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. Find and plot the instantaneous power p t . From v Ri, we have p t vi Ri2 1000 10 6 10 3 W 1 mW. See Fig. 10-2(b). EXAMPLE 10.2 The current in Example 10.1 passes through a 0.5-mF capacitor. Find the power p t entering the capacitor and the energy w t stored in it. Assume vC 0 0. Plot p t and w t . Figure 10-2(a) indicates that the current in the capacitor is a periodic function with a period T 2 ms. During one period the current is given by 1 mA 0 < t < 1 ms i 1 mA 1 < t < 2 ms

where we have used as an estimate of p the average proportion of cures in the two sample groups, given by (75 65) > 200 0.70, and where q 1 p 0.30. Then Z P1 P2 sP1 P2 0.750 0.650 0.0648 1.54

Fig. 10-2 The voltage across the capacitor is also a periodic function with the same period T [Fig. 10-3(a)]. period the voltage is 1 t 2000t V 0 < t < 1 ms v t i dt 4 2000t V 1 < t < 2 ms C 0 During one

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(a) On the basis of a one-tailed test at a 0.01 level of significance, we would reject the hypothesis H0 only if the z score were greater than 2.33. Since the z score is only 1.54, we must conclude that the results are due to chance at this level of significance. (b) On the basis of a one-tailed test at a 0.05 level of significance, we would reject H0 only if the z score were greater than 1.645. Hence, we must conclude that the results are due to chance at this level also. (c) If a one-tailed test at a 0.10 level of significance were used, we would reject H0 only if the z score were greater than 1.28. Since this condition is satisfied, we would conclude that the serum is effective at a 0.10 level of significance. (d) The P value of the test is P(Z 1.54) 0.0618, which is the probability that a z score of 1.54 or higher in favor of the user group would occur by chance if H0 were true. Note that our conclusions above depended on how much we were willing to risk being wrong. If results are actually due to chance and we conclude that they are due to the serum (Type I error), we might proceed to give the serum to large groups of people, only to find then that it is actually ineffective. This is a risk that we are not always willing to assume. On the other hand, we could conclude that the serum does not help when it actually does help (Type II error). Such a conclusion is very dangerous, especially if human lives are at stake.

7.14. Work Problem 7.13 if each group consists of 300 people and if 225 people in Group A and 195 people in Group B are cured.

Finally, the power entering the capacitor and the energy stored in it (both also periodic with period T) are 2000t mW 0 < t < 1 ms p t vi Fig. 10-3 b 2000t 4 mW 1 < t < 2 ms ( 1 t2 J 0 < t < 1 ms [Fig. 10-3 c w t Cv2 2 2 t 4 10 6 4 10 3 t J 1 < t < 2 ms Alternatively, w t may be obtained by integrating p t . Power entering the capacitor during one period is equally positive and negative [see Fig. 10-3(b)]. The energy stored in the capacitor is always positive as shown in Fig. 103(c). The maximum stored energy is Wmax 10 6 J 1 mJ at t 1; 3; 4; . . . ms.

If X and Y have joint density function f (x, y), then as we have seen in 2, the conditional density function of Y given X is f ( y u x) f (x, y) > f1 (x) where f1 (x) is the marginal density function of X. We can define the conditional expectation, or conditional mean, of Y given X by E(Y u X where X x is to be interpreted as x for conditional expectation. We note the following properties: 1. E(Y u X 2. E(Y) x)

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