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Is cross-fertilization good or bad?: An analysis of Darwin’s Zea Mays Data
By Jamie Chatman
and
Charlotte Hsieh
Outline
Short biography of Charles Darwin and Ronald Fisher
Description of the Zea Mays data Analysis of the data
Parametric tests (t-test, confidence intervals) Nonparametric test (i.e. Wilcoxon signed rank) Bootstrap tests
Conclusion
Short Biography of Charles Darwin
Darwin was born in 1809 in Shrewsbury, England At 16 went to Edinburgh University to study
medicine, but did not finish He went to Cambridge University, where he
received his degree studying to become a clergyman.
Darwin worked as an unpaid naturalist on a five-year scientific expedition to South America 1831.
Darwin’s research led to his book, On the Origin of Species by Means of Natural Selection, published in 1859.
1809-1882
Short Biography of Ronald Fisher
Fisher was born in East Finchley, London in 1890. Fisher went to Cambridge University and
received a degree in mathematics. Fisher made many discoveries in statistics
including maximum likelihood, analysis of variance, sufficiency, and was a pioneer for design of experiments.
1890-1962
Darwin’s Zea Mays Data
Hypothesis
Null Hypothesis: Ho: There is no difference in stalk height between
the cross-fertilized and self-fertilized plants.
Alternative Hypothesis: HA: Cross-fertilized stalk heights are not equal to
self-fertilized heights HA: Cross-fertilization leads to increased stalk
height
Galton’s Approach to the DataCrossed Self-Fert.
Pot I 23.500 17.375
12.000 20.375
21.000 20.000
Pot II 22.000 20.000
19.124 18.375
21.500 18.625
Pot III 22.125 18.625
20.375 15.250
18.250 16.500
21.625 18.000
23.250 16.250
Pot IV 21.000 18.000
22.125 12.750
23.000 15.500
12.000 18.000
Original DataCrossed Self-Fert. Difference
23.500 20.375 3.175
23.250 20.000 3.250
23.000 20.000 3.000
22.125 18.625 3.500
22.125 18.625 3.500
22.000 18.375 3.625
21.625 18.000 3.625
21.500 18.000 3.500
21.000 18.000 3.000
21.000 17.375 3.625
20.375 16.500 3.875
19.124 16.250 2.874
18.250 15.500 2.750
12.000 15.250 -3.250
12.000 12.750 -0.750
Galton’s Approach
Parametric Test Fisher made an assumption that the stalk heights
were normally distributed Crossed: X ~ Self-fertilized Y~ Difference: X-Y=d ~
p-value : 0.0497 Reject the null hypothesis that at the .05 level
),( 2XXN σµ
),( 2YYN σµ
),( 22XYxYN σσµµ +−
26.22
6166.22 =
=
ds
dd.f.= 14
148.206166.2
1526.22
=−=t
yx µµ =
Parametric Test
95% confidence interval
)15/7181.4*145.26167.215/7181.4*145.26167.2( +≤≤− d
))/()/(( 025.025. nstxdnstx +≤≤−
)2298.500364(. ≤≤ d
Since zero is not in the interval, the null hypothesis that the differences =0, (or that the means) are equal is rejected
Fisher’s Non-Parametric Approach If Ho is true, and the heights of the crossed and self-
fertilized are equal, then there should be an equal chance that each one of the pairs came from the self-fert. or the crossed If we look at all possible swaps in each pair there are
215 = 32,768 possibilities The sum of the differences is 39.25 But only 863 of these cases have sums of the difference as
great as 39.25 So the null hypothesis would be rejected at the
0526.
768,32
863*2 = level
Fisher’s Nonparametric Approach The results of the nonparametric test agreed with
the results of the t-test Fisher was happy with this However, Fisher believed that removing the
assumption of normality in the nonparametric test would result in a less powerful test than the t-test
“[Nonparametric tests] assume less knowledge, or more ignorance, of the experimental material than does the standard test…”
We disagree
Non-Parametric Test Wilcoxon Signed Rank Test
Diff.
6.125
-8.375
1
2
0.749
2.875
3.5
5.125
1.75
3.625
7
3
9.375
7.5
-6
Diff. Rank
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
721
== ∑=
n
iiRW
0.749
1
1.75
2
2.875
3
3.5
3.625
5.125
6
6.125
7
7.59.375
8.375
-
-
6
)12)(1(...21)(
02
1
2
1
2
1
2
1)(
222
1
1
++=+++=
=+
++
−=
nn
n
nRVar
nnRE
6
)12)(1()(
0)(1
++=
=
= ∑
=
nnnWVar
REWEn
ii
Non-Parametric Test
Wilcoxon Signed Rank Test When n is large W~N(0, Var(W))
This gives a p-value of 0.0409. Thus we reject the null hypothesis.
045.2072
)(
0
6)130)(115(15
=−=−++WVar
W
Bootstrap Methods
Introduced by Bradley Efron (1979) 44 years after Fisher’s analysis "If statistics had evolved at a time when computers existed,
it wouldn't be what it is today (Efron)." Uses repeated re-samples of the data Allows the use of computer sampling approaches
that are asymptotically equivalent to tests where exact significance levels require complicated manipulations
A sampling simulation approximation to Fisher’s nonparametric approach
The data “pull themselves up by their own bootstraps” by generating new data sets through which their reliability can be determined.
Bootstrap: Random Sign Change If Ho is true, there is an equal chance that the
plants in each pair are cross-fertilized or self-fertilized
Method: 1. Randomly shift from cross to self-fertilized in each
pair 2. Compute sum of differences 3. Repeat 5,000 times 4. Plot histogram of summed differences 5. Find the number of summed differences > 39.25
Bootstrap: Random Sign Change
-60 -40 -20 0 20 40 60
020
040
060
080
0
Histogram of 5000 Resampled Sums of (Sign) Randomized Zea Mays Differences
Total of Differences
Fre
quen
cy
Results 124/5000 are >39.25. The p-value is
2*(124/5000)=0.0496. Compare to exact
combinatorial p-value of 0.0526
Bootstrap: Resample Within Pots Experimenters will tend to present data in such a way as
to get significant results In order to be sure that pairings in each pot are random,
we can resample within pots We assume equality of heights in each pot Method:
1. Sample 3 crossed plants in pot 1 with replacement 2. Sample 3 self-fert. plants in pot 1 with replacement 3. Repeat for pots 2-4 4. Compute sum of differences 5. Repeat 5,000 times 6. Plot histogram of summed differences 5. Find the number of summed differences <0
Bootstrap: Resample Within Pots
-100 -50 0 50 100
050
010
0015
00
Histogram of Sums of Differences in 5000 Resamplings with Resampling Within Pots
Value of Sum of Differences
Freq
uenc
y
Results 27/5000 are <0 The p-value is
2*(27/5000)=0.0108
Resampling-Based Sign Test Disregard size of difference and look only at the sign of the
difference If Ho is true, the probability of any difference being positive or
negative is 0.5, and we can use a binomial approach, where we would expect half out of 15 pairs to have a positive difference and half to have a negative difference
We can count the number of positive differences in resampled pairs of size 15
Method: 1. Sample 3 crossed plants in pot 1 with replacement 2. Sample 3 self-fert. plants in pot 1 with replacement 3. Repeat for pots 2-4 4. Count the number of positive differences 5. Repeat 5,000 times
Resampling-Based Sign Test
Results Almost every time out of
5,000, we get over 8 positive differences out of 15.
#pos diff < 6: 0/5000 #pos diff < 8: 2/5000 p-value is essentially 0
6 8 10 12 14
050
010
0015
0020
00
Histogram of Number of Positive Differences Between Crossed and Self-Fertilized in 5000 Resamplings of Size 15 from the Zea Mays Data with Randomization
Within Pots
Number of Positive Differences
Freq
uenc
y
Randomization Within Pots Disregard information about cross or self-fertilized Find the distribution of summed differences by
resampling from pooled data Method:
1. Pool plants in pot 1 2. Sample 3 plants from the pool w/replacement, treat as crossed 3. Sample 3 plants from the pool w/replacement, treat as self-fert. 4. Repeat for pots 2-4 5. Compute sum of differences 6. Repeat 5,000 times 7. Plot histogram of summed differences (=distribution of null
hypothesis) 8. Find the number of summed differences >39.25
Randomization Within Pots
Results 38/5000 are >39.25 The p-value is
2*(38/5000)= 0.0152
-100 -50 0 50 100
05
001
000
150
0
Histogram of Null Hypothesis Randomization Test Distribution (resample of 5000)
Sum of Differences
Fre
que
ncy
Resampling Approach to Confidence Intervals Using Darwin’s original
differences: 1. Sample 15 differences
with replacement 2. Compute the sum of
differences 3. Repeat 5,000 times 4. Plot histogram of
summed differences 5. Take 125th and 4875th
summed difference Divide by sample size = 15
-100 -50 0 50 100
050
010
0015
00
Histogram of 5000 Sums of 15 Resampled Differences in Galton's Zea Mays Data
Sum of 15 Differences
Fre
quen
cy
We get 95% CI: (0.1749, 4.817), which is shorter than the t-interval (.0036, 5.230)
Resampling Approach to Confidence Intervals In the resampling approaches, “95% of the
resampled average differences were between 0.1749 and 4.817.”
This is not equivalent to the t- procedure, where “with probability 95%, the true value of the difference estimate lies between 0.0036 and 5.230.”
Conclusion
We can conclude from our tests that cross-fertilization leads to increased stalk heights
Despite Fisher’s concerns that removing normality assumptions was less intelligible than the t-test, nonparametric resampling-based methods are powerful and efficient
Is there anything else to consider?
Not using randomization, which might lead to environmental advantages and disadvantages Soil conditions or fertility Lighting Air currents Irrigation/evaporation
References Fisher, R.A.(1935). The Design of Experiments. Edinburgh:
Oliver & Boyd, 29-49. Thompson, J.R.(2000). Simulation: A Modeler’s Approach.
New York: Wiley-International Publication, 199-210. http://www.fact-index.com/r/ro/ronald_fisher.html http://www.lib.virginia.edu/science/parshall/darwin.html http://www.mste.uiuc.edu/stat/bootarticle.html http://www.psych.usyd.edu.au/difference5/scholars/galton.html
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