S
The challenge: from a sample of size n, what is the worst-case senario for the proportion below a given spec. limit?
The challenge involves potential that both the estimate of the average and the estimate of the standard deviation are worst-case estimates. This means (for the lower spec. situation) that the average derived from the sample is higher than the true process average, and that the standard deviation derived from the sample is lower than the true process standard deviation. The formulas that I have derived are as follows:
s(max)=sqrt(s^2(n-1)/Chi^2(1-alpha)
z=((xbar-z(alpha/2)(s(max)/sqrt
)-L.S.))/s(max)
I hope I got all of the parenthesis right.
s(max) is the worst-case estimate of the standard deviation for a given 1-alpha.
((xbar-z(alpha/2)(s(max)/sqrt
) is the lowest possible value for xbar based on the maximum possible error of the average >> z(alpha/2)(s(max)/sqrt
the final part is (this value (worst-case xbar), minus the lower spec.) divided by the worst-case estimate of the standard deviation.
Does this formula make sense?
The challenge involves potential that both the estimate of the average and the estimate of the standard deviation are worst-case estimates. This means (for the lower spec. situation) that the average derived from the sample is higher than the true process average, and that the standard deviation derived from the sample is lower than the true process standard deviation. The formulas that I have derived are as follows:
s(max)=sqrt(s^2(n-1)/Chi^2(1-alpha)
z=((xbar-z(alpha/2)(s(max)/sqrt
I hope I got all of the parenthesis right.
s(max) is the worst-case estimate of the standard deviation for a given 1-alpha.
((xbar-z(alpha/2)(s(max)/sqrt
the final part is (this value (worst-case xbar), minus the lower spec.) divided by the worst-case estimate of the standard deviation.
Does this formula make sense?