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Since the reference value of t is t0.005,18 2.88, the value of the test statistic is very far into the critical region, implying that H0: 1 0 should be rejected. The P-value for this test is P 1.23 10 9 . This was obtained manually with a calculator. Table 11-2 presents the Minitab output for this problem. Notice that the t-statistic value for the slope is computed as 11.35 and that the reported P-value is P 0.000. Minitab also reports the t-statistic for testing the hypothesis H0: 0 0. This statistic is computed from Equation 11-22, with 0,0 0, as t0 46.62. Clearly, then, the hypothesis that the intercept is zero is rejected.

x (a) (b)

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11-5.2

ko = ( H ( X ) L [ H ( A ) ] L [ H ( B ) ](mod 256), )

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A method called the analysis of variance can be used to test for signi cance of regression. The procedure partitions the total variability in the response variable into meaningful components as the basis for the test. The analysis of variance identity is as follows:

y 22

(3.4)

y 22

(11-24)

The two components on the right-hand-side of Equation 11-24 measure, respectively, the amount of variability in yi accounted for by the regression line and the residual variation left n unexplained by the regression line. We usually call SSE g i 1 1yi y i 2 2 the error sum of n 2 squares and SSR g i 1 1 yi y 2 the regression sum of squares. Symbolically, Equation 11-24 may be written as

The ORYX attack discussed here requires that some number of keystream bytes are known. Since ORYX is a stream cipher, known plaintext together with the corresponding ciphertext would yield the necessary keystream bytes. In practice about 25 known keystream bytes suffices to recover the entire key. The attack proceeds by trying each of the 2'' possible values for the pair ( H ( A ) , ( B ) ) in (3.4). Given a putative value for ( H ( A ) , ( B ) ) ,and H H assuming ko is known, we can solve for H ( X ) as

(11-25)

where SST g i 1 1 yi y2 2 is the total corrected sum of squares of y. In Section 11-2 we noted that SSE SST 1Sxy (see Equation 11-14), so since SST 1Sxy SSE, we note that the regression sum of squares in Equation 10-26 is SSR 1Sxy. The total sum of squares SST has n 1 degrees of freedom, and SSR and SSE have 1 and n 2 degrees of freedom, respectively. 2 2 2 2 , E1SSR 2 We may show that E3SSE 1n 22 4 and 1Sx x and that SSE 2 SSR are independent chi-square random variables with n 2 and 1 degrees of freedom, respectively. Thus, if the null hypothesis H0: 1 0 is true, the statistic

H ( X )= (ko - L [ H ( A ) ] L [ H ( B ) ](mod 256). )

(11-26)

follows the F1,n 2 distribution, and we would reject H0 if f0 f ,1,n 2. The quantities MSR SSR 1 and MSE SSE (n 2) are called mean squares. In general, a mean square is always computed by dividing a sum of squares by its number of degrees of freedom. The test procedure is usually arranged in an analysis of variance table, such as Table 11-3.

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