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10.3. The PQR Company claims that the lifetime of a type of battery that it manufactures is more than 250 hours. A consumer advocate wishing to determine whether the claim is justified measures the lifetimes of 24 of the company s batteries; the results are listed in Table 10-3. Assuming the sample to be random, determine whether the company s claim is justified at the 0.05 significance level. Table 10-3 271 253 264 230 216 295 198 262 211 275 288 252 282 236 294 225 291 243 284 253 272 219 224 268

The output v2 is a square pulse which switches between 5 V and 5 V with period of 2 =!. given by 5V 0 < t < =! v2 5 V =! < t < 2 =!

Let H0 be the hypothesis that the company s batteries have a lifetime equal to 250 hours, and let H1 be the hypothesis that they have a lifetime greater than 250 hours. To test H0 against H1, we can use the sign test. To do this, we subtract 250 from each entry in Table 10-3 and record the signs of the differences, as shown in Table 10-4. We see that there are 15 plus signs and 9 minus signs. Table 10-4

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Fig. 10-2

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EXAMPLE 5.24 The circuit of Fig. 5-31 is a parallel analog-to-digital converter. The Vcc and Vcc connections are omitted for simplicity. Let Vcc 5 V, vo 4 V, and vi t (V) for 0 < t < 4 s. Find outputs v3 ; v2 ; and v1 . Interpret the answer. The op amps have no feedback, and they function as comparators. The outputs with values at 5 or 5 V are given in Table 5-2.

Using a one-tailed test at the 0.05 significance level, we would reject H0 if the z score were greater than 1.645 (Fig. 10-2). Since the z score, using a correction for continuity, is z (15 0.5) (24)(0.5) 1.02

Table 5-2 time, s 0<t<1 1<t<2 2<t<3 3<t<4 input, V 0 < vi 1 < vi 2 < vi 3 < vi <1 <2 <3 <4 v3 v3 v3 v3 5 5 5 5 outputs, V v2 v2 v2 v2 5 5 5 5 v1 v1 v1 v1 5 5 5 5

!(24)(0.5)(0.5)

10.4. A sample of 40 grades from a statewide examination is shown in Table 10-5. Test the hypothesis at the 0.05 significance level that the median grade for all participants is (a) 66, (b) 75.

The binary sequences fv3 ; v2 ; v1 g in Table 5-2 uniquely specify the input voltage in discrete domain. However, in their present form they are not the binary numbers representing input amplitudes. Yet, by using a coder we could transform the above sequences into the binary numbers corresponding to the values of analog inputs.

Table 10-5 71 78 67 73 67 46 95 40 55 84 70 78 64 93 43 70 82 72 70 64 66 54 73 86 74 78 57 76 58 86 64 62 79 48 60 95 61 52 83 66

Fig. 14-3

CHAP. 5]

(a) Subtracting 66 from all the entries of Table 10-5 and retaining only the associated signs gives us Table 10-6, in which we see that there are 23 pluses, 15 minuses, and 2 zeros. Discarding the 2 zeros, our sample consists of 38 signs: 23 pluses and 15 minuses. Using a two-tailed test of the normal distribution with probabilities 1 0.025 in each tail (Fig. 10-3), we adopt the following decision rule: 2 (0.05) Accept the hypothesis if 1.96 Reject the hypothesis otherwise. Table 10-6 0 z 1.96.

Fig. 10-3

Fig. 5-31

0.5)

. Find (a) the Thevenin equivalent of the circuit seen by Rl and (b) v2 and the power dissipated in Rl for Rl 0:5, 1, 3, 5, 10, 100, and 1000

(38)(0.5)

!(38)(0.5)(0.5)

(a) The open-circuit voltage and short-circuit current at A B terminal are vo:c: 5v1 and is:c: 5v1 =3, respectively. We nd v1 by dividing vs between Rs and Ri . Thus, v1 Ri 990 20 19:8 V v Rs Ri s 10 990

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