Question

The upper specification limit of a certain chemical compound is set at 2.2% per unit. the product is shipped in lots of size 240 and inspection proceeds as per iso norm 3951 at general inspection level iii. the desired aql is 1.5%. find the parameters of the appropriate sampling plans according to the k-method and the m-method. suppose that a random sample of size 10 is drawn from an incoming lot and that the following results (expressed in percentage) are obtained:

1.2, 1.6, 0.8, 1.9, 1.8, 1.4, 0.6, 0.9, 1.5, 0.7.
a) Sentence this lot as per the k-method and the M-method, assuming a normal inspection regime.
b) Explain how the decision would change under either a tightened or reduced inspection regime.

Answer 1

To determine the parameters of the sampling plans, calculate the acceptance number (c) and the rejection number (r) for the k-method and the m-method. For the k-method, decide the sample size (n) and use cumulative binomial probability tables to find the values of c and r. For the m-method, determine the limiting quality ratio (LQR) and find the values of c and r. The decision would change under a tightened or reduced inspection regime by adjusting the acceptance and rejection numbers.

Step-by-step explanation:

To determine the parameters of the appropriate sampling plans according to the k-method and the m-method, we first need to calculate the acceptance number (c) and the rejection number (r) for each method. For the k-method, decide the sample size (n) and examine the cumulative binomial probability tables to find the value of c at the desired AQL.

For the m-method, determine the limiting quality ratio (LQR) and find the values of c and r at the desired AQL. In this case, the sample size is 10, and the given results are: 1.2, 1.6, 0.8, 1.9, 1.8, 1.4, 0.6, 0.9, 1.5, 0.7.

a) c and r for the k-method: With a sample size of 10, the AQL is 1.5%. Using cumulative binomial probability tables, we find that c = 0 and r = 3 at the desired AQL. For the m-method: Determine the LQR, which is the ratio of the upper specification limit to the AQL. In this case, the LQR is 2.2% / 1.5% = 1.47. From the m-tables, we find that c = 0 and r = 1 at the desired AQL.

b) Under a tightened inspection regime, both the k-method and m-method would have smaller acceptance numbers (c) and larger rejection numbers (r), making it more difficult for a lot to be accepted. Under a reduced inspection regime, both methods would have larger acceptance numbers (c) and smaller rejection numbers (r), making it easier for a lot to be accepted.