13) You perform a chi square test on a monohybrid cross (Aa x Aa) and determine

Question

13) You perform a chi square test on a monohybrid cross (Aa x Aa) and determine that chi square = 3.22. (HINT: Your df is the number of phenotypic groups minus 1). What is your p value based on this table? A. 3.22- B. 0.2 Probability (p) D. Between 0.5 and 0.2 0.90 | D..0 0.05 | 0.01 | 0.001 2 10.02 0.46 1.64 3.846.64 10.83 1.393.22 5.99 9.21 13.82 3 0.58 2.37 4.64 7.8211.316.27 4 1.06 3.36 5.99 9.49 13.2818.47 5 1.61 4.35 7.29 11.07 15.09 20.52 6 2.20 5.35 8.56 12.59 16.81 22.46 df 7 2.836.35 9.80 14.07 18.48 24.32 8 3.49 7.34 11.03 15.51 20.09 26.13 9 4.17 8.34 12.24 16.92 21.67 27.88 10 4.87 9.34 13.44 18.3123.21 29.59 15 8.55 14.34 19.31 25.00 30.58 37.30 25 16.47 24.34 30.68 37.65 44.31 52.62 50 37.69 49.34 58.16 67.51 76.15 86.60 E. Between 0.2 and 0.05 5 3 values

Explanation / Answer

CHI SQUARE

It is an alternative method of testing significance of difference between 2 propoertions. Here, Data is measured interms quality. It is non parametric ( not include maean , standard deviation method) It was developed by KARL PEARSON It is very important test in satistic & can be used for reasearch purpose with several variables on one go.

chi square ( x2) = sigma ( O-E)

WHERE

O= Observed frequency

E= Expected frequency

df ( degree cof freedom) - It is depend on number of columns & rouws in original table.

df = ( column -1 ) ( row - 1)

NOW AS GIVEN IN THE QUESTION,

1- chi square ( x2) = 3.22  

2- df = n-1

3- Here monohybrid sample is ued ( 3:1)( punnet square according to

mendel theory of genetics

therefore

n= 2 ( monohybride ) ( 2 samples are used to cross)

so, df = n-1

= 2-1

df = 1

so, when df=1 & x2= 3.22 then (p ) probability will be

df =1 ( 1st row ), x2= 3.22 p= 0.05

so option E = between 0.2 to 0.05 is correct

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