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(b) Plot the residuals on a normal probability scale. Do any points seem unusual on this plot (c) Delete the two points identi ed in part (b) from the sample and t the simple linear regression model to the remaining 18 points. Calculate the value of R2 for the new model. Is it larger or smaller than the value of R2 computed in part (a) Why (d) Did the value of 2 change dramatically when the two points identi ed above were deleted and the model t to the remaining points Why 11-51. Show that an equivalent way to de ne the test for signi cance of regression in simple linear regression is to base the test on R2 as follows: to test H0: 1 0 versus H1: 1 0, calculate F0 R2 1n 1 22 R

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and to reject H0: 1 0 if the computed value f0 f ,1,n 2. 11-52. Suppose that a simple linear regression model has been t to n 25 observations and R2 0.90. (a) Test for signi cance of regression at 0.05. Use the results of Exercise 11-51. (b) What is the smallest value of R2 that would lead to the conclusion of a signi cant regression if 0.05 11-53. Consider the rocket propellant data in Exercise 1112. Calculate the standardized residuals for these data. Does this provide any helpful information about the magnitude of the residuals 11-54. Studentized Residuals. Show that the variance of the ith residual is V1ei 2 Hint:

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previous chapter. As with the 511 in the WWII cipher machines. the W E P IV is necessary t o prevent messages from being sent in depth. Recall that two ciphertext messages are in depth if they were encrypted using the same key. Alessages in depth are a serious threat t o a stream cipher. In WEP. Trudy, the crypt analyst, knows many ciphertext messages (packets) and their corresponding IVs, and she would like t o recoaw the long-term key. The Fluher-Mantin-Shamir attack provides a clever. efficient, and elegant way to do just that. This attack has been successfully used t o break real WEP traffic [145]. Suppose that for a particular message, the three-byte initialization vector is of the form (3.8) where V can be any byte \-due. Then thcse three IV bytes become K O ,Kl and Kz in the RC-2 initialization algorithm of Table 3.9, while K3 is the first byte of the unknown long-term key. T h a t is. the message key is

40 400

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2. Press the Menu button and use the 4-Way Controller to navigate to Resize. 3. Press the Function/Set button. 4. Scroll through your images to nd one you want to resize and press the Function/Set button again. The Resize screen appears showing you the available resizing resolutions. 5. Use the 4-Way Controller to select a new resolution. 6. Press the Function/Set button to save it as a new image. 7. Select OK to save it as a new image le.

18 360

30 40

here V is known to Trudy. but K 3 , K4, K j . are unknown. To understand the attack. we need t o carefull\ consider what happens to the table S during the RC4 initialization phase when K is of the foiin in (3.9). In the RC4 initialization algorithm in Table 3.9 we fir5t set S t o the identity permutation, so that we have

10 40

Therefore, P1B A2 can be interpreted as the relative frequency of event B among the trials that produce an outcome in event A. EXAMPLE 2-17 Again consider the 400 parts in Table 2-3. From this table P1D F2 P1D F2 P1F2 10 40 ^ 400 400 10 40

Suppose that K is of the form in (3.9). Then at the i = 0 initialization step. we compute the index j = 0 + SO+ K O= 3 and elements i and j are swapped. resulting in the table

Note that in this example all four of the following probabilities are different: P1F2 P1D2 40 400 28 400 P1F D2 P1D F2 10 28 10 40

Here, P(D) and P1D F2 are probabilities of the same event, but they are computed under two different states of knowledge. Similarly, P(F) and P1F D2 are computed under two different states of knowledge. The tree diagram in Fig. 2-13 can also be used to display conditional probabilities. The rst branch is on surface aw. Of the 40 parts with surface aws, 10 are functionally defective and 30 are not. Therefore, P1D F2 10 40 and P1D F2 30 40

At the next step, i = 1 and j = 3 S1 t K1 = 3 1 255 = 3, since the addition is modulo 256. Elements i and j are again swapped, giving

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