1- Assuming this represents a random sample from the population, the sample mean is an unbiased estimator of the population mean. 2-Because they are robust, t procedures are justified in this case. 3- We would use z procedures here, since we are interested in the population mean.
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.
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.
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.
First you have chose an estimator for what you want to know about the population. In general the level of variability in the result that any estimator provides will depend on the variability in the population. Therefore, the greater the variability in the population the larger your sample size must be. You will also need to decide how much precision is required in your estimate. The more precision you require the greater your sample size will have to be.
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.
No, it is biased.
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 sample mean is an unbiased estimator of the population mean because the average of all the possible sample means of size n is equal to the population mean.
The best point estimator of the population mean would be the sample mean.
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.
1- Assuming this represents a random sample from the population, the sample mean is an unbiased estimator of the population mean. 2-Because they are robust, t procedures are justified in this case. 3- We would use z procedures here, since we are interested in the population mean.
It can get a bit confusing! The estimate is the value obtained from a sample. The estimator, as used in statistics, is the method used. There's one more, the estimand, which is the population parameter. If we have an unbiased estimator, then after sampling many times, or with a large sample, we should have an estimate which is close to the estimand. I will give you an example. I have a sample of 5 numbers and I take the average. The estimator is taking the average of the sample. It is the estimator of the mean of the population. The average = 4 (for example), this is my estmate.
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 standard deviation. There are many, and it's easy to construct one. The mean of a sample from a normal population is an unbiased estimator of the population mean. Let me call the sample mean xbar. If the sample size is n then n * xbar / ( n + 1 ) is a biased estimator of the mean with the property that its bias becomes smaller as the sample size rises.
No, the standard deviation is a measure of the entire population. The sample standard deviation is an unbiased estimator of the population. It is different in notation and is written as 's' as opposed to the greek letter sigma. Mathematically the difference is a factor of n/(n-1) in the variance of the sample. As you can see the value is greater than 1 so it will increase the value you get for your sample mean. Essentially, this covers for the fact that you are unlikely to obtain the full population variation when you sample.