80 1 Amorphous alloys cannot be used in practice without stabilization of the amorphous state—in particular, without heat treatment. Thus, annealed 71KNSR, 84KSR, and 10-020 amorphous alloys are used in the manufacture of broad-band magnetic heads, whose use in digital recording equipment increases the density and accuracy with which electroacoustic signals are recorded [1, 2]. Correct choice of the amorphous alloy and the annealing conditions is important in order to obtain magnetic and mechanical properties that correspond to the required operating parameters of the magnetic head and minimize the anisotropy of the amorphous tape and also so as to optimize the stability of the amorphous state with respect to time and temperature. As a rule, experimental data are used to select the optimal type of amorphous alloy for the recording head and optimal heat treatment, although it is also possible to use monofactorial dispersional analysis based on regression models, as will be shown in the present work. This method involves randomization of the 1 This article will be of interest primarily to those who develop and apply amorphous magnetic alloys. It employs statistical methods for optimization of the physical properties of the amorphous alloy and for harmonization of the alloy with a magnetic recording head. The article is practically overloaded with statistical calculations, but the editors have chosen to leave the text largely intact, because dispersional analysis is useful not only in the problem here presented but in many areas of metallurgy, physical metallurgy, and economics. For more details on these methods (including multifactorial dispersional analysis, extremal experiment planning, randomization strategies, and methods of information reduction), which were basically developed between 1960 and 1970, consult: Nalimov, V.V., Teoriya exksperimenta (Experiment Theory), Moscow: Nauka, 1971; Hutson, A., Dispersional Analysis (Russian translation), Moscow: Statistika, 1971. Such methods have been successfully employed in the analysis and monitoring of metals and alloys, the evaluation of chemical, metallurgical, and economic processes, and the development of mathematical models of these processes. See, for example: Shvarts, S.A., Prilozhenie matematicheskoi statistiki k analizu protsessov koksokhimicheskogo proizvodstva (Applying Statistics to the Analysis of Coke Production), Khar’kov: Metallurgizdat, 1962; Nalimov, V.V., Statisticheskie metody opisaniya khimicheskikh i metallurgicheskikh protsessov (Statistical Description of Chemical and Metallurgical Processes), Moscow: Metallurgiya, 1963. Mathematical methods prove extremely useful also in the development and prediction of new metallic materials. See, for example: Molotilov, B.V. and Matorin, V.I., Design Principles for New Functional Materials, Stal’ , 2004, no. 8, pp. 92‐94. results of annealing and their division into blocks, which allows the inclusion of the unknown systematic errors among the random errors, to which the laws of probability theory and statistics may be applied [3]. The results of annealing are divided into blocks so that all the levels of the given factor fall within each block, and the values within the block are randomized. The number of instances of annealing is determined by the number m of levels of the factor (magnetic permeability) and the number n of parallel measurements. To select three types of amorphous cobalt alloys that will ensure acceptable magnetic properties of the recording head, we consider annealing at 300 ° C in different media: air, vacuum, hydrogen. Thus, the number of blocks k = 3 (Table 1). Calculation of the sums required to determine the dispersion is based on the formulas Substituting in the numerical values, we obtain S 1 = 10 4 (0.011 + 0.0159 + 0.007056 + 0.0144 + 0.00746 + 0.016796 + 0.0096 + 0.0061 + 0.01383 + 0.054 + 0.007056 + 0.01016 + 0.0057 + 0.0045 + 0.0059 + 0.0092 + 0.0048 + 0.0175 + 0.00915 + 0.0039 + 0.00885 + 0.00348 + 0.0063 + 0.0089 + 0.00503 + 0.0397 + 0.005184 + 0.0081) = 10 4 (0.0042 + 0.0094 + 0.0054 + 0.00346 + 0.00778 + 0.003036 + 0.003036) = 0.312478 × 10 4 ; S 2 = 10 4 (0.168 + 0.187 + 0.135 + 0.107 + 0.119 + 0.0866 + 0.094 + 0.89685)/4 = 0.2242 × 10 4 ; S 3 = 10 4 (1.4617 + 0.936 + 0.822)/3 × 4