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2 2 2 where 0 w 1. Thus S p is a weighted average of the two sample variances S1 and S2, where the weights w and 1 w depend on the two sample sizes n1 and n2. Obviously, if n1 2 2 2 n2 n, w 0.5 and Sp is just the arithmetic average of S1 and S2. If n1 10 and n2 20 (say), w 0.32 and 1 w 0.68. The rst sample contributes n1 1 degrees of freedom 2 2 to Sp and the second sample contributes n2 1 degrees of freedom. Therefore, Sp has n1 n2 2 degrees of freedom. Now we know that

11 12 13 14 11 12 13 14 27 28 29 30 8 13 2 7

1 n2

1 22

15 15 31 12 47 2 63 9

1 n2

(10-13)

43 6 58 13

The use of this information to test the hypotheses in Equation 10-11 is now straightforward: simply replace 1 2 by 0, and the resulting test statistic has a t distribution with n1 n2 2 degrees of freedom under H0: 1 0 . Therefore, the reference distribu2 tion for the test statistic is the t distribution with n1 n2 2 degrees of freedom. The location of the critical region for both two- and one-sided alternatives parallels those in the one-sample case. Because a pooled estimate of variance is used, the procedure is often called the pooled t-test.

De nition:

The Two-Sample or Pooled t-Test*

44 9 60 4

Null hypothesis: Test statistic:

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Spot AE Point further re nes how the Spot metering system achieves an accurate reading. You can choose between the default setting, Center, or AF Point. When AF Point is selected, the exposure reading taken from the focus frame is locked. This is especially helpful when you have moved the focus point around in your frame; AF Point mimics the strengths of Evaluative and Spot metering by basing its reading on the information within the focus frame. When Spot AE Point is selected, a set of brackets displays within the focus frame.

(10-14)

45 12 61 11

Alternative Hypothesis H1: 1 2 0 H1: H1:

1 1 2 2 0 0

EXAMPLE 10-5

46 15 62 2

Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Speci cally, catalyst 1 is currently in use, but catalyst 2 is acceptable. Since catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield. A test is run in the pilot plant and results in the data shown in Table 10-1. Is there any difference between the mean yields Use 0.05, and assume equal variances. The solution using the eight-step hypothesis-testing procedure is as follows: 1. The parameters of interest are 1 and 2, the mean process yield using catalysts 1 and 2, respectively, and we want to know if 1 0. 2 2. H0: 1 0, or H0: 1 2 2

*While we have given the development of this procedure for the case where the sample sizes could be different, there is an advantage to using equal sample sizes n1 n2 n. When the sample sizes are the same from both populations, the t-test is more robust to the assumption of equal variances. Please see Section 10-3.2 on the CD.

function f . Figure 5.1 illustrates the general case. At this point we have the necessary notation and background to begin discussing the MD5 attack. The MD4 attack in the previous section finds two 512-bit blocks that have the same MD4 hash value. The MD5 attack presented here, which is due to Wang [157],finds a pair of 1024-bit messages that have the same MD5 hash value. We denote the two 1024-bit messages as M = (M, ,,Mi) and 1 1 = (Mo,M I ) ,where each Ml and Mi is a 512-bit 1 block. As with MD4, each 512-bit block consists of 16 words, where each word is 32 bits. Initially, Wang gave a collision [155, 1561 without providing any explanation of the underlying technique. This lack of information led to an impressive attempt to reverse engineer the Chinese method [64]. Incredibly, this reverse engineering was successful enough that it yielded a more efficient attack [81] than Wang s, and it has provided the basis for subsequent improvements in the attack. The Chinese team did eventually release some details of their attack [157]. Minor flaws and incremental improvements have been reported by various

Table 10-1 Catalyst Yield Data, Example 10-5 Observation Number 1 2 3 4 5 6 7 8 x1 s1 Catalyst 1 91.50 94.18 92.18 95.39 91.79 89.07 94.72 89.21 92.255 2.39 x2 s2 Catalyst 2 89.19 90.95 90.46 93.21 97.19 97.04 91.07 92.75 92.733 2.98

6. Reject H0 if t0 t0.025,14 2.145 or if t0 7. Computations: From Table 10-1 we have x1 s2 2.98, and n2 8. Therefore s2 p sp and t0 1n1 27.30 12s2 1n2 12s2 1 2 n1 n2 2 2.70

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