First, you must testing normality of your data distribution. You Can apply Anova one factor only if you data following normality.

In the other case, you must apply meta-analysis using the "random-effects model.

I Hope that this help you,

Good luck

You may try to do a meta-analysis using the "random-effects model". The heterogeneity variance and heterogeneity index are the measures of the difference between the groups (locations).

First, you must testing normality of your data distribution. You Can apply Anova one factor only if you data following normality.

In the other case, you must apply meta-analysis using the "random-effects model.

I Hope that this help you,

Good luck

Do your anova in regression form which automatically takes the difference in sample sizes into account.. Best regards, David Booth

I agree with David Eugene Booth , although the number of observations are very small particularly if there is a lot of natural heterogeneity in the phosphorous content.

David Eugene Booth wrote: "Do your anova in regression form which automatically takes the difference in sample sizes into account."

David (and others who recommended your reply): I do not understand what you mean when you say a regression approach "automatically takes the difference in sample sizes into account". Pasted below is an (SPSS) example (using made up data) in which the ONEWAY command and the REGRESSION command (using 2 group indicator variables) yield exactly the same ANOVA summary table (see attached image). I used SPSS because its ONEWAY (ANOVA) command has easy options for obtaining two different F-tests that are robust to heterogeneity of variance (the Welch and Brown-Forsythe F-tests).

My response to Louisa Habermann includes the following points:

HTH.

* SPSS syntax for the aforementioned example starts here.

* David Booth wrote, "Do your anova in regression form which

* automatically takes the difference in sample sizes into account".

* This implies that using an ANOVA command will somehow yield

* different results than using a REGRESSION command.

* That makes no sense to me.

* The results will be exactly the same with both approaches.

NEW FILE.

DATASET CLOSE ALL.

DATA LIST LIST / group y (2F5.0).

BEGIN DATA

1 60

1 50

1 55

2 60

2 62

2 69

2 74

2 57

2 61

2 68

2 56

2 65

2 58

2 55

2 65

2 69

2 62

2 63

2 62

2 57

3 44

3 64

3 26

3 63

END DATA.

* Generate indicator variables for Group.

COMPUTE g1 = group EQ 1.

COMPUTE g2 = group EQ 2.

COMPUTE g3 = group EQ 3.

FORMATS g1 to g3 (F1).

CROSSTABS g1 g2 g3 by group.

MEANS y by group.

* Estimate a one-way ANOVA using the REGRESSION command.

* Use group 2 (the largest group) as the reference category.

REGRESSION

/STATISTICS COEFF OUTS CI(95) R ANOVA

/DEPENDENT y

/METHOD=ENTER g1 g3.

* Estimate a one-way ANOVA using the ONEWAY command.

* Include both the Brown-Forsythe and Welch robust F-tests.

ONEWAY y BY group

/STATISTICS BROWNFORSYTHE WELCH

/MISSING ANALYSIS.

@BruceWeaver, Apologies it is called matrix arithmetic a subset of linear algebra, you know General Linear Model sort of thing. You might see, Mendenhall, Design and Analysis of experiment, or even George Seber's, General Linear Model. It would have saved you an awful lot

of work. David Booth

.Oh Bruce BTW

You did the calculation our way.in the old days AN0VA formuae were simpler for balanced designs a d hence balance was prefed. This is where she got the idea of equal sample size. Nowadays all packages

Use our approach but write it in the old style so some don't have to understand

May the Force be with you. David

Sorry for typoss it's hard to do on cell phone

If you have only One effect, there is no difference between means and least Square means. But if you have more than one effect, you can report least square means. See SAS example:

data soil;

input treat repeat Pho;

cards;

1 1 50

1 2 51

1 3 55

2 1 35

2 2 36

2 3 37

2 4 38

2 5 39

2 6 40

2 7 41

2 8 42

2 9 43

2 10 44

2 11 45

2 12 46

2 13 47

2 14 48

2 15 49

2 16 50

2 17 51

3 1 100

3 2 110

3 3 120

3 4 130

;

run;

proc glm;

class treat;

model pho =treat;

means treat ;

lsmeans treat/stderr;

run;

---

output

The SAS System 16:47 Monday,

The GLM Procedure

Class Level Information

Class Levels Values

treat 3 1 2 3

Number of Observations Read 24

Number of Observations Used 24

The SAS System 16:47 Monday,

The GLM Procedure

Dependent Variable: Pho

Sum of

Source DF Squares Mean Square F Value Pr > F

Model 2 16844.62500 8422.31250 191.83 F

treat 2 16844.62500 8422.31250 191.83 F

treat 2 16844.62500 8422.31250 191.83 |t|

1 52.000000 3.825561

Hello David Eugene Booth. I'm afraid I still do not see the distinction you are making. You said, "it is called matrix arithmetic a subset of linear algebra, you know General Linear Model sort of thing."

The reason I do not get your point is that if use the usual matrix equations for the OLS model in the MATRIX language in SPSS (which is far more work than using commands like ONEWAY, UNIANOVA or REGRESSION, by the way), I get exactly the same results. The attached image shows the same output I posted previously with selected parts of my MATRIX output added. Apologies if I'm just being obtuse.

Bruce Weaver All we are trying to say is to solve the problem enter the data and make one call to the appropriate regression method and you are finished. That's what SPSS does. Then for the Anova output it back calculates all the extra stuff that you don't need. I learned this in a summer course with the Mendenhall book. Compare that to Scheffe's ANOVA book from maybe 7 years before I took my course. The amazing thing is that people still mess with the ANOVA output. Please click on this link: Computer Age Statistical Inference: Algorithms, Evidence, and Data Science | Bradley Efron, Trevor Hastie | download (b-ok.cc)

Merry Xmas but a bit late. Dave Booth

Hello David, and belated season's greetings to you too. As I recall, you were facing some weather-related difficulties before Christmas. I hope all of that is now sorted out and all is well.

Re this discussion, my experience with SPSS, at least, is that it is far easier to estimate a one-way ANOVA model using either the ONEWAY command or the GLM/UNIANOVA than it is using the REGRESSION command. If there are 3 or more groups, REGRESSION (which does not accept "factor" variables) requires one to generate a set of k-1 indicator variables (or to use some other coding scheme with k-1 variables). As for getting a bunch of unwanted output, the default setting for the ONEWAY command is to return only the ANOVA summary table, which looks exactly the same (except for some labels) that REGRESSION would give you. And the REGRESSION output would include other stuff by default (e.g., a model summary table and a table of coefficients). So whether it is easier to use a regression command depends, I think, on what software one is using.

Anyway, I think my original point/question got lost. You wrote that doing a [one-way?] ANOVA "in regression form...automatically takes the difference in sample sizes into account" (emphasis added). I took that to imply that difference in sample sizes are not automatically taken into account if one uses an ANOVA form. But as shown in my previous posts, I get exactly the same results using ONEWAY, UNIANOVA and REGRESSION commands in SPSS. I also get the same results using the following 3 Stata commands:

oneway y group

anova y group

regress y i.group

It may be that I inferred something you did not imply. Apologies if so! ;-)

Cheers,

Bruce

This gives a nicely 'balanced' view!

https://www.theanalysisfactor.com/when-unequal-sample-sizes-are-and-are-not-a-problem-in-anova/