7.6.2 MSA Manual - Linearity - I can't get the "s" value

C

Clartsy

#1
Hi, I am a beginner for MSA. My calibration personnel just resigned and i had to pick up his work. (Some how I did not manage to ask him the questions below.....)

I just tried out the example for Linearity in MSA reference manual third edition (Pg92~96). But, I can't get the "s" value using the data and equation given in the manual. Because of the "s" value is not available, I can't calculate the confidence bands as well. Does anyone face this problem, or had I make some mistakes in my earlier calculation that cause this problem? :truce: I end up using the formula in Bias to calculate the confidence band. Not sure whether this is correct. Can anyone please help? :thanks:
 
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B

Barbara B

#2
Sorry, but I didn't understand your question. How could you calculate a confidence band without knowing the "s"? What measurement for variation did you use for the CB?

It would be helpful to see the formulas you used in your calculation, so we could help you in a more precisely way.

Barbara
 
C

Clartsy

#3
sorry for the late reply. coz i think i have get the 's' calculated correctly. :lmao:

This what i manage to work out so far, using Excel:
a= -0.131667
b= 0.736667
s= 0.239445
Sum (xi-xbar)^2 = 3.14
gm = 5 × 12 = 60

However, i got a new challenge. I can't figure out how the ta and tb are computed.

In pg 96, the book said that ta = -12.043 and tb=10.158
I use the equation in pg 93 (no. 8) for ta and pg 94 for tb, the answer i get is totally different. I was confused what do xj and xbar square in the equation refer to? :confused:

Ho: a=0 slope = 0

|t| = |a| ÷ [s/(Sum (xj-xbar)2)^0.5]

Ho: b = 0 intercept = 0

|t| = |b|/ [s[{(1/gm)+((xbar^2)/Sum (xi-xbar)^2)}^0.5]]

Can anyone help to explain on this?
 
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