• Variables Sampling Plans

• MIL-STD-414 (ISO 3951)

Variables Sampling Plans

• Variables sampling plans are based onthe numerical measurement of qualitycharacteristics

• Advantage: smaller samples arerequired than for attributes samplingplans.

• Disadvantages:

1. Precise measurements are costly.

2. If more than one characteristic ismeasured, a separate plan isneeded for each characteristic.

3. Measurements are assumed tofollow a normal distribution.

4. More computational effort isrequired.

• Example.

* Suppose a 1250 MHz Pentium chipis considered defective if its actualclockspeed is less than 1200 MHz.

* ⇒ LSL = 1200 MHz

* A lot is acceptable if no more than2 % of the chips in the lot aredefective.

* Reject such lots no more than 5 %of the time.

* ⇒ AQL = .02.

* If clockspeed is a normallydistributed random variable X withmean µ and variance σ2 (say 400),we can calculate the µ = µAQL thatcorresponds to the AQL:

.02 = P (X < LSL)

= P (Z <LSL− µAQL

σ)

* ⇒LSL− µAQL

σ= −2.05

so that

LSL− µAQL = −2.05σ.

Therefore,

µAQL = LSL + 2.05σ = 1241

* If the true mean clockspeed for alot is 1241 MHz or higher, the lot isconsidered acceptable.

* Problem: µ is unknown if we don’tinspect every item in the lot.

* Solution: Take a random sample ofchips from the lot and test the nullhypothesis that µ = 1241 versus thealternative hypothesis thatµ < 1241.

Ho : µ = 1241 acceptable lot

HA : µ < 1241 unacceptable lot

* Test statistic:

Z =X̄ − 1241

σ/√

n

where X̄ is the average of a randomsample of n clockspeedmeasurements. Reject Ho (i.e.reject the lot) if

Z < −zα = −1.645.

* ⇒ reject the lot if

X̄ − 124.1 < −1.645σ/√

n

or

X̄ < 124.1− 3.29/√

n.

* In general, reject the lot if

X̄ < (zAQL − z1−Pa(AQL)/√

n)σ + LSL

where zAQL is defined so that

P (Z > zAQL) = AQL.

* To determine the sample size n,use the LTPD.

* Suppose β = .1 and LTPD = 20 %

* ⇒ accept lots with 20 % defectiveno more than 10 % of the time.

* Let µLTPD be the mean clockspeedfor lots at the LTPD.

.2 = P (X < LSL) = P (Z <LSL− µLTPD

σ)

* ⇒LSL− µLTPD

σ= −.84

so

µLTPD = LSL + .84σ = 1216.8.

* ⇒ the mean clockspeed is 1216.8MHz in lots which are at the LTPD.

* We wish to reject 90 % of all suchlots.

* That is, we require

P (X̄ < 1241− 3.29/√

n) = .9.

* Standardize:

P (Z <1241− 3.29/

√n− 1216.8

2/√

n) = .9.

⇒

1.21√

n− 1.645 = 1.28.

We can solve for n:

n = 5.8

• In general,

n =

(

zα + zβ

zAQL − zLTPD

)2

• For one-sided tolerances of USL type,this is the formula for the sample size,but now reject the lot if

X̄ > USL− (zAQL − zα/√

n)σ.

• In both cases (USL or LSL), we canwrite the decision rule as reject thelot if

QL < k

or

QU < k

where

k = zAQL − zα/√

n

and

QL =X̄ − LSL

σand

QU =USL− X̄

σ.

• In other words, we reject a lot if thesample average is ’too close’ to theLSL (or USL).

• Example Find the sample size requiredfor a variables sampling plan with AQL= .1, LTPD = .2, α = .05 and β = .1.

*

n =

(

z.05 + z.1z.1 − z.2

)2

n =(1.645 + 1.28

1.28− .84

)2

n = 44.1

* ⇒ n = 45 should be used.

• If σ = 1, LSL = 10 and X̄ = 11, what isthe lot sentence under the aboveplan?

* k = 1.28− 1.645/√

45 = 1.03.

* QL = 11−101 = 1 < k. Therefore,

reject the lot.

MIL-STD-414

• aka ISO 3951.

• Another acceptance sampling system.It is actually comprised of severaltypes of plans. We will consider thecase in which σ is unknown, and isestimated using s.

• Example. Suppose LSL = 1200 MHz,AQL = 2 %, α = .05, β = .1 and lotsize is 1000. Find an appropriatesampling plan.

* From Table 11.1 (p. 351, Farnum),we convert AQL of 2 % to 2.5 %.

* From Table 11.2 (p. 352), with alot size of 1000, we use sample size

code letter K (under inspectionlevel IV).

* From Table 11.3 (p. 354), underAQL = 2.5 %, and across fromcode letter K, we find that thesample size n = 35, and M = 5.57,under normal inspection. Thismeans that if the estimatedproportion defective based on the35 measurements is greater than5.57 %, we reject the lot.

• Suppose 35 measurements are takengiving x̄ = 1230 and s = 20. What isthe lot sentence?

* Then QL = x̄−LSLs = 1230−1200

20 = 1.5

* Table 11.5 (p. 356) is used toconvert this number to an estimateof the fraction nonconforming:pL = 6.50% using QL = 1.5, n = 35).

* Therefore, we would reject the lot.

* If x̄ = 1235 and s = 20, we getQL = 1.75 so that pL = 3.77%. Inthis case, we would not reject thelot.