The QA 233 (Basic Business Statistics)

Solutions to Practice Problem Set III

Delta-Touch, a regional auditing firm, has been hired to audit the books for the Crystal Legends Marketing Research. Donna Rhodonda, Delta-Touch top auditor, has been given this assignment. Crystal Legends, which has twenty-three clients of their own, only wants a quick overview rather than a full-blown audit. Because of this, Ms. Rhodonda decides to only audit the accounts for five randomly selected Crystal Legends’ clients. The results of her audit are provided below:

Crystal Legends’ Client	Number of Correct Entries	Number of Incorrect Entries
Dunwoody Furniture Makers	1219	38
Hanson’s Appliance Production	2436	112
Molly’s Maid Service	844	11
Springfield Dry Cleaners	1856	39
Kozy Kennels	564	17

Before we start answering the proceeding questions, let’s agree on some notation for particular events in our data. We’ll let A = Correct Entry (so A = Incorrect Entry), B₁ = Dunwoody Furniture Makers, B₂ = Hanson’s Appliance Production, B₃ = Molly’s Maid Service, B₄ = Springfield Dry Cleaners, and B₅ = Kozy Kennels.

1. In how many different ways could Ms. Rhodonda have selected five Crystal Legends’ clients for her audit? Given these results, what is the probability that Ms. Rhodonda would select Dunwoody Furniture Makers, Hanson’s Appliance Production, Molly’s Maid Service, Springfield Dry Cleaners, and Kozy Kennels for her audit?

This is a simple question of combinations - we are asked in how many ways could we choose n = 5 clients from a population of N = 23 clients. Note that the order in which the clients are chosen does not matter, so we have

or 33,649 unique samples of five clients that could have been drawn from the population of twenty-three clients.

2. Using the data collected by Ms. Rhodonda, what would you estimate to be the probability of an incorrect entry? What approach(es) are you using to assess this probability? Please explain.

Here we wish to use our sample data to estimate Pr(A). We have

We have used raw data collected through an experiment to estimate this probability, thus we have used the Relative Frequency Approach to assign the probability to the event.

3. What would you estimate to be the probability of an incorrect entry in Kozy Kennels’ books? What kind of probability is this?

Here we wish to use our sample data to estimate Pr(A|B₅). We have

This is a Conditional Probability in which the conditioning argument is B₅ (i.e., Kozy Kennels).

4. What would you estimate to be the probability that an entry in Crystal legends’ books is drawn from a manufacturing organization (Dunwoody Furniture Makers or Hanson’s Appliance Production)?

Here we wish to use our sample data to estimate Pr(B₁ È B₂). Because B₁ (Dunwoody Furniture Makers) and B₂ (Hanson’s Appliance Production) are mutually exclusive (why?), we have

This is a Compound Probability.

5. What would you estimate to be the probability that an entry in Crystal legends’ books is incorrect and drawn from Molly’s Maid Service account entries?

Here we wish to use our sample data to estimate Pr(A Ç B₃). By the General Law of Addition, we have

or equivalently

This is a Joint Probability.

6. If an entry in Crystal legends’ books is drawn from Molly’s Maid Service account entries, what would you estimate to be the probability that the entry is incorrect?

Here we wish to use our sample data to estimate Pr(A|B₃). This is a Conditional Probability in which the conditioning argument is B₃ (i.e., Molly’s Maid Sevice). We have

7. What would you estimate to be the probability of an incorrect entry in the books of a service organization? Why is this an estimate? Hint: create a new contingency table with type of industry – service vs. manufacturing represented by the rows and status of entry – incorrect or correct represented by the columns. Use this resulting table to find the answer to the question.

We will first create a new contingency table to better reflect how we wish to analyze the data:

Type of Industry for   Crystal Legends’ Client	Number of Correct Entries	Number of Incorrect Entries
Manufacturing (Dunwoody Furniture Makers or Hanson’s Appliance Production)	3655	150
Service (Molly’s Maid Service, Springfield Dry Cleaners, or Kozy Kennels)	3264	67

Further, let’s define B = Manufacturing Industry Client (so B = Service Industry Client). Now we can more easily use use our sample data to estimate Pr(A|B). This is a Conditional Probability in which the conditioning argument is B (i.e., Service Industry Client). We have

8. Do the type of industry (service vs. manufacturing) and the status of an entry in the books (correct or incorrect) appear to be independent? Please explain. Hint: use the same table you created for question 7 to find the answer to the question.

In order for type of industry (service vs. manufacturing) and the status of an entry in the books (correct or incorrect) to be independent, the following condition must be met:

Referring to the table created for Question 7, we can see that

while our results from Question 7 indicated that

Since these two values are not equal, we conclude that type of industry (service vs. manufacturing) and the status of an entry in the books (correct or incorrect) are not independent. Note that these are sample results, so our assigned probabilities are estimates, so our results are far from conclusive. Also note that we could have attacked this problem in any number of ways, including by showing that any of the following conditions were met:

Deep-Six, Inc. is a manufacturer of SCUBA equipment. One of their most popular products is their Ultimate Depth Gauge, an important piece of equipment for preventing nitrogen narcosis (excess nitrogen in the bloodstream caused by spending too much time at too great a depth). Deep-Six claims that the Ultimate Depth Gauge has only a one in ten million chance of failure. To gain this level of reliability, the Ultimate Depth Gauge is actually comprised of two depth gauges (a primary gauge and a backup gauge). One of the safety features of the Ultimate Depth Gauge is a fail-safe indicator light that glows when at least one of the gauges (primary or backup) has failed. 

9. If the probability that each gauge fails is 0.001, and failures of the two gauges are independent, is Deep-Six's claim valid? If not, how low would the probability of failure for the first gauge have to fall before Deep-Six's claim would be valid?

Let's refer to a failure of the Ultimate Depth Gauge's primary gauge as event A and a failure of the Ultimate Depth Gauge's primary gauge as event B. Thus we have that

P(A) = P(B) = 0.001.

We are asked for the probability that both occur (which implies an intersection). Since these events are independent,  we can find the probability of their intersection by multiplying the probabilities of the two events directly

P(A Ç B) =  P(A)P(B) = 0.001² = 0.000001

which is a one in one million chance of total failure. Deep-Six's claim that the Ultimate Depth Gauge has only a one in ten million chance of failure is not correct under these conditions. To gain the level of reliability claimed by Deep-Six, the probability of failure for the primary gauge of the Ultimate Depth Gauge would have to satisfy this equation:

P(A Ç B) =  P(A)(0.001) = 0.0000001

thus P(A) (the probability of failure for the primary gauge of the Ultimate Depth Gauge would have to be so that Deep-Six's claim is correct) can be no more than

P(A) = 0.0001

10. What is the probability that the fail-safe indicator light will come on during a dive?

We again refer to a failure of the Ultimate Depth Gauge's primary gauge as event A and a failure of the Ultimate Depth Gauge's primary gauge as event B. Thus we have that

P(A) = P(B) = 0.001.

We are asked for the probability that either occurs (which implies a union). This is given by

P(A È B) =  P(A) + P(B) - P(A Ç B) = 0.001 + 0.001 -  0.000001 = 0.001999

which is almost a two in one thousand chance that the indicator light will come on.

 

11. Suppose that, given the first gauge fails, the probability that the second gauge fails is 0.01. Now what is the probability that  the  Ultimate Depth Gauge fails (i.e., both the primary and backup depth gauges that comprise the  Ultimate Depth Gauge fail)?

We'll again refer to a failure of the Ultimate Depth Gauge's primary gauge as event A and a failure of the Ultimate Depth Gauge's primary gauge as event B. Thus we have that

P(A) = 0.001 and P(B|A) = 0.01.

We are asked for the probability that both occur (which implies an intersection). Since these events are now not independent,  we can not find the probability of their intersection by multiplying the probabilities of the two events directly. We must use the General Law of Multiplication

P(A Ç B) =  P(A)P(B|A) = (0.001)(0.01) = 0.00001

which is a one in one hundred-thousand chance of total failure. Deep-Six's claim that the Ultimate Depth Gauge has only a one in ten million chance of failure is far from correct under these conditions.

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