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Since this value of z is greater than 2.33, we can reject the hypothesis at a 0.01 level of significance, i.e., we can conclude that the serum is effective with only a 0.01 probability of being wrong. Here the P value of the test is P(Z 2.67) 0.0038. This shows how increasing the sample size can increase the reliability of decisions. In many cases, however, it may be impractical to increase sample sizes. In such cases we are forced to make decisions on the basis of available information and so must contend with greater risks of incorrect decisions.

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7.15. A sample poll of 300 voters from district A and 200 voters from district B showed that 56% and 48%, respectively, were in favor of a given candidate. At a level of significance of 0.05, test the hypothesis that (a) there is a difference between the districts, (b) the candidate is preferred in district A. (c) Find the respective P values of the test.

Let p1 and p2 denote the proportions of all voters of districts A and B, respectively, who are in favor of the candidate. Under the hypothesis H0: p1 p2, we have mP1

See Fig. 2-12.

1 300

A capacitance of 60.0 mF has a voltage described as follows: 0 > t > 2 ms, v 25:0 103 t (V). Sketch i, p, and w for the given interval and nd Wmax .

1 R 200

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All problems will be reported to the appropriate lead. If appropriate, the lead will submit a problem report to the SQA lead, who will enter the problem report into the problem tracking system. Resolution of the problem will be reported to the SQA lead. A weekly problem tracking report will be submitted to the project lead.

where we have used as estimates of p and q the values [(0.56)(300) 1 0.528 0.472. Then Z P1 P2 sP1 P2 0.560 0.480 0.0456

For 0 > t > 2 ms,

(a) If we wish to determine only whether there is a difference between the districts, we must decide between the hypotheses H0: p1 p2 and H1: p1 2 p2, which involves a two-tailed test. On the basis of a two-tailed test at a 0.05 level of significance, we would reject H0 if Z were outside the interval 1.96 to 1.96. Since Z 1.75 lies inside this interval, we cannot reject H0 at this level, i.e., there is no significant difference between the districts. (b) If we wish to determine whether the candidate is preferred in district A, we must decide between the hypotheses H0: p1 p2 and H0: p1 p2, which involves a one-tailed test. On the basis of a one-tailed test at a 0.05 level of significance, we would reject H0 if Z were greater than 1.645. Since this is the case, we can reject H0 at this level and conclude that the candidate is preferred in district A. (c) In part (a), the P value is P(Z P(Z 1.75) 0.0401. 1.75) P(Z 1.75) 0.0802, and the P value in part (b) is

Fig. 2-12 dv d 60 10 6 25:0 103 t 1:5 A dt dt p vi 37:5 103 t W t wC p dt 1:875 104 t2 mJ i C

Tests involving student s t distribution 7.16. In the past a machine has produced washers having a mean thickness of 0.050 inch. To determine whether the machine is in proper working order a sample of 10 washers is chosen for which the mean thickness is 0.053 inch and the standard deviation is 0.003 inch. Test the hypothesis that the machine is in proper working order using a level of significance of (a) 0.05, (b) 0.01. (c) Find the P value of the test.

We wish to decide between the hypotheses H0: m 0.050, and the machine is in proper working order H1: m 2 0.050, and the machine is not in proper working order so that a two-tailed test is required. m 0.053 0.050 !n 1 !10 S 0.003 (a) For a two-tailed test at a 0.05 level of significance, we adopt the decision rule: Under the hypothesis H0, we have T (1) Accept H0 if T lies inside the interval interval 2.26 to 2.26. (2) Reject H0 otherwise. Since T 3.00, we reject H0 at the 0.05 level. t0.975 to t0.975, which for 10 1 # X 1 3.00.

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