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8/10/2019 Probablility- Chi Square
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Goodness of Fit Chi Square
By Rhondene Wint
CAPE BIOLOGY
Hypothesis Testing
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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).
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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.
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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
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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
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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
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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
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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.
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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
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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)
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Chi-Square Table
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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?