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Y. y. 5 T1 3 S 1 = S S 6 - S2 1 3 2 Then, the parameter estimates are: a = (S3S4-S2S5)/S6 (eq. 2 . 6 . 7 3 ) 3 = (S 1 S 5 -S 2 S i f )/S 6 (eq. 74) 2-50 - 53 - (n-2)S2 = E E (y, - âx,-§x2)2 1=1 (eq. 75) For a production item, the measured response is y0. is the solution of the equation The corresponding x value §XQ + axQ - y0 = 0 The solution is XQ = -0 (l- Vl+4ßy0/a2}/2ß (eq. 76) The estimated variances of the estimated parameters, and the estimated covariance between them are: V(e) = S^/Sg (eq.

64) Then, V(3) = — —— -————— ——2—— (eq. 65) V(o) = yiynV(3) (eq. 66) (n-i) (y^) V (x } = (eq 2>6 67) (Vn+yo2'2Vi)v(e) - s o The random error variance of x0 is zero since y0 is assumed to be measured without error in the cumulative model. In tank calibration applications, one is often interested in transfer amounts as determined by noting the difference between two measured responses, (yoz'^oi^' The transfer amount, 3(yo2~yoi) nas a systematic error variance given by - 2-6-68) 2-46 - 49 - For the total of m transfers, 2 V(3) 021 (eq.

45) Vs(xQt) - k2[V(C) + (XQ - x)2V(ß)]/ß2 (eq. 46) where x = Si/Se (eq. 7. 7 (a) given that the intercept is not zero, but is to be estimated. 7 (a) along with the w^. 2035 xQt = (76378 - 47562)72618 + (3) (5. 3 Single Point Calibration A problem commonly encountered in inspection may be called a single point calibration problem and is described as follows, in terms of a specific type of example. Suppose that the percent U-235 of items is to be measured by NDA, and further suppose that the expected range of percent U-235 values is quite narrow.