s=sample standard deviation s=square root (Sum(x-(xbar))2 /(n-1) Computing formula (so you don't have to find the mean and the distance from the mean over and over): square root(Sxx /(n-1)) Sxx= Sum(x2) - ((Sum(x))2/n)
The z score is calculated from the distance of a value from the distribution center divided by the standard dev. (x-xbar)/st. dev
Yes. The transform is z= (x-xbar)/s where x is the data value, xbar is the mean of the data and s is the standard deviation.
The simple answer is no. This depends on a lot of factors such as alpha which determines the critical value and the absolute value of the difference between the claim and sample data. Mathematically speaking, all things being equal, the larger the sample size the larger the absolute value of the test statistic. The formula for the test statistic mean with sigma known is shown below. You can substitute values in and perform the mathematics. The larger the sample size, the larger the Z value; but note if the numerator is small, even a small denominator will not produce a large Z value. In fact, the numerator could be zero which would make the test statistic zero. Z = (Xbar - μxbar)/(σ/√n) (formula from Elementary Statistics by Mario F. Triola)
z=x-mean / sd
s=sample standard deviation s=square root (Sum(x-(xbar))2 /(n-1) Computing formula (so you don't have to find the mean and the distance from the mean over and over): square root(Sxx /(n-1)) Sxx= Sum(x2) - ((Sum(x))2/n)
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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.
The z score is calculated from the distance of a value from the distribution center divided by the standard dev. (x-xbar)/st. dev
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.
There is no specific formula for measure - in statistics or any other branch of mathematics.
Yes. The transform is z= (x-xbar)/s where x is the data value, xbar is the mean of the data and s is the standard deviation.
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The simple answer is no. This depends on a lot of factors such as alpha which determines the critical value and the absolute value of the difference between the claim and sample data. Mathematically speaking, all things being equal, the larger the sample size the larger the absolute value of the test statistic. The formula for the test statistic mean with sigma known is shown below. You can substitute values in and perform the mathematics. The larger the sample size, the larger the Z value; but note if the numerator is small, even a small denominator will not produce a large Z value. In fact, the numerator could be zero which would make the test statistic zero. Z = (Xbar - μxbar)/(σ/√n) (formula from Elementary Statistics by Mario F. Triola)
(x value) - average
The formula for Z obtained is zobt equals the average of the sample (Xbar)minus the average of the population (u), divided b the standard deviation of the sample (s), divided be the square root of N. Example: Zobt= X-u ,divided by s ,divided by ,square root of N
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