X-bar represents the mean or average of the sample variable.
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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.
I believe you want the equation to calculate the standard deviation of a sample. The equation is: s = square root[ sum from i =1 to n of (xi- xbar)/(n-1)] where xbar is the average of values of the sample and n = size of sample.
I think, the estimate is a numerical value, wile the estimator is a function or operator, which can be generate more estimates according to some factors. For example (xbar) is estimator for (meu), which can be various when the sample size in various, the value that will be produced is an (estimate), but (xbar) is estimator.
Mean μ = 63.3 Standard deviation σ = 3.82 Standard error σ / √ n = 3.82 / √ 19 = 0.8763681 z = (xbar - μ) / (σ / √ n ) z = (61.6-63.3) / 0.876368 z = -1.9398
t= absolute value of ( sample 1 - sample two) THEN DIVIDED by the (standard error of sample one - standard error of sample 2) standard error = the standard deviation divided by (square root of the pop. sample number) You have to work in steps to get all info 1. mean ( REPRESENTED BY 'Xbar') 2. sum of squares ('SS') 3. Sample variance ('s^2') 4. standard deviation ('s') 5. standard error ('s subscript x') 6. pooled measure ('s^2p') 7. Standard error between means (s subscript mean one-mean two) 8. t test In other word finding the mean and having ht esample info leads you to each formula with the end formular being the t-test have fun, its easy but dumb