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where PI is the input plugboard permutation (when going from the input keyboard to the switches), PO is the output plugboard permutation (from

Table 9-2

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where ui is the probability that a randomly selected element falls in row class i and vj is the probability that a randomly selected element falls in column class j. Now, assuming independence, the estimators of ui and vj are

1 c n ja Oij 1 1 r n a Oij i 1 (9-40)

the output keyboard to the plugboard). Note that PI and PG1 follow the direction of the arrows in Figure 2.9. Also, PL, P ~ Iand PR are the left, , middle, and right twenties permutations, respectively, and Ps is the sixes permutation. As mentioned above, the Japanese always selected PI = PO. A major practical difference between Purple and Enigma is that Purple is not its own inverse (Problem 6 asks you to show that Enigma is its own inverse). This means that decryption with Purple is more complicated than with Enigma. Since decryption requires the inverse permutations, Purple can be decrypted by reversing the flow through the diagram in Figure 2.9, as illustrated in Figure 2.10. This implies that the output plugboard is used for input and the input plugboard for output. Of course, if the plugboards have identical permutations (as, apparently, was always the case with Purple), then it does not matter which plugboard is used for input and which is used for output.

(9-41)

(9-42)

P;lPo(y)

has an approximate chi-square distribution with (r 1)(c 1) degrees of freedom if the null hypothesis is true. Therefore, we would reject the hypothesis of independence if the observed value of the test statistic 2 exceeded 2 ,(r 1)(c 1). 0 EXAMPLE 9-14 A company has to choose among three pension plans. Management wishes to know whether the preference for plans is independent of job classi cation and wants to use 0.05. The opinions of a random sample of 500 employees are shown in Table 9-3. To nd the expected frequencies, we must rst compute u1 (340 500) 0.68, u2 1 2 3 (160 500) 0.32, v (200 500) 0.40, v (200 500) 0.40, and v (100 500) 0.20. The expected frequencies may now be computed from Equation 9-41. For example, the expected number of salaried workers favoring pension plan 1 is E11

50010.68210.402

Since the decryption formula works for any choice of PI and PO, why did the Japanese always select PI = PO If PI = PO, then a single plugboard

The expected frequencies are shown in Table 9-4. The eight-step hypothesis-testing procedure may now be applied to this problem. 1. 2. The variable of interest is employee preference among pension plans. H0: Preference is independent of salaried versus hourly job classi cation.

Table 9-3

Table 9-4

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