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I have the mean and standard deviation for body temperature ( in a group of people n=10 ), which measured using a thermometer in the left axilla, then in the right axilla 

Left  mean = 36.4; SD= 0.2
Right mean = 36.7; SD= 0.18

If I want to get one mean and one standard deviation that can express the whole body temperature. Can I average mean left and mean right then SD left and SD right.

i.e.

(mean left + mean right )/2
(SD left + SD right )/2

is this correct? if not . What is the correct methodology to get one mean and one SD ( if any)

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    $\begingroup$ "Standard divination" sounds like a procedure employed by the religiously inclined when they do not have data. $\endgroup$
    – whuber ♦
    Commented Apr 27, 2016 at 13:43
  • $\begingroup$ @whuber. Kindly, could you please clarify what you mean. I am sorry but I didn't get your idea! $\endgroup$
    – goro
    Commented Apr 27, 2016 at 13:47
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    $\begingroup$ google.com/search?q=divination+definition $\endgroup$
    – whuber ♦
    Commented Apr 27, 2016 at 13:50

2 Answers 2

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I am assuming you do not have the observed data, and just have the means and standard deviations. Let the data from the left axilla be $X_1, X_2, \dots, X_n$, and let the data from the right axilla be $Y_1, Y_2, \dots, Y_n$. I am assuming that you have equal sample sizes.

(Before I go on, I must warn you that your pooled/combined sample is no longer independent, because you have two data points on each individual. This is important if you are going to use the standard deviations to test a hypothesis.)

You are given the quantities $\bar{X}_n = 36.4$ and $sd(X_1) = 0.2$ and $\bar{Y}_n = 36.7$ and $sd(Y_1) = .18$.

(Here $\bar{X}_n$ denotes sample mean of $X$s similarly for $Y$).

If you were to pool all the data, you would have $X_1, \dots, X_n, Y_1, \dots, Y_n$. Let these be noted by $Z_1, \dots, Z_{2n}$. Then mean of $Z$, $$\bar{Z}_{2n} =\dfrac{X_1 + X_2 + \dots + X_n + Y_1 +Y_2 + \dots Y_n}{2n} = \dfrac{\bar{X}_n + \bar{Y}_n}{2} = 36.55$$

The standard deviation can be calculated using pooled estimators.

$$sd(Z_1)^2 = \dfrac{(n-1)sd(X_1)^2 + (n-1)sd(Y_1)^2}{2(n-1)} = \dfrac{sd(X_1)^2 + sd(Y_1)^2}{2}.$$

Thus the standard deviation is

$$sd(Z_1) = \sqrt{\dfrac{sd(X_1)^2 + sd(Y_1)^2}{2}}. $$

Like I mentioned before, since your sample is not independent, note that

$$sd(\bar{Z}_{2n})^2 \ne \dfrac{sd(Z_1)^2}{2n}$$.

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    $\begingroup$ Doesn't your pooled standard deviation calculation rely on the assumption of independence between $X_i$ and $Y_i$ which you stated is not valid? $\endgroup$
    – Mark L. Stone
    Commented Apr 27, 2016 at 14:38
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A suggestion, you are better off doing something like this:

$$E\left[T\right]=\overline{T}=E\left[\overline{T_{i}}\right] = \sum_{i=1}^{N}\frac{\overline{T_{i}}}{N}=\sum_{i=1}^{N}\frac{T_{right,i}+T_{left,i}}{2N}$$

Why? There's independence among subjects $i\in\left\{1,2,3,...,N\right\} $, while that's not true among samples from the same subject. Then the standard error follows.

$$\sigma_{\overline{T}}\approx{\frac{S}{\sqrt{N}}}=\sqrt{\sum_{i=1}^{N}\frac{\left(\overline{T_{i}}-E\left[\overline{T_{i}}\right]\right)^2}{N^{2}-N}} $$

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