While the anova is a generalization of the two-tailed t-test when the researcher has more than two treatments or populations to compare, with or without blocking. The t-test is more flexible because it can be one-tailed to the left, one-tailed to the right or two-tailed as the Anova. When the one-tailed test is adecuate, you have more power for the same alpha than with the alternative two tailed test. Besides, in the Anova you need homogeneity of variances for the data of each treatment, but in the t-test, when there is heterogeneity of variance, you may use the correction proposed by the Welch–Satterthwaite equation.

It is just about the number of treatments and nothing more. Because as you know, F=t^2. So both methods of inference are the same.

While the anova is a generalization of the two-tailed t-test when the researcher has more than two treatments or populations to compare, with or without blocking. The t-test is more flexible because it can be one-tailed to the left, one-tailed to the right or two-tailed as the Anova. When the one-tailed test is adecuate, you have more power for the same alpha than with the alternative two tailed test. Besides, in the Anova you need homogeneity of variances for the data of each treatment, but in the t-test, when there is heterogeneity of variance, you may use the correction proposed by the Welch–Satterthwaite equation.

Dear guillermo, thank you for your detailed answer. Is t-test more robust than ANOVA?

Dear Ehsan

I would not say that the t test is more robust, because it has problems with the assumptions similar to that of anova, but that it is more flexible. When you assume that there is homogeneity of variance between the data normally distributed of two independent groups and you use the pooled variance, this test is equivalente to the Anova with its assumptions, and it produces always the same conclusions in the two-tailed test that de ANOVA, because as you stated F=t^2. An outlier for example may result in invalid inferences in both tests. But when you detect a problem of heterogeneity of variances there is other t-test useful (with the correction proposed by the Welch–Satterthwaite equation) and you do not have a ANOVA test for doing the same. Then if you have more than two treatments and you want to use the ANOVA but there is heterogenity of variances, you have to desist and change for a suitable non parametric test or you may transform your data looking the verification of the anova assumptions on the transformed data.

Dear guillermo, can you describe a little about transformation in anova? Do you mean taking e.g. logarithm or etc.?

There are monotonic transformations of data from an experiment, that did not initially met the ANOVA assumptions, which cause that after applying on them, they adequately verify these assumptions.

See for example: http://en.wikipedia.org/wiki/Data_transformation_(statistics) and

http://en.wikipedia.org/wiki/Box-Cox_transformation

Dear Guillermo, thank you for the answer. I've read both pages, non of them mention in normalizing data by transformation in ANOVA as i supposed. Say we have 4 groups of weight data for ANOVA and one of them is not normally distributed. By, for instance, power transformation we normalize that group. The problem is, analyzing three groups with kg unit with one transformed kg unit is totally meaningless. I think we can only use transformation in ANOVA If and if all the groups are not normally distributed and all of them need only one type of transformation. Otherwise we should use non-parametric methods.

Dear Ehsan

All the data needs the same transformation. In your example each group is subjected to the same transformation. You may evaluate the lack of normality from the residuals of the anova. When you have succes in the transformation, the residual of the transformed data is more normal than before the transformation. The same is with heterogeneity of variances.

Dear Ehsan and Guillermo, Thank you for your responses.

I saw that in some papers used ANOVA with two treatments and they would prefer to use the ANOVA because of they believe that the power of test is more than in comparison with t-test. But I am not sure about it!

Any suggestion will be appreciate that why they would prefer to use the AVONA when there are tow groups, I could not find in some books that I had read anything about differences of ANOVA and t-test.

As I tried to say before: When you assume that the data is normally distributed and there is homogeneity of variances between the two independent groups and in the t test you use the pooled variance, if you have a two tailed test, This t-test is equal to the Anova test (which has the same assumptions), and it arrives always the same conclusions because the test in the ANOVA is also a two-tailed test, and F=t^2. Then each time that you reject the null hypotesis in the ANOVA, you reject it in the two-tailed t test.

Dear Behzad, as i said before the power of both methods are the same and the authors of that article are wrong about what they said. I think an instance is comprehensive here. Suppose we have two treatments with 8 replications each. So the degrees of freedom for t-test is 16-1=15, and for ANOVA is as 1 ( 2-1) for nominator and 15 (16- treatment DF). the tabular t at alpha level of 0.05 is 2.131 and for tabular F is 4.54. Simply you can see that F=t^2. So there is no difference in power of two methods.

Dear Ehsan

Your commentary is well but the degree of freedom are 16-2=14 in both tests, because we estimate the mean of each treatment and this estimation takes a degree of freedom for each.

Thanks for your correction Guillermo, I've mistaken with total DF.

Dear Guillermo Enrique Ramos and Ehsan Khedive , Thank you for this conversation. Vert helpful. I was dealing with the same question as I try to make sense of my research results. I have two treatments (burned & unburned) on litterfall within an agroforestry system. Samples were collected from both treatments in 10 replications for "unburned" and 5 replications for "burned". In SAS 9.4, we used a Glimmix Procedure with lsmeans/tukey.

The question was same as already asked but I need to confirm: Should report that "We used a one-way ANOVA" (with only 2 treatments) or more accurate to use t-test terminology? I simply need to know if ANOVA can be extended to a case like this one;

Dear Daniel,

If your test is bilateral, then the t test and Anova F test arrive to the same conclusions (under the assumptions of homogeneity of variance and normality). If you need an unilateral test, you should use the t test which is more flexible and has more power for the same signification level. Beside when there is lack of homogeneity of variance, the t test has a known correction (Welch–Satterthwaite equation).

Thank you Guillermo Enrique Ramos . Test is bilateral in our case. Thanks for suggesting this Welch-Satterthwaite equation in case I need it in the future. The Shapiro-Wilk test was used to test for Normality. Discovering this forum (even though it goes back to 2013) has just opened my interest in Stats.

Dear Daniel Syauswa

There is some issues with your analysis. Why did you use GLIMMIX? this procedure analyzes both fixed and random variables, so you should be cautious about determining what type is your treatment to get desired results. In case of normality, tests are not reasonable when the number of reps are lower than 12. Assume you have only 5 reps as is in your burned site, this means that every replicate have a 20% share of the Probability Density Function (aka PDF). So when a replicate changes randomly, it changes the PDF by 20%. One replicate in this way can change the PDF from normal to Severely skewed.

if you have heterogeneity of variances, first you should increase number of reps for the smaller group.

A t-test can best analyse both one-tailed and two-tailed analysis for you.

t-test or sometimes called Student's t is one of parametric statistical method of testing hypothesis for mean difference for one group and a maximum of two groups only. If you have more than two groups, then you need to employ one way ANOVA (Analysis of Variance) to assess the difference in mean score among the groups simultaneously. Once proven, then you can make pairwise comparison to identify which pair (a comparison between two group at a time) actually differ.