Download - Kul 3 Chi Square 2012
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Analisis Chi Square
The chi square analysis allows youto use statistics to determine if your
data good or not. In our fruit fly labs we are using laws of probability to
determine possible outcomes for genetic crosses.
How will we know if our fruit fly data is good?
xBlackbody,
eyeless
F1: all wild
wild
F1 x F1
5610
1881
1896
622
Jika F1 X F1 menghasilkan F2
dengan Rasio : 9:3:3:1
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The chi-square distribution can be used tosee whether or not an observed counts agree with an expected
counts.
Let
O = observed count and
E = Expected count
E
EO 2)(2
For testing significance of patterns in qualitative data
Test statistic is based on counts that represent the number of
items that fall in each category
Test statistics measures the agreement between actual counts
and expected counts assuming the null hypothesis
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The test statistic is compared to a theoretical
probability distribution
In order to use this distribution properly you need todetermine the degrees of freedom
Degrees of freedom is the number of phenotypic
possibilities in your cross minus one.
If the level of significance read from the table isgreater than .05 or 5% then your hypothesis is
accepted and the data is useful
The hypothesis is termed the null hypothesis
which states that there is no substantial
statistical deviation between observed and
expected data.
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Steps in hypothesis testing
1. State the hypotheses null
research Null hypothesis
Specifies a distribution of proportions
Research hypothesis
Specifies that the distribution will be different than that indicated in the null
hypothesis
1. Select an alpha level and determine the critical
value
2. Compute the test statistic
3. Make a decision (Kesimpulan)
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Calculating the test statistic Observed frequencies
the number of individuals from the samplewho are classified in a particular category
fo
Expected frequencies
the number of individuals from the samplewho are expectedto be classified in a
particular category fe
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Calculating the test statistic
Heads Tails
Percentages 50%50%
Proportions .5 .5
Coin flip: What percentage of people will predict heads? tails?
Expected Heads Tails
Proportions .5 .5
Frequencies 25 25
Expected frequency = fe =pn
n = 50 (sample size)
fe = .5 x 50 = 25
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Calculating the test statistic
x2 = (fo - fe)2fe
Heads Tails
Observed 35 15
Expected 25 25
Steps
1. find the difference between fo and fe foreach category
2. square the difference
3. divide the squared difference by fe
4. sum the values from all categories
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x2 = (fo - fe)2 = 4 + 4 = 8fe
Heads Tails
Observed (fo) 35 15
Expected (fe) 25 25
fo - fe 10 -10
(fo - fe)2 100 100
(fo - fe)2/fe 4 4
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Goodness of fit
Membuat Kesimpulan:
Critical value = 3.84 with df = 1 and = .05.
Observed chi square = 8.0
8.0 > 3.84
Observed chi square is greater than critical value
We reject the null hypothesis
Conclude that category frequencies are different
People were more likely to predict heads than tails
2
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Observed Expected
Frequency Frequency
H 40 50
T 60 50
sum 100 100
2
22
2 2
2 2
40 50
50
60 50
50
10
50
10
50
100
50
100
50
2 2
4
statistic formula
O E
E
( )
( ) ( )
( ) ( )
2 f l
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Observed Expected
Die Frequency Frequency
1 4 10
2 6 10
3 17 10
4 16 10
5 8 10
6 9 10
sum 60 60
2
22
2 2
2 2
2 2
4 10
10
6 10
10
17 10
10
16 10
10
8 1010
9 1050
14 2
statistic formula
O E
E
( )
( ) ( )
( ) ( )
( ) ( )
.
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Critical values for chi square distribution
Critical value (df= 1, = .05) = 3.84
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Chi square A chi square test is an inferential statistic
Analyzes proportions/frequency data
Uses proportions/distributions from a sample to test hypothesesabout proportions/ distributions in the population
Two tests Chi square test for goodness of fit
Chi square test for independence
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The Genetics Analysis:
To compute the hypothesis value take 10009/16
= 626
Phenotype Observed Hypothesis
Wild 5610 5634
Eyeless 1881 1878
Black body 1896 1878
Eyeless, black body 622 626
Total 10009
9/16 wild type: 3/16 normal body eyeless: 3/16
black body wild eyes: 1/16 black body eyeless.
Using the chi square formula
compute the chi square total
for this cross:
(5610 - 5630)2/ 5630 = .07
(1881 - 1877)2/ 1877 = .01
(1896 - 1877 )2/ 1877 = .20
(622 - 626) 2/ 626 = .02
2= .30
How many degrees offreedom? 3
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CHI-SQUARE DISTRIBUTION TABLE
Accept Hypothesis RejectHypothesis
Probability (p)
Degrees ofFreedom
0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001
1 0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83
2 0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82
3 0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27
4 0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.38 18.47
5 1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52
6 1.63 2.20 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46
7 2.17 2.83 3.82 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32
8 2.73 3.49 4.59 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.12
9 3.32 4.17 5.38 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88
10 3.94 4.86 6.18 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59