결합 된 표준 편차를 찾을 수 있습니까?

내가 2 세트가 있다고 가정하십시오.

세트 A : 항목 수 $n= 10$ , $\mu = 2.4$ , $\sigma = 0.8$

B 설정 : 항목 수 $n= 5$ , $\mu = 2$ , $\sigma = 1.2$

결합 평균 ( $\mu$ )을 쉽게 찾을 수 있지만 결합 표준 편차를 어떻게 구해야합니까?

따라서이 두 샘플 중 하나를 하나로 가져 오려면 다음을 수행하십시오.

$s_1 = \sqrt{\frac{1}{n_1}\Sigma_{i = 1}^{n_1} (x_i - \bar{y}_1)^2}$

$s_2 = \sqrt{\frac{1}{n_2}\Sigma_{i = 1}^{n_2} (y_i - \bar{y}_2)^2}$

여기서 및 ˉ y 2 는 표본 평균이며 s 1 및 s $\bar{y}_1$$\bar{y}_2$$s_1$$s_2$ 는 표본 표준 편차입니다.

그것들을 추가하려면 다음이 있습니다.

$s = \sqrt{\frac{1}{n_1 + n_2}\Sigma_{i = 1}^{n_1 + n_2} (z_i - \bar{y})^2}$

새로운 평균 가 ˉ y 1 및 ˉ y 2 와 다르기 때문에 그렇게 간단하지 않습니다 .$\bar{y}$$\bar{y}_1$$\bar{y}_2$

$\bar{y} = \frac{1}{n_1 + n_2}\Sigma_{i = 1}^{n_1 + n_2} z_i = \frac{n_1 \bar{y}_1 + n_2 \bar{y}_2}{n_1 + n_2}$

최종 공식은 다음과 같습니다.

$s = \sqrt{\frac{n_1 s_1^2 + n_2 s_2^2+ n_1(\bar{y}_1-\bar{y})^2 +n_2(\bar{y}_2-\bar{y})^2}{n_1 + n_2 }}$

일반적으로 사용되는 베셀 보정 ( " 분모") 버전의 표준 편차에 대한 평균 결과는 이전과 같지만$n-1$

$s = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2 + n_1(\bar{y}_1-\bar{y})^2 +n_2(\bar{y}_2-\bar{y})^2}{n_1+n_2 - 1} }$

You can read more info here: http://en.wikipedia.org/wiki/Standard_deviation

This obviously extends to $K$ groups:

$$s = \sqrt{ \frac{\sum_{k=1}^K (n_k-1)s_k^2 + n_k(\bar{y}_k-\bar{y})^2} {(\sum_{k=1}^K n_k) -1} }$$

I had the same problem: having the standard deviation, means and sizes of several subsets with empty intersection, compute the standard deviation of the union of those subsets.

I like the answer of sashkello and Glen_b ♦, but I wanted to find a proof of it. I did it in this way, and I leave it here in case it is of help for anybody.

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So the aim is to see that indeed:

$$s = \left(\frac{n_1 s_1^2 + n_2 s_2^2+ n_1(\bar{y}_1-\bar{y})^2 +n_2(\bar{y}_2-\bar{y})^2}{n_1 + n_2 }\right)^{1/2}$$

Step by step:

$$\left(\frac{n_1 s_1^2 + n_2 s_2^2+ n_1(\bar{y}_1-\bar{y})^2 +n_2(\bar{y}_2-\bar{y})^2}{n_1 + n_2 }\right)^{1/2} = \left(\frac{\sum_{i=1}^{n_1}(x_i - \bar{y_1})^2 + \sum_{i=1}^{n_2}(y_i - \bar{y_2})^2+ n_1(\bar{y}_1-\bar{y})^2 +n_2(\bar{y}_2-\bar{y})^2}{n_1 + n_2 }\right)^{1/2} = \left(\frac{\sum_{i=1}^{n_1}\left((x_i - \bar{y_1})^2 + (\bar{y}_1-\bar{y})^2\right) + \sum_{i=1}^{n_2}\left((y_i - \bar{y_2})^2 + (\bar{y}_2-\bar{y})^2\right)}{n_1 + n_2}\right)^{1/2} = \left(\frac{\sum_{i=1}^{n_1}\left(x_i^2 + \bar{y}^2 + 2\bar{y_1}^2 -2x_i\bar{y_1} -2\bar{y_1}\bar{y} \right)}{n_1 + n_2} + \frac{\sum_{i=1}^{n_2}\left(y_i^2 + \bar{y}^2 + 2\bar{y_2}^2 -2y_i\bar{y_2} -2\bar{y_2}\bar{y} \right)}{n_1 + n_2}\right)^{1/2} = \left(\frac{\sum_{i=1}^{n_1}\left(x_i^2 + \bar{y}^2 -2\bar{y}\sum_{j=1}^{n_1}\frac{x_j}{n_1}\right) + 2n_1\bar{y_1}^2 -2\bar{y_1}\sum_{i=1}^{n_1}x_i}{n_1 + n_2} + \frac{\sum_{i=1}^{n_2}\left(y_i^2 + \bar{y}^2 -2\bar{y}\sum_{j=1}^{n_2}\frac{y_j}{n_2}\right) + 2n_2\bar{y_2}^2 -2\bar{y_2}\sum_{i=1}^{n_2}y_i}{n_1 + n_2}\right)^{1/2} = \left(\frac{\sum_{i=1}^{n_1}\left(x_i^2 + \bar{y}^2 -2\bar{y}\sum_{j=1}^{n_1}\frac{x_j}{n_1}\right) + 2n_1\bar{y_1}^2 -2\bar{y_1}n_1\bar{y_1}}{n_1 + n_2} + \frac{\sum_{i=1}^{n_2}\left(y_i^2 + \bar{y}^2 -2\bar{y}\sum_{j=1}^{n_2}\frac{y_j}{n_2}\right) + 2n_2\bar{y_2}^2 -2\bar{y_2}n_2\bar{y_2}}{n_1 + n_2}\right)^{1/2} = \left(\frac{\sum_{i=1}^{n_1}\left(x_i^2 + \bar{y}^2 -2\bar{y}\sum_{j=1}^{n_1}\frac{x_j}{n_1}\right)}{n_1 + n_2} + \frac{\sum_{i=1}^{n_2}\left(y_i^2 + \bar{y}^2 -2\bar{y}\sum_{j=1}^{n_2}\frac{y_j}{n_2}\right)}{n_1 + n_2}\right)^{1/2}$$

Now the trick is to realize that we can reorder the sums: since each

$$-2\bar{y}\sum_{j=1}^{n_1}\frac{x_j}{n_1}$$ term appears $n_1$ times, we can re-write the numerator as $$\sum_{i=1}^{n_1}\left(x_i^2 + \bar{y}^2 -2\bar{y}x_i\right),$$

and hence, continuing with the equality chain:

$$= \left(\frac{\sum_{i=1}^{n_1}\left(x_i - \bar{y}\right)^2}{n_1 + n_2} + \frac{\sum_{i=1}^{n_2}\left(y_i - \bar{y}\right)^2}{n_1 + n_2}\right)^{1/2} = \left(\frac{\sum_{i=1}^{n_1 + n_2}\left(z_i - \bar{y}\right)^2}{n_1 + n_2}\right)^{1/2} = s \qquad \square$$

This been said, there is probably a simpler way to do this.

The formula can be extended to $k$ subsets as stated before. The proof would be induction on the number of sets. The base case is already proven, and for the induction step you should apply a similar equality chain to the latter.