In this context, ( s^2 ) would refer to the sample variance of the salaries of the 66 employees taken from the population of 820 employees. It is a measure of how much the salaries of these sampled employees deviate from their average salary. This sample variance provides an estimate of the variance of the population, assuming that the sample is representative.
You cannot prove it because it is not true.The expected value of the sample variance is the population variance but that is not the same as the two measures being the same.
The n-1 indicates that the calculation is being expanded from a sample of a population to the entire population. Bessel's correction(the use of n − 1 instead of n in the formula) is where n is the number of observations in a sample: it corrects the bias in the estimation of the population variance, and some (but not all) of the bias in the estimation of the population standard deviation. That is, when estimating the population variance and standard deviation from a sample when the population mean is unknown, the sample variance is a biased estimator of the population variance, and systematically underestimates it.
that you have a large variance in the population and/or your sample size is too small
The variance is: 1.6709957376e+13
The mean, by itself, does not provide sufficient information to make any assessment of the sample variance.
66
The proof that the sample variance is an unbiased estimator involves showing that, on average, the sample variance accurately estimates the true variance of the population from which the sample was drawn. This is achieved by demonstrating that the expected value of the sample variance equals the population variance, making it an unbiased estimator.
It means you can take a measure of the variance of the sample and expect that result to be consistent for the entire population, and the sample is a valid representation for/of the population and does not influence that measure of the population.
No, it is biased.
You cannot prove it because it is not true.The expected value of the sample variance is the population variance but that is not the same as the two measures being the same.
It is a biased estimator. S.R.S leads to a biased sample variance but i.i.d random sampling leads to a unbiased sample variance.
Yes, there is a mathematical proof that demonstrates the unbiasedness of the sample variance. This proof shows that the expected value of the sample variance is equal to the population variance, making it an unbiased estimator.
The n-1 indicates that the calculation is being expanded from a sample of a population to the entire population. Bessel's correction(the use of n − 1 instead of n in the formula) is where n is the number of observations in a sample: it corrects the bias in the estimation of the population variance, and some (but not all) of the bias in the estimation of the population standard deviation. That is, when estimating the population variance and standard deviation from a sample when the population mean is unknown, the sample variance is a biased estimator of the population variance, and systematically underestimates it.
The answer depends on the underlying variance (standard deviation) in the population, the size of the sample and the procedure used to select the sample.
yes, it can be smaller, equal or larger to the true value of the population varience.
The variance decreases with a larger sample so that the sample mean is likely to be closer to the population mean.
that you have a large variance in the population and/or your sample size is too small