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  GenStat  ~  REML meta-analysis

Kirsty Hassall
Posted: Fri Mar 15, 2019 3:53 pm Reply with quote
Guest
Hi David,
So the spatial models I’ve defined below, are the ones I selected as the best based on a single set of year by year analyses if that’s what you mean?
Before Genstat crashes, I see the following output, which implies the structure is well-defined …

REML variance components analysis

Response variate: GY
Fixed model: Constant + Genotype
Random model: Environment + Environment.Genotype + block@2007 + block@2007.subblock@2007 ([email]block@2007.subblock@2007[/email]) + block@2008R + block@2008R.subblock@2008R ([email]block@2008R.subblock@2008R[/email]) + block@2008W + block@2008W.subblock@2008W ([email]block@2008W.subblock@2008W[/email]) + block@2009 + block@2009.subblock@2009 ([email]block@2009.subblock@2009[/email]) + block@2010 + block@2010.subblock@2010 ([email]block@2010.subblock@2010[/email]) + block@2011 + block@2011.subblock@2011 ([email]block@2011.subblock@2011[/email]) + block@2012 + block@2012.row12F@2012 ([email]block@2012.row12F@2012[/email]) + block@2012.col12F@2012 ([email]block@2012.col12F@2012[/email]) + block@2014
Number of units: 4956

Separate residual terms for each level of experiment factor: Environment

Sparse algorithm with AI optimisation
Units with missing factor/covariate values included
- specific effect for term(s) omitted for units with missing values in block@2007, subblock@2007, block@2008R, subblock@2008R, block@2008W, subblock@2008W, block@2009, subblock@2009, block@2010, subblock@2010, block@2011, subblock@2011, block@2012, row12F@2012, col12F@2012, block@2014, row08rF, col08rF, row10F, col10F, row11F, col11F, row12F, col12F, row14F, col14F
Units with missing data values included


Residual models for multi-experiment analysis

Experiment factor: Environment

Experiment Term Factor Model Order Nrows
2007 Residual Whole term Identity 0 612
2008R row08rF.col08rF
row08rF Auto-regressive 1 12
col08rF Auto-regressive 1 52
2008W Residual Whole term Identity 0 612
2009 Residual Whole term Identity 0 612
2010 row10F.col10F row10F Auto-regressive 1 12
col10F Auto-regressive 1 52
2011 row11F.col11F row11F Identity 0 13
col11F Auto-regressive 1 48
2012 block.row12F.col12F
block Identity 0 3
row12F Identity 0 13
col12F Auto-regressive 1 16
2014 row14F.col14F row14F Identity 0 8
col14F Auto-regressive 1 78



Estimated variance components

Random term component s.e.
Environment 3.6181 2.0382
Environment.Genotype 0.2917 0.0134
block@2007 0.0026 0.0082
block@2007.subblock@2007 ([email]block@2007.subblock@2007[/email])
0.0593 0.0179
block@2008R 0.3577 0.3818
block@2008R.subblock@2008R ([email]block@2008R.subblock@2008R[/email])
0.0102 0.0127
block@2008W 2.3615 2.4477
block@2008W.subblock@2008W ([email]block@2008W.subblock@2008W[/email])
0.1523 0.0548
block@2009 0.1861 0.2042
block@2009.subblock@2009 ([email]block@2009.subblock@2009[/email])
0.1118 0.0434
block@2010 -0.0016 0.0155
block@2010.subblock@2010 ([email]block@2010.subblock@2010[/email])
-0.0008 0.0187
block@2011 0.6610 0.6562
block@2011.subblock@2011 ([email]block@2011.subblock@2011[/email])
0.0120 0.0067
block@2012 1.0578 1.0524
block@2012.row12F@2012 ([email]block@2012.row12F@2012[/email]) 0.0304 0.0140
block@2012.col12F@2012 ([email]block@2012.col12F@2012[/email]) 0.0008 0.0034
block@2014 0.0216 0.0251


Residual model for each experiment

Experiment factor: Environment

Experiment Term Factor Model(order) Parameter Estimate s.e.
2007 Residual Identity Variance 0.186 0.014
2008R row08rF.col08rF Variance 0.493 0.059
row08rF AR(1) phi_1 0.1481 0.0623
col08rF AR(1) phi_1 0.8055 0.0244
2008W Residual Identity Variance 0.453 0.031
2009 Residual Identity Variance 0.253 0.018
2010 row10F.col10F Variance 0.435 0.052
row10F AR(1) phi_1 0.2932 0.0518
col10F AR(1) phi_1 0.7423 0.0318
2011 row11F.col11F Variance 0.224 0.019
row11F Identity - - -
col11F AR(1) phi_1 0.4176 0.0459
2012 block.row12F.col12F Variance 0.162 0.015
block Identity - - -
row12F Identity - - -
col12F AR(1) phi_1 0.3769 0.0616
2014 row14F.col14F Variance 0.212 0.019
row14F Identity - - -
col14F AR(1) phi_1 0.6004 0.0354


Estimated covariance models


From: GENSTAT-Request <GENSTAT@jiscmail.ac.uk (GENSTAT@jiscmail.ac.uk)> On Behalf Of David Baird
Sent: 15 March 2019 10:13
To: GENSTAT@JISCMAIL.AC.UK (GENSTAT@JISCMAIL.AC.UK)
Subject: Re: REML meta-analysis



Dear Kirsty,

Have you tried VASERIES/VAMETA – this will try the various models and keep
the best fitting model for each experiment and then combines them.
If your structure fits within this framework, it’s definitely an easy way to go.

Regards, David.
______________________________________________
Dr David Baird Statistical Consultant and Genstat Developer
VSN (NZ) Limited (David@VSN.CO.NZ (David@VSN.CO.NZ))
8 Mariposa Crescent, Aidanfield, Christchurch 8025, New Zealand
Ph +64 3 3350588 Cell +64 21 1160803


From: GENSTAT-Request <GENSTAT@jiscmail.ac.uk (GENSTAT@jiscmail.ac.uk)> On Behalf Of Kirsty Hassall
Sent: 15 March 2019 9:59 PM
To: GENSTAT@JISCMAIL.AC.UK (GENSTAT@JISCMAIL.AC.UK)
Subject: REML meta-analysis



Dear Genstatters,

I am attempting to fit a meta-analysis across 8 different field experiments, each of which has a separate residual structure. The commands to set this up are:

vrmeta [terms=Environment*Genotype; experiments=Environment; random=random] \
'2007','2008R','2008W','2009','2010','2011','2012','2014';\
localterms=!f(block/subblock),!f(block/subblock),!f(block/subblock),\
!f(block/subblock),!f(block/subblock),!f(block/subblock), \
!f(block/(row12F + col12F)), !f(block)
vcomp [Genotype; Experiments=Environment] #random
vresid [experiment='2008R'; term=row08rF.col08rF] model=ar,ar; factor=row08rF,col08rF
vresid [experiment='2010'; term=row10F.col10F] model=ar,ar; factor=row10F,col10F
vresid [experiment='2011'; term=row11F.col11F] model=AR; order=1; factor=col11F
vresid [experiment='2012'; term=block.row12F.col12F] model=AR; order=1; factor=col12F
vresid [experiment='2014'; term=row14F.col14F] model=AR; order=1; factor=col14F
reml [mvincl=y,e; work=500] GY

Running the above, produces output for the random effects and residual terms which all seem to match the structure I would expect (i.e. no additional residual terms are added etc…). However, before reaching the output for the fixed effects I have the error “Genstat server applictaion has stopped working: A problem caused the program to stop working”.

Any advice on how to cope with this error would be greatly appreciated. Note, the above works if I remove the spatial structure over the residuals.

Thanks in advance,
Kirsty

-------------------------------------------------------------------------------------------------------------------------
Dr Kirsty Hassall
Scientific Statistician

Department of Computational and Analytical Sciences
Rothamsted Research
Harpenden, Herts
AL5 2JQ, UK

Tel: +44(0) 1582 763133 Ext: 2080
Email: kirsty.hassall@rothamsted.ac.uk (kirsty.hassall@rothamsted.ac.uk)
http://www.rothamsted.ac.uk


Rothamsted Research is a company limited by guarantee, registered in England at Harpenden, Hertfordshire, AL5 2JQ under the registration number 2393175 and a not for profit charity number 802038.


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Terry Koen
Posted: Tue Mar 26, 2019 4:43 am Reply with quote
Guest
Fellow Genstatters

I would like some advice on partitioning an ANOVA table into certain contrasts.

I have a Before / After design (sorry, no Control site(s)), with 7 equi-time spaced Surveys conducted - 3 Before an Event and 4 After; 4 Transects per Survey. N = 7 * 4 = 28.

I would like to expand the simple ANOVA model to partition the 6 df for Surveys into more meaningful comparisons, like to form separate linear contrasts for the 3 Before times and the 4 After times, to something resembling the following expansion

Survey 6
Pre v Post 1 Pre v Post 1
Survey / Pre 2 Survey / Pre Lin 1
Survey / Pre Dev 1
Survey / Post 3 Survey / Post Lin 1
Survey / Post Dev 2
Transects/Survey 21
Total 27

These statements give me the first basic partitioning
FACT [Levels=7; NV=28; VALUES=4(1,2,3,4,5,6,7)] Survey
FACT [Levels=4; NV=28; VALUES=(1,2,3,4)7] Transect
TREAT Survey
ANOVA [PR=aov,info,means; FPROB=yes] Bottles

These next statements give me the second level of partitioning
FACT [Levels=2; NV=28; LABELS=!t('Post','Pre'); VALUES=12(2),16(1)] PrePost
FACT [Levels=4; NV=28; LABELS=!t('2','3','4','Null'); VALUES=4(1,2,3),16(4)] SPre
FACT [Levels=5; NV=28; LABELS=!t('1','2','3','4','Null'); VALUES=12(5),4(1,2,3,4)] SPost
TREAT PrePost/(SPre + SPost)
ANOVA [PR=aov,info,means; FPROB=yes] Bottles

I had hoped that these statements would provide the linear contrasts …
VARIATE [Values = 2,3,4,*] PreLin
VARIATE [Values = 1,2,3,4,*] PostLin
BLOCK
TREATMENTS PrePost/(POL(SPre; 1; PreLin) + POL(SPost; 1; PostLin))
ANOVA [PR=aov,info,means,contrasts; FPROB=yes; PSE=diff] Bottles

… but alas, my efforts result in 2df for ‘Lin’ and zero Sums of Squares.
Source of variation d.f. s.s.
PrePost 1 4777.65
PrePost.SPre 2 715.17
PrePost.Lin 2 0.00
PrePost.SPost 3 8267.19
PrePost.Lin 2 0.00
Deviations 1 8267.19
Residual 21 1799.00
Total 27 15559.00

Is it correct to nullify the ‘Null’ level for SPre and SPost using a missing value indicator ‘*’? Or is there an alternative approach I could be using?

Many thanks, Terry

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david@vsn.co.nz
Posted: Tue Mar 26, 2019 5:41 am Reply with quote
Joined: 30 Jul 2009 Posts: 283
Dear Terry,

Wouldn’t the simplest thing be to apply contrasts to the factor Survey:

FACT [Levels=7; NV=28; VALUES=4(1,2,3,4,5,6,7)] Survey
FACT [Levels=4; NV=28; VALUES=(1,2,3,4)7] Transect
MATRIX [ROWS=!T('Pre vs Post','Pre linear','Pre Dev.','Post Linear', \
'Post Quad.', 'Post Dev.');COLUMNS=7] Contrasts; VALUES=\
!(-4,-4,-4,3,3,3,3, -1,0,1,4(0), 1,-2,1,4(0),\
3(0),-3,-1,1,3, 3(0),1,-1,-1,1, 3(0),-3,9,-9,3)
TREAT COMP(Survey;6;Contrasts)
BLOCK Transect
ANOVA [PR=aov,info,means; FPROB=yes; CONTRASTS=9] Bottles

Analysis of variance

Variate: Bottles

Source of variation d.f. s.s. m.s. v.r. F pr.

Transect stratum 3 0.990 0.330 0.20

Transect.*Units* stratum
Survey 6 136.123 22.687 13.94 <.001
Pre vs Post 1 84.079 84.079 51.65 <.001
Pre linear 1 10.556 10.556 6.48 0.020
Pre Dev. 1 0.020 0.020 0.01 0.912
Post Linear 1 37.000 37.000 22.73 <.001
Post Quad. 1 1.053 1.053 0.65 0.432
Post Dev. 1 3.414 3.414 2.10 0.165
Residual 18 29.303 1.628

Total 27 166.415

Note, I’ve split the 2 degrees of freedom in Post deviations into two named terms,
as Pre Dev. = Pre quadratic and Post Dev = Post cubic.

Regards, David.
______________________________________________
Dr David Baird Statistical Consultant and Genstat Developer
VSN (NZ) Limited (David@VSN.CO.NZ (David@VSN.CO.NZ))
8 Mariposa Crescent, Aidanfield, Christchurch 8025, New Zealand
Ph +64 3 3350588 Cell +64 21 1160803


From: GENSTAT-Request <GENSTAT@jiscmail.ac.uk> On Behalf Of Terry Koen
Sent: 26 March 2019 5:44 PM
To: GENSTAT@JISCMAIL.AC.UK
Subject: Re: REML meta-analysis



Fellow Genstatters

I would like some advice on partitioning an ANOVA table into certain contrasts.

I have a Before / After design (sorry, no Control site(s)), with 7 equi-time spaced Surveys conducted - 3 Before an Event and 4 After; 4 Transects per Survey. N = 7 * 4 = 28.

I would like to expand the simple ANOVA model to partition the 6 df for Surveys into more meaningful comparisons, like to form separate linear contrasts for the 3 Before times and the 4 After times, to something resembling the following expansion

Survey 6
Pre v Post 1 Pre v Post 1
Survey / Pre 2 Survey / Pre Lin 1
Survey / Pre Dev 1
Survey / Post 3 Survey / Post Lin 1
Survey / Post Dev 2
Transects/Survey 21
Total 27

These statements give me the first basic partitioning
FACT [Levels=7; NV=28; VALUES=4(1,2,3,4,5,6,7)] Survey
FACT [Levels=4; NV=28; VALUES=(1,2,3,4)7] Transect
TREAT Survey
ANOVA [PR=aov,info,means; FPROB=yes] Bottles

These next statements give me the second level of partitioning
FACT [Levels=2; NV=28; LABELS=!t('Post','Pre'); VALUES=12(2),16(1)] PrePost
FACT [Levels=4; NV=28; LABELS=!t('2','3','4','Null'); VALUES=4(1,2,3),16(4)] SPre
FACT [Levels=5; NV=28; LABELS=!t('1','2','3','4','Null'); VALUES=12(5),4(1,2,3,4)] SPost
TREAT PrePost/(SPre + SPost)
ANOVA [PR=aov,info,means; FPROB=yes] Bottles

I had hoped that these statements would provide the linear contrasts …
VARIATE [Values = 2,3,4,*] PreLin
VARIATE [Values = 1,2,3,4,*] PostLin
BLOCK
TREATMENTS PrePost/(POL(SPre; 1; PreLin) + POL(SPost; 1; PostLin))
ANOVA [PR=aov,info,means,contrasts; FPROB=yes; PSE=diff] Bottles

… but alas, my efforts result in 2df for ‘Lin’ and zero Sums of Squares.
Source of variation d.f. s.s.
PrePost 1 4777.65
PrePost.SPre 2 715.17
PrePost.Lin 2 0.00
PrePost.SPost 3 8267.19
PrePost.Lin 2 0.00
Deviations 1 8267.19
Residual 21 1799.00
Total 27 15559.00

Is it correct to nullify the ‘Null’ level for SPre and SPost using a missing value indicator ‘*’? Or is there an alternative approach I could be using?

Many thanks, Terry

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Terry Koen
Posted: Tue Mar 26, 2019 9:18 pm Reply with quote
Guest
Many thanks David

Yes, so simple when someone kindly opens my eyes for me, and reminds me of the COMPARISON function for estimating contrasts between treatments.

Regards, Terry

From: GENSTAT-Request <GENSTAT@jiscmail.ac.uk> On Behalf Of David Baird
Sent: Tuesday, 26 March 2019 4:41 PM
To: GENSTAT@JISCMAIL.AC.UK
Subject: Re: REML meta-analysis



Dear Terry,

Wouldn’t the simplest thing be to apply contrasts to the factor Survey:

FACT [Levels=7; NV=28; VALUES=4(1,2,3,4,5,6,7)] Survey
FACT [Levels=4; NV=28; VALUES=(1,2,3,4)7] Transect
MATRIX [ROWS=!T('Pre vs Post','Pre linear','Pre Dev.','Post Linear', \
'Post Quad.', 'Post Dev.');COLUMNS=7] Contrasts; VALUES=\
!(-4,-4,-4,3,3,3,3, -1,0,1,4(0), 1,-2,1,4(0),\
3(0),-3,-1,1,3, 3(0),1,-1,-1,1, 3(0),-3,9,-9,3)
TREAT COMP(Survey;6;Contrasts)
BLOCK Transect
ANOVA [PR=aov,info,means; FPROB=yes; CONTRASTS=9] Bottles

Analysis of variance

Variate: Bottles

Source of variation d.f. s.s. m.s. v.r. F pr.

Transect stratum 3 0.990 0.330 0.20

Transect.*Units* stratum
Survey 6 136.123 22.687 13.94 <.001
Pre vs Post 1 84.079 84.079 51.65 <.001
Pre linear 1 10.556 10.556 6.48 0.020
Pre Dev. 1 0.020 0.020 0.01 0.912
Post Linear 1 37.000 37.000 22.73 <.001
Post Quad. 1 1.053 1.053 0.65 0.432
Post Dev. 1 3.414 3.414 2.10 0.165
Residual 18 29.303 1.628

Total 27 166.415

Note, I’ve split the 2 degrees of freedom in Post deviations into two named terms,
as Pre Dev. = Pre quadratic and Post Dev = Post cubic.

Regards, David.
______________________________________________
Dr David Baird Statistical Consultant and Genstat Developer
VSN (NZ) Limited (David@VSN.CO.NZ (David@VSN.CO.NZ))
8 Mariposa Crescent, Aidanfield, Christchurch 8025, New Zealand
Ph +64 3 3350588 Cell +64 21 1160803


From: GENSTAT-Request <GENSTAT@jiscmail.ac.uk (GENSTAT@jiscmail.ac.uk)> On Behalf Of Terry Koen
Sent: 26 March 2019 5:44 PM
To: GENSTAT@JISCMAIL.AC.UK (GENSTAT@JISCMAIL.AC.UK)
Subject: Re: REML meta-analysis



Fellow Genstatters

I would like some advice on partitioning an ANOVA table into certain contrasts.

I have a Before / After design (sorry, no Control site(s)), with 7 equi-time spaced Surveys conducted - 3 Before an Event and 4 After; 4 Transects per Survey. N = 7 * 4 = 28.

I would like to expand the simple ANOVA model to partition the 6 df for Surveys into more meaningful comparisons, like to form separate linear contrasts for the 3 Before times and the 4 After times, to something resembling the following expansion

Survey 6
Pre v Post 1 Pre v Post 1
Survey / Pre 2 Survey / Pre Lin 1
Survey / Pre Dev 1
Survey / Post 3 Survey / Post Lin 1
Survey / Post Dev 2
Transects/Survey 21
Total 27

These statements give me the first basic partitioning
FACT [Levels=7; NV=28; VALUES=4(1,2,3,4,5,6,7)] Survey
FACT [Levels=4; NV=28; VALUES=(1,2,3,4)7] Transect
TREAT Survey
ANOVA [PR=aov,info,means; FPROB=yes] Bottles

These next statements give me the second level of partitioning
FACT [Levels=2; NV=28; LABELS=!t('Post','Pre'); VALUES=12(2),16(1)] PrePost
FACT [Levels=4; NV=28; LABELS=!t('2','3','4','Null'); VALUES=4(1,2,3),16(4)] SPre
FACT [Levels=5; NV=28; LABELS=!t('1','2','3','4','Null'); VALUES=12(5),4(1,2,3,4)] SPost
TREAT PrePost/(SPre + SPost)
ANOVA [PR=aov,info,means; FPROB=yes] Bottles

I had hoped that these statements would provide the linear contrasts …
VARIATE [Values = 2,3,4,*] PreLin
VARIATE [Values = 1,2,3,4,*] PostLin
BLOCK
TREATMENTS PrePost/(POL(SPre; 1; PreLin) + POL(SPost; 1; PostLin))
ANOVA [PR=aov,info,means,contrasts; FPROB=yes; PSE=diff] Bottles

… but alas, my efforts result in 2df for ‘Lin’ and zero Sums of Squares.
Source of variation d.f. s.s.
PrePost 1 4777.65
PrePost.SPre 2 715.17
PrePost.Lin 2 0.00
PrePost.SPost 3 8267.19
PrePost.Lin 2 0.00
Deviations 1 8267.19
Residual 21 1799.00
Total 27 15559.00

Is it correct to nullify the ‘Null’ level for SPre and SPost using a missing value indicator ‘*’? Or is there an alternative approach I could be using?

Many thanks, Terry


To unsubscribe from the GENSTAT list, click the following link:
https://www.jiscmail.ac.uk/cgi-bin/webadmin?SUBED1=GENSTAT&A=1

----------------------------------------------------------------------------------------------------------------------------------------------------------------------
This email is intended for the addressee(s) named and may contain confidential and/or privileged information.
If you are not the intended recipient, please notify the sender and then delete it immediately.
Any views expressed in this email are those of the individual sender except where the sender expressly and with authority states them to be the views of the NSW Office of Environment and Heritage.
PLEASE CONSIDER THE ENVIRONMENT BEFORE PRINTING THIS EMAIL

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