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Wi Ri 2 in Software Development ECC200 in Software Wi Ri 2

Wi Ri 2 using software tocompose ecc200 on asp.net web,windows application gs1 128 Ri = f i Software datamatrix 2d barcode f (a1 , . . .

, an , xi ). where Ri a re the residuals, and Wi is a weighting factor. For simple least squares, Wi = 1. For chi-squared, Wi = i 2 .

For relative least squares, Wi = f (a1 , . . .

, an , x) 2 . After computing 2 , Tables 5.3(a) and (b) can be used to determine the probability that that value or any larger value could have arisen by chance.

On the average, we expect 2 to be approximately equal to the number of degrees of freedom n f , and 2 n f is regarded as a good t of the model to the data.. Data analy Software 2d Data Matrix barcode sis and control Table 5.3(b) P(> 2 )/n f probability of exceeding 2 per degree of freedom P(> 2 ) 1.000,00 0.

900,00 0.800,00 0.700,00 0.

600,00 0.500,00 0.400,00 0.

300,00 0.200,00 0.100,00 0.

050,00 0.020,00 0.010,00 0.

005,00 0.002,00 0.001,00 0.

000,50 0.000,20 0.000,10 0.

000,05 0.000,02 0.000,01 n f = 30 0.

000,0 0.686,6 0.778,8 0.

850,3 0.914,7 0.977,9 1.

043,9 1.117,7 1.208,3 1.

341,9 1.459,1 1.598,7 1.

696,4 1.789,1 1.905,6 1.

990,1 2.072,1 2.177,1 2.

254,4 2.330,2 2.428,1 2.

500,8 50 0.000,0 0.753,8 0.

829,0 0.886,3 0.937,3 0.

986,7 1.037,8 1.094,5 1.

163,3 1.263,3 1.350,1 1.

452,3 1.523,1 1.589,8 1.

673,1 1.733,2 1.791,2 1.

865,2 1.919,4 1.972,3 2.

040,4 2.090,8 70 0.000,0 0.

790,4 0.855,7 0.904,9 0.

948,5 0.990,5 1.033,7 1.

081,3 1.138,8 1.221,8 1.

293,3 1.377,0 1.434,6 1.

488,8 1.556,1 1.604,5 1.

651,1 1.710,3 1.753,6 1.

795,8 1.850,0 1.890,0 100 0.

000,0 0.823,6 0.879,5 0.

921,3 0.958,1 0.993,3 1.

029,5 1.069,1 1.116,7 1.

185,0 1.243,4 1.311,4 1.

358,1 1.401,7 1.455,8 1.

494,5 1.531,7 1.578,8 1.

613,2 1.646,6 1.689,4 1.

721,0 150 0.000,0 0.855,2 0.

901,8 0.936,4 0.966,7 0.

995,6 1.025,0 1.057,2 1.

095,7 1.150,5 1.197,2 1.

251,2 1.288,1 1.322,4 1.

364,8 1.395,1 1.424,1 1.

460,8 1.487,4 1.513,3 1.

546,4 1.570,7 200 0.000,0 0.

874,2 0.915,0 0.945,2 0.

971,6 0.996,7 1.022,2 1.

049,9 1.083,0 1.130,1 1.

170,0 1.215,9 1.247,2 1.

276,3 1.312,2 1.337,7 1.

362,1 1.392,9 1.415,3 1.

437,0 1.464,7 1.485,0 300 0.

000,0 0.896,9 0.930,7 0.

955,6 0.977,3 0.997,8 1.

018,6 1.041,2 1.068,0 1.

106,0 1.138,0 1.174,7 1.

199,7 1.222,8 1.251,2 1.

271,4 1.290,7 1.314,9 1.

332,5 1.349,5 1.371,2 1.

387,1 500 0.000,0 0.919,9 0.

946,4 0.965,9 0.982,7 0.

998,7 1.014,8 1.032,2 1.

052,8 1.081,9 1.106,3 1.

134,1 1.153,0 1.170,4 1.

191,8 1.206,9 1.221,3 1.

239,4 1.252,5 1.265,1 1.

281,2 1.293,0. EXAMPLE 5. 6 Suppose we have a model function f (x) that is supposed to describe a measurable function and a series of 100 measurements f i and uncertainties i , each made at a corresponding xi . Applying Eq.

(5.5), we compute 2 = 150. What is the probability that the random uctuations in the measurements f i could have resulted in a 2 150 Using Table 5.

3, we look down the column for 100 degrees of freedom at nd that the probability of exceeding 2 = 149.4 is 0.001.

We may conclude that the model function does not adequately describe the measured data. EXAMPLE 5.7 Suppose we have m = 31 measurements ai , and using the equations in (5.

2), determine that the sample standard deviation = 1.38. Assuming that this is close to the true standard deviation 0 , the quantity (m 1) 2 /( 0 )2 = 30 2 /(1.

38)2 = 15.75 2 is distributed as 2 with 30 degrees of freedom. Using Table 5.

3(b), we nd that for 30 degrees of freedom, there is a 10% probability for 2 < 20.599 and a 10% probability for 2 > 40.256.

So the 80% con dence interval for is 20.599/15.75 = 1.

14 to 40.256/15.75 = 1.

60.. 5.6 Solving nonlinear equations Computing the probability of exceeding 2 The probab ECC200 for None ility of exceeding 2 by chance is given by:. P(> ) =. 2 2. x (n 2)/2 e x/2 dx 2n/2 (n/2). (5.7). Tables 5.3 (a) and (b) were computed by numerical integration of Eq. (5.

7), using adaptive quadrature and inverted using Newton s method (Appendix D). Tables 5.3(a) and (b) list the values of 2 corresponding to the probability factors in the rst column.

As a general rule, 2 n f indicates an acceptable t. For n f > 30, P(> 2 ) can be approximated as the Gaussian integral: P(> ) =. d x,.
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