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Consequently, we expect (3.11) to hold slightly more than 5% of the time. Then with a sufficient number of IVs of the form (3.8) Trudy can determine K3 from (3.11), assuming she knows the first keystrearn byte in each case. What is a sufficient number of IVs to recover K3 If we observe n encrypted packets, each with an IV of the form (3.8), then we expect to solve for the actual K3 using (3.11) for about 0 . 0 5 ~ ~ these. For the remainof ing 0.95n of the cases, we expect the result of (3.11) to be a random value in { 0 , 1 , 2 . . . . ,255}. Then the expected number of times that any particular value other than K3 appears is about 0.95n/256 and the correct value will have an expected count of 0.05n 0.95n/256 E 0.05n. We need to choose n large enough so that we can. with high probability, distinguish K3 from the random noise . If we choose n = 60, then we expect to see K3 three times, while it is unlikely that we will see any random value more than twice (see also Problem 7). This attack is easily extended to recover the remaining unknown key bytes. Wc illustrate the next step, that is, assuming that Trudy has recovered K3, we show that she can recover the key byte K4. In this case, Trudy will look for initialization vectors of the form

which is just the coef cient of determination. That is, the coef cient of determination R2 is just the square of the correlation coef cient between Y and X. It is often useful to test the hypotheses H0 : H1: 21 0 0 (11-45)

2 R2

IV = (4,255,V ) ,

(11-46)

which has the t distribution with n 2 degrees of freedom if H0: 0 is true. Therefore, we would reject the null hypothesis if t0 t 2,n 2. This test is equivalent to the test of the

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(3.12)

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hypothesis H0: 1 0 given in Section 11-6.1. This equivalence follows directly from Equation 10-51. The test procedure for the hypothesis is H0: H1: where 0 statistic

(11-48)

1 n 3

nhere V can be any value Then at the 1 = 0 step of t h e initialization. + So + I<" = 3 and elcnients 7 and j are snapped, resulting in

0 2 1n

321 2

i = 1 and j = 4 + S1 + A', = 4 (since tlie addition is mod 256) and elements S and 5 ~ swapped. giving 1 are

(11-49)

and reject H0: z 2. 0 if the value of the test statistic in Equation 11-49 is such that z0 It is also possible to construct an approximate 100(1 )% con dence interval for , using the transformation in Equation 10-55. The approximate 100(1 )% con dence interval is 1n z 1n z

1 2 1 2

(11-50)

where tanh u EXAMPLE 11-8

e u ).

... ....

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In 1 (Section 1-3) an application of regression analysis is described in which an engineer at a semiconductor assembly plant is investigating the relationship between pull strength of a wire bond and two factors: wire length and die height. In this example, we will consider only one of the factors, the wire length. A random sample of 25 units is selected and tested, and the wire bond pull strength and wire length are observed for each unit. The data are shown in Table 1-2. We assume that pull strength and wire length are jointly normally distributed. Figure 11-13 shows a scatter diagram of wire bond strength versus wire length. We have used the Minitab option of displaying box plots of each individual variable on the scatter diagram. There is evidence of a linear relationship between the two variables. The Minitab output for tting a simple linear regression model to the data is shown on the following page.

Figure 11-13 Scatter plot of wire bond strength versus wire length, Example 11-8.

4 5 1 5

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