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The problem is that in estimating the sample mean, the process has already made our estimate of the mean close to the value we sampled—identical, for n = 1. In the case of n = 1, the variance just cannot be estimated, because there is no variability in the sample. But consider n = 2. Suppose the sample were (0, 2).
Since the square root is a strictly concave function, it follows from Jensen's inequality that the square root of the sample variance is an underestimate. The use of n − 1 instead of n in the formula for the sample variance is known as Bessel's correction, which corrects the bias in the estimation of the population variance, and some, but not ...
Firstly, if the true population mean is unknown, then the sample variance (which uses the sample mean in place of the true mean) is a biased estimator: it underestimates the variance by a factor of (n − 1) / n; correcting this factor, resulting in the sum of squared deviations about the sample mean divided by n-1 instead of n, is called ...
This algorithm can easily be adapted to compute the variance of a finite population: simply divide by n instead of n − 1 on the last line.. Because SumSq and (Sum×Sum)/n can be very similar numbers, cancellation can lead to the precision of the result to be much less than the inherent precision of the floating-point arithmetic used to perform the computation.
An unbiased estimator for the variance is given by applying Bessel's correction, using N − 1 instead of N to yield the unbiased sample variance, denoted s 2: = = (¯). This estimator is unbiased if the variance exists and the sample values are drawn independently with replacement.
Mathematically, the variance of the sampling mean distribution obtained is equal to the variance of the population divided by the sample size. This is because as the sample size increases, sample means cluster more closely around the population mean.
Cochran's theorem then states that Q 1 and Q 2 are independent, with chi-squared distributions with n − 1 and 1 degree of freedom respectively. This shows that the sample mean and sample variance are independent.
The sample (2, 1, 0), for example, would have a sample mean of 1. If the statistician is interested in K variables rather than one, each observation having a value for each of those K variables, the overall sample mean consists of K sample means for individual variables.