Yes, it does. If the data are sample data, than the divisor is N. If the data are the entire population, than the divisor is N-1 is account for the loss of one degree of freedom in the calculation of both the mean and the standard deviation from the same data.
the sample standard deviation
Yes.
If the population standard deviation is sigma, then the estimate for the sample standard error for a sample of size n, is s = sigma*sqrt[n/(n-1)]
It can be.
Not a lot. After all, the sample sd is an estimate for the population sd.
The standard deviation of the population. the standard deviation of the population.
Yes
No.
The standard deviation if the data is a sample from a population is 7.7115; if it is the population the standard deviation is 7.0396.
the sample standard deviation
Yes.
If the population standard deviation is sigma, then the estimate for the sample standard error for a sample of size n, is s = sigma*sqrt[n/(n-1)]
It can be.
Not a lot. After all, the sample sd is an estimate for the population sd.
If the samples are drawn frm a normal population, when the population standard deviation is unknown and estimated by the sample standard deviation, the sampling distribution of the sample means follow a t-distribution.
The true / real standard deviation ("the mean deviation from the mean so to say") which is present in the population (everyone / everything you want to describe when you draw conclusions)
Sigma