The variance is: 1.6709957376e+13
The sample variance is considered an unbiased estimator of the population variance because it corrects for the bias introduced by estimating the population variance from a sample. When calculating the sample variance, we use ( n-1 ) (where ( n ) is the sample size) instead of ( n ) in the denominator, which compensates for the degree of freedom lost when estimating the population mean from the sample. This adjustment ensures that the expected value of the sample variance equals the true population variance, making it an unbiased estimator.
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 mean, by itself, does not provide sufficient information to make any assessment of the sample variance.
no
The sample variance (s²) is calculated using the formula ( s² = \frac{SS}{n - 1} ), where SS is the sum of squares and n is the sample size. For a sample size of n = 9 and SS = 72, the sample variance is ( s² = \frac{72}{9 - 1} = \frac{72}{8} = 9 ). The estimated standard error (SE) is the square root of the sample variance divided by the sample size, calculated as ( SE = \sqrt{\frac{s²}{n}} = \sqrt{\frac{9}{9}} = 1 ). Thus, the sample variance is 9 and the estimated standard error is 1.
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 considered an unbiased estimator of the population variance because it corrects for the bias introduced by estimating the population variance from a sample. When calculating the sample variance, we use ( n-1 ) (where ( n ) is the sample size) instead of ( n ) in the denominator, which compensates for the degree of freedom lost when estimating the population mean from the sample. This adjustment ensures that the expected value of the sample variance equals the true population variance, making it an unbiased estimator.
The variance of this data set is 22.611
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.
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.
No.
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.
The sample variance is 1.