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#### rschoi

the other a 95% confidence interval on an average result.

A 95/95 type requirement may be met by either an attribute or variables type plan ;

the confidence interval on an average is met by a variables plan.

95/95 - Variables

We wish to be 95% sure that at least 95% (of the items being sampled) fall within the tolerance. To do this we project a frequency distribution and observe that it falls well within the required tolerance.

Tolerance limits which are a function of the sample size, sample average, and standard deviation are calcuated. These tolerance limits must be within the specification limits for acceptance.

Acceptance calculations are made as follows. For the sample size an appropriate " K-factor" is found in a table. (An except from a table is shown below.)

n k

5 5.079

6 4.414

7 4.007

8 3.732

9 3.532

10 3.379

Let us assume that we are sampling 10 items to evaluate against a specification of 100 ¡¾10 pounds. An average of 104 pounds and a standard deviation of 2 pounds is calculated. The 95/95 limits are x-bar ¡¾Ks, or 104¡¾ (3.379) x (2) or 104¡¾6.758. The limits are 110.758 and 97.242 pounds. Since the 110.758 value is greater than the maximum requirement of 110, we would reject the lot.

As I know, The 95 limit is the confidence interval on an average.

The 95/95 limits are ......?

Thank you very much.

[This message has been edited by rschoi (edited 10 April 2000).]