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.
If a variable X, is distributed Normally with mean m and standard deviation s thenZ = (X - m)/s has a standard normal distribution.
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
Given a random variable X with mean M and standard deviation S, Z = (X - M)/S
A normal distribution is defined by two parameters: the mean, m, and the variance s2, (or standard deviation, s).The standard normal distribution is the special case of the normal distribution in which m = 0 and s = 1.
It is the estimate for s, the sample standard deviation.
To put your hand in a cup formation, and proceed to the breasts start under the nipple area and go up at a moderate speed.
Standard and Poors.
The force needed to lift a 50 kg mass would be approximately 490 newtons in a standard Earth's gravity of 9.81 m/s^2.
because their neck vertebrae are s shaped.
standard radial
Standard class.
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.
If a variable X, is distributed Normally with mean m and standard deviation s thenZ = (X - m)/s has a standard normal distribution.
You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).
"Eighty-eight Keys on a Standard Piano".
Standard Occupational Classification System.