- Jun 8, 2001
- 9,574
- 2
- 76
Maybe someone can explain this to me, because I sure as hell am confused. I am looking through past examples of work in my business stats class. It seems that two different but similar equations are being used for the same type of problem.
In one:
x-bar minus Mu
over
alpha divided by the square root of n
Where alpha is the error allowed.
In others I see
x-bar minus Mu
over
sigma divided by the square root of n
Where sigma is the standard deviation.
So I am confused. The standard deviation is not the same as the accepted error.
In one:
x-bar minus Mu
over
alpha divided by the square root of n
Where alpha is the error allowed.
In others I see
x-bar minus Mu
over
sigma divided by the square root of n
Where sigma is the standard deviation.
So I am confused. The standard deviation is not the same as the accepted error.
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