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How do I calculate the P value?

How do I calculate the P value?

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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How do I calculate the P value?
T Seed capsules of the Shepherd's pusse are either triangular or ovid. A cross blu a pure breeding plant with Ovoid seed capsules and a pure breeding plant with triangular seed Capsules viewed Fl hybrids that all have triangular seed capsules. The Fl hybrid self-cross produced 50 F2 plants, 72 of which had triangular seed Capsules and 8 of which had ovoid Seed Capswer You state your null hypothesis as follows: There is no significant differen blw the observed data and the expected ratio of 3:1 that would be predicted for a monohybrid cross. Any deviation abserved is simply due to random chance. a) Expected Ratio: 3:1 Fractions: 3/4 i b) Phena typic Class Observed # Expected # 72 8 80 Triangular Ovaid Total Expected Triangular: Expected Ovoid : 80 × 3/4 = 60 80 x 1/4= 20 Triangular: (72-60)² ( 60 ) Ovoid: (8-20)2 = 7.2 (20) d) P-value P-value: 0.00194577 = 2.4 60 20 80 P<0.05= reject the null hypothes P>0.05 fail to reject 1/4 (0-E)/E 2.4 7. 2 x² = 9.6 Total = 2.4+7.2 c) Degrees of Freedom : Degrees of = (# of phenotypic class - 1) = (2-1) = 1 Freedom 5 Nitrogenous Bases > U, T, C, A, G = 9.6 Chi square vake What is a P-valve = the probability that we would see the degree of deviation observed here given that the null hypothesis is true.
Transcribed Image Text:T Seed capsules of the Shepherd's pusse are either triangular or ovid. A cross blu a pure breeding plant with Ovoid seed capsules and a pure breeding plant with triangular seed Capsules viewed Fl hybrids that all have triangular seed capsules. The Fl hybrid self-cross produced 50 F2 plants, 72 of which had triangular seed Capsules and 8 of which had ovoid Seed Capswer You state your null hypothesis as follows: There is no significant differen blw the observed data and the expected ratio of 3:1 that would be predicted for a monohybrid cross. Any deviation abserved is simply due to random chance. a) Expected Ratio: 3:1 Fractions: 3/4 i b) Phena typic Class Observed # Expected # 72 8 80 Triangular Ovaid Total Expected Triangular: Expected Ovoid : 80 × 3/4 = 60 80 x 1/4= 20 Triangular: (72-60)² ( 60 ) Ovoid: (8-20)2 = 7.2 (20) d) P-value P-value: 0.00194577 = 2.4 60 20 80 P<0.05= reject the null hypothes P>0.05 fail to reject 1/4 (0-E)/E 2.4 7. 2 x² = 9.6 Total = 2.4+7.2 c) Degrees of Freedom : Degrees of = (# of phenotypic class - 1) = (2-1) = 1 Freedom 5 Nitrogenous Bases > U, T, C, A, G = 9.6 Chi square vake What is a P-valve = the probability that we would see the degree of deviation observed here given that the null hypothesis is true.
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