barcode font reporting services Tests of Hypotheses and Significance in Software

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CHAPTER 7 Tests of Hypotheses and Significance
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so that (3) (4) a1 n1P1, a2 n2P2, nA b1 np, n1(1 nB P1), nq b2 n2(1 P2)
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Using (3) and (4), we have from Problem 7.49 x2 n(a1b2 a2b1)2 n1n2nAnB n1n2(P1 P2)2 npq n[n1P1n2(1 P2) n2P2n1(1 n1n2npnq (since n P1)]2
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(P1 P2)2 pq(1>n1 1>n2)
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which is the square of the Z statistic given in (10) on page 217.
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Coefficient of contingency 7.52. Find the coefficient of contingency for the data in the contingency table of Problem 7.45.
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C x2 A x2 n 2.38 A 2.38 200 !0.01176 0.1084
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7.53. Find the maximum value of C for all 2
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2 tables that could arise in Problem 7.13.
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The maximum value of C occurs when the two classifications are perfectly dependent or associated. In such cases, all those who take the serum will recover and all those who do not take the serum will not recover. The contingency table then appears as in Table 7-21. Table 7-21 Recover Group A (using serum) Group B (not using serum) TOTAL 100 0 100 Do Not Recover 0 100 100 TOTAL 100 100 200
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Since the expected cell frequencies, assuming complete independence, are all equal to 50, x2 (100 50 50)2 (0 50)2 50 (0 50)2 50 (100 50 50)2 200
Then the maximum value of C is 2x2 >(x2 n) 2200>(200 200) 0.7071. In general, for perfect dependence in a contingency table where the numbers of rows and columns are both equal to k, the only nonzero cell frequencies occur in the diagonal from upper left to lower right. For such cases, Cmax 2(k 1)>k.
Miscellaneous problems 7.54. An instructor gives a short quiz involving 10 true-false questions. To test the hypothesis that the student is guessing, the following decision rule is adopted: (i) If 7 or more are correct, the student is not guessing; (ii) if fewer than 7 are correct, the student is guessing. Find the probability of rejecting the hypothesis when it is correct.
Let p probability that a question is answered correctly. The probability of getting x questions out of 10 correct is 10Cx pxq10 x, where q Then under the hypothesis p 0.5 (i.e., the student is guessing), 1 p.
CHAPTER 7 Tests of Hypotheses and Significance
P(7 or more correct)
P(7 correct)
10C7 2
Therefore, the probability of concluding that the student is not guessing when in fact he is guessing is 0.1719. Note that this is the probability of a Type I error.
1 3 2
P(8 correct) 1
10C8 2
1 2 2
P(9 correct)
10C9 2
P(10 correct)
1 2
10C10 2
7.55. In Problem 7.54, find the probability of accepting the hypothesis p
Under the hypothesis p P(less than 7 correct) 1 1 0.7, P(7 or more correct) [10C7(0.7)7(0.3)3
10C8(0.7)
0.5 when actually p
(0.3)2
10C9(0.7)
(0.3)
10C10(0.7)
7.56. In Problem 7.54, find the probability of accepting the hypothesis p 0.5 when actually (a) p (b) p 0.8, (c) p 0.9, (d) p 0.4, (e) p 0.3, (f) p 0.2, (g) p 0.1.
(a) If p 1 0.6, the required probability is given by [P(7 correct) 1 P(8 correct) P(9 correct)
10C8(0.6)
0.6,
P(10 correct)]
10C9(0.6)
[10C7(0.6)7(0.4)3
(0.4)2
(0.4)
10C10(0.6)
The results for (b), (c), . . . , (g) can be similarly found and are indicated in Table 7-22 together with the value corresponding to p 0.7 found in Problem 7.55. Note that the probability is denoted by b (probability of a Type II error). We have also included the entry for p 0.5, given by b 1 0.1719 0.828 from Problem 7.54. Table 7-22 p b 0.1 1.000 0.2 0.999 0.3 0.989 0.4 0.945 0.5 0.828 0.6 0.618 0.7 0.350 0.8 0.121 0.9 0.013
7.57. Use Problem 7.56 to construct the graph of b vs. p, the operating characteristic curve of the decision rule in Problem 7.54.
The required graph is shown in Fig. 7-14. Note the similarity with the OC curve of Problem 7.27.
Fig. 7-14
If we had plotted (1 b) vs. p, the power curve of the decision rule would have been obtained. The graph indicates that the given decision rule is powerful for rejecting p 0.5 when actually p
7.58. A coin that is tossed 6 times comes up heads 6 times. Can we conclude at (a) 0.05, (b) 0.01 significance level that the coin is not fair Consider both a one-tailed and a two-tailed test.
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