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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