# Help Required in calculating USL and LSL

#### ragav_in

##### New Member
Hi all,

I have a sample of 8 datapoints with a Std. Dev of 0.0195940953204932 and Mean of 0.10125 and Confidence Level(95%) value of 0.0163810736238815

I used the Descriptive Statistics Tools in Excel to arrive at this values. This exercise was carried out to calculate the USL and LSL for this mean. I used the Formula as below
USL = Mean + Confidence Level(95%) = 0.117631074
LSL = Mean - Confidence Level(95%) = 0.084868926

But my next step is to bring the USL and LSL to the following desired value of USL = 0.160032286
LSL = 0.042467714

Is this possible with this set of data? How can I modify the Confidence Level(95%) value in order to get this desired values?

Regards,
Subramanian

#### vinux

##### Dark Knight
Hi all,

I have a sample of 8 datapoints with a Std. Dev of 0.0195940953204932 and Mean of 0.10125 and Confidence Level(95%) value of 0.0163810736238815

I used the Descriptive Statistics Tools in Excel to arrive at this values. This exercise was carried out to calculate the USL and LSL for this mean. I used the Formula as below
USL = Mean + Confidence Level(95%) = 0.117631074
LSL = Mean - Confidence Level(95%) = 0.084868926

But my next step is to bring the USL and LSL to the following desired value of USL = 0.160032286
LSL = 0.042467714

Is this possible with this set of data? How can I modify the Confidence Level(95%) value in order to get this desired values?

Regards,
Subramanian
Ok. Actually you wanted to calculated the Confidence level for the new specification limits shakehead I am not commenting on sample size and how the specification limits are treated)

new USL = 0.160032286 = 0.10125 + talpha2 * 0.0195940953204932/sqrt(8)
and LSL = 0.10125 - talpha2 * 0.0195940953204932/sqrt(8)
ie talpha2 = 0.05878229*sqrt(8) /0.0195940953204932
=8.485282

Here we use students t distribution with 7 degress of freedom

So we have to calculate alpha such that
P [ -talpha2 < T < talpha2 ] = 1- alpha
So Confidence level will be 99.99374%

( Use any statistical package ... in excel the function is 1-TDIST(8.485282,7,2) )

Regards
VinuX Vinu CT

#### ragav_in

##### New Member
Hi Vinux,

Thank you for your reply. But still am confused about the details that you have mentioned below.

1. What is talpha2?
2. From where did the value of 0.10125 come in calculating the New USL and LSL?
3. From where did the value of 0.5878229 come in calculating the talpha2?

If you could clarify me on these points, I would be grateful to you. Eve nI read about using Students' t distribution for calculating the Confidence Level 95% value for a smaller sample, but I am not sure of how to use the formulae. Please help me out on this.

Regards,
Subramanian

Ok. Actually you wanted to calculated the Confidence level for the new specification limits shakehead I am not commenting on sample size and how the specification limits are treated)

new USL = 0.160032286 = 0.10125 + talpha2 * 0.0195940953204932/sqrt(8)
and LSL = 0.10125 - talpha2 * 0.0195940953204932/sqrt(8)
ie talpha2 = 0.05878229*sqrt(8) /0.0195940953204932
=8.485282

Here we use students t distribution with 7 degress of freedom

So we have to calculate alpha such that
P [ -talpha2 < T < talpha2 ] = 1- alpha
So Confidence level will be 99.99374%

( Use any statistical package ... in excel the function is 1-TDIST(8.485282,7,2) )

Regards
VinuX Vinu CT

#### vinux

##### Dark Knight
1. What is talpha2?
2. From where did the value of 0.10125 come in calculating the New USL and LSL?
3. From where did the value of 0.5878229 come in calculating the talpha2?
1. talpha2 is just a value such that P [ -talpha2 < T < talpha2 ] = 1- alpha ( I hope you know what is alpha )

2. I hope the new USL and LSL are equvidistant from mean ( ie 0.10125 )

3. 0.5878229 = 0.160032286 -0.10125 or 0.10125 - 0.084868926