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.


(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)

  • 8
    $\begingroup$ "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
  • 2
    $\begingroup$ google.com/search?q=divination+definition $\endgroup$
    – whuber
    Commented Apr 27, 2016 at 13:50

2 Answers 2


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

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

  • 2
    $\begingroup$ 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? $\endgroup$ Commented Apr 27, 2016 at 14:38

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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.