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Here's how you do it in Excel: use the function =STDEV(<range with data>). That function calculates standard deviation for a sample.

Q: How do you calculate sample standard deviation?

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in order to calculate the mean of the sample's mean and also to calculate the standard deviation of the sample's

The standard deviation of the population. the standard deviation of the population.

You calculate standard deviation the same way as always. You find the mean, and then you sum the squares of the deviations of the samples from the means, divide by N-1, and then take the square root. This has nothing to do with whether you have a normal distribution or not. This is how you calculate sample standard deviation, where the mean is determined along with the standard deviation, and the N-1 factor represents the loss of a degree of freedom in doing so. If you knew the mean a priori, you could calculate standard deviation of the sample, and only use N, instead of N-1.

Yes

Did you mean, "How do you calculate the 99.9 % confidence interval to a parameter using the mean and the standard deviation?" ? The parameter is the population mean μ. Let xbar and s denote the sample mean and the sample standard deviation. The formula for a 99.9% confidence limit for μ is xbar - 3.08 s / √n and xbar + 3.08 s / √n where xbar is the sample mean, n the sample size and s the sample standard deviation. 3.08 comes from a Normal probability table.

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Standard error of the sample mean is calculated dividing the the sample estimate of population standard deviation ("sample standard deviation") by the square root of sample size.

in order to calculate the mean of the sample's mean and also to calculate the standard deviation of the sample's

=stdev(...) will return the N-1 weighted sample standard deviation. =stdevp(...) will return the N weighted population standard deviation.

A single observation cannot have a sample standard deviation.

The standard deviation of the population. the standard deviation of the population.

You calculate standard deviation the same way as always. You find the mean, and then you sum the squares of the deviations of the samples from the means, divide by N-1, and then take the square root. This has nothing to do with whether you have a normal distribution or not. This is how you calculate sample standard deviation, where the mean is determined along with the standard deviation, and the N-1 factor represents the loss of a degree of freedom in doing so. If you knew the mean a priori, you could calculate standard deviation of the sample, and only use N, instead of N-1.

If I take 10 items (a small sample) from a population and calculate the standard deviation, then I take 100 items (larger sample), and calculate the standard deviation, how will my statistics change? The smaller sample could have a higher, lower or about equal the standard deviation of the larger sample. It's also possible that the smaller sample could be, by chance, closer to the standard deviation of the population. However, A properly taken larger sample will, in general, be a more reliable estimate of the standard deviation of the population than a smaller one. There are mathematical equations to show this, that in the long run, larger samples provide better estimates. This is generally but not always true. If your population is changing as you are collecting data, then a very large sample may not be representative as it takes time to collect.

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)]

Did you mean, "How do you calculate the 99.9 % confidence interval to a parameter using the mean and the standard deviation?" ? The parameter is the population mean μ. Let xbar and s denote the sample mean and the sample standard deviation. The formula for a 99.9% confidence limit for μ is xbar - 3.08 s / √n and xbar + 3.08 s / √n where xbar is the sample mean, n the sample size and s the sample standard deviation. 3.08 comes from a Normal probability table.

z=(x-mean)/(standard deviation of population distribution/square root of sample size) T-score is for when you don't have pop. standard deviation and must use sample s.d. as a substitute. t=(x-mean)/(standard deviation of sampling distribution/square root of sample size)

the sample standard deviation