# Standard deviation of the Mean

My data set focuses on the percentage cover of plants within quadrats at different locations. An example of my data looks like this:

Quadrat_Number Cover(%) Mean_Cover(%)
1          10         2.5
2          15         3.75
3           5         1.25
4          12         3


To calculate the mean cover at each location I've divided the Cover(%) by the number of quadrats in the sample (e.g. 10/4), and then summed the Mean_Cover(%). How would I then calculate my standard deviation of the mean? The answer may well be obvious but any help would be appreciated!

• What does the second column 'cover (%)' mean? How did you obtain/calculate the numbers in that column? – Sextus Empiricus Feb 2 '18 at 12:08
• That was the original observation, so in quadrat 1 there was 10 % cover. – Charlie Jones Feb 2 '18 at 12:15
• then what is the column 'Mean Cover' and how do you calculate it? Why is mean cover four times smaller than cover? – Sextus Empiricus Feb 2 '18 at 12:22
• I divided Cover(%) by the total number of quadrats (e.g 10/4). Doing so would enable me to add up Mean_Cover(%) to find the overall average cover in that location (e.g 2.5 + 3.75 + 1.25 + 3 = 10.5%). – Charlie Jones Feb 2 '18 at 12:24
• Okay well for the sake of argument, lets just ignore the Mean_Cover(%) column. I want to find out the mean cover of Cover(%), to do so is simple enough sum(Cover(%))/Number of quadrats. How do I then calculate my standard deviation? – Charlie Jones Feb 2 '18 at 12:29

To calculate the mean cover at each location I've divided the Cover(%) by the number of quadrats in the sample (e.g. 10/4), and then summed the Mean_Cover(%)

I dont know If I understand your words correctly but I feel that something like "Total_Cover(%)" would be better words to express that number. The mean_cover as in "average cover % per quadrat" should only be one number for your data set and should be estimated by

$$\sum_i^n (C_i)*1/n$$

where $C_i$ is the Cover_of_quadrat_i and $n$ is the number of samples. So $(10 + 15 +5+ 12)/4 = 10.5$

The standard deviation represents the idea of "average deviation from the mean" for your data. And it is estimated as:

$$\sqrt(\sum_i (C_i - \bar C )^2 * 1/n)$$

Where $\bar C$ is the mean of the sample.

• Hi, yeh you're absolutely right about that first part, that is what I have done, perhaps I didn't explain myself very well. As for the second part would I: sqrt(10 - 10.5)^2 * 1/(4), and do that for each quadrat? – Charlie Jones Feb 2 '18 at 12:06
• Or would it be sqrt(10 - 2.5)^2 * 1/(4)? And then calculate that for each quadrat? – Charlie Jones Feb 2 '18 at 12:17
• I don´t think I get what you are trying to accomplish, if you want to calculate the stndard deviation in the classical sense then you are doing it wrong. Please refer to, en.wikipedia.org/wiki/Standard_deviation – h3h325 Feb 2 '18 at 13:03

You can speak of the standard deviation of your population/sample, as a descriptive statistic:

$$\sigma = \frac{\sqrt{\sum{(\bar{x}-x_i})^2}}{n} = \frac{\sqrt{0.5^2 + 4.5^2 + 5.5^2 + 1.5^2}}{4} \simeq 1.8$$

and the $n$ can be replaced by $n-1$ depending on whether you just want to quantify the dispersion in your sample or whether you want to estimate the dispersion in the population from which you obtained the sample.

To express the 'deviation of the mean' is a bit ambiguous. You have only one single mean, namely 10.5 and the mean does not have a deviation in terms of a descriptive statistic.

However, you could view the mean, that you obtained in your single test, as a number that expresses just one of many other possible tests. In this case you can speak of the deviation of the multiple means from the population of different tests.

You just measured one of those means, but because you sampled multiple quadrats you can have an estimate of the variation of the population of means of quadrats just as you can have an estimate of the variation of the population of quadrats.

So, the estimate of the deviation of the mean cover is related to an estimate of the deviation of the cover, $\hat\sigma$:

$$\hat\sigma_{cover} = \frac{\sqrt{\sum{(\bar{x}-x_i})^2}}{n-1}$$

$$\hat\sigma_{mean\, of\, m\, covers} = \frac{\hat{\sigma}_{cover}}{\sqrt{m}} = \frac{\sqrt{\sum{(\bar{x}-x_i})^2}}{(n-1)\sqrt{m}}$$

Provided that the population has a finite deviation.