probablility- chi square

8/10/2019 Probablility- Chi Square

1/13

Goodness of Fit Chi Square

By Rhondene Wint

CAPE BIOLOGY

Hypothesis Testing


2/13

Chi-Square TestA fundamental problem is genetics is determining

whether the experimentally determined data fits

the results expected from theory (i.e. Mendels

laws as expressed in the Punnett square).


3/13

ProblemA geneticist is researching if a certain gene in a group

of offspring follows the Mendelian pattern ofinheritance, when two heterozygote parents were

crossed for the presence the dominant phenotype (A)

and the recessive phenotype (a).

A a Totals

A

10 42 52

a 33 15 48

Totals 43 57 100

Table 4. Results of a monohybrid cross between two heterozygotes for the 'a' gene.


4/13

Hypothesis Testing

Hypothesis testing is a systematic way to test claimsor ideas about a group or population based on a

sample

Null hypothesis HO: A claim that there is nodifference between an observation and an

expected/theoretical claim

Alternative Hypothesis Ha:A claim that the

observation and expected claims are different


5/13

How To: Hypothesis Testing

Step 1: State the hypotheses.

Step 2: Set the criteria for a decision.

Step 3: Compute the test statistic.

Step 4: Make a decision


6/13

Step 1:

State the Null Hypothesis: The gene Aa is inherited according to the Mendelian

heterozygote ratio 3 dominant:1 recessive

State the alternative hypothesis:

The gene Aa is not inherited according the

Mendelian heterozygote ratio of 3 dominant:1

recessive


7/13

Confidence Limit

To set the criteria for a decision, we state thelevel of significance for a test.

The level we set, called the alphalevel or p

value. The p value is the probability of rejecting the null,

when it is actually true


8/13

Goodness of Fit Chi Square

How "close" are the observed values tothose which would be expected under

the fitted- ideal model?

This test allows us to compare acollection of categorical data with some

theoretical expected distribution. This

test is often used in genetics to comparethe results of a cross with the theoretical

distribution based on genetic theory.


9/13

Observed

Expect

ed

(O

E)

(OE)2 (OE)2/ E

A-

type

85 75 10 100 1.33

a-typ

e

15 25 10 100 4.0

Tota 100 100 5.33


10/13


11/13

Teach by organising data into tables

Interpreting Chi squareconfidence limit degrees

of freedom

D.f(# of row1) x (# columns-1)

when the computed x2statistic exceeds the

critical value in the table for a 0.05 probability

level, then we can reject the null hypothesis of

equal distributions Since our x2statistic (3.418) did not exceed the

critical value for 0.05 probability level (3.841)


12/13

Chi-Square Table


13/13

3.A genetics engineer was attempting

to cross a tiger and a cheetah. She

predicted a phenotypic outcome of thetraits she was observing to be in the

following ratio 4 stripes only: 3 spots

only: 9 both stripes and spots. Whenthe cross was performed and she

counted the individuals she found 50

with stripes only, 41 with spots onlyand 85 with both. According to the

Chi-square test, did she get the

predicted outcome?

probablility- chi square

Documents