The 1st Hourly

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.

______________________________________________________________________________________

Name (PRINTED) Signature Date

Case One | Random Variables | Color Slot Machine

Probability and Random Variables

Pair of Dice

We have a pair of dice. We assume that the dice operate separately and independently of each other. The dice are not fair – here are their probability models:

1^st Face	Pr{1^st Face}	2^nd Face	Pr{2^nd Face}
1	2/10	2	1/15
7	3/10	3	2/15
8	5/10	4	3/15
		5	4/15
		6	5/15

Suppose that our experiment consists of tossing the dice, and noting the resulting face-pair.

List the possible face-pairs, and compute a probability for each pair.

Denote the face-value pairs as (d5 face value, d3 face value).

Pair→ Hyp	2	3	4	5	6
1	(2,1)	(3,1)	(4,1)	(5,1)	(6,1)
7	(2,7)	(3,7)	(4,7)	(5,7)	(6,7)
8	(2,8)	(3,8)	(4,8)	(5,8)	(6,8)

Then compute the pair probabilities as Pr{(d5 face value, d3 face value)} = Pr{d5 face value}*Pr{d3 face value}.

Pr{(2,1)} = Pr{2 from d5}*Pr{1 from d3) = (1/15)*(2/10) = 2/150

Pr{(2,7)} = Pr{2 from d5}*Pr{7 from d3) = (1/15)*(3/10) = 3/150

Pr{(2,8)} = Pr{2 from d5}*Pr{8 from d3) = (1/15)*(5/10) = 5/150

Pr{(3,1)} = Pr{3 from d5}*Pr{1 from d3) = (2/15)*(2/10) = 4/150

Pr{(3,7)} = Pr{3 from d5}*Pr{7 from d3) = (2/15)*(3/10) = 6/150

Pr{(3,8)} = Pr{3 from d5}*Pr{8 from d3) = (2/15)*(5/10) = 10/150

Pr{(4,1)} = Pr{4 from d5}*Pr{1 from d3) = (3/15)*(2/10) = 6/150

Pr{(4,7)} = Pr{4 from d5}*Pr{7 from d3) = (3/15)*(3/10) = 9/150

Pr{(4,8)} = Pr{4 from d5}*Pr{8 from d3) = (3/15)*(5/10) = 15/150

Pr{(5,1)} = Pr{5 from d5}*Pr{1 from d3) = (4/15)*(2/10) = 8/150

Pr{(5,7)} = Pr{5 from d5}*Pr{7 from d3) = (4/15)*(3/10) = 12/150

Pr{(5,8)} = Pr{5 from d5}*Pr{8 from d3) = (4/15)*(5/10) = 20/150

Pr{(6,1)} = Pr{6 from d5}*Pr{1 from d3) = (5/15)*(2/10) = 10/150

Pr{(6,7)} = Pr{6 from d5}*Pr{7 from d3) = (5/15)*(3/10) = 15/150

Pr{(6,8)} = Pr{6 from d5}*Pr{8 from d3) = (5/15)*(5/10) = 25/150

Define HYP = HIGHTIE² + LOWTIE². Compute the values and probabilities for HYP.

Define HYP as (d5 face value)² + (d3 face value)²

Pair→ Hyp	2	3	4	5	6
1	(2,1) →4+1=9	(3,1) →9+1=10	(4,1) →16+1=17	(5,1) →25+1=26	(6,1) →36+1=37
7	(2,7) →4+49=53	(3,7) →9+49=58	(4,7) →16+49=65	(5,7) →25+49=74	(6,7) →36+49=85
8	(2,8) →4+64=68	(3,8) →9+64=73	(4,8) →16+64=79	(5,8) →25+64=89	(6,8) →36+64=100

Pr{Hyp=9} = Pr{(2,1)} = 2/150

Pr{Hyp=10} = Pr{(3,1)} = 4/150

Pr{Hyp=17} = Pr{(4,1)} = 6/150

Pr{Hyp=26} = Pr{(5,1)} = 8/150

Pr{Hyp=37} = Pr{(6,1)} = 10/150

Pr{Hyp=9} = Pr{(2,7)} = 3/150

Pr{Hyp=10} = Pr{(3,7)} = 6/150

Pr{Hyp=17} = Pr{(4,7)} = 9/150

Pr{Hyp=26} = Pr{(5,7)} = 12/150

Pr{Hyp=37} = Pr{(6,7)} = 15/150

Pr{Hyp=9} = Pr{(2,8)} = 5/150

Pr{Hyp=10} = Pr{(3,8)} =10/150

Pr{Hyp=17} = Pr{(4,8)} = 15/150

Pr{Hyp=26} = Pr{(5,8)} = 20/150

Pr{Hyp=37} = Pr{(6,8)} = 25/150

Case Two | Long Run Argument and Perfect Samples | Plurality in US Resident Pregnancies

Plurality is the number of siblings born as the result of a single pregnancy. Singleton pregnancies yield one born infant, twin pregnancies yield two infants and triplet pregnancies yield three infants. Suppose that the probabilities tabled below apply to pregnancies to US resident mothers with pregnancies yielding one or more live births:

Plurality	Probability
Singleton (1)	0.9660
Twins (2)	0.0330
Triplets+ (3 or more)	0.0010
Total	1.00

1. Interpret each probability using the Long Run Argument.

In long runs of random sampling, approximately 96.6% of live births to US resident mothers were singleton births.

In long runs of random sampling, approximately 3.3% of live births to US resident mothers were twin births.

In long runs of random sampling, approximately 0.1% of live births to US resident mothers were triplet or higher order multiple births.

2. Compute and discuss Perfect Samples for n=3000.

The expected counts for n=3000 are

E_Singleton = 3000*P_Singleton = 3000*.9660 = 2898

E_Twin = 3000*P_Twin = 3000*.033 = 99

E_Triplets₊ = 3000*P_Triplets₊ = 3000*.001 = 3

In random samples of 3,000 live births to US resident mothers, approximately 2898 sampled live births are singleton live births.

In random samples of 3,000 live births to US resident mothers, approximately 99 sampled live births are twinned live births.

In random samples of 3,000 live births to US resident mothers, approximately 3 sampled live births are triplet or higher order multiple live births.

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
BBRRYRRR	.10
RGRGBRRB	.10
BBGGYGBR	.15
GRRYBRGG	.15
BGYGYGYY	.20
RRYYGBBY	.10
YYGBYYRR	.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

Sequence*	Probability
BBRRYRRR	.10
RGRGBRRB	.10
BBGGYGBR	.15
GRRYBRGG	.15
BGYGYGYY	.20
RRYYGBBY	.10
YYGBYYRR	.20
Total	1.00

Pr{Blue Shows} = Pr{One of BBRRYRRR, RGRGBRRB, BBGGYGBR, GRRYBRGG, BGYGYGYY, RRYYGBBY or YYGBYYRR Shows} =

Pr{BBRRYRRR}+Pr{RGRGBRRB}+Pr{BBGGYGBR}+Pr{GRRYBRGG}+Pr{BGYGYGYY}+Pr{RRYYGBBY}+Pr{YYGBYYRR} =

.10 + .10 + .15 + .15 + .20 + .10 + .20 = 1.00

Blue and “Yellow Shows Exactly Three Times”

Sequence*	Probability
RRYYGBBY	.10

Pr{Yellow Shows Exactly Three Times} = Pr{RRYYGBBY} = .10

Pr{ Yellow Shows Exactly Three Times | Blue Shows}= Pr{Yellow Shows Exactly Three Times and Blue Shows}/ Pr{Blue Shows} = .1/1 = .10

Pr{ Yellow Shows | “BR” Shows }

“BR” Shows

Sequence*	Probability
BBRRYRRR	.10
RGRGBRRB	.10
BBGGYGBR	.15
GRRYBRGG	.15
Total	.50

Pr{“BR” Shows} = Pr{One of BBRRYRRR, RGRGBRRB, BBGGYGBR, GRRYBRGG Shows} =

Pr{BBRRYRRR}+Pr{RGRGBRRB}+Pr{BBGGYGBR}+Pr{GRRYBRGG} =

.10 + .10 + .15 + .15 = .50

Yellow and “BR” Shows

Sequence*	Probability
BBRRYRRR	.10
BBGGYGBR	.15
GRRYBRGG	.15
Total	.40

Pr{Yellow and “BR” Shows} = Pr{One of BBRRYRRR, BBGGYGBR, GRRYBRGG Shows} =

Pr{BBRRYRRR}+ Pr{BBGGYGBR}+Pr{GRRYBRGG} =

.10 +.15 + .15 = .40

Pr{ Yellow Shows | “BR” Shows } = Pr{ Yellow and “BR” Show }/ Pr{ “BR” Shows } = .40/.50 = .80

Pr{ Green Shows | Red Shows}

Red Shows

Sequence*	Probability
BBRRYRRR	.10
RGRGBRRB	.10
BBGGYGBR	.15
GRRYBRGG	.15
RRYYGBBY	.10
YYGBYYRR	.20
Total	1.00

Pr{Red Shows} = Pr{One of BBRRYRRR, RGRGBRRB, BBGGYGBR, GRRYBRGG, BGYGYGYY, RRYYGBBY or YYGBYYRR Shows} =

Pr{BBRRYRRR}+Pr{RGRGBRRB}+Pr{BBGGYGBR}+Pr{GRRYBRGG}Pr{RRYYGBBY}+Pr{YYGBYYRR} =

.10 + .10 + .15 + .15 +.10 + .20 = .80

Green and Red Show

Sequence*	Probability
RGRGBRRB	.10
BBGGYGBR	.15
GRRYBRGG	.15
RRYYGBBY	.10
YYGBYYRR	.20
Total	1.00

Pr{Green and Red Shows} = Pr{One of BBRRYRRR, RGRGBRRB, BBGGYGBR, GRRYBRGG, RRYYGBBY or YYGBYYRR Shows} =

Pr{BBRRYRRR}+Pr{RGRGBRRB}+Pr{BBGGYGBR}+Pr{GRRYBRGG}Pr{RRYYGBBY}+Pr{YYGBYYRR} =

.10 + .10 + .15 + .15 +.10 + .20 = .80

Pr{ Green Shows | Red Shows}= Pr{Green and Red Show }/ Pr{Red Shows} = .80/.80 = 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

Sequence*	Probability
BBRRYRRR	.10
GRRYBRGG	.15
RRYYGBBY	.10
Total	.35

Pr{“RY” Shows} = Pr{One of BBRRYRRR, GRRYBRGG or RRYYGBBY Shows} =

Pr{BBRRYRRR}+Pr{GRRYBRGG}+Pr{RRYYGBBY} =

.10 + .15 + .10 = .35

2. Pr{ Green Shows 2^nd, 3^rd or 4^th }

Green Shows 2^nd, 3^rd or 4^th

Sequence*	Probability
RGRGBRRB	.10
BBGGYGBR	.15
BGYGYGYY	.20
YYGBYYRR	.20
Total	.80

Pr{Green Shows 2^nd, 3^rd or 4^th } = Pr{One of RGRGBRRB, BBGGYGBR, BGYGYGYY or YYGBYYRR Shows} =

Pr{RGRGBRRB}+Pr{BBGGYGBR}+Pr{BGYGYGYY}+ Pr{YYGBYYRR} =

.10 +.15 + .20 + .20 = .65

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

“GB” Does Not Show

Sequence*	Probability
BBRRYRRR	.10
GRRYBRGG	.15
BGYGYGYY	.20
Total	.45

Pr{“GB” Does Not Show} = Pr{One of BBRRYRRR, GRRYBRGG or BGYGYGYY Shows} =

Pr{BBRRYRRR+Pr{GRRYBRGG}+Pr{BGYGYGYY} =

.10 +.15 + .20 = .45

Pr{“GB” Does Not Show} = 1 – Pr{“GB” 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 | Non-small Cell Lung Cancer (NSCLC)

A key ability of malignant cells is the ability to induce angiogensis, the formation of new blood supply. These cells can release a substance that stimulates the formation of new blood vessels. This ability is key in the ability of malignant tumors to survive and grow.

Avastin is a monoclonal antibody that works by attaching to and inhibiting the action of vascular endothelial growth factor (VEGF) in laboratory experiments. VEGF is a substance that binds to certain cells to stimulate new blood vessel formation. When VEGF is bound to Avastin, it cannot stimulate the formation and growth of new blood vessels.

A number of cancers are driven by the derangement of cells composing the linings (epidermal cells) of various organs in the body. In particular, these cells lose control of their growth behaviors, leading to uncontrolled reproduction of cells. This deranged, accelerated cell reproduction is key to the ability of malignant tumors to grow.

Tarceva (erlotinib) is an oral anti-cancer drug under development by OSI Pharmaceuticals, Genentech and Roche. It is a member of the epidermal growth factor receptor (EGFR) inhibitor class of agents.

Lung Cancer

Two general types of lung cancer exist: Non-Small Cell Lung Cancer (NSCLC) and small-cell lung cancer (SCLC)

The most common type of lung cancer is NSCLC. Approximately 85% of all lung cancer cases are NSCLC.

Three main types of NSCLC - General treatment options for each of these are the same:

Squamous cell carcinoma. Most often related to smoking. These tumors may be found in the mucous membrane that lines the bronchi. Sometimes the tumor spreads beyond the bronchi. Coughing up blood may be a sign of squamous cell NSCLC.

Adenocarcinoma (including bronchioloalveolar carcinoma). Most often found in nonsmokers and women. Cancer is usually found near the edge of the lung. Adenocarcinoma can enter the chest lining. When that happens, fluid forms in the chest cavity. This type of NSCLC spreads (metastasizes) early in the disease to other body organs.

Large-cell undifferentiated carcinoma. Rare type of NSCLC. Tumors grow quickly and spread early in the disease. Tumors are usually larger than 1-1/2 inches.

First-line Treatments for NSCLC:

Surgery: Removes the tumor. This can be done if the tumor is small and has not spread to other areas of your body.

Radiation: Destroys any leftover cancer cells not removed by surgery. This may be done before surgery to make it easier to remove the tumor. Radiation can also be done after surgery.

Chemotherapy may help slow the growth of cancer cells and destroy them. Chemotherapy may be used with radiation to help shrink the tumor before surgery. It may be used after surgery or radiation to destroy any cancer cells that may have been left behind.

Consider patients with locally advanced or metastatic Non-Small Cell Lung Cancer (NSCLC) after failure of at least one previous chemotherapy regimen.

Consider two treatments: Avastin+Tarceva and Tarceva.

Sketch a comparative clinical trial for Avastin+Tarceva versus Tarceva in the treatment of patients with locally advanced or metastatic Non-Small Cell Lung Cancer (NSCLC) after failure of at least one previous chemotherapy regimen.

Solution

Purpose of Treatment: Treatment of patients with locally advanced or metastatic Non-Small Cell Lung Cancer (NSCLC) after failure of at least one previous chemotherapy regimen.

We recruit patients with locally advanced or metastatic Non-Small Cell Lung Cancer (NSCLC) after failure of at least one previous chemotherapy regimen. Those giving informed consent are then checked for study eligibity.

We employ a placebo version of Avastin. Those who are eligible and who give informed consent are then randomly assigned to either (Avastin and Tarceva) or (Placebo_Avastin and Tarceva).

Double blinding is employed – neither the subjects nor the clinical study workers know the actual individual treatment assignments.

Subjects are tracked for the following outcomes:

Safety: Side effects, adverse reactions, toxicity, fatal and severe complications, organ damage or failure, etc…

Effectiveness: Lung cancer status, survival status, survival time, quality of life.

Work all six (6) cases.