see the documents, there are six problems.

Acceptance Plan

Sample #

Sample size(ni)

Cum. Sample Size

AC

Re

1

50

50

*

2

2

50

100

1

3

3

50

150

2

3

* = acceptance not permitted on first sample, N=3000

Assuming, N>>>>>n, Use the appropriate method and the Poisson approximation to do the following:

Develop the acceptance scheme for this plan. Show all the components of the scheme

Compute the producer’s risk for a 1% nonconforming lot.

Compute the consumer’s risk for a 10% nonconforming lot.

What is the probability that plan in will pass both (i.e., a and b) inspection? (Assume this means passing either one or both inspections)

Compute the ATI for this plan if the lot is 10% nonconforming lot

Compute the ASN for this plan if the quality is 10% nonconforming.

2. In acceptance plan under MIL-STD-1916, Lot Size 2000, VL=IV. Determine

the Normal Plan

The Tightened Plan

The Reduced plan

Find: n, VL, CL for each plan (a-c)

For each plan what is the % nonconforming at α=0.05

For each plan what is the % nonconforming at β =0.1

For each plan, what is the AOQL, and the percent nonconforming at AOQL

What is the probability of instituting Tightened inspection, that is, switching from Normal inspection to Tightened inspection assuming the lot quality at AOQL is the bases of the analysis?

3. A double spec limits L and U study was implemented on the tensile strength of certain wire. The values for the Lower Spec and the Upper Spec are: 62 lb. and 76 lb. respectively. Variable sampling based on MIL-STD-1916 is used with Normal Inspection, Code letter B, and Verification Letter VL=III. Variability is unknown and standard deviation is to be used. A computation using the Form F criterion is to be made.

a). Determine the parameters of the plan (i.e., n, VL, CL, K, F) plan for Normal, Tightened, and Reduced Severity Levels

b). Should the lot be accepted?

b). If the upper limit is found to be 75 lb., what will be the decision based on the K criterion?

Tensile Strength of Certain Wires

i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

x 63 69 69 67 64 63 65 67 68 67 64 70 75 66 71 66 68 65 69 67

4. a). A certain firm claims that 99% its product meets specifications. To support this claim, an inspector draws a random sample of 10 items and ships the lot if the entire sample is in conformance. Find the probability of committing the following errors

i). Refusing to ship a lot even though 99% or more of the items are in conformance, Type I error

ii) Shipping a lot even though only 95% of the items are in conformance, Type II error

5. Devise a sampling plan for which the producer’s risk (P. R.) = 5% (Pa=0.95) with AQL of 0.01 and for which the consumer’s risk (C.R.) = 1.15% with LTPD = 0.1. Determine the plan that exactly meets P. R. and for which C. R < 1.15%

6. Two parts are assembled as shown. Assume that the dimensions X, and Y are normally distributed with means X and X and variance X, Y respectively. The parts are produced on different machines and are assembled at random. Control charts are maintained on each dimension for the range of each sample (n=5). Both range charts are in control.

(a). Given that twenty (20) samples on the range chart for X and ten(10) sample on the range chart for Y yielded the following , estimate X, and Y .

Note that each sample consists of 5 units, i.e., n=5.

If it is desired that the probability of a smaller clearance (.i.e, D=X-Y) than 0.09 should be 0.0006, what separation between the average dimensions (X – Y ) should be specified

Suppose μ_D=μ_(X-Y)=1.95.0 cm,and σ_D=0.42 cm, what is the probability of a smaller clearance than 0.09 cm.

How many parts out-of-one million will have a clearance less than or equal to this minimum clearance value.

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Requirements: Answer all the problem | .doc file

LASP LOT acceptance sampling plan