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It can be shown that, if the population follows the hypothesized distribution, X 2 has, approx0 imately, a chi-square distribution with k p 1 degrees of freedom, where p represents the number of parameters of the hypothesized distribution estimated by sample statistics. This approximation improves as n increases. We would reject the hypothesis that the distribution of the population is the hypothesized distribution if the calculated value of the test statistic 2 2 0 ,k p 1. One point to be noted in the application of this test procedure concerns the magnitude 2 of the expected frequencies. If these expected frequencies are too small, the test statistic X0 will not re ect the departure of observed from expected, but only the small magnitude of the expected frequencies. There is no general agreement regarding the minimum value of expected frequencies, but values of 3, 4, and 5 are widely used as minimal. Some writers suggest that an expected frequency could be as small as 1 or 2, so long as most of them exceed 5. Should an expected frequency be too small, it can be combined with the expected frequency in an adjacent class interval. The corresponding observed frequencies would then also be combined, and k would be reduced by 1. Class intervals are not required to be of equal width. We now give two examples of the test procedure. EXAMPLE 9-12 A Poisson Distribution The number of defects in printed circuit boards is hypothesized to follow a Poisson distribution. A random sample of n 60 printed boards has been collected, and the following number of defects observed.

The mean of the assumed Poisson distribution in this example is unknown and must be estimated from the sample data. The estimate of the mean number of defects per board is the sample average, that is, (32 0 15 1 9 2 4 3) 60 0.75. From the Poisson

5.4 11105

distribution with parameter 0.75, we may compute pi, the theoretical, hypothesized probability associated with the ith class interval. Since each class interval corresponds to a particular number of defects, we may nd the pi as follows: p1 p2 p3 p4 P1X P1X P1X P1X 02 12 22 32 e e e 1

10.752 0 0! 10.752 1 1!

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Figure 11: Encryption and decryption not know any of the underlying plaintext A basic assumption is that the ciphertext is always available to an attacker After all if the ciphertext is not available to the attacker, why bother to encrypt In a known plaintext attack, Trudy has the ciphertext as well as some of the corresponding plaintext This might give the attacker some advantage over the ciphertext only scenario-certainly the attacker is no worse off with known plaintext If Trudy knows all of the plaintext, there is probably not much point in bothering to attack the system, so the implicit assumption is that the amount of known plaintext is relatively limited As the name implies, in a chosen plaintext attack, an adversary can choose the plaintext and then obtain the corresponding ciphertext This can only help the attacker, as compared to a known plaintext scenario.

The expected frequencies are computed by multiplying the sample size n probabilities pi. That is, Ei npi. The expected frequencies follow:

Number of Defects 0 1 2 3 (or more)

Similarly, in a chosen ciphertext attack, the cryptanalyst chooses ciphertext and gets to see the corresponding plaintext There are also related key attacks, where the attacker can break the system if two keys are used that happen to be related in some very special way While this may seem somewhat esoteric, we will see an example of a real-world related key attack in 3 In most cases, recovering the key is Trudy s ultimate goal, but there are attacks that recover the plaintext without revealing the key A cipher is generally not considered secure unless it is secure against all plausible attacks Cryptographers are, by nature, a paranoid bunch, so plausible is usually defined very broadly Kerckhoffs Principle is one of the fundamental concepts underlying cryptography.

Since the expected frequency in the last cell is less than 3, we combine the last two cells:

Number of Defects 0 1 2 (or more)

The chi-square test statistic in Equation 9-39 will have k p 1 3 1 1 1 degree of freedom, because the mean of the Poisson distribution was estimated from the data. The eight-step hypothesis-testing procedure may now be applied, using 0.05, as follows: 1. 2. 3. 4. 5. The variable of interest is the form of the distribution of defects in printed circuit boards. H0: The form of the distribution of defects is Poisson. H1: The form of the distribution of defects is not Poisson. 0.05 The test statistic is

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