Error In Estimation
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the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error in estimation routine error of the estimate based on a sample Figure 1 shows two
Error Estimation Calculus
regression examples. You can see that in Graph A, the points are closer to the line than they error estimation taylor series are in Graph B. Therefore, the predictions in Graph A are more accurate than in Graph B. Figure 1. Regressions differing in accuracy of prediction. The standard error of the error propagation estimate is a measure of the accuracy of predictions. Recall that the regression line is the line that minimizes the sum of squared deviations of prediction (also called the sum of squares error). The standard error of the estimate is closely related to this quantity and is defined below: where σest is the standard error of the estimate, Y is an
Estimation Error Formula
actual score, Y' is a predicted score, and N is the number of pairs of scores. The numerator is the sum of squared differences between the actual scores and the predicted scores. Note the similarity of the formula for σest to the formula for σ.  It turns out that σest is the standard deviation of the errors of prediction (each Y - Y' is an error of prediction). Assume the data in Table 1 are the data from a population of five X, Y pairs. Table 1. Example data. X Y Y' Y-Y' (Y-Y')2 1.00 1.00 1.210 -0.210 0.044 2.00 2.00 1.635 0.365 0.133 3.00 1.30 2.060 -0.760 0.578 4.00 3.75 2.485 1.265 1.600 5.00 2.25 2.910 -0.660 0.436 Sum 15.00 10.30 10.30 0.000 2.791 The last column shows that the sum of the squared errors of prediction is 2.791. Therefore, the standard error of the estimate is There is a version of the formula for the standard error in terms of Pearson's correlation: where ρ is the population value of Pearson's
the quantity to be estimated generally differs from its estimator , thus implying an error of estimation . When the mean grade of a vertical bore-hole through the middle of sampling error a block is used to estimate the true mean grade of the error calculation block, the error involved is . Just as the regionalized variable has been interpreted as a particular realization
Error Analysis
of the random function , so the error also appears as a particular realization of the random variable at the point . Suppose now that the entire deposit is http://onlinestatbook.com/2/regression/accuracy.html divided into blocks of equal size , and that each block is intersected by a vertical bore-hole passing through its center. If the deposit is an homogeneous mineralization, i.e., if the random function can be considered as stationary, then the error is also stationary and any, two errors and can be considered as two different realizations of the same stationary http://www.statistik.tuwien.ac.at/public/dutt/vorles/geost_03/node93.html random function . If the histogram of experimental errors is available in a control zone, then it will be possible, because of the stationary assumption, to infer the complete distribution function of . Even if such a histogram is not available, it will still be possible to calculate the stationary expectation and variance of the distribution function of the error. The particular error involved when estimating the block remains unknown, but the mean and variance of the errors (or the complete distribution function if it is known) will provide a measure of the quality of the estimation. In case of the assumption of second-order stationarity of the random function we have (i) the mathematical expectation, (ii) a variance, called the ``estimation variance'', The expectation characterizes the mean error, while the variance is a dispersion index of the error. Thus, a good estimation procedure must be such that it ensures (i) a mean error close to zero, this property of the estimator is known as unbiasedness; (ii) a dispersion of errors very concentrated around this mean
Finite Math Everything for Finite Math & Calculus Español Note To understand this topic, you will need to be familiar with derivatives, as discussed in Chapter 3 of Calculus Applied to the Real World. If you like, you http://www.zweigmedia.com/RealWorld/calctopic1/linearapprox.html can review the topic summary material on techniques of differentiation or, for a more detailed study, the on-line tutorials on derivatives of powers, sums, and constant multipes. We start with the observation that if you zoom in to a portion of a smooth curve near a specified point, it becomes indistinguishable from the tangent line at that point. In other words: The values of the function are close to the values of the linear function whose graph is error in the tangent line. For this reason, the linear function whose graph is the tangent line to $y = f(x)$ at a specified point $(a, f(a))$ is called the linear approximation of $f(x)$ near $x = a.$ Q What is the formula for the linear approximation? A All we need is the equation of the tangent line at a specified point $(a, f(a)).$ Since the tangent line at $(a, f(a))$ has slope $f'(a),$ we can write down its error in estimation equation using the point-slope formula: $y= y_0 + m(x - x_0)$ $= f(a) + f'(a)(x - a)$ Thus, the the linear approximation to $f(x)$ near $x = a$ is given by $L(x) = f(a) + f'(a)(x - a).$ Q The above argument is based on geometry: the fact that the tangent line is close to the original graph near the point of tangency. Is there an algebriac way of seeing why this is true? A Yes. This links to an algebraic derivation of the linear approximation. Linear Approximation of $f(x)$ Near $x = a$ If $x$ is close to a, then $f(x) \approx f(a) + (x-a)f'(a).$ The right-hand side, $L(x) = f(a) + (x-a)f'(a),$ which is a linear function of $x,$ is called the linear approximation of $f(x)$ near $x = a.$ Example 1 Linear Approximation of the Square Root Let $f(x) = x^{1/2}.$ Find the linear approximation of $f$ near $x = 4$ (at the point $(4, f(4)) = (4, 2)$ on the graph), and use it to approximate $\sqrt{4.1.}$ Solution Since $f'(x) = 1/(2x^{1/2}),$ $f'(4) = 1/(2 \cdot 4^{1/2}) = 1/4.$ so the linear approximation is $L(x) = f(4) + (x-4)f'(4)$ $ = 2 + (x-4)/4$ $ = 0.25x + 1.$ We can use $L(x)$ to approximate the square root of any number close to $4$ very easily without using a ca