p-Value of mean of sample means I suspect this question is related to:
When combining p-values, why not just averaging?
Let me explain my problem: 


*

*Suppose you have a set of sample means and associated p-values, standard errors and degrees of freedom. Let’s denote them as: $\mu_i$, $p_i$, $\text{SE}_i$, $df_i$ where $i=\{1,n\}$

*It’s straight forward to figure out the mean of sample means. But what is the resultant standard error and p-value of the mean of sample means? I suspect p-values multiply in some manner… 
 A: You can use fixed effects meta-analysis to do this. Here's an example, in R.
library(metafor)

#set up means
df <- as.data.frame(rnorm(1000))
names(df) <- "y"


#create 10 groups and calculate mean and se for each
df$group <- sample(1:10, 1000, replace=TRUE)

summaryStats <- data.frame(1:10)
summaryStats$m <- NA
    summaryStats$se <- NA

summaryStats$m <- tapply(df$y, df$group, mean)
    summaryStats$se <- tapply(df$y, df$group, function(x) sd(x)/sqrt(length(x)))
summaryStats$v <- summaryStats$se^2
summaryStats <- as.data.frame(summaryStats)


#Run meta-analysis to get combined mean and se for 10 groups
rma(yi=summaryStats$m, vi=summaryStats$v, method="FE")
#Run t-test to get mean and SE for original sample
t.test(df$y)

Here's the meta-analysis output:
> rma(yi=summaryStats$m, vi=summaryStats$v, method="FE")

Fixed-Effects Model (k = 10)

Test for Heterogeneity: 
Q(df = 9) = 4.3262, p-val = 0.8887

Model Results:

estimate       se     zval     pval    ci.lb    ci.ub          
 -0.0034   0.0316  -0.1078   0.9142  -0.0653   0.0585          

Here's the t-test output:

t.test(df$y)

    One Sample t-test

data:  df$y
t = -0.0551, df = 999, p-value = 0.9561
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 -0.06380216  0.06031609
sample estimates:
   mean of x 
-0.001743035 

They're not exactly the same, but they're very close.
