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is the t distribution with n1 n2 2 degrees of freedom. Therefore P( t 2,n1 n2 2 T t 2,n1 n2 2) 1 . Now substituting Equation 10-18 for T and manipulating the quantities inside the probability statement will lead to the 100(1 )% con dence interval on . 1 2 De nition If x1, x2 , s2 and s2 are the sample means and variances of two random samples of 1 2 sizes n1 and n2, respectively, from two independent normal populations with unknown but equal variances, then a 100(1 )% con dence interval on the difference in means 1 2 is x1 x2 t

Figure 2.21: Merged paths causal and thereby reduce the number of random cases. Note also that only 1 in 26 of the random paths survive the first test, and many more vanish at later steps. Suppose we are in a random case and the first plaintext letter happens to encrypt to the first ciphertext letter. Then the probability that none of the 30 possible steps yields a match for the second letter is (25/26)" z 0.31, z and this holds for each subsequent letter. While the total number of paths increases, the number of distinct merged paths decreases; see Problems 31 and 32. The results in Table 2.7 illustrate typical numbers for the random case, while the results in Table 2.8 illustrate typical numbers for the causal case. For both cases, we have merged paths, as discussed above (and illustrated in Figures 2.20 and 2.21). Table 2.7 indicates that using 30 known plaintext letters, we expect only a fraction of about 0.000427 of the random cases to survive, and each of these survivors will have expanded to an average of about 16.5 paths, with a maximum for the cases tested of 84. In contrast, Table 2.8 shows that with 30 known plaintext letters, we expect the causal path to have generated about 29.6 consistent branches, with, for the 10.000 cases tested, a maximum of 151 and a minimum of one consistent path.

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23 1n1 12 s2 1n2 12 s2 4 1n1 n2 22 is the pooled estimate where sp 1 2 of the common population standard deviation, and t 2, n1 n2 2 is the upper 2 percentage point of the t distribution with n1 n2 2 degrees of freedom.

1 n2

(10-19)

Table 2.7: Random Case Steps 10 20 30 40 50 60 70 80 90 100 Average 6.5 11.8 16.5 20.8 28.4 38.8 47.1 71.3 77.6 100.5 Maximum 27 56 84 105 194 163 415 524 486 1005 Tests Nonzero 763 100,000 516 100,000 427 100,000 324 100,000 290 100,000 275 100,000 269 100,000 212 100,000 100,000 216 203 100.000

EXAMPLE 10-8

An article in the journal Hazardous Waste and Hazardous Materials (Vol. 6, 1989) reported the results of an analysis of the weight of calcium in standard cement and cement doped with lead. Reduced levels of calcium would indicate that the hydration mechanism in the cement is blocked and would allow water to attack various locations in the cement structure. Ten samples of standard cement had an average weight percent calcium of x1 90.0, with a sample standard deviation of s1 5.0, while 15 samples of the lead-doped cement had an average weight percent calcium of x2 87.0, with a sample standard deviation of s2 4.0. We will assume that weight percent calcium is normally distributed and nd a 95% confidence interval on the difference in means, 1 2, for the two types of cement. Furthermore, we will assume that both normal populations have the same standard deviation. The pooled estimate of the common standard deviation is found using Equation 10-12 as follows: s2 p 1n1 12 s2 1n2 12 s2 1 2 n1 n2 2 119.52

Table 2.8: Causal Case Steps 10 20 30 40 50 60 70 80 90 100 Average 10.2 19.6 29.6 40.1 54.1 69.2 85.0 105.0 130.4 161.1 Maximum 51 94 151 237 404 566 689 829 1152 1926 Minimum 1 1 1 1 1 1 1 2 1 1 Tests 10,000 10,000 10,000 10,000 10,000 10,000 5,000 5,000 3,000 3,000

915.02 2 1414.02 2 10 15 2 19.52 Therefore, the pooled standard deviation estimate is sp dence interval is found using Equation 10-19: x1 x2 t0.025,23 sp 1 n1 B 1 n2 1 B 10 x1

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