Your task is to perform a chi-square goodness of fit test on the F2 data and make a decision, based on your statistical analysis, as to whether you reject or do not reject the computer generated mode of inheritance as being consistent with the observed data. The mode of inheritance you are to test on the observed data is autosomal recessive. PARENTAL CROSS Parental cross: Father with disease phenotype, Mother with wild-type phenotype. F1 DATA Phenotype Gender Disease Wild-Type Male 18 23 Female 22 17 F2 DATA Phenotype Gender Disease Wild-Type Male 22 23 Female 15 20 Gender Phenotype Observed counts (O) Expected Proportions Expected Counts (E) (O-E) (O-E)^2/E Male WT 23 Male Disease 22 Female WT 20 Female Disease 15 Total 80 DegFrdm P-value type it with the calculation please
Your task is to perform a chi-square goodness of fit test on the F2 data and make a decision, based on your statistical analysis, as to whether you reject or do not reject the computer generated mode of inheritance as being consistent with the observed data. The mode of inheritance you are to test on the observed data is autosomal recessive. PARENTAL CROSS Parental cross: Father with disease phenotype, Mother with wild-type phenotype. F1 DATA Phenotype Gender Disease Wild-Type Male 18 23 Female 22 17 F2 DATA Phenotype Gender Disease Wild-Type Male 22 23 Female 15 20 Gender Phenotype Observed counts (O) Expected Proportions Expected Counts (E) (O-E) (O-E)^2/E Male WT 23 Male Disease 22 Female WT 20 Female Disease 15 Total 80 DegFrdm P-value type it with the calculation please
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Your task is to perform a chi-square goodness of fit test
on the F2 data and make a decision, based on your statistical analysis, as to whether
you reject or do not reject the computer generated
with the observed data.
The mode of inheritance you are to test on the observed data is autosomal recessive.
PARENTAL CROSS
Parental cross: Father with disease phenotype, Mother with wild-type phenotype.
F1 DATA
Phenotype
Gender Disease Wild-Type
Male 18 23
Female 22 17
F2 DATA
Phenotype
Gender Disease Wild-Type
Male 22 23
Female 15 20
Gender | Phenotype | Observed counts (O) |
Expected Proportions |
Expected Counts (E) |
(O-E) | (O-E)^2/E |
Male | WT | 23 | ||||
Male | Disease | 22 | ||||
Female | WT | 20 | ||||
Female | Disease | 15 | ||||
Total | 80 | |||||
DegFrdm | ||||||
P-value |
type it with the calculation please
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