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Glen_b
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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}$$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.

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 $x_{ij}$.

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.

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.

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Glen_b
  • 290.5k
  • 37
  • 652
  • 1.1k

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 $x_{ij}$.

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.

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!

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.

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 $x_{ij}$.

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.

Source Link
Glen_b
  • 290.5k
  • 37
  • 652
  • 1.1k

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!

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.