NIST Technical Note 1337: Characterization of Clocks and Oscillators

12 FREQUENCY AND TIME MEASUREMENT213allocation. By rderring to tables of the chi-squared distribution, one findsthat for 10 degrees of freedom (df = 10) the 5% and 95~~ points correspond to(12-37)Thus, with 90~~ probability the calculated sample variance s; = 3 satisfies theinequalityand this inequality can be rearranged in the form3.94 < (df)s;/CT; < 18.3, (12-38)1.64 < a; < 7.61. (12-39)The estimate s; = 3 is a point estimate. The estimate 1.64 < a; < 7.61 isan interval estimate and should be interpreted to mean that 90% of the timethe interval calculated in this manner will contain the true 11;.12.1.8 Efficient Use of the Data and Determination of theDegrees of FreedomTypically, the sample variance is calculated from a data set using therelation1 NS2 ::::::-- L (~ - Z)2 (12-40)- N - 1 11=1 -II ,where it is implicitly assumed that the z..'s are random and uncorrelated (i.e.,white) and where zis the sample mean calculated from the same data set. Ifallof this is true, then S2 is chi-squared distributed and has N - 1 degrees offreedom.Consider the case of two oscillators being compared in phase with N valuesof the phase difference obtained at equally spaced intervals To' From these Nphase values one obtains N - 1 consecutive values of average frequency, andfrom these one can compute N - 2 individual sample Allan variances (not allindependent) for r = To. These N - 2 values can be averaged to obtain anestimate of the Allan variance at T = To'The variance of this Allan variance has been calculated (Lesage andAudoin, 1973; Yoshimura, 1978). This approach is less versatile than themethod of the previous section since it yields only symmetric error limits.However, it is simple and easy to use. let 6(N) be the relative differencebetween the sample Allan variance and the true value. Thus,(12-41)IN-83