When you have an apriori alternative hypothesis, you can look directly at doing a hypothesis test specifically for that, that is, an hypothesis test with increased power for the alternative you are interested in. You have an ordered alternative, so could use a test specifically constructed for that.
Ordered alternatives also named monotone alternatives. There is a section about that in the old book by Miller: "Beyond ANOVA".
The likelihood ratio test for monotone alternatives was developed by Bartholomew (1959, 1961, refs in Miller), and is connected to isotone regression and the "pool adjacent violators" algorithm. The problem is that the null distribution is complicated, and needs special tables, Miller's book has references. And, there is a complete book:
There is an alternative, maybe easier to implement (following Miller). This is due to Abelson and Tukey (1963), and summarized by: If the monotone alternative is $\mu_1 \le \mu_2 \le \dots \le \mu_I$ then base the test on the contrast with coefficients $c_1, c_2, \dots, c_I$. Write $R$ for the region in parameter space satisfying $\mu_1 \le \mu_2 \le \dots \le \mu_I$
and choose a maximin approach by choosing the contrast by maximizing the minimum power over $R$. Detail in Miller, above. The paper by Abelson and Tukey: "Efficient Utilization of Non-Numerical Information in Quantitative Analysis: General Theory and the Case of Simple Order", the Annals of Mathematical Statistics, 1963, have a table of maximin contrasts, which we report a part of below:
$$
\begin{array}{crrrrrrr}
j & n=1 & n=2& n=3 & n=4 & n=5 & n=6& n=7&n=8 \\ \hline
1 &-.707&-.816&-.866&-.894&-.913&-.926&-.935 \\
2 & .707&.000&-.134&-.201&-.242&-.269&-.289 \\
3 & & .816& .134&.000&-.070&-.114&-.144& \\
4 & & & .866&.201&.070& .000&-.045 \\
5 & & & &.894&.242&.114&.045 \\
6 & & & & &.913&.269&.144 \\
7 & & & & & &.926&.289 \\
8 & & & & & & &.935 \\ \hline
\end{array}
$$
By using such an approach, there is only one hypothesis test and obviously no multiplicity problems!
A formula for the contrasts in the simple order case $\mu_1 \le \dots \le \mu_n$ is given by:
$$
c_j = \{(j-1)[1-((j-1)/n)]\}^{1/2} - \{j(1-j/n)\}^{1/2}
$$
and then one can simply use a $t$-test with this contrast. This do not need any special code (or package) in R.
There do seem to exist some implementations in R for tests for ordered alternatives, not the ones I wrote about, but a nonparametric alternative, the Jonckheere-Terpstra test. See https://en.wikipedia.org/wiki/Jonckheere's_trend_test . I will show an implementation in CRAN package clinfun, code below:
library(clinfun) ### From CRAN
set.seed(7*11*13) ### for reproducibility
mu <- c(10, 11, 12)
g <- length(mu)
n <- 10 ### number of obs in each group
gg <- list()
for (j in seq(along=mu)) gg[[j]] <- rnorm(n, mu[j], sd=1.0)
nn <- sapply(gg, length)
ind <- rep(1:g, nn)
xx <- unlist(gg)
test.ordered <- jonckheere.test(xx, ind, alternative="increasing")
Output when run is:
> test.ordered
Jonckheere-Terpstra test
data:
JT = 219, p-value = 0.00407
alternative hypothesis: increasing
Other implementations in coin, npsm and the package lawstat has analogous tests for equality of variances against ordered alternatives. I will come back extending this when I get hold of the referenced papers!
