The "problem" is that you seem to have fixed the desired family-wise type-I error-rate at 5%. When this is fixed, then your data simply des not allow you to reject the null hypothesis that the means of groups 1 and 2 are equal.

The two p-values are 0.045 and 0.069 - they are also quite similar (as the differences in means are very similar, too). One is slightly above the "magical 5% level" and the other is slightly below. Nothing strange about that.

I think your "problem" is caused by a flawed use of hypothesis tests. Let us assume the test is justified and sensible, and that the assumptions are met (normal residuals, equal variances). In this case you simply must not reject the null that m1=m2. Full stop. There is nothing more to discuss about. The interpretation depends a little on the question if the power was defined (a-priori). I assume that this was not the case. Then the interpretation is: do as if m1 and m3 were different and call the difference between the other groups inconclusive.

But since you think that there is some "problem", you may think about following questions:

why are you testing?

are the null hypotheses sensible?

what are the alternative hypotheses (relevant effect sizes)?

does the test have a resonable power?

what is the justification of the 5% significance level?

The "problem" is that you seem to have fixed the desired family-wise type-I error-rate at 5%. When this is fixed, then your data simply des not allow you to reject the null hypothesis that the means of groups 1 and 2 are equal.

The two p-values are 0.045 and 0.069 - they are also quite similar (as the differences in means are very similar, too). One is slightly above the "magical 5% level" and the other is slightly below. Nothing strange about that.

I think your "problem" is caused by a flawed use of hypothesis tests. Let us assume the test is justified and sensible, and that the assumptions are met (normal residuals, equal variances). In this case you simply must not reject the null that m1=m2. Full stop. There is nothing more to discuss about. The interpretation depends a little on the question if the power was defined (a-priori). I assume that this was not the case. Then the interpretation is: do as if m1 and m3 were different and call the difference between the other groups inconclusive.

But since you think that there is some "problem", you may think about following questions:

why are you testing?

are the null hypotheses sensible?

what are the alternative hypotheses (relevant effect sizes)?

does the test have a resonable power?

what is the justification of the 5% significance level?

Re Jochen's question about whether the test has reasonable power, note that when you have 3 groups, Fisher's LSD preserves the family-wise alpha, and is more powerful than alternatives such as Tukey's HSD.  If you need references, see the Multiple Comparisons chapter in Dave Howell's book Statistical Methods for Psychology.  I believe Thom Baguley's book Serious Stats also makes the same point.  And finally, see the article linked below.  HTH.

http://www.ncbi.nlm.nih.gov/pubmed/17128424

http://www.palgrave.com/page/detail/?k=9780230577183

http://www.cengage.com/search/productOverview.do;jsessionid=1E81FE48BA368F14BF322B3542DE5AAA?N=11%204294963777&Ntk=P_Isbn13&Ntt=9781111835484&Ntx=mode+matchallpartial

Yuting -

As Jochen indicated, setting a level of 5% for all tests is a mistake that is often made. The p-value is a function of sample size. (See the link below.)

I think you will find confidence intervals to be far more useful. You can probably assume normality, but the Chebyshev inequality is always available to you. You might be required to produce a p-value, but without a power analysis or other sensitivity analysis, it does not tell you much, and is not well interpreted at any rate. If you produce confidence intervals for each of your three point estimates, that will be far more useful.

You can make the same pairwise comparisons by estimating a confidence interval around the difference in estimated means. (This can be done for independent samples, which i suspect you have, or when there is covariance.)  I just did an Internet search on "confidence interval for difference in means," and the first link that popped up is attached below. However, I recommend when finding information that way that you check more than one source to (1) make certain there was no error on the first site you found, and (2) see if one site explains it to you better than another. 

Cheers - Jim  

http://onlinestatbook.com/2/estimation/difference_means.html

Article Practical Interpretation of Hypothesis Tests - letter to the...

James, just a note:

The 2nd page you linked (about CI between means) contains several mistakes, or better: a quite flawed logic. Just some points:

- Likert data is not normally distributed. It can not be, and this is well known. So this assumption is not met. But it is not discussed if the violation of this assumption is relevant (what will, in turn, question the whole testing procedure...)

- They use 17 values per group (each "value" a number between 1 and 7) and give the means with 4 significant digits. The means that can be calculated from 17 of such values have a "granularity" of 0.06. Claiming a precision higher than this is somehow unsound.

- a CI is a random inteval. It is a frequentist's tool, and as such it is not intended for estimation and does not provide a measure for precision. The procedure of generation of (say, 95%-)CIs will make sure that not more than 5% of such intervals miss the population difference. This is "error-rate control" and not "estimation". However, it is often abused of being an interval for estimation. This is reasonable only for likelhood-based CI's (one can invent other valid rules for constructing such CIs that are incompatible with estimation).

- worst: the conclusion went really wrong: "This analysis provides evidence that the mean for females is higher than the mean for males, and that the difference between means in the population is likely to be between 0.29 and 2.65." - CI's do not per se give a "range of likely values". Likelihood-based CIs give a range of parameter values (here: population difference) for which the observed data is not too unlikely (the "likely-statement" is about the data, not about the population parameter!). A correct conclusion would be: based on the observed CI hypotheses about mean differences of less than 0.29 and more than 2.56 can be rejected witout exceeding a long-run type-I error-rate of 5%.

- the results are valid only when the assumed model is correct (e.g. the errors may not be normally distributed for differences but possibly for log-ratios; just to mention a possible alternative). This is not mentioned.

If estimation is the aim, CIs make not much sense logically - although they may be sensible practically in sime situations. It would be more direct to simply provife likelihood intervals instead, although they still are about the data (and not about the parameter). There is only one way to get a statement about the parameter, and this is via Bayes theorem.

In the case above, what is the dependent variable? Is interval or ratio level data at least the measure for the dependent variable? Why are there three age groups and not two age groups for the independent variable age? Would reducing the degrees of freedom improve the model? Would the theory you are testing support two, rather than the three age groups? For example, if you are testing a theory based on differences among Boomers, Gen-Xe-rs, and Traditionalist on some scale, perhaps the groups cannot be merged. On the other hand, if you conduct a survey and some respondents claim their ethnicity to be African American while others claim Black American, it would be safe to merge those responses. Since you have three groups of ages, I assume they are age ranges, and thus, subject to subjective classification.

Nonetheless, the theory should justify the age groups you have established. In your case, two larger (samples) age groups is better than three smaller groups if your theory can justify two age groups. 

Regarding Bruce's point, there is an online video that illustrates (and includes the reference) why comparisons of 3 groups, which is your situation, will NOT increase family-wise error.  It is easy to understand and using Fisher's LSD should give you more statistical power as said.  

Here is the link of the video

https://www.youtube.com/watch?v=o5IOyr-7tVY