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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.