometer data or Eq. (16) for Cassia flask data. Plot tp versus Vm. Determine the slope dcf)ldy[m and the intercept at w = 0 from the best straight line through these data points. This should be done with a linear least-squares fitting procedure. 

Plot die isotherm (u versus x) at 77 K and compare it qualitatively with the one shown in Fig. 1. Finally, calculate x/v(l x) for each point and plot that quantity versus x see Eq. (3). Draw the best straight hne through the points betweenx = 0.05 and 0.3, and determine the slope and intercept of this hne. This evaluation can be made graphically or by a linear least-squares fit of the data with Eq. (3) (see Chapter XXI). From Eq. (4), calculate and c for the sample studied. 

The procedure described for calibration of K and a is laborious because of the required fractionation process. The two constants are derived as described from the intercept and slope of a linear least squares fit to values for a series... 

Note that this equation is again in a form that gives a zero intercept. Thus, a plot and linear least squares analysis of ... 

Numerous kinetic expressions can be placed into a form that would yield a zero y-intercept when using the linear least-squares method. A survey of a few of these models is provided in Table B.3.1. Given that the y-intercept is a known value (i.e., zero), if a perfect correlation could be achieved, the hypothesis that the true value of the parameter, Si, is equal to the specified value, a, could be tested by referring the quantity ... 

The correlation coefficient of the best linear least squares regression model should be between 0.98 and 1.00 or greater than 0.999 with the slope and intercept reported. However, there is no rule stating that the... 

The continuum model has been applied to an experimental study of the solvent effect on the 6-chloro-2-hydroxypyridine/6-chloro-2-pyridone equilibrium in a variety of essentially non-hydrogen-bonding solvents (Beak et al., 1980). In this study, a plot of log A nh/oh) versus (e - 1)/ (2e + 1), the solvent dielectric term, yielded a linear least-squares fit with a slope of 2.5 0.2, an intercept of -1.71, and a correlation coefficient of 0.9944. This result was used to estimate the gas phase free-energy difference of 9.2 kJ mole-1, which compares favorably with the observed value of 8.8 kJ mole-1 for this system. The authors also reported that alcohol solvents are correlated fairly well in this study but that other solvents seem to be divided into two classes, those that are electron-pair donors and those that are electron-pair acceptors in a hydrogen bond. The hydrogen bonding effect is assumed to be independent from the reaction field effect and is included in the continuum model by means of the Kamlet and Taft (1976) empirical parameters. The interested reader is referred to the original paper for a detailed discussion of the method and its application. 

Some of the Eql(l/1) compounds are predicted to have a second region at lower temperatures. If one or two data points exist in the second region, the slope will be lower and the intercept higher than the average value. In the case of benz[e]pyrene the linear least-squares fit gives an intercept of 15.10 1.60, which is higher than... 

Figure 5.5 Plots of ECD data as In KT3/2 versus 1,000/7 for acetophenone taken at different times 1,000, 1,000, 200, 100, and 50 ps. The lines were calculated by linear least squares through the data. The two intercepts at 1,000 ps are different because of the different amounts of data. The parameters are given in Table 5.4.

Solid line linear least squares fit to the data, Equation 21 in the text. Dashed lines one standard deviation ( 0.168) of the intercept (0.749) of log 1/K. Calculated from April-Oc-tober 1966, temperature-depth profiles in Lake Tiberias. Sources of data and computation methods discussed in Ref. 

The linear least-squares line gives a slope of 0.861 and an intercept of —0.002 (using Options under Chart, Add Trendline, when highlighting the chart or line). Hence, the concentration of the unknown is equal to (0.463 — 0.002)70.861, as given by the formula in the spreadsheet (below). The sample concentration is 0.540 ppm. We will now perform the same calculation without charting the calibration curve, and including the standard deviation of the sample concentration. 

If the linear least-squares method is used to find the line that best represents a set of data, this line can be used to predict a value for the dependent variable corresponding to any given value of the independent variable. We now consider the probable error in such a prediction. Since the dependent variable y is a function of the slope m and the intercept b, we might try to apply Eq. (11.28) ... 

Linear Least Squares with Fixed Slope or Intercept... 

At times it is necessary to do a linear least-squares fit with the constraint that the slope or the intercept must have a specific value. For example, the Bouguer-Beer law states that the absorbance of a solution is proportional to the concentration of the colored substance. In fitting the absorbance of several solutions to their concentrations, one would specify that the intercept of the least-squares line had to be zero. 

Using Excel or by hand calculation carry out a linear least squares fit on the following data, once with the intercept fixed at zero and one without specifying the intercept ... 

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