Key

The 1st Hourly

Math 1107

Summer Term 2007

 

Protocol

 

You will use only the following resources: Your individual calculator; individual tool-sheet (single 8.5 by 11 inch sheet); your writing utensils; blank paper (provided by me) and this copy of the hourly. Do not share these resources with anyone else.

 

In each case, show complete detail and work for full credit. Follow case study solutions and sample hourly keys in presenting your solutions. Work all six cases. Using only one side of the blank sheets provided, present your work. Do not write on both sides of the sheets provided, and present your work only on these sheets. All of your work goes on one side each of the blank sheets provided. Space out your work. Do not share information with any other students during this hourly. Sign and Acknowledge:

 

I agree to follow this protocol.

 

______________________________________________________________________________________

Name (PRINTED)                                       Signature                                         Date

 

Case One

Pair of Dice

Random Variable

25 Points Maximum

 

We have a pair of dice – a fair d3 {faces 1, 3, 5} and a loaded d4 {faces 2, 4, 6, 7} – note the probability model for the d4 below. 

 Face

Probability

2

0.40

4

0.25

6

0.25

7

0.10

Total

1.00

We assume that the dice operate separately and independently of each other. Suppose that our experiment consists of tossing the dice, and noting the resulting face-pair.

1. List the possible pairs of face values, and compute a probability for each pair of face values.

25 Points Maximum

 

15 Points Maximum

 (d4,d3)

2

4

6

7

1

(2,1)

(4,1)

(6,1)

(7,1)

3

(2,3)

(4,3)

(6,3)

(7,3)

5

(2,5)

(4,5)

(6,5)

(7,5)

 

The pairs are: (2,1), (4,1), (6,1), (7,1), (2,3), (4,3), (6,3), (7,3), (2,5), (4,5), (6,5) and (7,5).

 

Pr{(2,1)} = Pr{2 from d4}*Pr{1 from d3} = (4/10)*(1/3) = 4/30 = 16/120                                                              

Pr{(2,3)} = Pr{2 from d4}*Pr{3 from d3} = (4/10)*(1/3) = 4/30 = 16/120                                                             

Pr{(2,5)} = Pr{2 from d4}*Pr{5 from d3} = (4/10)*(1/3) = 4/30 = 16/120 [subtotal = 48/120]

 

Pr{(4,1)} = Pr{4 from d4}*Pr{1 from d3} = (1/4)*(1/3) = 1/12 = 10/120                                                             

Pr{(4,3)} = Pr{4 from d4}*Pr{3 from d3} = (1/4)*(1/3) = 1/12 = 10/120                                                             

Pr{(4,5)} = Pr{4 from d4}*Pr{5 from d3} = (1/4)*(1/3) = 1/12 = 10/120 [subtotal = 30/120]

 

Pr{(6,1)} = Pr{6 from d4}*Pr{1 from d3} = (1/4)*(1/3) = 1/12 = 10/120                                                              

Pr{(6,3)} = Pr{6 from d4}*Pr{3 from d3} = (1/4)*(1/3) = 1/12 = 10/120                                                             

Pr{(6,5)} = Pr{6 from d4}*Pr{5 from d3} = (1/4)*(1/3) = 1/12 = 10/120 [subtotal = 30/120]

 

Pr{(7,1)} = Pr{7 from d4}*Pr{1 from d3} = (1/10)*(1/3) = 1/30 = 4/120                                                              

Pr{(7,3)} = Pr{7 from d4}*Pr{3 from d3} = (1/10)*(1/3) = 1/30 = 4/120                                                             

Pr{(7,5)} = Pr{7 from d4}*Pr{5 from d3} = (1/10)*(1/3) = 1/30 = 4/120 [subtotal = 12/120]

 

Check: (48/120)+(30/120)+(30/120)+(12/120) = (78+42)/120 = 120/120 = 1

2. When both faces in the pair are odd, define THING as the product of the face values in the pair. When exactly one face in the pair is odd, define THING as the sum of the face values in the pair. Compute and list the possible values for the variable THING, and compute a probability for each value of THING.

Show all work and full detail for full credit.

10 Points Maximum

 

(d4,d3)

2

4

6

7

1

(2,1) One Odd Face: THING = (2+1) = 3

(4,1) One Odd Face: THING = (4+1) = 5

(6,1) One Odd Face: THING = (6+1) = 7

(7,1) Both Faces Odd: THING = 7*1=7

3

(2,3) One Odd Face: THING = (2+3) = 5

(4,3) One Odd Face: THING = (4+3) = 7

(6,3) One Odd Face: THING = (6+3) = 9

(7,3) Both Faces Odd: THING = 7*3=21

5

(2,5) One Odd Face: THING = (2+5) = 7

(4,5) One Odd Face: THING = (4+5) = 9

(6,5) One Odd Face: THING = (6+5) = 11

(7,5) Both Faces Odd: THING = 7*5=35

The pairs are: (2,1), (4,1), (6,1), (7,1), (2,3), (4,3), (6,3), (7,3), (2,5), (4,5), (6,5) and (7,5).

THING{(2,1)} = 2 + 1 = 3

THING{(2,3)} = 2 + 3 = 5

THING{(2,5)} = 2 + 5 = 7

THING{(4,1)} = 4 + 1 = 5

THING{(4,3)} = 4 + 3 = 7

THING{(4,5)} = 4 + 5 = 9

THING{(6,1)} = 6 + 1 = 7

THING{(6,3)} = 6 + 3 = 9

THING{(6,5)} = 6 + 5 = 11

THING{(7,1)} = 7*1 = 7

THING{(7,3)} = 7*3 = 21

THING{(7,5)} = 7*5 = 35

 

Pr{THING=3} = Pr{(2,1)} = 4/30 = 16/120

Pr{THING=5} = Pr{One of (2,3), (4,1) Shows} = Pr{(2,3)} + Pr{(4,1)} = (4/30) + (1/12) = (16+10)/120 = 26/120

 

Pr{THING=7} = Pr{One of (2,5), (4,3), (6,1), (7,1) Shows} = Pr{(2,5)} + Pr{(4,3)} + Pr{(6,1)} + Pr{(7,1)} =

(16/120) + (10/120) + (10/120) + (4/120) = 40/120

 

Pr{THING=9} = Pr{One of (4,5), (6,3) Shows} = Pr{(4,5)} + Pr{(6,3)} = (10/120) + (10/120) = 20/120

Pr{THING=11} = Pr{(6,5)} = 10/120

Pr{THING=21} = Pr{(7,3)} = 4/120

Pr{THING=35} = Pr{(7,5)} = 4/120

 

Check: (16/120) + (26/120) + (40/120) + (20/120) + (10/120) + (4/120) + (4/120) = (42 + 60 + 18)/120 = (120)/120 =1

Case Two

Color Slot Machine

Conditional Probabilities

25 Points Maximum

Here is our slot machine – on each trial, it produces a six color sequence, using the table below:

Sequence*

Probability

BGBYYG

0.10

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

BBBBBY

0.01

Total

1.00

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st to 6th , from left to right: (1st 2nd 3rd 4th 5th6th )

Compute the following conditional probabilities.

 

1. Pr{“BB” Shows | Red Shows}

9 Points Maximum

Sequence*

Probability

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

Total

0.89

 

Pr{Red Shows} = Pr{One of GBBRRG, YRBGGR, YRGGBB, YBYYRR, BBYYRY Shows} =

Pr{GBBRRG} + Pr{YRBGGR} + Pr{YRGGBB} + Pr{YBYYRR} + Pr{BBYYRY} =

0.15 + 0.11 + 0.35 + 0.23 + 0.05 = 0.89

 

 

Sequence*

Probability

GBBRRG

0.15

YRGGBB

0.35

BBYYRY

0.05

Total

0.55

 

Pr{“BB” and Red Show} = Pr{One of GBBRRG, YRGGBB, BBYYRY Shows} =

Pr{GBBRRG} + Pr{YRGGBB} + Pr{BBYYRY} = 0.15 + 0.35 + 0.05 = 0.55

 

Pr{“BB” Shows | Red Shows} = Pr{“BB” and Red Show}/ Pr{Red Shows} = 0.55/0.89

 

2. Pr{Red Shows  | “BY” Shows }

8 Points Maximum

Sequence*

Probability

BGBYYG

0.10

YBYYRR

0.23

BBYYRY

0.05

BBBBBY

0.01

Total

0.39

 

Pr{“BY” Shows} = Pr{One of BGBYYG, YBYYRR,  BBYYRY, BBBBBY Shows} =

Pr{BGBYYG} + Pr{YBYYRR} + Pr{BBYYRY} + Pr{BBBBBY} =

0.10 + 0.23 + 0.05 + 0.01 = 0.39

 

 

Sequence*

Probability

YBYYRR

0.23

BBYYRY

0.05

Total

0.28

 

Pr{“BY” and Red Show} = Pr{One of YBYYRR,  BBYYRY Shows} =

Pr{YBYYRR} + Pr{BBYYRY} = 0.23 + 0.05 = 0.28

 

Pr{Red Shows | “BY” Shows} = Pr{“BY” and Red Show}/ Pr{“BY” Shows} = 0.28/0.39

 

 

3. Pr{Green Shows | Red Shows}

9 Points Maximum

Sequence*

Probability

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

Total

0.89

 

Pr{Red Shows} = Pr{One of GBBRRG, YRBGGR, YRGGBB, YBYYRR, BBYYRY Shows} =

Pr{GBBRRG} + Pr{YRBGGR} + Pr{YRGGBB} + Pr{YBYYRR} + Pr{BBYYRY} =

0.15 + 0.11 + 0.35 + 0.23 + 0.05 = 0.89

 

Sequence*

Probability

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

Total

0.61

 

Pr{Green and Red Show} = Pr{One of GBBRRG, YRBGGR, YRGGBB Shows} =

Pr{GBBRRG} + Pr{YRBGGR} + Pr{YRGGBB} = 0.15 + 0.11 + 0.35 = 0.61

 

Pr{Green Shows | Red Shows} = Pr{Green and Red Show}/ Pr{Red Shows} = 0.61/0.89

 

Show all work and full detail for full credit.

 

Case Three

Long Run Argument and Perfect Samples

Traumatic Brain Injury (TBI) and Glasgow Coma Scale (GCS)

25 Points Maximum

 

Traumatic brain injury (TBI) is an insult to the brain from an external mechanical force, possibly leading to permanent or temporary impairments of cognitive, physical, and psychosocial functions with an associated diminished or altered state of consciousness. The Glasgow Coma Scale (GCS) is the most widely used system for scoring the level of consciousness of a patient who has had a traumatic brain injury. GCS is based on the patient's best eye-opening, verbal, and motor responses. Each response is scored and then the sum of the three scores is computed. The total score varies from 3 to 15. The GCS categories are Mild (for GCS scores between 13 and 15), Moderate (for GCS scores between 9 and 12) and Severe (for GCS scores between 3 and 8). It is not unusual for people to die with TBI before they can be treated or evaluated. We can augment the GCS categories by adding a PAD (Pre-admission Death, TBI Noted) category. Suppose that the probabilities tabled below apply to TBI cases:

 

TBI Severity

Probability

Mild

0.10

Moderate

0.15

Severe

0.50

PAD

0.25

Total

1.00

 

1. Interpret each probability using the Long Run Argument.

10 Points Maximum

 

In long runs of sampling with replacement, approximately 10% of sampled TBI cases are mild.

In long runs of sampling with replacement, approximately 15% of sampled TBI cases are moderate.

In long runs of sampling with replacement, approximately 50% of sampled TBI cases are severe.

In long runs of sampling with replacement, approximately 25% of sampled TBI cases are pre-admission deaths(PAD).

 

2. Compute and discuss Perfect Samples for n=1200.

15 Points Maximum

 

9 Points Maximum

EMild = 1200*PMild = 1200*0.10 = 120

EModerate = 1200*PModerate = 1200*0.15 = 180

ESevere = 1200*PSevere = 1200*0.50 = 600

EPAD = 1200*PPAD = 1200*0.25 = 300

 

Check: EMild + EModerate + ESevere + EPAD = 120 + 180 + 600 + 300 = 1200

 

6 Points Maximum

Random samples of TBI cases of size 1200 yield approximately 120 mild cases.

Random samples of TBI cases of size 1200 yield approximately 180 moderate cases.

Random samples of TBI cases of size 1200 yield approximately 600 severe cases.

Random samples of TBI cases of size 1200 yield approximately 300 PAD cases.

 

Show all work and full detail for full credit. Provide complete discussion for full credit.

 

Case Four

Color Slot Machine

Probability Rules

25 Points Maximum

 

Here is our slot machine – on each trial, it produces a six color sequence, using the table below:

Sequence*

Probability

BGBYYG

0.10

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

BBBBBY

0.01

Total

1.00

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st to 6th , from left to right: (1st 2nd 3rd 4th 5th6th )

Compute the following probabilities. If a rule is specified, you must use that rule.

 

1. Pr{“YR” Shows }

9 Points Maximum

Sequence*

Probability

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

Total

0.74

 

Pr{“YR” Shows } = Pr{One of YRBGGR, YRGGBB, YBYYRR, BBYYRY Shows} =

Pr{YRBGGR} + Pr{YRGGBB} + Pr{YBYYRR} + Pr{BBYYRY} = 0.11 + 0.35 + 0.23 + 0.05 = 0.46 + 0.28 = 0.74

 

2. Pr{ Green Shows 2nd or 3rd }

8 Points Maximum

Sequence*

Probability

BGBYYG

0.10

YRGGBB

0.35

Total

0.45

 

Pr{ Green Shows 2nd or 3rd } = Pr{One of BGBYYG, YRGGBB Shows} =

Pr{BGBYYG} + Pr{YRGGBB} = 0.10 + 0.35 = 0.45

 

3. Pr{ Green Shows } – Use the Complementary Rule

8 Points Maximum

Failure to Use CR: 6 points off

Sequence*

Probability

BGBYYG

0.10

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

BBBBBY

0.01

Total

1.00

 

Other Event = “Green does not Show”

Pr{Green does not Show} = Pr{YBYYRR} + Pr{BBYYRY} + Pr{BBBBBY} = 0.23 + 0.05 + 0.01 = 0.29

 

Pr{Green Shows} = 1 – Pr{Green does not Show} = 1 – 0.29 = 0.71

 

Check:

 

Pr{Green Shows} = Pr{One of BGBYYG, GBBRRG, YRBGGR, YRGGBB Shows} =

Pr{BGBYYG} + Pr{GBBRRG} + Pr{YRBGGR} + Pr{YRGGBB} =

0.10 + 0.15 + 0.11 + 0.35 = 0.25 + 0.46 = 0.71

 

Show all work and full detail for full credit.

 

Case Five

Design Fault Spot

25 Points Maximum

 

In each of the following a brief description of a design is presented. Briefly identify faults present in the design. Use the information provided. Be brief and complete.

 

1. A sample of college students is needed for a sample survey. The people running the study decide on the

following: they divide the population of colleges and universities into groups based upon enrollment size and whether the college or university is private or public; next, they used judgment to select one school from each group. Then, a random sample of students was selected from each selected school.

7 Points Maximum

 

The first stage of sampling is non-random. Both stages of sampling should be random.

 

2. In a comparative clinical trial, treatment methods are compared in the treatment of Condition Z, which when left untreated leads to severe complications and possibly death. Suppose we have a new candidate treatment, and further suppose that a standard treatment for a similar (but different) disease is available. A comparative clinical trial is proposed that would compare these treatments in patients with condition Z.

6 Points Maximum

 

The new treatment is not necessarily appropriate for Condition Z.

 

3. In a comparative clinical trial, treatment methods are compared in the treatment of Condition X, which when left untreated leads to severe complications and possibly death. A new surgical method is compared to a standard surgical method. Study physicians classify subjects by the severity of their disease, and assign only the "moderate" subjects to the new surgical method. Only the "severe" subjects are assigned to the standard surgical method.

6 Points Maximum

 

Patients must be randomly assigned to treatment.

 

4. Sample survey planning is under way to study voter support levels for a proposed (federal) constitutional amendment. The proposal is to randomly sample US resident adults, aged 18 years or older.

6 Points Maximum

 

The survey should target registered or likely voters. Many adults rarely or never vote, and even registered voters might choose to not vote.

 

Case Six

Clinical Trial Sketch

Behcet’s Syndrome

25 Points Maximum

 

Inflammation is the body’s response to irritation, infection, injury or other insults. Inflammation is mediated by the immune system, and includes redness, heat, swelling and pain. Behcet’s sndrome is a medical condition in which the patient suffers chronic inflammation of blood vessels. Patients with Behcet’s syndrome suffer complications when organs are affected by inflammatory flare-ups. One source of complications is uveitis (inflammation of the uvea (which includes the iris)). In some cases, complications of Behcet’s-induced uveitis can threaten vision. A similar condition, retinitis, affects the retina. Vision may be threatened by certain forms of uveitis and retinitis.

 

The uvea is the pigmented, middle lining of the eye. The term derives from the Latin word for grape, because the uvea can resemble the outer skin of a grape. The uvea includes the iris up front and includes the ciliary body and choroid in back. The uvea lies behind the retina.

 

The symptoms of Behcet's syndrome are usually treated with corticosteroids to suppress inflammation. Other medicines such as methotrexate, cyclophosphamide, or azathioprine may also be used. These drugs all can have serious side effects, including liver or kidney damage. As a result, some patients forego treatment with these medications.

 

Repeated inflammatory attacks can damage the uvea and/or retina. Accumulated damage can affect, impair and ultimately cause vision loss. The standard treatment is the use of anti-inflammatories, such as cortico-steroids, which treat inflammation. Successful response to these treatments includes fewer attacks, attacks of shorter duration and less severe attacks. Ultimately, the objective is the preservation of sight.

 

Other treatments attempt to prevent inflammation. Zenapax is an immuno-suppressive drug that basically blocks the signaling of lymphocytes by locking IL-2 receptors on the lymphocytes. As a result, the blocked lymphocytes do not initiate inflammatory processes. In this trial, we focus on patients with Behcet’s syndrome. More specifically, we focus on those patients with uveitis/retinitis and whose vision is threatened.

Sketch a comparative clinical trial comparing the effect of Anti-inflammatories versus Anti-inflammatories + Zenapax in the treatment of Behcet’s syndrome with vision-threatening uveitis/retinitis.

Make your sketch concise and complete, following the style demonstrated in class, in the sample second hourlies and in case study summaries.

We seek to recruit adult human subjects with Behcet’s syndrome with vision-threatening uveitis/retinitis. We inform potential study volunteers of the potential risks and benefits of study participation, as well as the details of the study protocol. Those who qualify and volunteer under informed consent are then enrolled in the study. Enrolled subjects present Behcet’s syndrome with vision-threatening uveitis or retinitis. The standard treatment is an anti-inflammatory regime plus a placebo version of Zenapax. The candidate treatment is an anti-inflammatory regime plus Zenapax. Subjects are randomly assigned to either the standard or the candidate treatement protocol. Double-blinding is employed, so that neither the subjects nor the clinical personnel know the individual treatment assignments.

 

Treated subjects are then followed for safety outcomes, including routine or minor side effects. Treated subjects are followed for toxic events, including kidney/liver damage, organ failure and the like. Treated subjects are followed for treatment response, including frequency, duration and severity of inflammatory flare-ups in the uvea and/or retina, as well as visual acuity and preservation of sight.

 

Work all six (6) cases.