Variance, t-value, sample mean
The sample variance is obtained by dividing SS by the degrees of freedom (n-1). In this case, the sample variance is SS/(n-1) = 300/(4-1) = 300/3 = 100 In order to get the standard error, you can do one of two things: a) divide the variance by n and get the square root of the result: square.root (100/4) = square.root(25) = 5, or b) get the standard deviation and divide it by the square root of n. 10/square.root(4) = 10/2 = 5
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.
No.
I believe you are interested in calculating the variance from a set of data related to salaries. Variance = square of the standard deviation, where: s= square root[sum (xi- mean)2/(n-1)] where mean of the set is the sum of all data divided by the number in the sample. X of i is a single data point (single salary). If instead of a sample of data, you have the entire population of size N, substitute N for n-1 in the above equation. You may find more information on the interpretation of variance, by searching wikipedia under variance and standard deviation. I note that an advantage of using the standard deviation rather than variance, is because the standard deviation will be in the same units as the mean.
Variance, t-value, sample mean
The sample variance is obtained by dividing SS by the degrees of freedom (n-1). In this case, the sample variance is SS/(n-1) = 300/(4-1) = 300/3 = 100 In order to get the standard error, you can do one of two things: a) divide the variance by n and get the square root of the result: square.root (100/4) = square.root(25) = 5, or b) get the standard deviation and divide it by the square root of n. 10/square.root(4) = 10/2 = 5
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.
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.
No, it is biased.
The variance is: 1.6709957376e+13
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.
No.
I believe you are interested in calculating the variance from a set of data related to salaries. Variance = square of the standard deviation, where: s= square root[sum (xi- mean)2/(n-1)] where mean of the set is the sum of all data divided by the number in the sample. X of i is a single data point (single salary). If instead of a sample of data, you have the entire population of size N, substitute N for n-1 in the above equation. You may find more information on the interpretation of variance, by searching wikipedia under variance and standard deviation. I note that an advantage of using the standard deviation rather than variance, is because the standard deviation will be in the same units as the mean.
The mean, by itself, does not provide sufficient information to make any assessment of the sample variance.
no
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.