1. Is cross-fertilization good or bad?: An
analysis of Darwin’s Zea Mays Data
By Jamie Chatman
and
Charlotte Hsieh

2. 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

3. 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

4. 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

5. Darwin’s Zea Mays Data

6. 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

7. Galton’s Approach to the Data
Crossed 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 Data
Crossed 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

8. 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
),(
2
XXN σµ
),(
2
YYN σµ
),(
22
XYxYN σσµµ +−
26.22
6166.2
2
=
=
d
s
d
d.f.= 14
148.2
06166.2
15
26.22
=
−
=t
yx µµ =

9. 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

10. 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

11. 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

12. 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
72
1
== ∑=
n
i
iRW
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
)(
0
2
1
2
1
2
1
2
1
)(
222
1
1
++
=
+++
=
=
+






+
+






−=
nn
n
n
RVar
nn
RE
6
)12)(1(
)(
0)(
1
++
=
=





= ∑=
nnn
WVar
REWE
n
i
i

13. 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.
045.2
072
)(
0
6
)130)(115(15
=
−
=
−
++
WVar
W

14. 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.

15. 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

16. Bootstrap: Random Sign Change
-60 -40 -20 0 20 40 60
0200400600800
Histogram of 5000 Resampled Sums of (Sign) Randomized
Zea Mays Differences
Total of Differences
Frequency
Results
 124/5000 are >39.25.
 The p-value is
2*(124/5000)=0.0496.
 Compare to exact
combinatorial p-value of
0.0526

17. 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

18. Bootstrap: Resample Within Pots
-100 -50 0 50 100
050010001500
Histogram of Sums of Differences in 5000
Resamplings with Resampling Within Pots
Value of Sum of Differences
Frequency
Results
 27/5000 are <0
 The p-value is
2*(27/5000)=0.0108

19. 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

20. 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
0500100015002000
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
Frequency

21. 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

22. Randomization Within Pots
Results
 38/5000 are >39.25
 The p-value is
2*(38/5000)= 0.0152
-100 -50 0 50 100
050010001500
Histogram of Null Hypothesis Randomization
Test Distribution (resample of 5000)
Sum of Differences
Frequency

23. 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
050010001500
Histogram of 5000 Sums of 15 Resampled
Differences in Galton's Zea Mays Data
Sum of 15 Differences
Frequency
We get 95% CI: (0.1749, 4.817),
which is shorter than the t-interval
(.0036, 5.230)

24. 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.”

25. 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

26. 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

27. 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