barcode font reporting services Tests of Hypotheses and Significance in Software

Creation QR-Code in Software Tests of Hypotheses and Significance

CHAPTER 7 Tests of Hypotheses and Significance
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Then x2 (315 312.75)2 312.75 (108 104.25)2 104.25 (101 104.25)2 104.25 (32 34.75)2 37.75 4 1 3. 0.470
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Since there are four categories, k (a) For n (b) For n 3, x2 0.99 3, x2 0.95
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4 and the number of degrees of freedom is n
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11.3 so that we cannot reject the theory at the 0.01 level. 7.81 so that we cannot reject the theory at the 0.05 level.
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We conclude that the theory and experiment are in agreement. Note that for 3 degrees of freedom, x2 0.352 and x2 0.470 0.352. Therefore, although the 0.05 agreement is good, the results obtained are subject to a reasonable amount of sampling error. (c) The P value is P(x2 P 0.93. 0.470). The table in Appendix E shows that 0.9 P 0.95. By computer software,
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7.40. An urn consists of a very large number of marbles of four different colors: red, orange, yellow, and green. A sample of 12 marbles drawn at random from the urn revealed 2 red, 5 orange, 4 yellow, and 1 green marble. Test the hypothesis that the urn contains equal proportions of the differently colored marbles, and find the P value of the sample results.
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Under the hypothesis that the urn contains equal proportions of the differently colored marbles, we would expect 3 of each kind in a sample of 12 marbles. Since these expected numbers are less than 5, the chi-square approximation will be in error. To avoid this, we combine categories so that the expected number in each category is at least 5. If we wish to reject the hypothesis, we should combine categories in such a way that the evidence against the hypothesis shows up best. This is achieved in our case by considering the categories red or green and orange or yellow, for which the sample revealed 3 and 9 marbles, respectively. Since the expected number in each category under the hypothesis of equal proportions is 6, we have x2 (3 6 6)2 (9 6 6)2 3
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For n 2 1 1, x2 3.84. Therefore, we cannot reject the hypothesis at the 0.05 level of significance 0.95 (although we can at the 0.10 level). Conceivably the observed results could arise on the basis of chance even when equal proportions of the colors are present. The P value is P(x2 3) 0.083. Another method Using Yates correction, we find x2 (u 3 6u 6 0.5)2 (u 9 6u 6 0.5)2 (2.5)2 6 (2.5)2 6 2.1
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which leads to the same conclusion given above. This is of course to be expected, since Yates correction always reduces the value of x2. Here the P value is P(x2 2.1) 0.15. It should be noted that if the x2 approximation is used despite the fact that the frequencies are too small, we would obtain x2 (2 3 3)2 (5 3 3)2 (4 3 3)2 (1 3 3)2 3.33
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with a P value of 0.34. Since for n 4 1 3, x2 7.81, we would arrive at the same conclusions as above. Unfortunately 0.95 the x2 approximation for small frequencies is poor; hence when it is not advisable to combine frequencies we must resort to exact probability methods involving the multinomial distribution.
7.41. In 360 tosses of a pair of dice, 74 sevens and 24 elevens are observed. Using a 0.05 level of significance, test the hypothesis that the dice are fair, and find the P value of the observed results.
A pair of dice can fall in 36 ways. A seven can occur in 6 ways, an eleven in 2 ways.
CHAPTER 7 Tests of Hypotheses and Significance
6 36 1 6 2 36 1 18 .
Then P(seven) 1 sevens and 18 (360)
and P(eleven) 20 elevens, so that x2
Therefore, in 360 tosses we would expect 1 (360) 6 (24 20 20)2
(74 60
60)2
with a P value of 0.044. For n 2 1 1, x2 3.84. Then since 4.07 3.84, we would be inclined to reject the hypothesis 0.95 that the dice are fair. Using Yates correction, however, we find x2 (corrected) (u74 60u 60 0.5)2 (u 24 20u 20 0.5)2 (13.5)2 60 (3.5)2 20 3.65
with a P value of 0.056. Therefore, on the basis of the corrected x2, we could not reject the hypothesis at the 0.05 level. In general, for large samples such as we have here, results using Yates correction prove to be more reliable than uncorrected results. However, since even the corrected value of x2 lies so close to the critical value, we are hesitant about making decisions one way or the other. In such cases it is perhaps best to increase the sample size by taking more observations if we are interested especially in the 0.05 level. Otherwise, we could reject the hypothesis at some other level (such as 0.10).
7.42. A survey of 320 families with 5 children each revealed the distribution of boys and girls shown in Table 7-12. (a) Is the result consistent with the hypothesis that male and female births are equally probable (b) What is the P value of the sample results
(a) Number of Boys and Girls Number of Families 5 boys 0 girls 18 4 boys 1 girl 56 Table 7-12 3 boys 2 girls 110 2 boys 3 girls 88 1 boy 4 girls 40 0 boys 5 girls TOTAL 8 320
Let p probability of a male birth, and q 1 p probability of a female birth. Then the probabilities of (5 boys), (4 boys and 1 girl), . . . , (5 girls) are given by the terms in the binomial expansion (p If p q
1 2,
q)5 1 5 2
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