The 1st Hourly

Math 1107

Summer Term 2004

Protocol

You will use only the following resources: Your individual calculator, Your individual tool-sheet (single 8.5 by 11 inch sheet), Your writing utensils, Blank Paper (provided by me ) and this copy of the hourly. Share these resources with no-one else. 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. 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

Long Run Argument

Perfect Samples

Landsteiner's Human Blood Types

In the early 20th century, an Austrian scientist named Karl Landsteiner classified blood according to chemical molecular differences – surface antigen proteins. Landsteiner observed two distinct chemical molecules present on the surface of the red blood cells. He labeled one molecule "A" and the other molecule "B." If the red blood cell had only "A" molecules on it, that blood was called type A. If the red blood cell had only "B" molecules on it, that blood was called type B. If the red blood cell had a mixture of both molecules, that blood was called type AB. If the red blood cell had neither molecule, that blood was called type O. The Rh(esus) factor is a blood protein isolated in Rhesus monkeys by K. Landsteiner and A. S. Wiener that is also present in some humans. Humans with this protein are said to be Rh positive; those without are said to be Rh negative. The introduction of Rh positive blood in Rh negative individuals causes a strong immune response, hence individuals who are Rh negative cannot safely receive Rh positive donor blood. Safe blood transfusing requires careful matching of ABO and Rh blood types.

Suppose that the probabilities for blood types for US residents are noted below:

Blood Type (Rh Factor)	Probability
O(+)	0.38
O(-)	0.07
A(+)	0.34
A(-)	0.06
B(+)	0.09
B(-)	0.02
AB(+)	0.03
AB(-)	0.01
Total	1.00

1.1) Interpret each probability using the Long Run Argument.

In large random samples of US residents, approximately 38% of sampled US residents present blood type O+, approximately 7% of sampled US residents present blood type O-, approximately 34% of sampled US residents present blood type A+, approximately 6% of sampled US residents present blood type A-, approximately 9% of sampled US residents present blood type B+, approximately 2% of sampled US residents present blood type B-, approximately 3% of sampled US residents present blood type AB+- and approximately 1% of sampled US residents present blood type AB-.

1.2) Compute the perfect sample of n=1500 US residents, and describe the relationship of this perfect sample to real random samples of US residents.

Blood Type (Rh Factor)	Probability	Perfect Count (n=1500)
O(+)	0.38	1500*0.38 =570
O(-)	0.07	1500*0.07=105
A(+)	0.34	1500*0.34=510
A(-)	0.06	1500*0.06=90
B(+)	0.09	1500*0.09=135
B(-)	0.02	1500*0.02=30
AB(+)	0.03	1500*0.03=45
AB(-)	0.01	1500*0.01=15
Total	1	1500

Perfect Count = 1500 * Probability

In random samples of 1500 US residents, approximately 570 sampled US residents present blood type O+, approximately 105 sampled US residents present blood type O-, approximately 510 sampled US residents present blood type A+, approximately 90 sampled US residents present blood type A-, approximately 135 sampled US residents present blood type B+, approximately 30 sampled US residents present blood type B-, approximately 45 sampled US residents present blood type AB+- and approximately 15 sampled US residents present blood type AB-.

Case Two

Clinical Trial Sketch

Non-small Cell Lung Cancer (NSCLC)

Key abilities of malignant cells are to induce the formation of new blood supply, and to grow aggressively. Certain malignant 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.

Iressa (Gefitinib) belongs to a group of anticancer drugs called epidermal growth factor receptor-tyrosine kinase inhibitors (EGFR-TKI). It may be of use in the treatment for non-small cell lung cancer (NSCLC). It may be useful in patients whose cancer has gotten worse despite treatment with platinum-based and docetaxel chemotherapy, two drugs that are currently the standard treatment of NSCLC. Iressa blocks growth signals in cancer cells. These signals are caused by an enzyme called tyrosine kinase (TK). IRESSA blocks several of these tyrosine kinases, including one associated with Epidermal Growth Factor Receptor (EGFR). EGFR is found on the cell surface of many normal cells and cancer cells. IRESSA works by binding to the tyrosine kinase of the EGFR to directly block growth signals turned on by triggers outside or inside the cell. In theory, this should impair the growth of malignant cells.

Lung Cancer

There are two general types of lung cancer: 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. There are 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.

Sketch a basic clinical trial for Iressa in the treatment of patients with locally advanced or metastatic Non-Small Cell Lung Cancer (NSCLC) after failure treatment with platinum-based and docetaxel chemotherapy.

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 Iressa. Those who are eligible and who give informed consent are then randomly assigned to either Iressa or Placebo. 

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.

Case Three

Random Variables

Pair of Dice

We have a pair of special dice, each fair:

1st Die		2^nd Die
Face Value (FV1)	Probability	Face Value (FV2)	Probability
0	¼	0	½
90	¼	90	½
180	¼
270	¼
Total	1	Total	1

3.a) List the possible pairs, and compute a probability for each.

(0,0),(0,90), (90,0),(90,90), (180,0),(180,90), (270,0),(270,90)

Pr{(0,0)} = Pr{0 from 1^st die}*Pr{0 from 2^nd die} = (1/4)*(1/2) = 1/8 Pr{(0,90)} = Pr{0 from 1^st die}*Pr{90 from 2^nd die} = (1/4)*(1/2) = 1/8 Pr{(90,0)} = Pr{90 from 1^st die}*Pr{0 from 2^nd die} = (1/4)*(1/2) = 1/8 Pr{(90,90)} = Pr{90 from 1^st die}*Pr{90 from 2^nd die} = (1/4)*(1/2) = 1/8 Pr{(180,0)} = Pr{180 from 1^st die}*Pr{0 from 2^nd die} = (1/4)*(1/2) = 1/8 Pr{(180,90)} = Pr{180 from 1^st die}*Pr{90 from 2^nd die} = (1/4)*(1/2) = 1/8 Pr{(270,0)} = Pr{270 from 1^st die}*Pr{0 from 2^nd die} = (1/4)*(1/2) = 1/8 Pr{(270,90)} = Pr{270 from 1^st die}*Pr{90 from 2^nd die} = (1/4)*(1/2) = 1/8

3.b) Define the random variable THING as THING= COSINE(FV1)+SINE(FV2). List the possible values for THING, and compute a probability for each value of THING.

THING{(0,0)}=COSINE(0) + SINE(0) = 1 + 0 = 1; THING{(0,90)}=COSINE(0) + SINE(90) = 1 + 1 = 2; THING{(90,0)}=COSINE(90) + SINE(0) = 0 + 0 = 0; THING{(90,90)}=COSINE(90) + SINE(90) = 0 + 1 = 1; THING{(180,0)}=COSINE(180) + SINE(0) = -1 + 0 = -1; THING{(180,90)}=COSINE(180) + SINE(90) = -1 + 1 = 0; THING{(270,0)}=COSINE(270) + SINE(0) = 0 + 0 = 0; THING{(270,90)}=COSINE(270) + SINE(90) = 0 + 1 = 1;

Pr{THING=-1}=Pr{(180,0)} = 1/8 Pr{THING=0}=Pr{one of (90,0),(180,90),(270,0) shows} = Pr{ (90,0) }+ Pr{ (180,90) }+ Pr{ (270,0) } = (1/8)+ (1/8)+ (1/8) = 3/8 Pr{THING=1}=Pr{one of (0,0),(90,90),(270,90) shows} = Pr{ (0,0) }+ Pr{ (90,90) }+ Pr{ (270,90) } = (1/8)+ (1/8)+ (1/8) = 3/8 Pr{THING=2}=Pr{(0,90)} = 1/8

Case Four

Design Spot Fault

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.

4.1) 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. A standard treatment is available. Suppose we have a new candidate treatment, and further suppose that a basic clinical trial is proposed to evaluate this new candidate treatment. Inappropriate denial of treatment.

4.2) A clinical trial of a new Hepatitis C treatment is designed as follows: subjects are screened for Hepatitis C infection. Those who test positive for Hepatitis C infection are then told of their status, and are offered treatment for Hepatitis C at no cost, and are given no further information. Those who accept the free treatment offer are then randomly assigned to either a Placebo, or to the New Treatment Plan. Denial of informed consent. Inappropriate denial of treatment in the placebo group.

4.3) A survey of parents and legal guardians of US high school students is used to study the sexual habits, knowledge and attitudes of US high school students. An appropriate random sample is used, and there are no problems with the wording and delivery of the survey instrument. Proxy responses may not be reliable when used in the study of illicit or sensitive subject matter.

4.4) The objective of a sample survey is to study the attitudes of urban residents of the United States regarding federal programs, taxation and spending. A random sample of urban business owners is employed in this design. Assume that there are no problems with the wording and delivery of the survey instrument. A random sample of urban residents is required for this design. Owners of urban businesses need not be residents, and not all residents own businesses.

Case Five

Probability Computational Rules

Pair of Dice

In this experiment we have a weird pair of dice – they are linked and do not operate independently. In fact, the dice produce the following face-pairs with the following probabilities:

(d2face,d4face)@Pr{(d2face,d4face)} d2 face Þ d4 face ß	3	4
1	(3,1) @ .10	(4,1) @ .10
2	(3,2) @ .10	(4,2) @ .20
5	(3,5) @ .05	(4,5) @ .20
6	(3,6) @ .15	(4,6) @ .10

In this experiment we toss this weird pair of dice and note the resulting pair of faces. In each of the following, show your intermediate steps and work. If a rule is specified, you must use that rule for your computation.

5.1) Compute Pr{ d2 face shows even } using the Additive Rule.

Pr{ d2 shows even } = Pr{ one of (4,1),(4,2),(4,5),(4,6) shows} = Pr{(4,1) } + Pr{ (4,2) } + Pr{ (4,5) } + Pr{ (4,6) } = .1 + .2 + .2 + .1 = .60

5.2) Compute Pr{ Sum of the faces < 7 } using the Complementary Rule.

Pr{SUM ³ 7} = Pr{ one of (4,5),(3,5),(4,5),(4,6) shows} = Pr{(4,5)}+ Pr{(4,6)}+ Pr{(3,5)}+ Pr{(3,6)}= .2+.1+.05+.15+= .50

Pr{SUM < 7} = 1 - Pr{SUM ³ 7} = 1 - .50 = .50

Check: Pr{SUM < 7} = Pr{ one of (3,1),(3,2),(4,1),(4,2) shows}= Pr{(3,1)}+ Pr{(3,2)} + Pr{(4,1)}+ Pr{(4,2)} = .1+.1+.1+.2 = .50 Ö

Case Six

Color Slot Machine

Computation of Conditional Probabilities.

Here is our slot machine – on each trial, it produces a 6-color sequence, using the table below:

Sequence*	Probability
BBBBBB	.10
BBGGBG	.10
RRYYGG	.15
GBBGGY	.10
RGYYYR	.25
BBYYGR	.10
YYGRYY	.20
Total	1.00

* B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1^st to 6^th , from left to right: (1^st 2^nd 3^rd 4^th 5^th6^th )

Compute the following conditional probabilities:

6.1) Pr{Blue Shows 3^rd | Green Shows 4^th }

Pr{Blue Shows 3^rd and Green Shows 4^th } = Pr{GBBGGY} = .10

Pr{ Green Shows 4^th } = Pr{ GBBGGY} + Pr{ BBGGBG} = .10 + .10 = .20

Pr{Blue Shows 3^rd | Green Shows 4^th } = Pr{Blue Shows 3^rd and Green Shows 4^th }/ Pr{ Green Shows 4^th } = .1/.2 = .50

6.2) Pr{Yellow Shows | Green Shows }

Pr{Yellow Shows and Green Shows }= Pr{ RRYYGG }+ Pr{ GBBGGY }+ Pr{ RGYYYR }+ Pr{ BBYYGR }+Pr{ YYGRYY}=.15+.10+.25+.10+.20=.80

Pr{ Green Shows }= Pr{ BBGGBG }+ Pr{ RRYYGG }+ Pr{ GBBGGY }+ Pr{ RGYYYR }+ Pr{ BBYYGR }+Pr{ YYGRYY}=.1+.15+.1+.25+.1+.2=.90

Pr{Yellow Shows | Green Shows }= Pr{Yellow Shows | Green Shows }/ Pr{Yellow Shows | Green Shows }= .8/.9 = 8/9 ≈ .8889

6.3) Pr{Blue Shows | Yellow Shows}

Pr{Blue Shows and Yellow Shows}= Pr{BBYYGR} + Pr{ GBBGGY } = .10 + .10 = .20

Pr{ Yellow Shows}=Pr{RRYYGG}+ Pr{GBBGGY}+ Pr{RGYYYR}+ Pr{BBYYGR}+ Pr{YYGRYY}= .15+.10+.25+.10+.20=.80

Pr{Blue Shows | Yellow Shows}= Pr{Blue Shows and Yellow Shows}/ Pr{ Yellow Shows}= .20/.80 = 2/8 = .250