# Is the mean of equal length subsets always equal to the mean of the set?

I've been fooling around with some numbers while learning R and would like to know if the following is generalizable.

When I calculated the mean of numbers 1 through 100, I got this:

> mean(1:100)
 50.5


Then when I calculated the mean of equal length subsets, I got the same:

> mean25 <- mean(1:25)
> mean50 <- mean(26:50)
> mean75 <- mean(51:75)
> mean100 <- mean(76:100)
> mean(c(mean25, mean50, mean75, mean100))
 50.5


As an aside, when calculating the mean of different length subsets, the mean is expectedly not the same:

> mean5 <- mean(1:5)
> mean37 <- mean(6:37)
> mean91 <- mean(38:91)
> mean100 <- mean(92:100)
> mean(c(mean5, mean37, mean91, mean100))
 46.25


My question:

Would it be accurate to say the mean of equal length subsets is always equal to the mean of the set? Are there any counterexamples to this?

Would it be accurate to say the mean of equal length subsets is always equal to the mean of the set?

It's always true!

Consider $n=mk$ observations, where you take $k$ mutually-exclusive groups of size $m$.

Label the observations in group $i$ as $x_{ij},\: j=1,2,...,m$.

The individual means are $\bar{x}_i = \frac{1}{m}\sum_{j=1}^m x_{ij}$.

The mean-of-means is

$\overline{\bar{x}_i} = \frac{1}{k}\sum_{i=1}^k \bar{x}_{i}$

$\hspace{.5cm}=\frac{1}{k}\sum_{i=1}^k (\frac{1}{m}\sum_{j=1}^m x_{ij})$

$\hspace{.5cm}=\frac{1}{km}\sum_{i=1}^k \sum_{j=1}^m x_{ij}$

$\hspace{.5cm}=\frac{1}{n}\sum_{i,j} x_{ij}$

which is just the overall mean of the data.

For the unequal-length case, they're also equal (that is, the mean of means equals the overall mean) if you do an appropriately weighted average when taking the mean of means.

• Thank you for such a tremendously thorough answer, and I appreciate the bit about weighted means at the end as well. I'm off to explore that now! – user60305 Dec 4 '14 at 16:27
• could you please clarify this last sentence ("For the unequal-length case, they're also equal (that is, the mean of means equals the overall mean) if you do an appropriately weighted average when taking the mean of means.")? The first part (equal lengths) makes sense to me. But in case of unequal lengths, they are in fact NOT the same...right? – Tilen Jan 5 '20 at 11:24
• I don't follow you. It's possible you have misunderstood the intent of the phrase "appropriately-weighted" – Glen_b Jan 6 '20 at 10:43
• That's very likely. So, could you clarify the intent of that phrase please? Many thanks. – Tilen Jan 6 '20 at 13:20
• If you take weights proportional to OP calls "length" (i.e. number of values in each subset), then the weighted average of those subset means will be the overall mean. – Glen_b Jan 6 '20 at 22:54