T-value used for MSA - t-value used in MSA Bias Example is different from Excel

[clcheng]

I noticed that we can get the t-value from Excel. However, the t-value used in MSA for Bias Example is different from the Excel. Instead, it was tabulated based on Student T Distribution Table using proportionate method coz it's not an interger value.

If the degree of freedom is an integer, t-value obtained from Excel and Student T Distribution Table are the same. However if the df is in decimal, the t-value from Excel is different from table (proportionate method).

Which one is the correct t-value to use for Bias study???

Anyone willing to help me?

[Atul Khandekar]

clcheng,
I don't think there is a function in Excel that will give you the t-value for 'decimal' DF. You can always use the same math in Excel that you used for calculating t from tables. For DF=10.8, get t for 10 and 11, then interpolate.

Hope this helps... and welcome to the Cove!

[Atul Khandekar]

The latest addition to the AIAG MSA FAQ, answers this exact question:

Q: I can't determine how to calculate the significant t value (two-tailed) in Table 3 on page 88 (value shown as 2.206) and Table 4 on page 90 (value shown as 1.993).

Ans:...
For Table 3: Given that the df = 10.8, look in a t-table in a standard statistical reference book. Look for the value in the t-table for df=10 and df=11, in the t-sub-0.975 column (this column represents the two-sided values for 95% confidence). The value at 10 df = 2.2281 and the value for 11 df = 2.2010. You must interpolate between these values to get the answer for this problem.

Take the difference between 2.2281 and 2.2010. This equals 0.0271. Since the df we are after is 10.8, we need to interpolate by either adding 20% of this value to the 2.2010, or by subtracting 80% of this from 2.2281. 20% of .0271 = 0.00542; adding that to the df for 11 = 2.2010 + .00542 = 2.206, rounded off to 3 places as shown in MSA3. Or you may find 80% of .0271 = .02168. Subtracting that from the df for 10 = 2.2281 - .02168 = 2.206, the same value we found using the alternate method.

[Zunaidy]

there is a function in EXCEL namely TINV(x,y)
x = you just input the default value of 0.05
y = it is the degree of freedom. for the value in table page 90, refer to Appendix C d2 table page 195, look at g=20 and m=5. you will have y=72.7

the result of TINV(0.05,72.7) equals to 1.993
i had compared to the table 3 page 88, it has the correct value.

i'm just using excel to compute bias study & linearity study.

[Atul Khandekar]

Originally Posted by Zunaidy

there is a function in EXCEL namely TINV(x,y)
x = you just input the default value of 0.05
y = it is the degree of freedom. for the value in table page 90, refer to Appendix C d2 table page 195, look at g=20 and m=5. you will have y=72.7

the result of TINV(0.05,72.7) equals to 1.993
i had compared to the table 3 page 88, it has the correct value.

i'm just using excel to compute bias study & linearity study.

A word of caution: In TINV, if degrees_freedom is not an integer, it is truncated. So if you use 72.1,72.2...72.9 Excel returns the same value as TINV(0.05,72).

In this case (I think you are refering to Table 4, page 90),
TINV(0.05,72)= 1.99346231966047 and
TINV(0.05,73)= 1.99299847736256,
They are very close, so interpolated value for 72.7 will be correct (1.993) upto 3 decimal places. You may run into trouble with lower value of DF.