Calculating scale mean from item means I'm working on a review paper and need to collect the means and standard deviations of a given measure (such as a measure of depression) from papers of interest. However, some authors report means and standard deviations for each item on the measure, but do not calculate the overall mean and standard deviation. For example, for a measure with a 1-5 Likert-type response scale, rather than reporting a mean depression score of 2.5, sd = .34, for the whole scale, they'll report: item 1 mean 2.4, sd=.41, item 2 mean 2.5, sd=.38, etc.
1) Can I calculate the scale mean by simply taking the average of the item means?
2) Is it possible to calculate the accompanying scale standard deviation? And if so, how? It's probably not as simple as taking the average of the item standard deviations....
If it isn't possible, I can just contact the authors, but I hate to bother them if the needed stats can be calculated from the provided information.
Any advice would be greatly appreciated! Thank you!
 A: If the scale score is an unweighted mean of ratings that we're pretending are numeric (not ordinal, which Likert ratings actually are), then: 


*

*Yes, a sample's mean score is equal to the grand mean of the sample's mean ratings across all items.
Here's a demonstration in r:
d8a=data.frame(replicate(10,rbinom(100,4,.5)))                    #simulates Likert data
for(i in 1:100){d8a$score[i]=mean(as.numeric(d8a[i,1:10]))}    #appends unweighted means
colMeans(d8a)                                                  #means of items and score
mean(colMeans(d8a)[1:10])                                           #grand mean of items
colMeans(d8a)[11]==mean(colMeans(d8a)[1:10])                     #usually prints "TRUE"*

*Oddly enough, with set.seed(2), colMeans(d8a)[11]= 1.985, as does mean(colMeans(d8a)[1:10]), but colMeans(d8a)[11]==mean(colMeans(d8a)[1:10]) prints FALSE. Not sure what's up with that. Might have to ask about this on Stack Overflow later.
If scale scores are produced by other means, such as by factor scoring using weights based on sample covariance, then these won't be equivalent to the unweighted mean of item ratings.

*I can't say for sure that there's no way of doing it, but as you suspect, the simple way won't work. 
sd(d8a$score)                                               #standard deviation of score
x=c();for(i in 1:10){x=append(x,sd(d8a[,i]))};mean(x)  #mean of item standard deviations

The score's $SD$ will be smaller in general as a consequence of regression to the mean.
