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(b) For a two-tailed test at a 0.01 level of significance, we adopt the decision rule: (1) Accept H0 if T lies inside the interval interval 3.25 to 3.25. (2) Reject H0 otherwise. Since T 3.00, we accept H0 at the 0.01 level. Because we can reject H0 at the 0.05 level but not at the 0.01 level, we can say that the sample result is probably significant (see terminology at the end of Problem 7.5). It would therefore be advisable to check the machine or at least to take another sample. (c) The P value is P(T 3) P(T 3). The table in Appendix D shows that 0.01 computer software, we find P 0.015. P 0.02. Using t0.995 to t0.995, which for 10 1 9 degrees of freedom is the

Fig. 8-13.

12.1 FREQUENCY RESPONSE The response of linear circuits to a sinusoidal input is also a sinusoid, with the same frequency but possibly a di erent amplitude and phase angle. This response is a function of the frequency. We have already seen that a sinusoid can be represented by a phasor which shows its magnitude and phase. The frequency response is de ned as the ratio of the output phasor to the input phasor. It is a real function of j! and is given by H j! Re H j Im H jHje j 1a

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7.17. A test of the breaking strengths of 6 ropes manufactured by a company showed a mean breaking strength of 7750 lb and a standard deviation of 145 lb, whereas the manufacturer claimed a mean breaking strength of 8000 lb. Can we support the manufactur r s claim at a level of significance of (a) 0.05, (b) 0.01 (c) What is the P value of the test

where Re [H] and Im [H] are the real and imaginary parts of H j! and jHj and are its magnitude and phase angle. Re H , Im H , |H|, and are, in general, functions of !. They are related by jHj2 jH j! j2 Re2 H Im2 H H j! tan 1 Im H Re H 1b 1c

We must decide between the hypotheses H0: m H1: m 8000 lb, and the manufacturer s claim is justified 8000 lb, and the manufacturer s claim is not justified

7750 8000 !6 145

The frequency response, therefore, depends on the choice of input and output variables. For example, if a current source is connected across the network of Fig. 12-1(a), the terminal current is the input and the terminal voltage may be taken as the output. In this case, the input impedance Z V1 =I1 constitutes the frequency response. Conversely, if a voltage source is applied to the input and

(a) For a one-tailed test at a 0.05 level of significance, we adopt the decision rule: (1) Accept H0 if T is greater than (2) Reject H0 otherwise. Since T 3.86, we reject H0. t0.95, which for 6 1 5 degrees of freedom means T 2.01.

Fig. 12-1

(b) For a one-tailed test at a 0.01 level of significance, we adopt the decision rule: (1) Accept H0 if T is greater than (2) Reject H0 otherwise. Since T 3.86, we reject H0. We conclude that it is extremely unlikely that the manufacturer s claim is justified. (c) The P value is P(T P 0.006. 3.86). The table in Appendix D shows 0.005 P 0.01. By computer software, t0.99, which for 5 degrees of freedom means T 3.36.

7. Write scenarios for the B&B problem (see Problem 4.3).

7.18. The IQs (intelligence quotients) of 16 students from one area of a city showed a mean of 107 with a standard deviation of 10, while the IQs of 14 students from another area of the city showed a mean of 112 with a standard deviation of 8. Is there a significant difference between the IQs of the two groups at a (a) 0.01, (b) 0.05 level of significance (c) What is the P value of the test

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