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7.46. Work Problem 7.45 by using Yates correction.
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x2(corrected) (u75 70u 70 0.5)2 (u65 70u 70 0.5)2 (u 25 30u 30 0.5)2 (u 35 30u 30 0.5)2 1.93
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with a P value of 0.16. Therefore, the conclusions given in Problem 7.45 are valid. This could have been realized at once by noting Yates correction always decreases the value of x2 and increases the P value.
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7.47. In Table 7-16 are indicated the numbers of students passed and failed by three instructors: Mr. X, Mr. Y, and Mr. Z. Test the hypothesis that the proportions of students failed by the three instructors are equal.
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Table 7-16 Frequencies Observed Mr. X Passed Failed TOTAL 50 5 55 Mr. Y 47 14 61 Mr. Z 56 8 64 TOTAL 153 27 180
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Under the hypothesis H0 that the proportions of students failed by the three instructors are the same, they would have failed 27 > 180 15% of the students and passed 85% of the students. The frequencies expected under H0 are shown in Table 7-17.
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CHAPTER 7 Tests of Hypotheses and Significance
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Then x2 (50 46.75)2 46.75 (47 51.85)2 51.85 (56 54.40)2 54.40 (5 8.25)2 8.25 (14 9.15)2 9.15 (8 9.60)2 9.60 4.84
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Table 7-17 Frequencies Expected under H0 Mr. X Passed Failed TOTAL 85% of 55 15% of 55 55 46.75 8.25 Mr. Y 85% of 61 15% of 61 61 51.85 9.15 Mr. Z 85% of 64 15% of 64 64 54.40 9.60 TOTAL 153 27 180
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To determine the number of degrees of freedom, consider Table 7-18, which is the same as Tables 7-16 and 7-17 except that only totals are shown. It is clear that we have the freedom of placing only one number into an empty cell of the first column and one number into an empty cell of the second or third column, after which all numbers in the remaining cells will be uniquely determined from the indicated totals. Therefore, there are 2 degrees of freedom in this case. Table 7-18 Mr. X Passed Failed TOTAL 55 61 64 Mr. Y Mr. Z TOTAL 153 27 180
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Since x2 5.99, we cannot reject H0 at the 0.05 level. Note, however, that since x2 4.61, we can 0.95 0.90 reject H0 at the 0.10 level if we are willing to take the risk of 1 chance in 10 of being wrong. The P value of the observed frequencies is P(x2 4.84) 0.089.
7.48. Show that for an h (h 1)(k 1).
There are h k
k contingency table (h
1, k
1), the number of degrees of freedom is given by
1 independent totals of the hk entries. It follows that the number of degrees of freedom is hk (h k 1) (h 1)(k 1)
as required. The result holds if the population parameters needed in obtaining theoretical frequencies are known; otherwise adjustment is needed as described in (b), page 220.
7.49. Table 7-19 represents a general 2
2 contingency table. Show that
x2 n(a1b2 a2b1)2 n1n2nAnB
Table 7-19 Results Observed I A B TOTAL a1 b1 n1 II a2 b2 n2 TOTAL nA nB n
CHAPTER 7 Tests of Hypotheses and Significance
Table 7-20 Results Expected I A B TOTAL n1nA> n n 1 nB > n n1 II n2nA> n n2nB> n n2 TOTAL nA nB n
As in Problem 7.45, the results expected under a null hypothesis are shown in Table 7-20. Then x2 (a1 n1nA >n)2 n1nA >n n1nA n n2nA n a1 b1 (a2 n2nA >n)2 n2nA >n b1)(a1 b1 a2 b2 (b1 n1nB >n)2 n1nB >n a1b2 n a1b2 n n a1b2 a2b1 2 n2nB n a2b1 a2b1 (b2 n2nB >n)2 n2nB >n
a1 a2
(a1 a1 n1nB n
a2) b2 n2nB n
Similarly We can therefore write x2
which simplifies to (1)
n a1b2 a2b1 2 n1nA n x2
n a1b2 a2b1 2 n2nA n n(a1b2 a2b1)2 n1n2nAnB a1
n 2 n1n2nAnB b1, n2
n a1b2 a2b1 2 n1nB n
where a1b2 a2b1, n a1 a2 b1 b2, n1 Yates correction is applied, (1) is replaced by (2) x2(corrected)
b2, nA
a2, nB
b2. If
n A u u 1n B 2 n1n2nAnB
7.50. Illustrate the result of Problem 7.49 for the data of Problem 7.45.
In Problem 7.45, a1 75, a2 25, b1 then (1) of Problem 7.49 gives x2 65, b2 35, n1 140, n2 60, nA 100, nB 100, and n 200;
200[(75)(35) (25)(65)]2 (140)(60)(100)(100)
Using Yates correction, the result is the same as in Problem 7.46: x2(corrected) n A u a1b2 a2b1 u n1n2nAnB
1 2 2n
200[u (75)(35) (25)(65)u 100]2 (140)(60)(100)(100)
7.51. Show that a chi-square test involving two sample proportions is equivalent to a significance test of differences in proportions using the normal approximation (see page 217).
Let P1 and P2 denote the two sample proportions and p the population proportion. With reference to Problem 7.49, we have (1) (2) P1 a1 n1 , P2 p a2 n2 , nA n, 1 1 P1 p b1 n1 , q nB n 1 P2 b2 n2
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