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Step 3. Since the alloy I sample has the smaller size, N1 ranks are R1 106 and R2 65. Then U mU Thus sU N1N2 2 N1N2 (8)(10) 2 N1(N1 2 40 1) R1 (8)(10)

. Find the gains G v2 =vs in Fig. 5-4 and G v2 =vs in Fig. 5-5 for k 1, 2, 4, 6, 8, 10, 100, 1000, and 1. Compare the results.

(8)(9) 106 10 2 1) (8)(10)(19) 12

1. All are okay except for pre:self.a.b = self.c.a, which should be pre: self.a.b = self.c.d. 2. Given the object model shown in Fig. 14-4, explain each OCL statement. What does it specify Is the OCL invariant reasonable Is it always true familytree a) self.person = self.marriage.child marriage b) self.child.birthfam = self c) self.husband.birthdate < self.wife.birthdate

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The gains G and G are calculated for nine values of k in Table 5-4. As k becomes very large, G and G approach the limit gain of 5, which is the negative of the ratio R2 =R1 and is independent of k. The circuit of Fig. 5-5 (with negative feedback) is always stable and its gain monotonically approaches the limit gain. However, the circuit of Fig. 5-4 (with positive feedback) is unstable. The gain G becomes very large as k approaches six. At k 6, G 1. Table 5-4 k 1 2 4 6 8 10 100 1000 1 G 1:0 2:5 10:0 1 20:0 12:5 5:32 5:03 5:00 G 0:71 1:25 2:00 2:50 2:86 3:12 4:72 4:97 5:00

Table 10-9 Alloy I Cable Strength 18.3 16.4 22.7 17.8 18.9 25.3 16.1 24.2 Rank 12 10 16 11 13 18 9 17 Sum 106 Alloy II Cable Strength 12.6 14.1 20.5 10.7 15.9 19.6 12.9 15.2 11.8 14.7 Rank 3 5 15 1 8 14 4 7 2 6 Sum 65

CHAP. 5]

Since the hypothesis H0 that we are testing is whether there is no difference between the alloys, a two-tailed test is required. For the 0.05 significance level, we have the decision rule: Accept H0 if 1.96 Reject H0 otherwise. Because z z 1.96.

(a) Since samples 1 and 2 yield values for U given by U1 U2 N1N2 N1N2 N1(N1 1) 2 N2(N2 1) 2 R1 R2 (8)(10) (8)(10) (8)(9) 106 2 (10)(11) 65 2 10 70

in the circuit of Fig. 5-33. Find v2 =vs for k 1, 10, 100, 1000, 1 and compare the results with the values of G in Table 5-4.

we have U1 U2 10 70 N(N 2 80, and N1N2 1) (N1 106 65 N2 1) 171 and (18)(19) 2 171

Fig. 5-33 This problem is solved by application of KCL at node A (another approach which uses the Thevenin equivalent is suggested in Problem 5.30). Thus, v1 vs v1 v2 v1 0 1 5 50 From the ampli er we obtain v2 kv1 or v1 v2 =k 34 33

For the alloy II sample, U so that Z U mU sU 70 40 11.25 2.67 N1N2 N2(N2 2 1) R2 (8)(10) (10)(11) 2 65 70

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This value of z is the negative of the z in Problem 10.5, and the right-hand tail of the normal distribution is used instead of the left-hand tail. Since this value of z also lies outside 1.96 z 1.96, the conclusion is the same as that for Problem 10.5.

Replacing v1 in (34) into (33) and rearranging terms, we obtain v2 50k 5k vs 61 10k 6:1 k 35

10.8. A professor has two classes in psychology: a morning class of 9 students, and an afternoon class of 12 students. On a final examination scheduled at the same time for all students, the classes received the grades shown in Table 10-10. Can one conclude at the 0.05 significance level that the morning class performed worse than the afternoon class Table 10-10 Morning class Afternoon class 73 86 87 81 79 84 75 88 82 90 66 85 95 84 75 92 70 83 91 53 84

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