A one-way Analysis of Variance will tell you if one (or more) of the group means is significantly different from any (or all) of the others. However, it will not tell you exactly which group(s). For that, you need to perform an exhaustive post hoc comparison of all possible unique combinations (where order of combination of the two groups being exhaustively compared is not considered.

Example Groups A, B, C, D

Post Hoc Exhaustive Comparisons:

A vs B, A vs C, A vs D, B vs C, B vs D, C vs D.

There are many post hoc methods, and only a few are valid, depending on the nature of of data being analyzed.

Also, beware that all of the General Linear Model tests (t-test, ANOVA, etc.) require that all of your data and data subgroups being tested, be normally and uniformly distributed. Normal distribution is a primary assumption of the GLM tests.

If your data is not normally and uniformly distributed (use the Kolmogorov-Smirnov and Uniform Distribution tests) then use a non-parametric method of ANOVA known as the Kruskal-Wallis Test and perform exhaustive subgroup post hoc tests using the Mann-Whitney U test (non-parametric equivalent of the MWU test.

Richard E. Gilder, MS RN-C

Bioinformatics Scientist

The Gilder Company

TTUHSC Adjunct Faculty

[email protected]

T-Test is used when the independent variable doesn't exceed two values when ANOVA accepts more than two values. You'll choose according to the nature of your independent variables.

In general it is better to use a single model that takes into account all of the data. That is, it is better to use anova than to use pairwise t-tests. Likewise, for post-hoc tests after the omnibus anova analysis: for example, Tukey HSD takes into account the variability in all groups, while pairwise t-tests looks only at two groups at a time while ignoring the other groups. (Side note to reader: This also applies to Kruskal-Wallis: Dunn (1964) is a better post-hoc test than pairwise Mann-Whitney tests)

There is some discussion about sample size, anova, and non-parametric approaches, for a similar study design here. I think you will find the discussion helpful.

https://www.researchgate.net/post/What_alternative_for_ANOVA_test_when_levene_test_shows_a_violation_of_the_homogeneity_assumption

Richard E Gilder brings up some of the assumptions that go in to using analysis of variance. That is, for the test to be really valid, the underlying population of the sampled data needs to meet some conditions. There is an awful lot that can be said here, as there is some judgement, subtlety, and sometimes outright disagreements about how things should be approached.

In any case, find a good source for the assumptions about general linear models. There's a lot of bad advice and misinformation on the internet, so, really, use a real textbook. For example, Montgomery, Design and Analysis of Experiments, is a good one. At the following presentation, see slide 8, and then slides 19 - 22.

https://staff.emu.edu.tr/adhammackieh/Documents/courses/ieng581/lecture-notes/ch03.pdf .

That being said, I would like to offer a couple of disagreements with some of the advice of Richard E Gilder . (See, I told you there would be disagreements !) Just to focus on one issue, I advise against using a test to assess if data meet the normality assumption. No real data is truly normal, so if a test tells you your data are not normal, how does that help you? And these tests become increasingly sensitive to minor deviations from normality as the sample size becomes large.

It's better to use plots to assess if your data suggest conditional normality. To do this, plot the residuals from the analysis. (This is the approach described in the presentation linked above, with excerpts from the Montgomery book.) And the operative criterion is if the data approximates a conditionally normal distribution well enough to make the test close enough to valid. Note also that this approach isn't about if the data are normal. It's about the conditional distribution, approximated by the residuals.

As practical advice, heteroscedasticity may be more of a problem for anova than a modest violation of the normality assumption. (Again, don't use a test! In Montgomery, a plot of residuals vs. predicted values is used.) For a one-way anova there are approaches that can address differences in variance among the groups (like Welch's anova).

Tests on mean or means, when we have one sample , we use one sample t-test, for two independent samples, we use two-sample t-test. For more than two independent sample, one way ANOVA is used.

Use anova.

I concur with Marilou Lastrollo

ANOVA is the solution to the problem since the number of the group is more than one.

Anova is a great way.

Dear Denis,

from your question, I understand that, in the end, you want to know which of the four reactants, compared to the other three, scores ‘best’. That is, which of the four reactants gives the ‘best’ recovery yields, compared to ‘the rest’. That question does not call for ‘post hoc’, ‘a posteriori’, or so called ‘unplanned’ (pairwise) contrasts such as those of Tukey or Bonferroni etc.

Instead, comparing each reactant to the other three is specifically the purpose of the ‘a priori’ or ‘planned’ contrast called the ‘deviation contrast’. Below, I show with a concrete example how this contrast can be computed and tested in SPSS (but SAS, Stata, and R etc. have similar commands, and also similar results). I also show that this contrast can not only be done with ‘ANOVA’, but also with regression (using the right effect code variables). Note however, that, in general, the deviation contrast compares the mean of each reactant to the UNweighted mean of the other three reactants. I also show how to compare each reactant with the WEIGHTED mean of the other three reactants. Further details are given in the comments (‘*’) within the following SPSS-syntax:

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

* To illustrate, we create a hypothetical dataset:.

data list free /Reactant, Yield, N.

begin data.

1 0 99

1 1 11

2 0 21

2 1 09

3 0 08

3 1 02

4 0 08

4 1 12

end data.

weight by N.

value labels Reactant

1 'A (N=110, Yield=11/110=0.10)'

2 'B (N=30, Yield=9/30=0.30)'

3 'C (N=10, Yield=2/10=0.20)'

4 'D (N=20, Yield=12/20=0.60)'/.

* In SPSS, the UNweighted deviation contrast within ANOVA

is done with just a few keywords:.

glm Yield by Reactant /print=descriptive /contrast(Reactant)=Deviation(1).

glm Yield by Reactant /print=descriptive /contrast(Reactant)=Deviation(4).

* The UNweighted grand mean is (0.10+0.30+0.20+0.60)/4=0.30.

* The SPSS-results of this UNweighted deviation contrast show that:

mean ‘A’=0.10 is sign. < than the UNweighted grand mean=0.30, p than the UNweighted grand mean=0.30, p than the WEIGHTED grand mean=0.20, p than the WEIGHTED mean of the rest=0.15, p

When comparison of more than two populations' means is to be carried out, ANOVA is the best analysis to be used; OR ANOVA is used for testing for equality of three or more populations' means.

Thus, in this your problem...ANOVA would be appropriate.