Key

The 1st Hourly

Math 1107

Summer Term 2009

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. Show all work and full detail for full credit. Provide complete discussion for full credit. Sign and Acknowledge: I agree to follow this protocol.

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Name (PRINTED) Signature Date

Case One | Random Variables | Pair of Dice

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

Face	Probability
1	0.10
2	0.20
3	0.30
4	0.40
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.

Denote the face-value pairs as (d4 face value, d3 face value).

	1	2	3	4
1	(1,1)	(2,1)	(3,1)	(4,1)
2	(1,2)	(2,2)	(3,2)	(4,2)
3	(1,3)	(2,3)	(3,3)	(4,3)

Then compute the pair probabilities as Pr{(d4 face value, d3 face value)} = Pr{d4 face value}*Pr{d3 face value}.

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

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

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

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

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

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

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

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

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

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

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

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

2. When both faces in the pair are odd, define THING as the product of the face values in the pair. When both faces in the pair are even, define THING as the product of the face values in the pair.When one and only 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.

When both face values in the pair are odd, THING=(d4 face value)*(d3 face value).

When both face values in the pair are even, THING=(d4 face value)*(d3 face value).

When exactly one face value in the pair are odd, THING=(d4 face value) + (d3 face value).

Pair→Thing	1	2	3	4
1	(1,1) → 1*1=1	(2,1) → 2+1=3	(3,1) → 3*1=3	(4,1) → 4+1=5
2	(1,2) → 1+2=3	(2,2) → 2*2=4	(3,2) → 3+2=5	(4,2) → 4*2=8
3	(1,3) → 1*3=3	(2,3) → 2+3=5	(3,3) → 3*3=9	(4,3) → 4+3=7

Pr{Thing=1} = Pr{(1,1)} = 1/30

Pr{Thing=3} = Pr{One of (1,2), (1,3), (2,1), (3,1) Shows} = Pr{(1,2)} + Pr{(1,3) } + Pr{(2,1)} + Pr{(3,1)} = (1/30)+(1/30)+(2/30)+(3/30) = 7/30

Pr{Thing=4} = Pr{One of (2,2) Shows} = Pr{(2,1)} = (2/30)

Pr{Thing=5} = Pr{One of (2,3), (3,2) , (4,1) Shows} = Pr{(2,3)} + Pr{(3,2)} + Pr{(4,1)} = (2/30)+(3/30) +(4/30) = 9/30

Pr{Thing=7} = Pr{(4,3)} = 4/30

Pr{Thing=8} = Pr{(4,2)} = 4/30

Pr{Thing=9} = Pr{(3,3)} = 3/30

Case Two | Long Run Argument and Perfect Samples | Alzheimer’s Disease Survival Time

Alzheimer’s Disease (AD) is progressive and fatal, involving the progressive loss of higher cognitive functions, including memory, speech, language and emotional control. Survival time is the time, in years, from initial diagnosis to death. Suppose that the following probability model correctly describes survival time (in years) for people diagnosed with Alzheimer’s Disease.

Survival Category (years past diagnosis)	Probability
5 or less	.05
Strictly more than 5, but less than or equal to 10	.50
Strictly more than 10, but less than or equal to 15	.27
Strictly more than 15, but less than or equal to 20	.13
Strictly more than 20	.05
Total	1.00

1. Interpret each probability using the Long Run Argument.

In long runs of random sampling, approximately 5% of Alzheimer’s disease patients live 5 years or less after diagnosis.

In long runs of random sampling, approximately 50% of Alzheimer’s disease patients live strictly more than 5, but less than or equal to 10 years or less after diagnosis.

In long runs of random sampling, approximately 27% of Alzheimer’s disease patients live strictly more than 10, but less than or equal to 15 years or less after diagnosis.

In long runs of random sampling, approximately 13% of Alzheimer’s disease patients live strictly more than 15, but less than or equal to 20 years or less after diagnosis.

In long runs of random sampling, approximately 5% of Alzheimer’s disease patients live strictly more than 20 after diagnosis.

2. Compute and discuss Perfect Samples for n=5,000.

The expected counts for n=5000 are

E_[0,5] = 5000*P_[0,5] = 5000*.05 = 25

E_(5,10] = 5000*P_(5,10] = 5000*.50 = 2500

E_(10,15] = 5000*P_(10,15] = 5000*.27 = 1350

E_(15,20] = 5000*P_(5,10] = 5000*.13 = 650

E_(20+] = 5000*P_(20+] = 5000*.05 = 250

In random samples of 5,000 Alzheimer’s disease patients, approximately 25 sampled patients live 5 years or less after diagnosis.

In random samples of 5,000 Alzheimer’s disease patients, approximately 2500 sampled patients live strictly more than 5, but less than or equal to 10 years or less after diagnosis.

In random samples of 5,000 Alzheimer’s disease patients, approximately 1350 sampled patients live strictly more than 10, but less than or equal to 15 years or less after diagnosis.

In random samples of 5,000 Alzheimer’s disease patients, approximately 650 sampled patients live strictly more than 15, but less than or equal to 20 years or less after diagnosis.

In random samples of 5,000 Alzheimer’s disease patients, approximately 250 sampled patients live strictly more than 20 after diagnosis.

Case Three | Color Slot Machine | Conditional Probabilities

Here is our slot machine – on each trial, it produces a 8-color sequence, using the table below:

Sequence*	Probability
GYYRYRRR	.10
RGRBBRRB	.10
BGYBBGGY	.15
GRRYBRGG	.15
GBRGYGYY	.20
RYRYGBBY	.10
YYGBYRYY	.20
Total	1.00

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered from left to right: (1^st 2^nd 3^rd 4^th 5^th6^th7^th 8^th )

Compute the following conditional probabilities:

Pr{ Yellow Shows Exactly Three Times | Blue Shows}

Blue Shows

RGRBBRRB	.10
BGYBBGGY	.15
GRRYBRGG	.15
GBRGYGYY	.20
RYRYGBBY	.10
YYGBYRYY	.20

Pr{Blue Shows} =

Pr{Exactly One of RGRBBRRB, BGYBBGGY, GRRYBRGG, GBRGYGYY, RYRYGBBY or YYGBYRYY Shows} =

Pr{ RGRBBRRB} + Pr{BGYBBGGY} + Pr{ GRRYBRGG} + Pr{ GBRGYGYY} +Pr{ RYRYGBBY } + Pr{YYGBYRYY} =

.10 + .15 + .15 + .20 + .10 + .20 = .90

Yellow Shows Exactly Three Times and Blue Shows

GBRGYGYY	.20
RYRYGBBY	.10

Pr{Yellow Shows Exactly Three Times and Blue Shows} =

Pr{Exactly One of GBRGYGYY or RYRYGBBY } =

Pr{ GBRGYGYY} +Pr{ RYRYGBBY } = .10 + .20 = .30

Pr{Yellow Shows Exactly Three Times|Blue Shows} = Pr{Yellow Shows Exactly Three Times and Blue Shows}/ Pr{ Blue Shows} = .30/.90 = 1/3

Pr{ Yellow Shows | “BR” Shows }

“BR” Shows

RGRBBRRB	.10
GRRYBRGG	.15
GBRGYGYY	.20

Pr{“BR” Shows} =

Pr{Exactly One of RGRBBRRB, GRRYBRGG, GBRGYGYY Shows} =

Pr{ RGRBBRRB} + Pr{ GRRYBRGG} + Pr{ GBRGYGYY} =

.10 + .15 +.20 = .45

Yellow and “BR” Shows

GRRYBRGG	.15
GBRGYGYY	.20

Pr{“BR” Shows} =

Pr{Exactly One of GRRYBRGG, GBRGYGYY Shows} =

Pr{ GRRYBRGG} + Pr{ GBRGYGYY} =

.15 +.20 = .35

Pr{Yellow Shows Exactly Three Times|”BR” Shows} = Pr{Yellow Shows Exactly Three Times and “BR” Shows}/ Pr{ Blue Shows} = .35/.45 = 7/9

Pr{ Green Shows | Red Shows}

Red Shows

GYYRYRRR	.10
RGRBBRRB	.10
GRRYBRGG	.15
GBRGYGYY	.20
RYRYGBBY	.10
YYGBYRYY	.20

Pr{Red Shows} =

Pr{Exactly One of GYYRYRRR, RGRBBRRB, GRRYBRGG, GBRGYGYY, RYRYGBBY, YYGBYRYY Shows} =

Pr{ GYYRYRRR } + Pr{RGRBBRRB } + Pr{GRRYBRGG } + Pr{ GBRGYGYY } +Pr{RYRYGBBY } + Pr{YYGBYRYY } =

.10 + .10 + .15 + .20 + .10 + .20 = .85

Red and Green Shows

GYYRYRRR	.10
RGRBBRRB	.10
GRRYBRGG	.15
GBRGYGYY	.20
RYRYGBBY	.10
YYGBYRYY	.20

Pr{Red and Green Shows} =

Pr{Exactly One of GYYRYRRR, RGRBBRRB, GRRYBRGG, GBRGYGYY, RYRYGBBY, YYGBYRYY Shows} =

Pr{ GYYRYRRR } + Pr{RGRBBRRB } + Pr{GRRYBRGG } + Pr{ GBRGYGYY } +Pr{RYRYGBBY } + Pr{YYGBYRYY } =

.10 + .10 + .15 + .20 + .10 + .20 = .85

Pr{Green Shows | Red Shows} = Pr{Red and Green Show}/ Pr{ red Shows} = .85/.85 = 1

Case Four | Color Slot Machine | Probability Rules

Using the color slot machine from Case Three, compute the following probabilities. If a rule is specified, you must use that rule.

1. Pr{“RY” Shows }

“RY” Shows

GYYRYRRR	.10
GRRYBRGG	.15
RYRYGBBY	.10
YYGBYRYY	.20

Pr{“RY” Shows} =

Pr{Exactly One of GYYRYRRR, GRRYBRGG, RYRYGBBY or YYGBYRYY Shows} =

Pr{ GYYRYRRR } + Pr{ GRRYBRGG} + Pr{ RYRYGBBY } + Pr{YYGBYRYY} =

.10 + .15 + .10 + .20 = .55

2. Pr{ Green Shows In 1^st or 2^nd slot}

Green Shows 1^st or 2^nd

GYYRYRRR	.10
RGRBBRRB	.10
BGYBBGGY	.15
GRRYBRGG	.15
GBRGYGYY	.20

Pr{Green Shows 1^st or 2^nd} =

Pr{Exactly One of GYYRYRRR, RGRBBRRB, BGYBBGGY, GRRYBRGG, GBRGYGYY Shows} =

Pr{ GYYRYRRR } + Pr{RGRBBRRB } + Pr{BGYBBGGY } + Pr{GRRYBRGG } +Pr{GBRGYGYY } =

.10 + .10 + .15 + .15 +.20 = .70

3. Pr{ “RY” Shows } – Use the Complementary Rule

Other Event = “RY” Does Not SHow

“RY” Does Not Show

RGRBBRRB	.10
BGYBBGGY	.15
GBRGYGYY	.20

Pr{“RY” Does Not Show} =

Pr{Exactly One of RGRBBRRB, BGYBBGGY, GBRGYGYY Shows} =

Pr{ RGRBBRRB } + Pr{ BGYBBGGY } + Pr{ RGRBBRRB } =

.10 +.15 +.20 = .45

Pr{“RY” Shows} = 1 – Pr{“RY” Does Not Show} = 1 – .45 = .55

Case Five | Design Fault Spot

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. WidgetCorpsä is conducting an Employee Satisfaction Survey. They hire a third party to conduct the survey, and a random sample of employees is employed in the survey. The interviews are conducted after the annual performance and salary reviews, and while the third party identifies the identities of the respondents on the surveys, it tells the respondents that these will be removed before the results are given to WidgetCorpsä.

Widget Corps Survey – If the survey is given too close in time to the annual review, responses may be distorted. Confidentiality is at risk.

2. A sample survey of US University and College undergraduates is used to study the sexual habits, knowledge and attitudes of their parents and legal guardians. Appropriate random sampling of students is used, and there are no problems with the wording and delivery of the survey instrument.

Parental Sexual Mores Survey – survey the parents directly.

3. Disease W is a disease caused by an infection. Left untreated, disease W produces severe and occasionally fatal symptoms and complications. Suppose that no effective, standard treatments are available. Suppose further that a new treatment, ugorbitx is available for evaluation. A clinical trial is proposed to evaluate ugorbitx by giving all trial subjects ugorbitx. Their results would be compared to similar groups of untreated patients who are not enrolled in the trial.

Disease W Clinical Trial – Use a placebo group within the trial itself.

4. A clinical trial of a new Hepatitis C (HepC) treatment is designed as follows: subjects are screened for HepC

infection. Untreated HepC can lead to liver disease, liver failure, liver cancer and death. Those who test positive for HepC infection are then told of their status, and are offered treatment for HepC at no cost, and are offered enrollment in a comparative clinical trial for the treatment of HepC infection. Those who qualify and who give informed consent are then stratified by risk of progression to symptoms of hepatitis. Those judged to be at high risk are assigned to the new treatment, those judged to be at moderate risk of progression to symptoms of hepatitis are assigned to standard treatment, and those judged to be at minimal risk of progression to symptoms of hepatitis symptoms are assigned to placebo only.

Hepatitis C Clinical Trial – Use informed consent in the recruitment of subjects, and randomly assign subject volunteers to treatment.

Case Six | Clinical Trial Sketch | Congenital Nephrogenic Diabetes Insipidus

Diabetes insipidus (DI) is a condition that occurs when the kidneys are unable to conserve water as they perform their function of filtering blood. The amount of water conserved is controlled by antidiuretic hormone (ADH), also called vasopressin. ADH is a hormone produced in a region of the brain called the hypothalamus. It is then stored and released from the pituitary gland, a small gland at the base of the brain. DI caused by a lack of ADH is called central diabetes insipidus. Nephrogenic Diabetes Insipidus is a disease in which the patient’s kidneys are resistant to the diuretic hormone vasopressin. Vasopressin is a hormone produced by the hypothalamus, and among other things, stimulates the kidneys to preserve water and concentrate urine. In NDI, the kidneys are not responsive to normal amounts of vasopressin. When DI is caused by a failure of the kidneys to respond to ADH, the condition is called nephrogenic diabetes insipidus (NDI). Nephrogenic DI involves a defect in the parts of the kidneys that reabsorb water back into the bloodstream. It occurs less often than central DI. Congenital NDI (CNDI) occurs as an inherited disorder in which male children receive the abnormal gene that causes the disease from their mothers.

Standard Treatments for NDI include hydrochlorothiazide/amiloride (moduretic) and indomethacin. Moduretic helps to prevent excessive fluid build-up in the kidneys, reduces excessive blood pressure and preserves potassium levels that might otherwise fall due to use of a diuretic. Diuretics reduce urine volume in people with NDI. Indomethacin is a nonsteriodal anti-inflammatory drug (NSAID), used to treat pain and inflammation in people with NDI, also helps to reduce urine volume.

Experimental Treatments in this trial are calcitonin and sildenafil. Calcitonin reduces the amount of calcium and phosphates in the blood – use of thiazide diuretics may lead to elevated calcium in the blood. Sildenafil may enhance water reabsorption in people with NDI.

Inclusion Criteria: Known diagnosis of CNDI, Age 5 to 25 years, Normal kidney function, Post-void residual urine < 200 ml (determined by bladder ultrasound).

Exclusion Criteria: Impaired kidney function, Known urinary retention or bladder dysfunction, High blood pressure, Other significant chronic medical disease (e.g., heart failure, liver disease, etc.), Allergy to study drugs.

Primary Outcome: Urine volume – amount of urine cleared.

Secondary Outcomes: Frequency of urination (How often), Urine osmolality (Concentration of waste in urine)

Other considerations: Excessive Thirst – polydipsia, Acute Hyperosmolar Dehydration – excessively high blood plasma concentration, Low Blood Pressure – hypotension, Shock, Poor Nutrition and Growth

Sketch a comparative clinical trial for the treatment of people with congenital nephrogenic diabetes insipidus, comparing {Moduretic+Indomethacin} to {Moduretic+Indomethacin+Calcitonin+Sildenafil}. Make your sketch concise and complete, following the style demonstrated in class, in the first and second hourlies and in case study summaries. 

http://clinicaltrials.gov/ct2/show/NCT00478335

Population – Subjects with congenital nephrogenic diabetes insipidus(CNDI). Inclusion/exclusion criteria further restrict the population to males aged 5-25, diagnosed with CNDI, with normal renal function.

Disease – Congenital Nephrogenic Diabetes Insipidus

This is a therapeutic trial – the treatments are intended to actively alter the course of disease in subjects with CNDI.

We recruit subjects who are diagnosed with congenital nephrogenic diabetes insipidus. Eligible subjects are male, aged 5-25, have normal renal and bladder function, normal blood pressure range and no other major medical conditions.

Eligible adult subjects who give informed consent, and eligible pediatric subjects whose medical guardians give informed consent by proxy are enrolled in the trial.

Enrolled subjects are randomly assigned to either (Moduretic+Indomethacin) + (PlaceboCalcitonin + PlaceboSildenifil) or to (Moduretic+Indomethacin) + (Calcitonin + Sildenifil). Double blinding is employed, so that neither the subjects nor their clinical personnel know individual treatment assignments during the trial.

Treated Subjects are followed for: urine volume, frequency of urination, concentration of urine, excessive thirst, water consumption, blood plasma concentration, blood pressure, nutrition and growth status.

Treated Subjects are followed for drug safety (kidney and liver, in particular), side effects and quality of life.

Work all six (6) cases.