Transcript Statistical Analysis: Chi

Statistical Analysis: Chi Square
AP Biology
Ms. Haut
“Goodness of Fit Test”
Chi Square (χ2)
• Statistical analysis used to determine
whether data obtained experimentally
provides a “good fit” to the expected data
• Used to determine if any deviations from
the expected results are due to random
chance alone or to other circumstances
Chi Square
2
(χ )
• Use the equation to test the “null” hypothesis
– The prediction that data from the experiment will match
the expected results
• χ2 = Σ (observed results – expected results)2
expected results
• When all else is equal, the value of χ2 increases as
the difference between the observed and expected
values increase
Chi Square
2
(χ )
• Once you have calculated the value of χ2,
you must determine the probability that the
difference between the observed and
expected values (χ2) occurred simply by
chance (sample error)
• You compare the calculated value to the
appropriate value in a “degrees of freedom”
table
Degrees of Freedom
• Degrees of freedom = # of categories – 1
• Takes into account the natural increase in χ2
as the number of categories increases
df
1
2
3
4
5
6
7
P = 0.05 P = 0.01 P = 0.001
3.84
5.99
7.82
9.49
11.07
12.59
14.07
6.64
9.21
11.35
13.28
15.09
16.81
18.48
10.83
13.82
16.27
18.47
20.52
22.46
24.32
• Scientists usually say that if their
probability of getting the observed deviation
from the expected results by chance is
greater that 0.05 (5%), then we can accept
the null hypothesis.
– The differences we see between what we
expected and what we actually see happened by
chance sampling error
• If the probability is less than 0.05, then we
reject the null hypothesis
– The differences between expected and observed
results did not just happen by chance sampling
error
Practice
•
A newly identified fruit fly mutant, cyclops eye
(large and single in the middle of the head), is
hypothesized to be autosomal dominant. The
experimenter started with homozygous wild type
females and homozygous cyclops males. The
data from the F2 generation was 44 wild type
males, 60 wild type females, 110 cyclops males
and 150 cyclops females. Does this data support
or reject the hypothesis? Use chi square to prove
your position.
E = cyclops
e = normal
Expected:
P = EE x ee
F1 = all Ee
F2 = Ee x Ee
3: cyclops:1 normal
Out of 364 offspring we should get an expected ratio of
273 cyclops: 91 normal
What we actually got was 260 cyclops: 104 normal
2
2
(104-91)
(260-273)
=
+
91
273
= 169/273 + 169/91
= 0.61 + 1.86
= 2.48
2
Test of Significance: Chi-Square Distribution of X2 Probability, Abridged* Probability
Accept
Reject
0.99
0.90
0.70
0.50
0.30
0.10
0.05
0.01
0.001
1
0.0002
0.016
0.15
0.46
1.1
2.7
3.8
6.6
10.8
2
0.02
0.21
0.71
1.39
2.4
4.6
6.0
9.2
13.8
3
0.12
0.58
1.42
2.37
3.7
6.3
7.8
11.3
16.3
4
0.30
1.06
2.02
3.36
4.9
7.8
9.5
13.3
18.5
5
0.55
1.61
3.00
4.35
6.1
9.2
11.1
15.1
20.5
6
0.87
2.20
3.83
5.35
7.2
10.6
12.6
16.8
22.5
7
1.24
2.83
4.67
6.35
8.4
12.0
14.1
18.8
24.3
8
1.65
3.49
5.53
7.34
9.5
13.4
15.5
20.1
26.1
9
2.09
4.17
6.39
8.34
10.6
14.7
16.9
21.7
27.9
10
2.56
4.87
7.27
9.34
11.8
16.0
18.3
23.2
29.6
Degrees
of
freedom
2 = 2.48 Do we accept the hypothesis?
Accepted: pattern of inheritance is autosomal dominant
Null Hypothesis:
If the Mars Co. M & M sorters are doing their job correctly,
then there should be no difference in M & M color ratios
between actual store-bought bags of M &Ms and what the Mars
Co. claims are the actual ratios.