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3.84, we reject the hypothesis
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(b) The critical value x2 for 1 degree of freedom is 6.63. Then since 4.50 6.63, we cannot reject the 0.99 hypothesis that the coin is fair at a 0.01 level of significance. We conclude that the observed results are probably significant and the coin is probably not fair. For a comparison of this method with previous methods used, see Method 1 of Problem 7.36. (c) The P value is P(x2 P 0.039. 4.50). The table in Appendix E shows 0.025 P 0.05. By computer software,
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7.35. Work Problem 7.34 using Yates correction.
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x2 (corrected) (u x (u115 1 np1 u np1 100u 100 0.5)2 0.5)2 (u x2 ( u 85 np2 u np2 100 u 100 0.5)2 0.5)2 (14.5)2 100 (14.5)2 100
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The corrected P value is 0.04 Since 4.205 3.84 and 4.205 6.63, the conclusions arrived at in Problem 7.34 are valid. For a comparison with previous methods, see Method 2 of Problem 7.36.
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7.36. Work Problem 7.34 by using the normal approximation to the binomial distribution.
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Under the hypothesis that the coin is fair, the mean and standard deviation of the number of heads in 200 tosses of a coin are m np (200)(0.5) 100 and s !(200)(0.5)(0.5) 7.07. !npq Method 1 115 heads in standard units (115 100) > 7.07 2.12.
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Using a 0.05 significance level and a two-tailed test, we would reject the hypothesis that the coin is fair if the z score were outside the interval 1.96 to 1.96. With a 0.01 level the corresponding interval would be 2.58 to 2.58. It follows as in Problem 7.34 that we can reject the hypothesis at a 0.05 level but cannot reject it at a 0.01 level. The P value of the test is 0.034. Note that the square of the above standard score, (2.12)2 4.50, is the same as the value of x2 obtained in Problem 7.34. This is always the case for a chi-square test involving two categories. See Problem 7.60. Method 2 Using the correction for continuity, 115 or more heads is equivalent to 114.5 or more heads. Then 114.5 in standard units (114.5 100) > 7.07 2.05. This leads to the same conclusions as in the first method. The corrected P value is 0.04. Note that the square of this standard score is (2.05)2 4.20, agreeing with the value of x2 corrected for continuity using Yates correction of Problem 7.35. This is always the case for a chi-square test involving two categories in which Yates correction is applied, again in consequence of Problem 7.60.
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7.37. Table 7-10 shows the observed and expected frequencies in tossing a die 120 times. (a) Test the hypothesis that the die is fair, using a significance level of 0.05. (b) Find the P value of the test.
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CHAPTER 7 Tests of Hypotheses and Significance
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(a) Face 1 25 20 Table 7-10 2 17 20 3 15 20 4 23 20 5 24 20 6 16 20
Observed Frequency Expected Frequency
(x1 (25 20 np1)2 np1 20)2 (x2 (17 20 np2)2 np2 20)2
(x3 (15
np3)2 np3 20)2 20
(x4 (23
np4)2 np4 20)2 20
(x5 (24
np5)2 np5 20)2 20
(x6 (16
np6)2 np6 20)2 20 5.00
Since the number of categories or classes (faces 1, 2, 3, 4, 5, 6) is k 6, n k 1 6 1 5. The critical value x2 for 5 degrees of freedom is 11.1. Then since 5.00 11.1, we cannot reject the 0.95 hypothesis that the die is fair. For 5 degrees of freedom x2 1.15, so that x2 5.00 1.15. It follows that the agreement is not so 0.05 exceptionally good that we would look upon it with suspicion. (b) The P value of the test is P(x2 software, P 0.42. 5.00). The table in Appendix E shows 0.25 P 0.5. By computer
7.38. A random number table of 250 digits had the distribution of the digits 0, 1, 2, . . . , 9 shown in Table 7-11. (a) Does the observed distribution differ significantly from the expected distribution (b) What is the P value of the observation (a)
Digit Observed Frequency Expected Frequency (17 25 0 17 25 1 31 25 2 29 25 Table 7-11 3 18 25 4 14 25 5 20 25 6 35 25 7 30 25 8 20 25 9 36 25
25)2
(31 25
25)2
(29 25
25)2
(18 25
25)2
(36 25
25)2
The critical value x2 for n k 1 9 degrees of freedom is 21.7, and 23.3 21.7. Hence we 0.99 conclude that the observed distribution differs significantly from the expected distribution at the 0.01 level of significance. Some suspicion is therefore upon the table. (b) The P value is P(x2 23.3). The table in Appendix E shows that 0.005 software, P 0.0056. P 0.01. By computer
7.39. In Mendel s experiments with peas he observed 315 round and yellow, 108 round and green, 101 wrinkled and yellow, and 32 wrinkled and green. According to his theory of heredity the numbers should be in the proportion 9:3:3:1. Is there any evidence to doubt his theory at the (a) 0.01, (b) 0.05 level of significance (c) What is the P value of the observation
The total number of peas is 315 108 101 32 556. Since the expected numbers are in the proportion 9:3:3:1 (and 9 3 3 1 16), we would expect 9 (556) 16 3 (556) 16 312.75 round and yellow 104.25 round and green 3 (556) 16 1 (556) 16 104.25 wrinkled and yellow 34.75 wrinkled and green
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