B

I don't know if this is the t-value riosimbolon mentioned, but the 1.662 can be found in your "MSA 3rd Ed with ANOVA.xls" (Intro to Measurement System Analysis (MSA) of Continuous Data – Part 5b: R&R), sheet "Stability", cell U43 (called "t_critical alpha=.05" in U42).

According to the MSA manual the critical t-value should be derived using

t^(-1)_{nu, 1-alpha/2}

(1-alpha/2)-quantile of the t-distribution with "nu" degrees of freedom, "nu" from table C1 (appendix C) in MSA 3rd (formula: see p.89 in MSA 3rd and p94 in MSA4th, resp.)

Table C1 does provide "nu"-numbers for a maximum of m=20 readings within a subgroup and a maximum of g=20 subgroups. At the bottom of the table it is stated that further "nu"-values could be build by using cd (constant difference). For a subgroup sample size of m=5 readings, the "nu"-value for g=20 subgroups is 72.7 and the cd=3.623. To obtain the "nu"-value for g=25 subgroups cd have to be added 5 times to the g=20-subgroups-"nu"-value (because 25 = 20 + 5 subgroups):

nu = 72.7 + 5*3.623 = 90.815

1-alpha/2 = 0.975 (for alpha=5%)

The critical t-value for the bias test with m=5 readings, g=20 subgroups and alpha=5% is given by

t^(-1)_{nu, 1-alpha/2} = t^(-1)_{90.815, 0.975} = 1.9864

But in the Excel file 1.662 is presented as the critical t-value. If alpha is set to 10%, this would be the correct value:

t^(-1)_{nu, 1-alpha/2} = t^(-1)_{90.815, 0.95} = 1.6618

(checked with R and Minitab)

Maybe this occurs due to Excel's use of alpha-values within the function TINV (see "A one-tailed t-value..." on TINV function) which differs from the usual behaviour of statistical software packages?

Best regards,

Barbara

According to the MSA manual the critical t-value should be derived using

t^(-1)_{nu, 1-alpha/2}

(1-alpha/2)-quantile of the t-distribution with "nu" degrees of freedom, "nu" from table C1 (appendix C) in MSA 3rd (formula: see p.89 in MSA 3rd and p94 in MSA4th, resp.)

Table C1 does provide "nu"-numbers for a maximum of m=20 readings within a subgroup and a maximum of g=20 subgroups. At the bottom of the table it is stated that further "nu"-values could be build by using cd (constant difference). For a subgroup sample size of m=5 readings, the "nu"-value for g=20 subgroups is 72.7 and the cd=3.623. To obtain the "nu"-value for g=25 subgroups cd have to be added 5 times to the g=20-subgroups-"nu"-value (because 25 = 20 + 5 subgroups):

nu = 72.7 + 5*3.623 = 90.815

1-alpha/2 = 0.975 (for alpha=5%)

The critical t-value for the bias test with m=5 readings, g=20 subgroups and alpha=5% is given by

t^(-1)_{nu, 1-alpha/2} = t^(-1)_{90.815, 0.975} = 1.9864

But in the Excel file 1.662 is presented as the critical t-value. If alpha is set to 10%, this would be the correct value:

t^(-1)_{nu, 1-alpha/2} = t^(-1)_{90.815, 0.95} = 1.6618

(checked with R and Minitab)

Maybe this occurs due to Excel's use of alpha-values within the function TINV (see "A one-tailed t-value..." on TINV function) which differs from the usual behaviour of statistical software packages?

Best regards,

Barbara

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