Question

Given the data 28.65 26.55 26.65 27.65 27.35 28.35 26.85 28.65 29.65 27.85 27.05 28.25 28.85 26.75 27.65 28.45 28.65 28.45 31.65 26.35 27.75 29.25 27.65 28.65 27.65 28.55 27.65 27.25 Determine (a) the mean, (b) the standard deviation, (c) the variance, (d) the coefficient of variation, and (e) the 90% confidence interval for the mean. (f) Construct a histogram. Use a range from 26 to 32 with increments of 0.5. (g) Assuming that the distribution Thus, the improved estimates of the parameters are a0 = 0.7286 and a1 = 1.5019. The new parameters result in a sum of the squares of the residuals equal to 0.0242. Equation (17.36) can be used to compute ε0 and ε1 equal to 37 and 33 percent, respectively. The computation would then be repeated until these values fell below the prescribed stopping criterion. The final result is a0 = 0.79186 and a1 = 1.6751. These coefficients give a sum of the squares of the residuals of 0.000662. Apotential problem with the Gauss-Newton method as developed to this point is that the partial derivatives of the function may be difficult to evaluate. Consequently, many computer programs use difference equations to approximate the partial derivatives. One method is ∂ fi ∂ak ∼= f(xi ; a0, . . . , ak + δak, . . . , am) − f(xi ; a0, . . . , ak, . . . , am) δak (17.37) where δ = a small fractional perturbation. The Gauss-Newton method has a number of other possible shortcomings: 1. It may converge slowly. 2. It may oscillate widely, that is, continually change directions. 3. It may not converge at all. Modifications of the method (Booth and Peterson, 1958; Hartley, 1961) have been developed to remedy the shortcomings. In addition, although there are several approaches expressly designed for regression, a more general approach is to use nonlinear optimization routines as described in Part Four. To do this, a guess for the parameters is made, and the sum of the squares of the residuals is computed. For example, for Eq. (17.31) it would be computed as Sr = n i=1 [yi − a0(1 − e−a1xi )]2 (17.38) Then, the parameters would be adjusted systematically to minimize Sr using search techniques of the type described previously in Chap. 14. We will illustrate how this is done when we describe software applications at the end of Chap. 19. 484 LEAST-SQUARES REGRESSION PROBLEMS PROBLEMS 485 is normal and that your estimate of the standard deviation is valid, compute the range (that is, the lower and the upper values) that encompasses 68% of the readings. Determine whether this is a valid estimate for the data in this problem. 17.3 Use least-squares regression to fit a straight line to x 0 2 4 6 9 11 12 15 17 19 y 5 6 7 6 9 8 7 10 12 12 Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the regression line. Then repeat the problem, but regress x versus y— that is, switch the variables. Interpret your results. 17.4 Use least-squares regression to fit a straight line to x 6 7 11 15 17 21 23 29 29 37 39 y 29 21 29 14 21 15 7 7 13 0 3 Along with the slope and the intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the regression line. If someone made an additional measurement of x = 10, y = 10, would you suspect, based on a visual assessment and the standard error, that the measurement was valid or faulty? Justify your conclusion. 17.5 Using the same approach as was employed to derive Eqs. (17.15) and (17.16), derive the least-squares fit of the following model: y = a1x + e That is, determine the slope that results in the least-squares fit for a straight line with a zero intercept. Fit the following data with this model and display the result graphically: x 2 4 6 7 10 11 14 17 20 y 1 2 5 2 8 7 6 9 12 17.6 Use least-squares regression to fit a straight line to x 1 2 3 4 5 6 7 8 9 y 1 1.5 2 3 4 5 8 10 13 (a) Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the straight line. Assess the fit. (b) Recompute (a), but use polynomial regression to fit a parabola to the data. Compare the results with those of (a). 17.7 Fit the following data with (a) a saturation-growth-rate model, (b) a power equation, and (c) a parabola. In each case, plot the data and the equation.

Answer 1

Mean , Standard deviations, Variance and coefficient of variance has been shown as below.

Variance and Standard deviation measure the distribution of data around the mean. The more consistent or precise the data the least is Standard deviation and variance.