using permutation tests to study infant handling by female baboons thomas l. moore vicki...

Using Permutation Tests to Study Infant Handling by

Female Baboons

Thomas L. Moore

Vicki Bentley-Condit

Grinnell College

Infant handling examples

Plan for talk

• The data & problem

• The use of permutation tests

• Interpret results

• The stability of results

• The choice of test statistics

• Summary

The data (handout) HANDLERS ranks

INFANTS/ KM KN NQ PO HQ LL NY PS SK ST WK AL CO DD LS LY MH ML MM PA PH PT RSMothers 1 1 1 1 | 2 2 2 2 2 2 2 | 3 3 3 3 3 3 3 3 3 3 3 3 ranks KG/KM 1 0 0 4 1 | 1 0 0 0 3 1 0 | 0 0 0 0 0 0 0 0 0 0 2 1 HZ/HQ 2 13 23 7 5 | 0 2 1 1 5 6 18 | 1 6 3 0 1 4 1 0 9 0 10 1 LC/LL 2 4 0 1 4 | 3 0 2 1 1 5 3 | 1 0 0 1 0 2 1 1 1 0 1 6 NK/NY 2 12 4 10 5 | 9 1 0 2 3 11 7 | 8 6 3 1 0 2 1 1 5 3 2 3 PZ/PS 2 1 3 4 1 | 0 0 0 0 0 0 2 | 0 2 0 0 0 3 0 1 1 0 3 0 CY/CO 3 2 2 7 3 | 1 1 2 0 3 12 16 | 3 0 2 0 0 2 0 0 1 0 0 2 LZ/LS 3 1 0 3 2 | 1 1 0 0 2 0 5 | 2 2 2 0 1 9 2 0 0 0 3 2 MQ/ML 3 0 1 5 2 | 2 4 2 2 2 4 5 | 7 5 2 1 1 7 0 4 4 1 0 2 MW/MH 3 3 0 7 4 | 2 3 0 5 2 8 13 | 7 14 2 0 0 0 4 0 8 0 13 6 MX/MM 3 2 3 4 5 | 0 0 0 0 0 5 2 | 9 3 1 0 0 2 0 0 1 2 2 3 PK/PH 3 2 0 6 4 | 3 4 1 0 0 15 10 | 8 5 1 0 3 1 1 6 3 0 7 5

HANDLERS ranks KM KN NQ PO HQ LL NY PS SK ST WK AL CO DD LS LY MH ML MM PA PH PT RS 1 1 1 1 | 2 2 2 2 2 2 2 | 3 3 3 3 3 3 3 3 3 3 3 3 INFANTS/ Mothers ranks KG/KM 1 0 0 4 1 | 1 0 0 0 3 1 0 | 0 0 0 0 0 0 0 0 0 0 2 1 HZ/HQ 2 13 23 7 5 | 0 2 1 1 5 6 18 | 1 6 3 0 1 4 1 0 9 0 10 1 LC/LL 2 4 0 1 4 | 3 0 2 1 1 5 3 | 1 0 0 1 0 2 1 1 1 0 1 6 NK/NY 2 12 4 10 5 | 9 1 0 2 3 11 7 | 8 6 3 1 0 2 1 1 5 3 2 3 PZ/PS 2 1 3 4 1 | 0 0 0 0 0 0 2 | 0 2 0 0 0 3 0 1 1 0 3 0 CY/CO 3 2 2 7 3 | 1 1 2 0 3 12 16 | 3 0 2 0 0 2 0 0 1 0 0 2 LZ/LS 3 1 0 3 2 | 1 1 0 0 2 0 5 | 2 2 2 0 1 9 2 0 0 0 3 2 MQ/ML 3 0 1 5 2 | 2 4 2 2 2 4 5 | 7 5 2 1 1 7 0 4 4 1 0 2 MW/MH 3 3 0 7 4 | 2 3 0 5 2 8 13 | 7 14 2 0 0 0 4 0 8 0 13 6 MX/MM 3 2 3 4 5 | 0 0 0 0 0 5 2 | 9 3 1 0 0 2 0 0 1 2 2 3 PK/PH 3 2 0 6 4 | 3 4 1 0 0 15 10 | 8 5 1 0 3 1 1 6 3 0 7 5

High-ranked female handles mid-ranked infant: Female NQ handles Infant NK 10 times

NK’s mother is NY

KM KN NQ PO HQ

1 1 1 1 2

KG/KM 1 0 0 4 1 1

HZ/HQ 2 13 23 7 5 0

LC/LL 2 4 0 1 4 3

NK/NY 2 12 4 10 5 9

PZ/PS 2 1 3 4 1 0


NK’s mother is NY

The variables

• Handler rank: high(1), mid(2), low(3)

• Infant rank: high(1), mid(2), low(3)

• The number of interactions between a given infant-handler pair

NOTE: Rank is determined from a dominance hierarchy score measured independently of infant handling.

3 kinds of handling behavior

• Passive: movement to within 1m. of the mother-infant pair with no attempt to handle,

• Unsuccessful: movement to within 1m. of the mother-infant pair with attempted (but not successful) handle, or

• Successful: a successful handle.

• NOTE: Each count in the matrix above is the sum of counts from the 3 categories listed on this slide.

Research hypotheses

1. Females will tend to handle the infants of females who are ranked the same as or lower than themselves. (RH1)

2. Females will tend to handle the infants of females who are ranked directly below them (or same rank if female is low-ranked). (RH2)

The data (handout) HANDLERS ranks

INFANTS/ KM KN NQ PO HQ LL NY PS SK ST WK AL CO DD LS LY MH ML MM PA PH PT RSMothers 1 1 1 1 | 2 2 2 2 2 2 2 | 3 3 3 3 3 3 3 3 3 3 3 3 ranks KG/KM 1 0 0 4 1 | 1 0 0 0 3 1 0 | 0 0 0 0 0 0 0 0 0 0 2 1 HZ/HQ 2 13 23 7 5 | 0 2 1 1 5 6 18 | 1 6 3 0 1 4 1 0 9 0 10 1 LC/LL 2 4 0 1 4 | 3 0 2 1 1 5 3 | 1 0 0 1 0 2 1 1 1 0 1 6 NK/NY 2 12 4 10 5 | 9 1 0 2 3 11 7 | 8 6 3 1 0 2 1 1 5 3 2 3 PZ/PS 2 1 3 4 1 | 0 0 0 0 0 0 2 | 0 2 0 0 0 3 0 1 1 0 3 0 CY/CO 3 2 2 7 3 | 1 1 2 0 3 12 16 | 3 0 2 0 0 2 0 0 1 0 0 2 LZ/LS 3 1 0 3 2 | 1 1 0 0 2 0 5 | 2 2 2 0 1 9 2 0 0 0 3 2 MQ/ML 3 0 1 5 2 | 2 4 2 2 2 4 5 | 7 5 2 1 1 7 0 4 4 1 0 2 MW/MH 3 3 0 7 4 | 2 3 0 5 2 8 13 | 7 14 2 0 0 0 4 0 8 0 13 6 MX/MM 3 2 3 4 5 | 0 0 0 0 0 5 2 | 9 3 1 0 0 2 0 0 1 2 2 3 PK/PH 3 2 0 6 4 | 3 4 1 0 0 15 10 | 8 5 1 0 3 1 1 6 3 0 7 5

[,1] [,2] [,3] [1,] 5 5 3 [2,] 97 83 95 [3,] 68 138 184

X=handler rank; Y=Infant rank

Handler's rank Hi Mid Low Infant Hi 5 5 3 Rank Mi 97 83 95 Lo 68 138 184 Totals: 170 226 282 (A)Counts

Handler's rank Hi Mid Low Infant Hi 2.9% > 2.2% > 1.1% Rank Mi 57.1% > 36.7% > 33.9% Lo 40.0% < 61.1% < 65.0% (B)Column%

Handler's rank Hi Mid Low

Infant Hi 1.12 0.40 -1.37

Rank Mi 5.06 -1.44 -3.08

Lo -5.34 1.32 3.43

Adjusted residuals

Adjusted residuals

Handler's rank Hi Mid Low

Infant Hi 1.12 0.40 -1.37

Rank Mi 5.06 -1.44 -3.08

Lo -5.34 1.32 3.43

Is the relationship statistically significant?

The Null Model

The female handlers interacted with infants as given in the data set. These interactions involved a variety of complex causes, but none of this complexity had anything to do with ranks. That is, ranks can be viewed as meaningless labels attached to infants and females.

Computing a permutation test

• Choose a test statistic, C.• (1) Assign ranks at random to infants and females using

the rank distributions of the data set. That is, assign ranks at random so that infants are assigned, in this case, 1 High, 4 Mid, and 6 Low and so that females are assigned 4 High’s, 7 Mid’s, and 12 Low’s. This assignment leads to the original data table but with permuted ranks.

• (2) Re-form the 3-by-3 table.• (3) Compute the value of C for this table.• Iterate (1)-(3) many times for empirical null distribution.• )( CC D

PvalueP

Test statistic for Research hypothesis 1

472

18413868

958397

355

*

LTE


160

18413868

958397

355

*

LT

A sample permutation (handout)

HANDLERS ranks KM KN NQ PO HQ LL NY PS SK ST WK AL CO DD LS LY MH ML MM PA PH PT RS 1 3 1 3 3 3 2 2 3 1 3 2 2 3 3 1 3 3 3 2 2 3 2 INFANTS/ Mothers ranks KG/KM 1 0 0 4 1 | 1 0 0 0 3 1 0 | 0 0 0 0 0 0 0 0 0 0 2 1 HZ/HQ 3 13 23 7 5 | 0 2 1 1 5 6 18 | 1 6 3 0 1 4 1 0 9 0 10 1 LC/LL 3 4 0 1 4 | 3 0 2 1 1 5 3 | 1 0 0 1 0 2 1 1 1 0 1 6 NK/NY 2 12 4 10 5 | 9 1 0 2 3 11 7 | 8 6 3 1 0 2 1 1 5 3 2 3 PZ/PS 2 1 3 4 1 | 0 0 0 0 0 0 2 | 0 2 0 0 0 3 0 1 1 0 3 0 CY/CO 2 2 2 7 3 | 1 1 2 0 3 12 16 | 3 0 2 0 0 2 0 0 1 0 0 2 LZ/LS 3 1 0 3 2 | 1 1 0 0 2 0 5 | 2 2 2 0 1 9 2 0 0 0 3 2 MQ/ML 3 0 1 5 2 | 2 4 2 2 2 4 5 | 7 5 2 1 1 7 0 4 4 1 0 2 MW/MH 3 3 0 7 4 | 2 3 0 5 2 8 13 | 7 14 2 0 0 0 4 0 8 0 13 6 MX/MM 3 2 3 4 5 | 0 0 0 0 0 5 2 | 9 3 1 0 0 2 0 0 1 2 2 3 PK/PH 2 2 0 6 4 | 3 4 1 0 0 15 10 | 8 5 1 0 3 1 1 6 3 0 7 5

[,1] [,2] [,3] [1,] 5 1 7 [2,] 85 60 119 [3,] 81 117 203


424

20311781

1196085

715

*

LTE

Null distribution: 1000 resamples

C

Fre

qu

en

cy

100 200 300 400 500

05

01

00

15

02

00

Conclusion

• P-value ≈ 15/1000 = .015

• Observed pattern is unlikely the result of chance alone.

Summary by type of interaction

LTE LT n

PA 0.015 0.038 678

Pass 0.013 0.071 377

Un 0.012 0.119 189

Succ 0.372 0.017 112

---------------------------

p-values for two test statistics (LTE and LT) for 4 datasets of counts.

Look at Successful interactions

hi mid lo

hi 0 3 2

mid 19 6 22

lo 3 14 43

hi mid lo

hi 0% 13% 3%

mid 86% 26% 33%

lo 14% 61% 64%

hi mid lo

hi -1.13 2.23 -0.92

mid 4.71 -1.73 -2.39

lo -4.19 0.79 2.75

Counts

Resids

Column%

Stability of results

• Suggested by Clifford Lunneborg (Stats 2002)

• Stable description: “finding of the study … is not unduly influenced by the inclusion in the study of one particular source of observations.”

Four views of stability

Remove infants

Remove handlers

Test statistic ???? ????

P-value ???? ????

For example …

• Remove infants, one at a time, recompute the test statistic.

• Use a normalized test statistic = LTE*LTE / Sum of table entries;

• LTE* and LTE have same permutation distribution, …

• But LTE* accounts for sub-table count variation.

• LTE* values are stable (plot below)

Remove infant, LTE*, PA

2 4 6 8 10

0.64

0.68

0.72

0.76

Remove infant, p-value, Pass

Infant ID2 4 6 8 10

0.02

0.06

0.10

Remove handler, p-value, Pass

Handler ID5 10 15 20

0.01

0.03

0.05

0.07

Stability summary

Remove infants

Remove handlers

Test statistic Stable for both LTE and LT

Stable for both LTE and LT

P-value Slight instability for Infant #1 and Passive handling

Slight instability for Handler #1 and Passive handling

Choice of test statistics

• LTE and LT ad hoc, but intuitive• Power analysis to compare LTE and LT to

some other statistics– Correlation-based statistics

• M = Pearson’s correlation putting scores on ranks (Agresti,88)

• GK = Goodman and Kruskal’s gamma (Agresti,58)

– Beta = the asymmetry parameter in an ordinal quasi-symmetric log-linear model (Agresti,202)

Simulation

• Product multinomial model where J-th handler generates nJ interactions

• 2^(7-3) design to estimate main effects for 7 factors, including– Total sample size– Table dimension– RH1 vs RH2– Strength of research hypothesis– Patterns of non-homogeneity of counts– Infant rank distribution– Handler rank distribution

Results of power study

• LTE and LT outperform others;

• LTE does best under RH1

• LT does best under RH2

• Good news!

Summary of talk

• There is evidence for both research hypotheses, depending on the type of handling behavior;

• This suggests a nuanced description of how infant-handling behavior works.

• Permutation tests gave a way of analyzing a messy data set;

• We assessed stability through a simple remove-one-at-a-time strategy;

• We compared the power of test statistics and found simple ones to perform well.

Final slide

• Thank you for listening.

• Email: [email protected]

• Slides and handout: http://www.math.grinnell.edu/~mooret/reports/reports.html

• Note: this info is on your handout.

http://www.math.grinnell.edu/~mooret/reports/reports.html

http://www.math.grinnell.edu/~mooret/reports/reports.html

using permutation tests to study infant handling by female baboons thomas l. moore vicki...

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