Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Precalculus Mathematics Homework Help
Calculating Variance: Sample vs Population
Reply to thread
Message
[QUOTE="statdad, post: 5099966, member: 136993"] The only time you would have [itex] n^2 [/itex] in the denominator is if you were (for some reason) considering mathematically a single observation [itex] X_1 [/itex] and calculate [tex] Var\left( \dfrac{X_1}{n}\right) = \dfrac{Var(X)}{n^2} [/tex] "is this the same for all cases as in, calculating the variance of a sample and estimating a population variance?" I'm not exactly sure what you mean by this, so if my response is off-target that's why. If you've talked about sampling distributions for the sample mean, the expression [itex] \frac{\sigma^2}n [/itex] is the population variance for that sampling distribution. It will never have denominator [itex] n^2 [/itex], since, as long as the distribution being sampled has a variance, the steps shown above apply. The (sample) variance of a sample is a different beast. Essentially * if the population variance is [itex] \sigma^2 [/itex], then the sample variance [tex] s^2 = \dfrac 1 {n-1} \sum_{i=1}^n \, \left(x_i - \bar x\right)^2 [/tex] is an unbiased estimator of the population variance * If you refer to the variance of the sampling distribution of [itex] \bar x [/itex] - which is given above - then to estimate that you have two options a) If you have a single sample, use the sample variance [tex] s^2 [/tex] to estimate the sampling distribution's variance by calculating [tex] \dfrac{s^2}{n} [/tex] b) If you have a large number of samples, all the same sample size, from the same population, then calculate each sample mean and treat those sample means as a new sample. The sample variance of those (call it [itex] s_{\bar x}^2[/itex]) is the estimate of [itex] \dfrac{\sigma^2}{n}[/itex] So there are several subtleties to wade through, but in none but the one unusual and unrealistic comment I made at the start will [itex] \frac{\sigma^2}{n^2} [/itex] play a role. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Precalculus Mathematics Homework Help
Calculating Variance: Sample vs Population
Back
Top