Timeline for Intuitive Explanation Why the Standard Error of the Sample Proportions is not Divided by $n - 1$
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 20, 2021 at 4:20 | comment | added | Glen_b | @Dan It's $1/\sqrt{n}$ because the variance of the mean of independent identically distributed variables is $1/n$ times the variance of the individual observations. $\text{var}(\bar{X})=\text{var}(\frac{1}{n}\sum_i X_i)=\frac{1}{n^2}\text{var}(\sum_i X_i)$ (multiplication by a constant) $=\frac{1}{n^2}\sum_i\text{var}(X_i)=\frac{1}{n^2}\,n\,\text{var}(X_i)$ (independence), $=\frac{1}{n}\text{var}(X_i)$. See en.wikipedia.org/wiki/Variance#Basic_properties | |
| Sep 20, 2021 at 4:05 | comment | added | Alexis | @Dan I could, but you already have it in your question! For a sample mean (say, continuous and normalish-ly distributed), you use information calculating the mean, but then you use the mean to calculate the (variance and the) standard deviation: $s = \sqrt{\frac{\sum_{i}^{n} \left(x_{i}-\overline{x}\right)^{2}}{n-1}} = \sqrt{\frac{\sum_{i}^{n} \left(x_{i}-\frac{\sum_{i}^{n}x_{i}}{n}\right)^{2}}{n-1}}$: that $\overline{x}$ is using $\frac{1}{n}^{\text{th}}$ of the information in each observation… and since you have $n$ observations that's 1 degree of freedom lost from the denominator. | |
| Sep 20, 2021 at 2:46 | comment | added | Dan | Thank you for this explanation. This makes complete sense. Can you comment on the standard error for the sample mean calculation? Why divide by sqrt(n) and not sqrt(n-1)? Thanks! | |
| Sep 20, 2021 at 2:45 | vote | accept | Dan | ||
| Sep 20, 2021 at 1:38 | history | answered | Alexis | CC BY-SA 4.0 |