Explanation Given info: A random sample of 469 teenage girls has been taken to test the rate of twin births. Calculation: The test hypotheses are given below: Null hypothesis: H 0 : p = 0.03 _ That is, the proportion of twin births for young mothers is 3%. Alternative hypothesis: H A : p ≠ 0.03 _ That is, the proportion of twin births for young mothers is differs from 3%. Verify the condition: Independence assumption: The twin birth given by one mother does not affect the birth given by another. The samples taken from the mother is independent. Randomization condition: The sample is not random, it is more reasonable to consider the mothers giving twin birth as the representative of all the teen mothers. Success/failure condition : n p 0 ≥ 10 , n q 0 ≥ 10 Condition 1: n p 0 = ( 469 ) ( 0.03 ) = 14.07 ( > 10 ) Condition 2: n q 0 = ( 469 ) ( 0.97 ) = 454.93 ( > 10 ) Here, n p 0 and n q 0 is greater than 10. Hence, the success/failure condition is satisfied 10% condition: size of 469 is less than the10% population of all mothers

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Chapter 17, Problem 31E

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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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Whether the proportion of twin births for young mothers is different.

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