Are these formulas for transforming P, LSD, MSD, HSD, CI, to SE as an exact or inflated/conservative estimate of $\hat{\sigma}$ correct?

Question

Background

I am conducting a meta-analysis that includes previously published data. Often, differences between treatments are reported with P-values, least significant differences (LSD), and other statistics but provide no direct estimate of the variance.

In the context of the model that I am using, an overestimate of variance is okay.

Problem

Here is a list of transformations to $SE$ where $SE=\sqrt{MSE/n}$ (Saville 2003) that I am considering, feedback appreciated; below, I assume that $\alpha=0.05$ so $1-^{\alpha}/_2=0.975$ and variables are normally distributed unless otherwise stated:

Questions:

1. given $P$, $n$, and treatment means $\bar X_1$ and $\bar X_2$ $$SE=\frac{\bar X_1-\bar X_2}{t_{(1-\frac{P}{2},2n-2)}\sqrt{2/n}}$$

2. given LSD (Rosenberg 2004), $\alpha$, $n$, $b$ where $b$ is number of blocks, and $n=b$ by default for RCBD $$SE = \frac{LSD}{t_{(0.975,n)}\sqrt{2bn}}$$

3. given MSD (minimum significant difference) (Wang 2000), $n$, $\alpha$, df = $2n-2$ $$SE = \frac{MSD}{t_{(0.975, 2n-2)}\sqrt{2}}$$

4. given a 95% Confidence Interval (Saville 2003) (measured from mean to upper or lower confidence limit), $\alpha$, and $n$ $$SE = \frac{CI}{t_{(\alpha/2,n)}}$$

5. given Tukey's HSD, $n$, where $q$ is the 'studentized range statistic', $$SE = \frac{HSD}{q_{(0.975,n)}}$$

An R function to encapsulate these equations:

1. Example Data:

   data <- data.frame(Y=rep(1,5), 
                      stat=rep(1,5), 
                      n=rep(4,5), 
                      statname=c('SD', 'MSE', 'LSD', 'HSD', 'MSD') 
   

2. Example Use:

   transformstats(data)    
   

3. The transformstats function:

   transformstats <- function(data) {
     ## Transformation of stats to SE
     ## transform SD to SE
     if ("SD" %in% data$statname) {
       sdi <- which(data$statname == "SD")
       data$stat[sdi] <- data$stat[sdi] / sqrt(data$n[sdi])
       data$statname[sdi] <- "SE"
         }
     ## transform MSE to SE
     if ("MSE" %in% data$statname) {
       msei <- which(data$statname == "MSE")
       data$stat[msei] <- sqrt (data$stat[msei]/data$n[msei])
       data$statname[msei] <- "SE"
     }
     ## 95%CI measured from mean to upper or lower CI
     ## SE = CI/t
     if ("95%CI" %in% data$statname) {
       cii <- which(data$statname == '95%CI')
       data$stat[cii] <- data$stat[cii]/qt(0.975,data$n[cii])
       data$statname[cii] <- "SE"
     }
     ## Fisher's Least Significant Difference (LSD)
     ## conservatively assume no within block replication
     if ("LSD" %in% data$statname) {
       lsdi <- which(data$statname == "LSD")
       data$stat[lsdi] <- data$stat[lsdi] / (qt(0.975,data$n[lsdi]) * sqrt( (2 * data$n[lsdi])))
       data$statname[lsdi] <- "SE"
     }
     ## Tukey's Honestly Significant Difference (HSD),
     ## conservatively assuming 3 groups being tested so df =2
     if ("HSD" %in% data$statname) {
       hsdi <- which(data$statname == "HSD" & data$n > 1)
       data$stat[hsdi] <- data$stat[hsdi] / (qtukey(0.975, data$n[lsdi], df = 2))
       data$statname[hsdi] <- "SE"
     }              
     ## MSD Minimum Squared Difference
     ## MSD = t_{\alpha/2, 2n-2}*SD*sqrt(2/n)
     ## SE  = MSD*n/(t*sqrt(2))
     if ("MSD" %in% data$statname) {
       msdi <- which(data$statname == "MSD")
       data$stat[msdi] <- data$stat[msdi] * data$n[msdi] / (qt(0.975,2*data$n[lsdi]-2)*sqrt(2))
       data$statname[msdi] <- "SE"
     }
     if (FALSE %in% c('SE','none') %in% data$statname) {
       print(paste(trait, ': ERROR!!! data contains untransformed statistics'))
     }
     return(data)
   }
   

References

Saville 2003Can J. Exptl Psych. (pdf)

Rosenberg et al 2004 (link)

Wang et al. 2000 Env. Tox. and Chem 19(1):113-117 (link)

Answer 1

Your LSD equation looks fine. If you want to get back to variance and you have a summary statistic that says something about variability or significance of an effect then you can almost always get back to variance—-you just need to know the formula. For example, in your equation for LSD you want to solve for MSE, MSE = (LSD/t_)^2 / 2 * b

Answer 2

I can only agree with John. Furthermore, perhaps this paper by David Saville helps you with some formula to recalculate variability measures from LSDs et al.:
Saville D.J. (2003). Basic statistics and the inconsistency of multiple comparison procedures. Canadian Journal of Experimental Psychology, 57, 167–175

UPDATE:
If you are looking for more formulas to convert between various effect sizes, books on meta-analysis should provide a lot of these. However, I am not an expert in this area and can't recommend one.
But, I remember that the book by Rosenthal and Rosnow once helped with some formula:
Essentials of Behavioral Research: Methods and Data Analysis
Furthermore, I have heard a lot of good things about the formulas in this book by Rosenthal, Rosnow & Rubin (although I have never used it):
Contrasts and Effect Sizes in Behavioral Research: A Correlational Approach (You should definitely give it a try if a nearby library has it).

If this is not enough, perhaps ask another question on literature for converting effect sizes for meta-analyses. Perhaps someone more into meta-analysis has more grounded recommendations.

Answer 3

You may consider trying the R package compute.es. There are several functions for deriving effect size estimates and the variance of the effect size.