Download Chi Square Analyses

Chi Square Analyses:
Comparing Frequency Distributions
Chi-Square Tests
• test probability distributions from nominal,
ordinal, or discrete data
• Compare data to a theoretical distribution.
• Compare two sets of data
Chi Square Tests for Goodness of Fit
• Two types – extrinsic and intrinsic
• Assumptions of both tests
– Measurement on at least a nominal scale
– Observations are independent
– The expected frequencies for each category must
be specified
– The sample size must be sufficiently large so that
no category has an expected frequency of < 5.
Chi Square Tests for Goodness of Fit
• Hypotheses
– Null – the observed frequency distribution is the
same as the hypothesized frequency distribution
– Alternative - the observed and hypothesized
distributions are different
Chi Square Tests for Goodness of Fit
• Test Statistic
– The test statistic is based on the difference between the
observed and expected frequencies. It is calculated by:
(O  E)
 
E
2
2
Chi Square Test for Goodness of Fit
• In an extrinsic test, no population parameters need
to be estimated from the data.
• An intrinsic test requires an estimation of a
population parameter from the data collected.
– Technically, the degrees of freedom should be reduced by
1 for each parameter estimated
– However, this is a minor effect and not always considered
(we won’t worry about it).
– An intrinsic test is commonly used when comparing a
sample to a derived distribution such as the poisson or
binomial distribution
Chi Square Test for Goodness of Fit
(Extrinsic)
• Example
– Cross of two pea plants with purple flowers.
– When you do the cross, you get 80 plants with
round seeds, and 20 with wrinkled.
– Your biological hypotheses are that:
• the parents were heterozygous (since some white
flowered offspring were produced)
• P is completely dominant to p
• genes segregate correctly
• fertilization is random
• zygotes have the same probability of survival with
respect to this gene.
• Example
– Your biological hypotheses are that:
• the parents were heterozygous (since some white
flowered offspring were produced)
• P is completely dominant to p
• genes segregate correctly
• fertilization is random
• zygotes have the same probability of survival with
respect to this gene.
GAMETES of
PARENTS in =
Frequency
P
p
Expected Ratio under
THESE hypotheses:
P
PP
Pp
¾ Purple offspring
¼ White offspring
p
Pp
pp
Chi Square Test for Goodness of Fit
(Extrinsic)
So, we want to see how close our observed results are to what we expect under
our hypothesis. Maybe the “total difference” would be a good measure…
Offspring
OBSERVED
Phenotype
EXPECTED
O-E
by
HYPOTHESIS
(O-E)2
Purple
80
75 (3/4)
5
25
White
20
25 (1/4)
-5
25
100
100
SUM = 0
bummer
SUM = 25
Hmmm…
But sample size matters….
Chi Square Test for Goodness of Fit
(Extrinsic)
So, we want to see how close our observed results are to what we expect under
our hypothesis. Maybe the “total difference” would be a good measure…
Offspring
OBSERVED
Phenotype
EXPECTED
O-E
by
HYPOTHESIS
(O-E)2
Purple
7505
7500 (3/4)
5
25
White
2495
2500 (1/4)
-5
25
10000
10000
SUM = 0
bummer
SUM = 25
same
But sample size matters….these results are a lot closer to
the expected values, but give the same total. So we need to
evaluate the “sum of Squares” in relation to sample size…
“mean square”
Chi Square Test for Goodness of Fit
(Extrinsic)
So, we want to see how close our observed results are to what we expect under
our hypothesis. Maybe the “total difference” would be a good measure…
Offspring
OBSERVED
Phenotype
EXPECTED
O-E
by
HYPOTHESIS
(O-E)2
(O-E)2/E
Purple
80
75 (3/4)
5
25
0.33
White
20
25 (1/4)
-5
25
1.00
100
100
This = your calculated Chi-Square value, and
you compare it to a Chi-Square table with df =
Categories (P or W = 2) – 1 = 2-1 = 1.
1.33
The critical value is associated with a probability; in
this case p = 0.05. This is the probability that
results as deviant as yours could have occurred by
chance if your null hypothesis was true. You only
reject the null hypothesis if you observe a more
deviant pattern. (This would make your calculated
value greater than the threshold critical value).
Chi Square Test for Goodness of Fit
(Intrinsic)
• Example
– In the 98 year period from 1900-1997, there were 159
U.S. landfalling hurricanes. Does the number of
landfalling hurricanes per year follow a Poisson
distribution?
– Calculate the expected frequencies
– Calculate the expected number by multiplying the
frequency by the number of categories (here, years = 98)
Formula:
p(x) = Xxe-x
x!
Chi Square Test for Goodness of Fit
(Intrinsic)
Hurricanes
per year
0
1
2
3
4
5
6
Observed # Expected
freq
18
0.198
34
0.320
24
0.260
16
0.140
3
0.057
1
0.018
2
0.007
159
Expected #
19.4
31.36
25.48
13.72
5.59
1.76
0.69
Chi Square Test for Goodness of Fit
(Intrinsic)
Hurricanes
per year
0
1
2
3
>4
Observed # Expected
freq
18
0.198
34
0.320
24
0.260
16
0.140
6
8.04
Expected #
19.4
31.36
25.48
13.72
0.518
Since we had an expected value <5, we combined
categories to fix this problem.
Chi Square Test for Goodness of Fit
(Intrinsic)
• Calculate the chi square statistic in the same
way as before, and look up on table.
• Here:
– X2 = 1.306
– Tabled value for  = 0.05 = 7.81
– Thus, we fail to reject the null hypothesis,
supporting the claim that the annual number of
landfalling U.S. hurricanes follows a Poisson
distribution (rare, independent, random).
Chi Square Test of Independence
• Also called the Chi Square Test for
Contingency Tables
• This test is performed to see if two
variables, both measured on a nominal
scale, are related in some way.
• The question asked here is if there is a
relationship between the variables; the
null hypothesis is that no relationship
exists – they are “independent”.
Chi Square Test of Independence
• Steps in doing the test
– 1. Form a table, or matrix, from the data collected
– 2. Calculate row, column, and grand totals for the
matrix
– 3. Use these totals to calculate expected values
(frequencies) for each cell in the matrix
• Calculated by: [(row total) x (column
total)]/grand total
• Based on the product rule – the probability of
two independent events occurring together is
the product of their independent probabilities.
Chi Square Test of Independence
Classic Example: Testing for Linkage or Independent Assortment between
two loci
Suppose we cross two pea plants: PpTt x pptt
- Purple is completely dominant to white
- Tall is completely dominant to short
Produce the following results in the offspring:
PT = 32
Pt = 22
pT = 23
Pt = 36
113
ARE THE GENES ASSORTING INDEPENDENTLY, OR ARE THEY LINKED?
Chi Square Test of Independence
ARE THE GENES ASSORTING INDEPENDENTLY, OR ARE THEY LINKED?
PT = 32
Pt = 22
pT = 23
Pt = 36
113
CONTINGENCY TABLE
T
t
P
32
22
54
p
23
36
59
55
58
113
IF these events (flower color and plant height) are inherited independently,
THEN the frequency of any combined outcome should be = to the product of
their independent probabilities:
IF IA, THEN f(PT) = f(P) x f(T) x N = 54/113 x 55/113 x 113 = 26.28318
Reduces to: f(PT) = f(P) x f(T) x N = 54 x 55/113 = 26.28318 = RT x CT/GT
Chi Square Test of Independence
ARE THE GENES ASSORTING INDEPENDENTLY, OR ARE THEY LINKED?
PT = 32
Pt = 22
pT = 23
Pt = 36
113
CONTINGENCY TABLE
T
exp
t
exp
P
32
26.28
22
27.72
54
p
23
28.72
36
30.28
59
55
58
113
IF these events (flower color and plant height) are inherited independently,
THEN the frequency of any combined outcome should be = to the product of
their independent probabilities:
IF IA, THEN f(PT) = f(P) x f(T) x N = 54/113 x 55/113 x 113 = 26.28318
Reduces to: f(PT) = f(P) x f(T) x N = 54 x 55/113 = 26.28318 = RT x CT/GT
T
exp
t
exp
P
32
26.28
22
27.72
54
p
23
28.72
36
30.28
59
55
58
113
Obs
Exp
O-E
(O-E)2/E
PT
32
26.28
5.72
1.24
Pt
22
27.72
-5.72
1.18
pT
23
28.72
-5.72
1.14
pt
36
30.28
5.72
1.08
4.64
Df = (R-1)(C-1) in contingency table
(1)(1) = 1, p = 0.05, critical = 3.84…. Reject Ho.