14. The formula for Chi Square (x) is: * = chi squared x² = E (0; – E;)? E; O; = observed value E¡ = expected value Chi-square Table. Probabilities a 0.95 0.90 0.70 0.50 0.30 0.20 0.10 0.05 0.01 Df 1 0.004 0.016 0.15 0.46 1.07 1.64 2.71 3.84 6.64 0.10 0.21 1.39 1.39 2.41 3.22 4.61 5.99 9.21 3 0.35 0.58 1.42 2.37 3.67 4.64 6.25 7.82 11.35 4 0.71 1.06 2.20 3.36 4.88 5.99 7.78 9.49 13.28 1.15 1.61 3.00 4.35 6.06 7.29 9.24 11.07 15.09 1.64 2.20 3.83 5.35 7.23 8.56 10.65 12.59 16.81 7 2.17 2.83 4.67 6.35 8.38 9.80 12.02 14.07 18.48 8 2.73 3.49 5.53 7.34 9.52 11.03 13.36 15.51 20.09 3.33 4.17 6.39 8.34 10.66 12.24 14.68 16.92 21.67 Chi-Square Value (X²). If the observed had equaled the expected, the value would have been zero. Thus, a small X value indicates that the observed and expected ratios are in close agreement. Into a container put 270 corn kernels, 90 rice grains, 90 mongo seeds and 30 squash seeds. Mix them together by stirring or shaking. Supposing that we are crossing two double heterozygotic pea plants (AaBb x AaBb). A=tall; a=Dwarf; B=Normal; b=wrinkled. The four types of seeds represent the four possible phenotypes of the cross: Tall- normal (Corn), Tall-wrinkled (Rice), Dwarf-normal (Mongo) and Dwarf-wrinkled (Squash). I. Chi-Square Test d? d/E Observed Frequency (0) Seed Phenotype Class Expected Frequency (E) Deviation (d=0-E) Corr Rice Mongo Squash What can you conclude based on the value of the computed Chi-square? How can you relate the two principles of Mendel to Chi-Square Values?
14. The formula for Chi Square (x) is: * = chi squared x² = E (0; – E;)? E; O; = observed value E¡ = expected value Chi-square Table. Probabilities a 0.95 0.90 0.70 0.50 0.30 0.20 0.10 0.05 0.01 Df 1 0.004 0.016 0.15 0.46 1.07 1.64 2.71 3.84 6.64 0.10 0.21 1.39 1.39 2.41 3.22 4.61 5.99 9.21 3 0.35 0.58 1.42 2.37 3.67 4.64 6.25 7.82 11.35 4 0.71 1.06 2.20 3.36 4.88 5.99 7.78 9.49 13.28 1.15 1.61 3.00 4.35 6.06 7.29 9.24 11.07 15.09 1.64 2.20 3.83 5.35 7.23 8.56 10.65 12.59 16.81 7 2.17 2.83 4.67 6.35 8.38 9.80 12.02 14.07 18.48 8 2.73 3.49 5.53 7.34 9.52 11.03 13.36 15.51 20.09 3.33 4.17 6.39 8.34 10.66 12.24 14.68 16.92 21.67 Chi-Square Value (X²). If the observed had equaled the expected, the value would have been zero. Thus, a small X value indicates that the observed and expected ratios are in close agreement. Into a container put 270 corn kernels, 90 rice grains, 90 mongo seeds and 30 squash seeds. Mix them together by stirring or shaking. Supposing that we are crossing two double heterozygotic pea plants (AaBb x AaBb). A=tall; a=Dwarf; B=Normal; b=wrinkled. The four types of seeds represent the four possible phenotypes of the cross: Tall- normal (Corn), Tall-wrinkled (Rice), Dwarf-normal (Mongo) and Dwarf-wrinkled (Squash). I. Chi-Square Test d? d/E Observed Frequency (0) Seed Phenotype Class Expected Frequency (E) Deviation (d=0-E) Corr Rice Mongo Squash What can you conclude based on the value of the computed Chi-square? How can you relate the two principles of Mendel to Chi-Square Values?
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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Transcribed Image Text:14. The formula for Chi Square (x) is:
X* = chi squared
x² =E
(0; - E;)?
O: = observed value
E;
E = expected value
Chi-square Table.
Probabilities
a
0.95
0.90
0.70
0.50
0.30
0.20
0.10
0.05
0.01
Df
1
0.004
0.016
0.15
0.46
1.07
1.64
2.71
3.84
6.64
2
0.10
0.21
1.39
1.39
2.41
3.22
4.61
5.99
9.21
3
0.35
0.58
1.42
2.37
3.67
4.64
6.25
7.82
11.35
4
0.71
1.06
2.20
3.36
4.88
5.99
7.78
9.49
13.28
1.15
1.61
3.00
4.35
6.06
7.29
9.24
11.07
15.09
1.64
2.20
3.83
5.35
7.23
8.56
10.65
12.59
16.81
7
2.17
2.83
4.67
6.35
8.38
9.80
12.02
14.07
18.48
8
2.73
3.49
5.53
7.34
9.52
11.03
13.36
15.51
20.09
9
3.33
4.17
6.39
8.34
10.66
12.24
14.68
16.92
21.67
Chi-Square Value (X*). If the observed had equaled the expected, the value would
have been zero. Thus, a small X value indicates that the observed and expected
ratios are in close agreement.
Into a container put 270 corn kernels, 90 rice grains, 90 mongo seeds and 30 squash
seeds. Mix them together by stirring or shaking.
Supposing that we are crossing two double heterozygotic pea plants (AaBb x AaBb).
A=tall; a=Dwarf; B=Normal; b=wrinkled.
The four types of seeds represent the four possible phenotypes of the cross: Tall-
normal (Corn), Tall-wrinkled (Rice), Dwarf-normal (Mongo) and Dwarf-wrinkled
(Squash).
I.
Chi-Square Test
d?
d/E
Observed
Frequency
(0)
Seed
Phenotype
Class
Expected
Frequency
Deviation
(d=0-E)
Corn
Rice
Mongo
Squash
What can you conclude based on the value of the computed Chi-square?
How can you relate the two principles of Mendel to Chi-Square Values?
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