Key | The 1st Hourly | Math 1107 | Summer Term 2010

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– note the probability models for the dice below. 

1^st d4		2^nd d4
Face	Probability	Face	Probability
1	1/10	3	4/10
2	2/10	4	3/10
5	3/10	7	2/10
6	4/10	8	1/10
Total	10/10=1	Total	10/10=1

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-value-pair.

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

(1^stÞ, 2^nd ß)	1(1/10)	2(2/10)	5(3/10)	6(4/10)
3(4/10)	(1,3)	(2,3)	(5,3)	(6,3)
4(3/10)	(1,4)	(2,4)	(5,4)	(6,4)
7(2/10)	(1,7)	(2,7)	(5,7)	(6,7)
8(1/10)	(1,8)	(2,8)	(5,8)	(6,8)

(1,3), (2,3), (5,3), (6,3), (1,4), (2,4), (5,4), (6,4), (1,7), (2,7), (5,7), (6,7), (1,8), (2,8), (5,8), (6,8)

Pr{(1,3)} = Pr{1 from 1^st }*Pr{3 from 2^nd } = (1/10)*(4/10) = 4/100 = 0.04

Pr{(1,4)} = Pr{1 from 1^st }*Pr{4 from 2^nd } = (1/10)*(3/10) = 3/100 = 0.03

Pr{(1,7)} = Pr{1 from 1^st }*Pr{7 from 2^nd } = (1/10)*(2/10) = 2/100 = 0.02

Pr{(1,8)} = Pr{1 from 1^st }*Pr{8 from 2^nd } = (1/10)*(1/10) = 1/100 = 0.01

Pr{(2,3)} = Pr{2 from 1^st }*Pr{3 from 2^nd } = (2/10)*(4/10) = 8/100 = 0.08

Pr{(2,4)} = Pr{2 from 1^st }*Pr{4 from 2^nd } = (2/10)*(3/10) = 6/100 = 0.06

Pr{(2,7)} = Pr{2 from 1^st }*Pr{7 from 2^nd } = (2/10)*(2/10) = 4/100 = 0.04

Pr{(2,8)} = Pr{2 from 1^st }*Pr{8 from 2^nd } = (2/10)*(1/10) = 2/100 = 0.02

Pr{(5,3)} = Pr{5 from 1^st }*Pr{3 from 2^nd } = (3/10)*(4/10) = 12/100 = 0.12

Pr{(5,4)} = Pr{5 from 1^st }*Pr{4 from 2^nd } = (3/10)*(3/10) = 9/100 = 0.09

Pr{(5,7)} = Pr{5 from 1^st }*Pr{7 from 2^nd } = (3/10)*(2/10) = 6/100 = 0.06

Pr{(5,8)} = Pr{5 from 1^st }*Pr{8 from 2^nd } = (3/10)*(1/10) = 3/100 = 0.03

Pr{(6,3)} = Pr{6 from 1^st }*Pr{3 from 2^nd } = (4/10)*(4/10) = 16/100 = 0.16

Pr{(6,4)} = Pr{6 from 1^st }*Pr{4 from 2^nd } = (4/10)*(3/10) = 12/100 = 0.12

Pr{(6,7)} = Pr{6 from 1^st }*Pr{7 from 2^nd } = (4/10)*(2/10) = 8/100 = 0.08

Pr{(6,8)} = Pr{6 from 1^st }*Pr{8 from 2^nd } = (4/10)*(1/10) = 4/100 = 0.04

Define the random variable SIGNDIFF as:

SIGNDIFF = +1 if 1^st Die > 2^nd Die

and

SIGNDIFF = -1 if 1^st Die < 2^nd Die.

1.2) Compute and list the possible values for the variable SIGNDIFF, and compute a probability for each value of SIGNDIFF.

SIGNDIFF{(1,3)} = -1

SIGNDIFF{(1,4)} = -1

SIGNDIFF{(1,7)} = -1

SIGNDIFF{(1,8)} = -1

SIGNDIFF{(2,3)} = -1

SIGNDIFF{(2,4)} = -1

SIGNDIFF{(2,7)} = -1

SIGNDIFF{(2,8)} = -1

SIGNDIFF{(5,3)} = +1

SIGNDIFF{(5,4)} = +1

SIGNDIFF{(5,7)} = -1

SIGNDIFF{(5,8)} = -1

SIGNDIFF{(6,3)} = +1

SIGNDIFF{(6,4)} = +1

SIGNDIFF{(6,7)} = -1

SIGNDIFF{(6,8)} = -1

Pr{SIGNDIFF = +1} = Pr{One of (5,3), (5,4), (6,3) or (6,4) Shows} =

Pr{(5,3)} + Pr{(5,4) } + Pr{(6,3) } + Pr{(6,4) } = (12/100) + (9/100) + (16/100) +(12/100) = 49/100 = .49

Complementary Rule:

Pr{SIGNDIFF = -1} = 1 - Pr{ SIGNDIFF = +1} = 1 - .49 = .51

Or Directly:

Pr{SIGNDIFF = -1} =

Pr{One of (1,3), (2,3), (1,4), (2,4), (1,7), (2,7), (5,7), (6,7), (1,8), (2,8), (5,8) or (6,8) Shows} =

Pr{(1,3)} + Pr{(2,3)} + Pr{(1,4)} + Pr{(2,4)} + Pr{(1,7)} + Pr{(2,7)} + Pr{(5,7)} + Pr{(6,7)} +

Pr{(1,8)} + Pr{(2,8)} + Pr{(5,8)} + Pr{(6,8)} =

0.04 + 0.08 + 0.03 + 0.06 + 0.02 + 0.04 + 0.06 + 0.08 + 0.01 + 0.02 + 0.03 + 0.04 =

0.12 + 0.09 + 0.06 + 0.14 + 0.03 + 0.07 = 0.21 + 0.20 + 0.10 = 0.51

Case Two | Long Run Argument and Perfect Samples | Birthweight

Low birthweight is a strong marker of complications in liveborn infants. Low birthweight is strongly associated with a number of complications, including infant mortality, incomplete and impaired organ development and a number of birth defects. Suppose that the following probability model applies to year 2005 United States Resident Live Births:

Birthweight Status	Probability
Very Low Birthweight (Birthweight < 1500g)	0.016
Low Birthweight (1500g ≤ Birthweight < 2500g)	0.067
Normal Birthweight (2500 ≤ Birthweight < 4000g)	0.900
Macrosomia (Birthweight ≥ 4000g)	0.017
Total	1.00

2.1) Interpret each probability using the Long Run Argument.  

In long runs of random sampling of year 2005 US Resident live births, approximately 1.6% of sampled births were very low birthweight (birthweight < 1500 grams).

In long runs of random sampling of year 2005 US Resident live births, approximately 6.7% of sampled births were low birthweight (1500 grams ≤ Birthweight < 2500 grams).

In long runs of random sampling of year 2005 US Resident live births, approximately 90.0% of sampled births were normal birthweight (2500 grams ≤ Birthweight < 4000 grams).

In long runs of random sampling of year 2005 US Resident live births, approximately 1.7% of sampled births were macrosomic (birthweight > 4000 grams).

2.2) Compute and discuss Perfect Samples for n=3000.

Expected Count, Very Low Birthweight = 3000*Pr{ Very Low Birthweight} = 3000*0.016 = 48

Expected Count, Low Birthweight = 3000*Pr{ Low Birthweight} = 3000*0.067 = 201

Expected Count, Normal Birthweight = 3000*Pr{ Normal Birthweight} = 3000*0.900 = 2700

Expected Count, Macrosomia = 3000*Pr{ Macrosomia } = 3000*0.017 = 51

In random samples of 3000 year 2005 US Resident live births, approximately 48 of 3000 of sampled births were very low birthweight (birthweight < 1500 grams).

In random samples of 3000 year 2005 US Resident live births, approximately 201 of 3000 sampled births were low birthweight (1500 grams ≤ Birthweight < 2500 grams).

In random samples of 3000 year 2005 US Resident live births, approximately 2700 of 3000 sampled births were normal birthweight (2500 grams ≤ Birthweight < 4000 grams).

In random samples of 3000 year 2005 US Resident live births, approximately 51 of 3000 of sampled births were macrosomic (birthweight > 4000 grams).

Case Three | Probability Rules, Conditional Probability | Color Slot Machine

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

Color Sequence*	Color Sequence Probability
GBRB	0.300
GBBR	0.060
RGYB	0.300
YYYY	0.160
BBYY	0.060
YBGR	0.120
Total	1.000

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

Compute the following probabilities. In each of the following, show your intermediate steps and work. If a rule is specified, you must use that rule for your computation.

3.1) Pr{Green Shows} - Use the Additive Rule.

Color Sequence*	Color Sequence Probability
GBRB	0.300
GBBR	0.060
RGYB	0.300
YBGR	0.120
Total	0.780

Pr{Green Shows} = Pr{One of GBRB, GBBR, RGYB or YBGR Shows} =

Pr{GBRB} + Pr{GBBR} + Pr{RGYB} + Pr{YBGR} =

0.300 + 0.060 + 0.300 + 0.120 = 0.780

3.2) Pr{Green or Blue Show} - Use the Complementary Rule.

Other Event = “Neither Green Nor Blue Shows”

Color Sequence*	Color Sequence Probability
YYYY	0.160
Total	0.160

Pr{“Neither Green Nor Blue Shows”} = Pr{YYYY} = 0.160

Pr{Green or Blue Show} = 1 – Pr{“Neither Green Nor Blue Shows”} = 1 – 0.160 = 0.84

3.3) Pr{Green Shows | Blue Shows} - This is a Conditional Probability.

Prior Event = “Blue Shows”

Color Sequence*	Color Sequence Probability
GBRB	0.300
GBBR	0.060
RGYB	0.300
BBYY	0.060
YBGR	0.120
Total	0.840

Pr{Blue Shows} = Pr{One of GBRB, GBBR, RGYB, BBYY or YBGR Shows} =

Pr{GBRB} + Pr{GBBR} + Pr{RGYB} + Pr{BBYY} + Pr{YBGR} =

0.300 + 0.060 + 0.300 + 0.060 + 0.120 = 0.840

Joint Event = “Green and Blue Show”

Color Sequence*	Color Sequence Probability
GBRB	0.300
GBBR	0.060
RGYB	0.300
YBGR	0.120
Total	0.78

Pr{Green and Blue Show} = Pr{One of GBRB, GBBR, RGYB or YBGR Shows} =

Pr{GBRB} + Pr{GBBR} + Pr{RGYB} + Pr{YBGR} =

0.300 + 0.060 + 0.300 + 0.120 = 0.78

Pr{Green Shows | Blue Shows} = Pr{Green and Blue Shows}/ Pr{ Blue Shows} = 0.78/0.84

Case Four | Conditional Probability | Color Bowl, Draws without Replacement

We have a bowl containing the following colors and counts of balls (color @ count):

White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

Each trial of our experiment consists of five draws without replacement from the bowl. Compute the following conditional probabilities. Compute these directly. In each of the following, show your intermediate steps and work. Compute the following conditional probabilities:

4.1) Pr{ Yellow shows 3^rd | Blue Shows 2^nd and Yellow shows 1^st}

Before 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

After 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 1

After 2^nd Draw: White @ 2, Black @ 3, Blue @ 4, Green @ 5, Red @ 4, Yellow @ 1

Pr{ Yellow shows 3^rd | Blue Shows 2^nd and Yellow shows 1^st} = 1/(2+3+4+5+4+1) = 1/19

4.2) Pr{ Blue shows 5^th | Black shows 1^st, Blue shows 2^nd, Blue shows 3^rd, and Green shows 4^th }  

Before 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

After 1^st Draw: White @ 2, Black @ 2, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

After 2^nd Draw: White @ 2, Black @ 2, Blue @ 4, Green @ 5, Red @ 4, Yellow @ 2

After 3^rd Draw: White @ 2, Black @ 2, Blue @ 3, Green @ 5, Red @ 4, Yellow @ 2

After 4^th Draw: White @ 2, Black @ 2, Blue @ 3, Green @ 4, Red @ 4, Yellow @ 2

Pr{ Blue shows 5^th | Black shows 1^st, Blue shows 2^nd, Blue shows 3^rd, and Green shows 4^th } =

2/(2+2+3+4+4+2) = 3/17

4.3) Pr{ Blue shows 3^rd | Yellow shows 1^st, Blue shows 2^nd }

Before 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 2

After 1^st Draw: White @ 2, Black @ 3, Blue @ 5, Green @ 5, Red @ 4, Yellow @ 1

After 2^nd Draw: White @ 2, Black @ 3, Blue @ 4, Green @ 5, Red @ 4, Yellow @ 1

Pr{ Blue shows 3^rd | Yellow shows 1^st, Blue shows 2^nd } = 4/(2+3+4+5+4+1) = 4/19

Case Five | Clinical Trial Sketch | Ocular Hypertension, Glaucoma and Intraocular Stents.

Ocular Hypertenion and Open-angle Glaucoma

The eye consists of two fluid-filled chambers. The front(anterior) chamber is filled with aqueous humour, and maintaining proper aqueous fluid pressure is mediated by the trabecular network. Glaucoma is loss of vision due to optic nerve damage associated with excessive fluid pressure, which is commonly caused by drainage problems. A variety of drugs treat or delay glaucoma by reducing the production of aqueous humour – but the problem of drainage persists. A stent is a tube – stents are now used in cardiovascular procedures to restore the ability of blocked blood vessels to permit blood flow. A similar approach may work in glaucoma and ocular hypertension.

Ocular hypertension is commonly defined as a condition with the following criteria: an intraocular pressure of greater than 21 mm Hg is measured in one or both eyes on two or more occasions; the optic nerve appears normal; no signs of glaucoma are evident on visual field testing; no signs of any ocular disease are present. Glaucoma occurs when increased intraocular pressure, optic nerve damage, and vision loss are present.

Treatments:

Brinzolamide reduces intraocular pressure by reducing production of aqueous humour. Timolol maleate lowers blood pressure by promoting the relaxation of the smooth muscle liningsof the blood vessels. It also reduces the production of aqueous humour, which then lowers intraocular pressure. These drugs are administered as eye drops.

Case Five | Clinical Trial Sketch | Ocular Hypertension, Glaucoma and Intraocular Stents.

Treatments: Trabecular Micro-Bypass Stent: The trabecular network allows the drainage of aqueous humour from the eye. The basis for intraocular hypertension and for most types of glaucoma is excess fluid pressure in the eye, usually due to partial or total blockage of the trabecular network. The implantation of a stent may restore normal or otherwise improve drainage.

Subjects: Consider those adults with ocular hypertension without any glaucoma, as well as those recently diagnosed with open angle glaucoma.

Endpoints: Ocular Hypertension Only Subjects: Intraocular Pressure, Optic Nerve Status, Visual Field Loss, Visual Acuity, Progression to Glaucoma (among the ocular hypertensive only subjects), Time to Progression to Glaucoma

Recently Diagnosed Open Angle Glaucoma Subjects: Intraocular Pressure, Optic Nerve Status, Visual Field Loss, Visual Acuity

Suppose that previous surgical trials have established the trabecular micro-bypass stent as safe and effective in the treatment of ocular hypertension and glaucoma.

Sketch a comparative clinical trial comparing the effect of Stent + Brinzolamide + Timolol versus Stent in the treatment of Ocular Hypertension or Recently diagnosed Open Angle Glaucoma.

We recruit volunteers who have been diagnosed with ocular hypertension or recently diagnosed with open angle glaucoma. Those meeting inclusion and exclusion requirements and who give informed consent are enrolled in the study.

The treatments are either trabecular micro stent + brinzolamide + timolol or trabecular micro stent + placebo_brinzolamide + placebo_timolol.

Enrolled subjects are randomly assigned to either trabecular micro stent + brinzolamide + timolol or to trabecular micro stent + placebo_brinzolamide + placebo_timolol. Double-blinding is employed, so that neither the subjects nor the study personnel know the individual treatment assignments.

Treated subjects diagnosed with ocular hypertension only are tracked for Intraocular Pressure, Optic Nerve Status, Visual Field Loss, Visual Acuity, Progression to Glaucoma (among the ocular hypertensive only subjects), Time to Progression to Glaucoma.

Treated subjects recently diagnosed with open angle glaucoma are tracked for Intraocular Pressure, Optic Nerve Status, Visual Field Loss, Visual Acuity.

All treated subjects are tracked for side effects and toxicity, including reactions at the sites of dosage and stent injection, as well as kidney or liver toxicity.

Case Six | 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.

6.1) 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.

Urban business owners might not be residents, and urban residents who are not business owners are excluded from the survey.

6.2) A survey design targets a nation-wide (US) population of Public High School biology teachers by mailing survey instruments to a random sample of these teachers. Assume that there are no problems with the wording of the survey instrument. A total of 200 (out of 20,000 mailed) surveys are returned. Among other things, an analysis of the 200 responses indicated that 19% of the respondents thought that "dinosaurs and humans lived at the same time.." The researchers claim that their results and sample reliably reflect the nation-wide population of HS Biology Teachers.

The response rate is very low. Teachers from private high schools are excluded from the survey.

6.3) In a comparative clinical trial, two new treatment methods (TM1 and TM2) are compared in the treatment of Condition X, which when left untreated leads to severe complications and possibly death. A standard treatment is available. A comparative trial is proposed in which appropriately recruited subjects with condition X are randomly assigned with informed consent either to TM1 or to TM2. Double blinding is employed.

If there is a standard treatment, study subjects are denied appropriate treatment, and the standard for comparison should be the standard treatment. If there is no standard treatment, the basis of comparison is placebo – there should be a placebo group.

6.4) The cervix serves as the gateway between the uterus and vagina. Cervical cancer (or cancer of the uterine cervix) is cancer that originates from the cells of the cervix. Taxol is a potent anticancer natural product (with activity against a number of leukemia's and solid tumors in the breast, ovary, brain, and lung in humans) has stimulated an intense research effort over recent years. Consider cervical patients with an advanced (but not terminal) prognosis. In a clinical trial for Taxol as a possible treatment for patients with advanced cervical cancer it is proposed that Taxol be compared to a placebo-only group.

The study subjects’ cancers are advanced, not necessarily terminal – the subjects may benefit from standard treatment.

Work all six (6) cases.