Question

Problem 2 A firm manufacturing memory chips found that if everything was going right, 10% of them were defective. If the production process was in trouble then 40% were defective. Firm's quality control office tests four memory chips each hour. If two or more of four were defective, production would be shut down to look for trouble. (a) What is Type I and Type II errors here? (b) What is probability that production will be shut down if everything was going right? (c) What is the probability of the missed alarm? (d) Suppose, you have additional information: I. Production goes "out of control" about 10% of hours. 2. Testing of one chip costs $10 3. Missed alarm costs $10000. 4. False alarm costs $2000. Calculate the expected total cost associated with faulty production. Is it better to use another decision rule for detecting the trouble (shut the production if at least one of four tested chips is defective)? (e) New manager suggested testing 100 chips and shutting down production if more than 25 were defective. Calculate expected total cost associated with faulty production for this (number 3) decision rule. Compare with the first two.

Answer 1

A. For Type 1 Error to happen, everything should be going right, and yet the production should stop. In this case, Probability of defective , P(D) = 0.1, Probability of non Defective, P(ND) = 0.9. Now , for Type 1 error to happen, atleast 2 out of 4 chips that are picked are to be defective.

Probability of 2 chips defective, 2 chips non defective is given by, P(2) = $P(D)^{2}*P(ND)^{2} = 0.1^{2}*0.9^{2}$ = 0.81%

Probability of 3 chips defective, 1 chips non defective is given by, P(3) = $P(D)^{3}*P(ND)^{1} = 0.1^{3}*0.9^{1}$ = 0.09%

Probability of 4 chips defective , 0 chips non defective is given by, P(4) = $P(D)^{4} = 0.1^{4}$ = 0.01%

Total Probability of Type 1 error is given by , P(2)+ P(3)+P(4) = 0.81% +0.09% +0.01% = 0.91%

For Type 2 Error to happen, production process has to be in trouble and yet the production shouldnt stop. In this case, Probability of defective , P(D) = 0.4, Probability of non Defective, P(ND) = 0.6. Now , for Type 2 error to happen, only 1 out of 4 chips that are picked are to be defective.

Probability of 1 chips defective, 3 chips non defective is given by, P(1) = $P(D)^{1}*P(ND)^{3} = 0.4^{1}*0.6^{3}$ = 8.6%

So, there is a 8.6% probability of committing a Type 2 Error.

B. The probability that everything will be shutdown if everyhting was going right is given by the Type 1 Error, which is 0.91%.( The entire explanation is given in the first part).

C. The probability of missed alarm is given by Type 2 Error = 8.6%. (The explanation of which is given in the first part)

D. Cost of Testing a chip = $10

Cost of Missed Alarm = $10000

Cost of False Alarm = $2000

Total cost of Missed Alarm = ($10000+ 4* $10 )*8.6%= $863.44

Total cost of False Alarm = ( $2000 + 4 *$10 )*0.91%= $18.56

Sice, the cost of Missed Alarm is more, it is better to avoid the missed alarm, which means it is better to stop the production, if there is one defective in the lot of 4 chips.