Explanation Given info: The data shows that the first 30 decimal places of pi. Calculation: From the given sequence the number of odd numbers is n 1 = 17 and the number of even numbers is n 2 = 13 . The hypotheses are given below: Null hypothesis: H 0 : The sequence of digits is random. Alternative hypothesis: H 1 : The sequence of digits is not random. Value of G: Let G be the number of runs in the sequence. A run is the sequence of the same observations. Then, the runs are calculated as follows: Run 1 = { O } Run 2 = { E } Run 3 = { O, O, O } Run 4 = { E, E } Run 5 = { O, O, O } Run 6 = { E } Run 7 = { O, O, O , O } Run 8 = { E }

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Chapter 13.7, Problem 6BSC

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To test: The randomness of odd (O) and eve (E) digits.

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Chapter 13.7, Problem 5BSC

Chapter 13.7, Problem 7BSC

Chapter 13 Solutions

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Ch. 13.2 - In Exercises 9-12, use the sign test for the data...Ch. 13.2 - Prob. 12BSC Ch. 13.2 - Prob. 13BSC Ch. 13.2 - In Exercises 13-16, use the sign test for the...Ch. 13.2 - Prob. 15BSC Ch. 13.2 - In Exercises 13-16, use the sign test for the...Ch. 13.2 - Appendix B Data Sets.In Exercises 13-16, refer to...Ch. 13.2 - Prob. 18BSC Ch. 13.2 - Appendix B Data Sets.In Exercises 17-20, refer to...Ch. 13.2 - Appendix B Data Sets.In Exercises 17-20, refer to...Ch. 13.2 - Procedures for Handling Ties In the sign test...Ch. 13.2 - Prob. 22BB Ch. 13.3 - Wilcoxon Signed-Ranks Test for Freshman 15 The...Ch. 13.3 - Prob. 2BSC Ch. 13.3 - Prob. 3BSC Ch. 13.3 - Prob. 4BSC Ch. 13.3 - Prob. 5BSC Ch. 13.3 - Prob. 6BSC Ch. 13.3 - Using the Wilcoxon Signed-Ranks Test.In Exercises...Ch. 13.3 - Prob. 8BSC Ch. 13.3 - Prob. 9BSC Ch. 13.3 - Prob. 10BSC Ch. 13.3 - Appendix B Data Sets.In Exercises 9-12, refer to...Ch. 13.3 - Prob. 12BSC Ch. 13.3 - Prob. 13BB Ch. 13.4 - Prob. 1BSC Ch. 13.4 - Prob. 2BSC Ch. 13.4 - Prob. 3BSC Ch. 13.4 - Prob. 4BSC Ch. 13.4 - Prob. 5BSC Ch. 13.4 - Prob. 6BSC Ch. 13.4 - Prob. 7BSC Ch. 13.4 - Prob. 8BSC Ch. 13.4 - Prob. 9BSC Ch. 13.4 - Prob. 10BSC Ch. 13.4 - Prob. 11BSC Ch. 13.4 - Prob. 12BSC Ch. 13.4 - Prob. 13BB Ch. 13.4 - Prob. 14BB Ch. 13.5 - Effect of Lead on IQ Score Listed below are full...Ch. 13.5 - Prob. 2BSC Ch. 13.5 - Notation For the data given in Exercise 1,...Ch. 13.5 - Prob. 4BSC Ch. 13.5 - Prob. 5BSC Ch. 13.5 - Prob. 6BSC Ch. 13.5 - Prob. 7BSC Ch. 13.5 - Using tho Kruskal-Wallis Test.In Exercises 5-8,...Ch. 13.5 - Prob. 9BSC Ch. 13.5 - Prob. 10BSC Ch. 13.5 - Prob. 11BSC Ch. 13.5 - Car Crash Measurements Refer to Data See 13 in...Ch. 13.5 - Prob. 13BB Ch. 13.6 - Regression If the methods of this section arc used...Ch. 13.6 - Level of Measurement Which of the levels of...Ch. 13.6 - Prob. 3BSC Ch. 13.6 - Prob. 4BSC Ch. 13.6 - Prob. 5BSC Ch. 13.6 - Prob. 6BSC Ch. 13.6 - Prob. 7BSC Ch. 13.6 - Testing for Rank Correlation. In Exercises 7-12,...Ch. 13.6 - Prob. 9BSC Ch. 13.6 - Testing for Rank Correlation. In Exercises 7-12,...Ch. 13.6 - Prob. 11BSC Ch. 13.6 - Prob. 12BSC Ch. 13.6 - Appendix B Data Sets. In Exercises 13-16, use the...Ch. 13.6 - Prob. 14BSC Ch. 13.6 - Appendix B Data Sets. In Exercises 13-16, use the...Ch. 13.6 - Prob. 16BSC Ch. 13.6 - Prob. 17BB Ch. 13.7 - Prob. 1BSC Ch. 13.7 - Prob. 2BSC Ch. 13.7 - Prob. 3BSC Ch. 13.7 - Prob. 4BSC Ch. 13.7 - Prob. 5BSC Ch. 13.7 - Prob. 6BSC Ch. 13.7 - Prob. 7BSC Ch. 13.7 - Prob. 8BSC Ch. 13.7 - Prob. 9BSC Ch. 13.7 - Baseball World Series Victories Test the claim...Ch. 13.7 - Prob. 11BSC Ch. 13.7 - Prob. 12BSC Ch. 13 - Prob. 1CQQ Ch. 13 - Prob. 2CQQ Ch. 13 - Prob. 3CQQ Ch. 13 - Prob. 4CQQ Ch. 13 - Prob. 5CQQ Ch. 13 - Prob. 6CQQ Ch. 13 - Prob. 7CQQ Ch. 13 - Prob. 8CQQ Ch. 13 - Prob. 9CQQ Ch. 13 - Sign Test Identify three different applications of...Ch. 13 - Prob. 1RE Ch. 13 - Prob. 2RE Ch. 13 - Prob. 3RE Ch. 13 - Prob. 4RE Ch. 13 - Prob. 5RE Ch. 13 - Using Nonparametric Tests. In Exercises 110, use a...Ch. 13 - Prob. 7RE Ch. 13 - Prob. 8RE Ch. 13 - Prob. 9RE Ch. 13 - Prob. 10RE Ch. 13 - Prob. 1CRE Ch. 13 - Please be aware that some of the following...Ch. 13 - Prob. 3CRE Ch. 13 - Prob. 4CRE Ch. 13 - Prob. 5CRE Ch. 13 - Prob. 6CRE Ch. 13 - Prob. 7CRE Ch. 13 - Prob. 8CRE Ch. 13 - Prob. 9CRE Ch. 13 - Prob. 10CRE Ch. 13 - Prob. 11TP Ch. 13 - Prob. 10FDD

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Population Genetics In the study of population genetics, an important measure of inbreeding is the proportion of homozygous genotypesthat is, instances in which the two alleles carried at a particular site on an individuals chromosomes are both the same. For population in which blood-related individual mate, them is a higher than expected frequency of homozygous individuals. Examples of such populations include endangered or rare species, selectively bred breeds, and isolated populations. in general. the frequency of homozygous children from mating of blood-related parents is greater than that for children from unrelated parents Measured over a large number of generations, the proportion of heterozygous genotypesthat is, nonhomozygous genotypeschanges by a constant factor 1 from generation to generation. The factor 1 is a number between 0 and 1. If 1=0.75, for example then the proportion of heterozygous individuals in the population decreases by 25 in each generation In this case, after 10 generations, the proportion of heterozygous individuals in the population decreases by 94.37, since 0.7510=0.0563, or 5.63. In other words, 94.37 of the population is homozygous. For specific types of matings, the proportion of heterozygous genotypes can be related to that of previous generations and is found from an equation. For mating between siblings 1 can be determined as the largest value of for which 2=12+14. This equation comes from carefully accounting for the genotypes for the present generation the 2 term in terms of those previous two generations represented by for the parents generation and by the constant term of the grandparents generation. a Find both solutions to the quadratic equation above and identify which is 1 use a horizontal span of 1 to 1 in this exercise and the following exercise. b After 5 generations, what proportion of the population will be homozygous? c After 20 generations, what proportion of the population will be homozygous?

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To test: The randomness of odd (O) and eve (E) digits.

Solution Summary: The author explains that the sequence of digits is random.

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To test: The randomness of odd (O) and eve (E) digits.

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Chapter 13 Solutions