Usually the sum of squared deviations from the mean is divided by n-1, where n is the number of observations in the sample.
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.
The proof of sample variance involves calculating the sum of squared differences between each data point and the sample mean, dividing by the number of data points minus one, and taking the square root. This formula is derived from the definition of variance as the average of the squared differences from the mean.
Variance, t-value, sample mean
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.
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.
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.
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.