Standard deviation of the Mean - Cross Validated most recent 30 from stats.stackexchange.com 2019-08-24T10:08:33Z https://stats.stackexchange.com/feeds/question/326438 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://stats.stackexchange.com/q/326438 0 Standard deviation of the Mean Charlie Jones https://stats.stackexchange.com/users/191180 2018-02-02T11:41:26Z 2018-02-02T14:34:17Z <p>My data set focuses on the percentage cover of plants within quadrats at different locations. An example of my data looks like this:</p> <pre><code>Quadrat_Number Cover(%) Mean_Cover(%) 1 10 2.5 2 15 3.75 3 5 1.25 4 12 3 </code></pre> <p>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!</p> https://stats.stackexchange.com/questions/326438/-/326440#326440 1 Answer by h3h325 for Standard deviation of the Mean h3h325 https://stats.stackexchange.com/users/165179 2018-02-02T11:58:00Z 2018-02-02T13:02:18Z <blockquote> <p>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(%)</p> </blockquote> <p>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</p> <p>$$\sum_i^n (C_i)*1/n$$</p> <p>where $C_i$ is the Cover_of_quadrat_i and $n$ is the number of samples. So $(10 + 15 +5+ 12)/4 = 10.5$ </p> <p>The standard deviation represents the idea of "average deviation from the mean" for your data. And it is estimated as:</p> <p>$$\sqrt(\sum_i (C_i - \bar C )^2 * 1/n)$$</p> <p>Where $\bar C$ is the mean of the sample.</p> https://stats.stackexchange.com/questions/326438/-/326445#326445 1 Answer by Martijn Weterings for Standard deviation of the Mean Martijn Weterings https://stats.stackexchange.com/users/164061 2018-02-02T12:54:23Z 2018-02-02T14:34:17Z <p>You can speak of the <em>standard deviation of your population/sample</em>, as a descriptive statistic:</p> <p>$$\sigma = \frac{\sqrt{\sum{(\bar{x}-x_i})^2}}{n} = \frac{\sqrt{0.5^2 + 4.5^2 + 5.5^2 + 1.5^2}}{4} \simeq 1.8$$</p> <p>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 <em>estimate</em> the dispersion in the population from which you obtained the sample. </p> <hr> <p>To express the 'deviation of the mean' is a bit ambiguous. You have only <em>one single</em> mean, namely 10.5 and the mean does not have a deviation in terms of a descriptive statistic.</p> <p>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 <strong>deviation of the <em>multiple</em> means</strong> from the population of different tests. </p> <p>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. </p> <p>So, the estimate of the deviation of the mean cover is related to an estimate of the deviation of the cover, $\hat\sigma$:</p> <p>$$\hat\sigma_{cover} = \frac{\sqrt{\sum{(\bar{x}-x_i})^2}}{n-1}$$ </p> <p>$$\hat\sigma_{mean\, of\, m\, covers} = \frac{\hat{\sigma}_{cover}}{\sqrt{m}} = \frac{\sqrt{\sum{(\bar{x}-x_i})^2}}{(n-1)\sqrt{m}}$$ </p> <p>Provided that the population has a finite deviation.</p>