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Data result of Testing Hardy-Weinberg Equilibrium with natural selection Chi-square of results from bean model for F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 3.1 The resulting chi-squared value is 3.1. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 3.1 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis. Chi-square of results from bean model for F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 6.5 The resulting chi-squared value is 6.5. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 6.5 is greater than the critical value (5.99), Thus, we reject the null hypothesis. c. The population is not in Hardy-Weinberg Equilibrium, and evolution is indeed taking place or does not conform to null hypothesis.

Data result of Testing Hardy-Weinberg Equilibrium with natural selection Chi-square of results from bean model for F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 3.1 The resulting chi-squared value is 3.1. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 3.1 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis. Chi-square of results from bean model for F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 6.5 The resulting chi-squared value is 6.5. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 6.5 is greater than the critical value (5.99), Thus, we reject the null hypothesis. c. The population is not in Hardy-Weinberg Equilibrium, and evolution is indeed taking place or does not conform to null hypothesis.

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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2. What would have happened over a long period of time?
Data result of Testing Hardy-Weinberg Equilibrium with natural selection Chi-square of results from bean model for F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 3.1 The resulting chi-squared value is 3.1. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 3.1 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis. Chi-square of results from bean model for F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 6.5 The resulting chi-squared value is 6.5. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 6.5 is greater than the critical value (5.99), Thus, we reject the null hypothesis. c. The population is not in Hardy-Weinberg Equilibrium, and evolution is indeed taking place or does not conform to null hypothesis.
Transcribed Image Text:Data result of Testing Hardy-Weinberg Equilibrium with natural selection Chi-square of results from bean model for F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 3.1 The resulting chi-squared value is 3.1. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 3.1 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis. Chi-square of results from bean model for F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 6.5 The resulting chi-squared value is 6.5. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 6.5 is greater than the critical value (5.99), Thus, we reject the null hypothesis. c. The population is not in Hardy-Weinberg Equilibrium, and evolution is indeed taking place or does not conform to null hypothesis.
Data result of Testing Hardy-Weinberg Equilibrium without natural selection Chi-square of results from bean model F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 0.21 The resulting chi-squared value is 0.21. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 0.21 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis Chi-square of results from bean model F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 2.52 The resulting chi-squared value is 2.52. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 2.52 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis
Transcribed Image Text:Data result of Testing Hardy-Weinberg Equilibrium without natural selection Chi-square of results from bean model F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 0.21 The resulting chi-squared value is 0.21. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 0.21 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis Chi-square of results from bean model F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 2.52 The resulting chi-squared value is 2.52. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 2.52 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis
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