Explanation Calculation: From the given information the value of α is 0.01, and to test the claim that the distributions of growths are different for root stocks A and B . The null and alternative hypothesis is, Null hypothesis: H 0 : The distributions of growths are same for root stocks A and B . Alternative hypothesis: H 1 : The distributions of growths are different for root stocks A and B . Test statistic : The z value for the sample test statistic R is, z = R − μ R σ R In the formula R is the sum of ranks from the sample of size n 1 (smaller sample), μ R = n 1 ( n 1 + n 2 + 1 ) 2 σ R = n 1 n 2 ( n 1 + n 2 + 1 ) 12 n 1 be the sample size of the smaller sample and n 2 be the sample size of the larger sample Procedure for assigned rank to data values: First combine both the samples. Arrange the data values in ascending order. Rank each of the data value in sequential order. The rank for each of tied data is computed as, First assign the sequential position ranks for all the values that are same. Take the mean of all the position ranks that are assigned for same data values. Assign the mean rank as the rank position for all the tied data. The assigned mean rank is used in calculating the test statistic. The ranks are, Meters Rank Stock 1.58 1 B 1.94 2 A 1.96 3 B 2.1 4 A 2.13 5 B 2.26 6 A 2.37 7 A 2.44 8 B 2.52 9 B 2.58 10 A 2.71 11 B 2.74 12 A 2.77 13 B 2.78 14 B 2.81 15 A 2.86 16 B 2.88 17 A 2.91 18 B 2.94 19 A 3.02 20 B 3.11 21 A 3.41 22 A The smaller sample is 11 which corresponds the both ‘ A ’, and ‘ B ’. That is, n 1 = 11 , n 2 = 11 . The value of R is, R = 2 + 4 + 6 + 7 + 10 + 12 + 15 + 17 + 19 + 21 + 221 = 135 The value of R is 135

Understandable Statistics: Concepts and Methods

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ISBN: 9781337119917

Author: Charles Henry Brase, Corrinne Pellillo Brase

Publisher: Cengage Learning

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Understandable Statistics: Concepts and Methods

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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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Identify whether the distributions of growths are different for root stocks A and B . Interpret the results.

Solution Summary: The author concludes that there is no evidence that the distributions of growths are different for root stocks A and B

Videos

Identify whether the distributions of growths are different for root stocks A and B.

Interpret the results.

Chapter 11 Solutions

Identify whether the distributions of growths are different for root stocks A and B . Interpret the results.

Solution Summary: The author concludes that there is no evidence that the distributions of growths are different for root stocks A and B

Videos

Identify whether the distributions of growths are different for root stocks A and B. Interpret the results.

Chapter 11 Solutions

Identify whether the distributions of growths are different for root stocks A and B.

Interpret the results.