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x?
That's the average.
sum of all x divided by nx
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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