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2.22. If the random variables X and Y have joint density function f (x, y) e xy>96 0 0 x 4, 1 otherwise XY 2, V y 5

Consider u xy2, v x2y. Dividing these equations, we obtain y >x u >v so that y ux >v. This leads to the simultaneous solution x v2>3 u 1>3, y u2>3 v 1>3. The image of 0 x 4, 1 y 5 in the uv-plane is given by 0 which are equivalent to v2 This region is shown shaded in Fig. 2-15. The Jacobian is given by

A phasor current 25 408 A has complex frequency s 2 j3 s 1 . What is the magnitude of i t at t 0:2 s Ans: 4:51 A Calculate the impedance Z s for the circuit shown in Fig. 8-32, at (a) s 0; b s j1 rad/s, (c) s j2 rad/s, (d) jsj 1. Ans: a 1

(b) If the level of significance is 0.01, the z1 value in Fig. 7-5 is (1) Reject H0 if Z is less than 2.33.

; b 1:58 18:438

2.33. Hence we adopt the decision rule:

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(2) Accept H0 (or withhold any decision) otherwise. Since, as in Problem 7.7(a), the z score is 2.50, which is less than 2.33, we reject H0 at a 0.01 level of significance. Note that this decision is not the same as that reached in Problem 7.7(b) using a two-tailed test. It follows that decisions concerning a given hypothesis H0 based on one-tailed and two-tailed tests are not always in agreement. This is, of course, to be expected since we are testing H0 against a different alternative in each case. (c) The P value of the test is P(Z 1570) 0.0062, which is the probability that a mean lifetime of less than 1570 hours would occur by chance if H0 were true.

; c 1:84 12:538

4 2 3 2 1

7.9. The breaking strengths of cables produced by a manufacturer have mean 1800 lb and standard deviation 100 lb. By a new technique in the manufacturing process it is claimed that the breaking strength can be increased. To test this claim, a sample of 50 cables is tested, and it is found that the mean breaking strength is 1850 lb. (a) Can we support the claim at a 0.01 level of significance (b) What is the P value of the test

Fig. 8-32

(a) We have to decide between the two hypotheses H0: m H1: u 1800 lb, and there is really no change in breaking strength 1800 lb, and there is a change in breaking strength

A one-tailed test should be used here (see Fig. 7-4). At a 0.01 level of significance the decision rule is: (1) If the z score observed is greater than 2.33, the results are significant at the 0.01 level and H0 is rejected. (2) Otherwise H0 is accepted (or the decision is withheld). Under the hypothesis that H0 is true, we find Z # X m s> !n 1850 1800 100> !50 3.55

Fig. 8-33

which is greater than 2.33. Hence we conclude that the results are highly significant and the claim should be supported. (b) The P value of the test is P(Z 3.55) 0.0002, which is the probability that a mean breaking strength of 1850 lb or more would occur by chance if H0 were true.

The voltage source in the s-domain circuit shown in Fig. 8-33 has the time-domain expression vi t 10e t cos 2t Obtain io t . Ans: 7:07e cos 2t 98:138

Tests involving differences of means and proportions 7.10. An examination was given to two classes consisting of 40 and 50 students, respectively. In the first class the mean grade was 74 with a standard deviation of 8, while in the second class the mean grade was 78 with a standard deviation of 7. Is there a significant difference between the performance of the two classes at a level of significance of (a) 0.05, (b) 0.01 (c) What is the P value of the test

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