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OB 0²A=x0² ₁A= m-1 6²B = 02²/2 X ½-1 =

OB 0²A=x0² ₁A= m-1 6²B = 02²/2 X ½-1 =

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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IS Q.9. The following results were obtained from a laboratory of Pune from two samples. Sample Size Sample average Standard deviation A 10 B 10 Does the result indicate they are identical (F= 9.9 df 3.18). Solution: Null Hypothesis (Ho): The sample population were identical Hozo = 02 0 Calculation: A: n₁ = 10 X = 2.01 0A = 0.50 А B: 12 = 10 Y = 1.96 OB = 0.031 0²₁A = 1 × 0²/² m-1 B = Since o B > of A mx 02/²2 12-1 2.01 1.96 F = 10× (0.050)² 10-1 10 × (0.031)² 10-1 0.050 0.031 10 ×.0025 0.025 9 9. 10×.00096 9 = : 0.357 = 0.38 = 0.0028 0.0096 9 0.0010 0.0028 • Critical value: The tabulated value of F at 0.05 level is (9, 9) df is 3.18. Decision: Since the calculated value of F is 0.38<table value of F i.e., 3.18 at df (9, 9) so the null hypothesis is accepted. = 0.0010
Transcribed Image Text:IS Q.9. The following results were obtained from a laboratory of Pune from two samples. Sample Size Sample average Standard deviation A 10 B 10 Does the result indicate they are identical (F= 9.9 df 3.18). Solution: Null Hypothesis (Ho): The sample population were identical Hozo = 02 0 Calculation: A: n₁ = 10 X = 2.01 0A = 0.50 А B: 12 = 10 Y = 1.96 OB = 0.031 0²₁A = 1 × 0²/² m-1 B = Since o B > of A mx 02/²2 12-1 2.01 1.96 F = 10× (0.050)² 10-1 10 × (0.031)² 10-1 0.050 0.031 10 ×.0025 0.025 9 9. 10×.00096 9 = : 0.357 = 0.38 = 0.0028 0.0096 9 0.0010 0.0028 • Critical value: The tabulated value of F at 0.05 level is (9, 9) df is 3.18. Decision: Since the calculated value of F is 0.38<table value of F i.e., 3.18 at df (9, 9) so the null hypothesis is accepted. = 0.0010
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