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The hypothesis H0 that the coin is fair, i.e., p 0.5, is accepted when the number of heads in 100 tosses lies between 39.5 and 60.5. The probability of rejecting H0 when it should be accepted (i.e., the probability of a Type I error) is represented by the total area a of the shaded region under the normal curve to the left in Fig. 7-6. As computed in Problem 7.2(a), this area a, which represents the level of significance of the test of H0, is equal to 0.0358.

Fig. 12-15

Fig. 7-6

If the probability of heads is p 0.7, then the distribution of heads in 100 tosses is represented by the normal curve to the right in Fig. 7-6. The probability of accepting H0 when actually p 0.7 (i.e., the probability of a Type II error) is given by the cross-hatched area b. To compute this area, we observe that the distribution under the hypothesis p 0.7 has mean and standard deviation given by m np (100)(0.7) 70 s !npq 60.5 70 4.58 39.5 70 4.58 !(100)(0.7)(0.3) 2.07 6.66 4.58

Fig. 14-30

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Then b area under the standard normal curve between z 6.66 and z 2.07 0.0192. Therefore, with the given decision rule there is very little chance of accepting the hypothesis that the coin is fair when actually p 0.7. Note that in this problem we were given the decision rule from which we computed a and b. In practice two other possibilities may arise:

2. Why does Gunter restrict the terms/events that can be used in a speci cation What is the di erence between a user s requirements and a speci cation 3. The proposed system is a face recognition based on image processing. The system will have a camera and is intended to prevent nonemployees from entering the company s secret facilities by controlling the lock on the door. When a person tries to turn the door handle, the system takes an image and compares it with a set of images of current employees. Classify each of the following events as to whether the events are in the environment or in the system and whether the events are hidden or visible: 1. A person tries to turn the door handles. 2. The door is unlocked by the system. 3. An employee lets a nonemployee through the door. 4. An employee has an identical twin. 5. An image has the minimal number of similarities for the matching algorithm.

The impedances in Fig. 14-32 are selected to give the identical Z-matrix. Thus, since I1 and I2 pass through the common impedance, Zb , in opposite directions, Z12 in the matrix is Zb . Then, Zb 3 j2 . Since Z11 is to include all impedances through which I1 passes, 3 j1 Za 3 j2 from which Za j1 . Similarly, Z22 8 j6 Zb Zc and Zc 5 j4 .

(1) We decide on a (such as 0.05 or 0.01), arrive at a decision rule, and then compute b. (2) We decide on a and b and then arrive at a decision rule.

0.6, the distribution of heads has mean and standard deviation given by m np (100)(0.6) 60 s !npq 60.5 60 4.90 39.5 60 4.90 !(100)(0.6)(0.4) 0.102 4.18 4.90

Fig. 14-32

Then b area under the standard normal curve between z 4.18 and z 0.102 0.5405 Therefore, with the given decision rule there is a large chance of accepting the hypothesis that the coin is fair when actually p 0.6. (b) If p 0.8, then m np (100)(0.8) 80 and s !npq 60.5 4 39.5 4 80 80 !(100)(0.08)(0.2) 4.88 10.12 4.88 0.0000, very closely. 4.

14.15 For the transformer circuit of Fig. 14-11(b), k 0:96, R1 1:2 , R2 0:3 , X1 20 , X2 5 , ZL 5:0 36:878 , and V2 100 08 V. Obtain the coil emfs E1 and E2 , and the magnetizing current I .

(c) From comparison with (b) or by calculation, we see that if p (d) By symmetry, p 0.4 yields the same value of b as p

7.25. Represent the results of Problems 7.23 and 7.24 by constructing a graph of (a) b vs. p, (b) (1 Interpret the graphs obtained.

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