A manufacturer of computer memory chips produces chips in lots of 1000. If nothi

Question

A manufacturer of computer memory chips produces chips in lots of 1000. If nothing has gone wrong in the manufacturing process, at most 5 chips each lot would be defective, but if something does go wrong, there could be far more defective chips. If something goes wrong with a given lot, they discard the entire lot. It would be prohibitively expensive to test every chip in every lot, so they want to make the decision of whether or not to discard a given lot on the basis of the number of defective chips in a simple random sample. They decide they can afford to test 100 chips from each lot. You are hired as their statistician.

There is a tradeoff between the cost of eroneously discarding a good lot, and the cost of warranty claims if a bad lot is sold. The next few problems refer to this scenario.

Suppose that whether or not a lot is good is random, that the long-run fraction of lots that are good is 98%, and that whether each lot is good is independent of whether any other lot or lots are good. Assume that the sample drawn from a lot is independent of whether the lot is good or bad. To simplify the problem even more, assume that good lots contain exactly 5 defective chips, and that bad lots contain exactly 70 defective chips.

The expected number of lots the manufacturer must make to get one good lot that is not rejected by the test is _____?

With this test and this mix of good and bad lots, among the lots that pass the test, the long-run fraction of lots that are actually bad is _____?

Explanation / Answer

Ans:

1. First let's determine the hypotheses:

H0: The given lot is good,
H1: The given lot is not good.

The above hypotheses are equivalent to:

H0: The number of defectives in the given lot is less than or equal to 7,
H1: The number of defectives in the given lot is greater to 7.

Now let P be the proportion of defectives in the given lot (or the population). Therefore, the above hypotheses can be written as

H0: P<=7/1000
H1: P>7/1000,

since the #of chips of the given lot (population size) is 1000. Now let X be 0 if a given chip is defective and 1 if not. So X has a binomial distribution with parameters 1000 and P. Now we can test the hypotheses by the common test method available for proportion. As Glenn Curth said, we want that the type two error be .1. Here we will reject H0 if and only if:

(p-.007)/(p(1-p)/n)^.5>Z(.01)=2.33,

where p is the sample proportion of defectives and n=100. So if we show the number of defectives in the sample with x, it can be found via the inequality below:

(x/100-.007)/(x/100(1-x/100)/100)^.5>2...

Now it's easy to find x.

2. If the lot has the real number of 70 defectives it concludes that P=70/1000=.07. So the chance of rejecting the lot if it really has 70 defective chips is

P((p-.007)/(p(1-p)/n)^.5>2.33|P=.07).

We know that (p-.007)/(p(1-p)/n)^.5 has a standard normal distribution if H0 is true, but here H0 isn't true. So we must find the above probability using the fact that P=.07. We have:

(p-.007)/(p(1-p)/n)^.5>2.33
===> (p-.007)>2.33(p(1-p)/n)^.5
===> p>2.33(p(1-p)/n)^.5+.007
===> p-.07>2.33(p(1-p)/n)^.5+.007-.07=2.33(p(...
===> (p-.07)/ (p(1-p)/n)^.5>2.33-.063/(p(1-p)/n)^.5.
===>P((p-.007)/(p(1-p)/n)^.5>2.33|P=.0... (p(1-p)/n)^.5>2.33-.063/(p(1-p)/n)^.5).

Now because the sample size is large, we can approximate the term (p(1-p)/n)^.5 by (P(1-P)/n)^.5 where P=.07 and say that (p-.007)/(p(1-p)/n)^.5 has a standard normal distribution. So we have (P(1-P)/n)^.5=(.07*.93/100)^.5=.025. Thus the favorite probability can be calculated as:

P((p-.07)/ (p(1-p)/n)^.5>2.33-.063/(p(1-p)/n)^.5)=P...
P(Z>-.19),

where Z is a standard normal variable. So the above probability can be easily found using a normal table.

3. This is exactly the definition of type one error! In the part 1, we said that type one error is .01, so the fraction is 1/100, or %.01.

4. We know that Z(.01)=2.33. So set p=.007 and find the smallest value of n form the inequality below:

(.007*.093/n)^.5<2.33.

5. You're right about geometric distribution. The parameter of this distribution, theta, is the probability that a good lot isn't rejected by the test. It is

theta = P(Accept H0 | H0 is true) = 1- P(Reject H0 | H0 is true),

and you know that P(Reject H0 | H0 is true) is the type one error of the test. If, as the part 1, we take it .01 we have theta=1-.01=.99.

6. Since # of lots the manufacturer must make to get one good lot that is not rejected has a geometric distribution with parameter theta=.99, the expected value of # of lots the manufacturer must make to get one good lot that is not rejected is:

1/theta=1/.99.

7. P(Accept H0 | H1 is true) = P(Reject H1 | H1 is true) = the type two error of the test.

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