Key

Comprehensive Final Examination: Monday Version

Math 1107

Fall Semester 2007

Protocol

You will use only the following resources: Your individual calculator; Your individual tool-sheets (two (2) 8.5 by 11 inch sheet); Your writing utensils; Blank Paper (provided by me); This copy of the hourly and

the tables provided by me. Do not share these resources with anyone else. Show complete detail and work for full credit. Follow case study solutions and sample hourly keys in presenting your solutions.

Work all four 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. When you’re done: Print your name on a blank sheet of paper. Place your toolsheet, test and work under this sheet, and turn it all in to me.

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 | Random Variables | Color Slot Machine

Here is our slot machine – on each trial, it produces a 5-color sequence, using the table below:

Sequence*	Probability
GBRBY	.25
BBGGB	.10
GBBRG	.25
RGYBG	.10
BBYYG	.10
RRYYB	.10
YBGRY	.10
Total	1.00

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

Define REDCOUNT as the number of red slots in the sequence. Compute the probability model for

REDCOUNT. Compute the values of REDCOUNT, showing in detail how these values are computed from each color sequence. Compute the probability for each value of REDCOUNT, showing full detail.

Sequence*	Probability	REDCOUNT
GBRBY	.25	1
BBGGB	.10	0
GBBRG	.25	1
RGYBG	.10	1
BBYYG	.10	0
RRYYB	.10	2
YBGRY	.10	1
Total	1.00

Pr{REDCOUNT=0} =

Pr{One of BBGGB or BBYYG Shows } =

Pr{BBGGB}+ Pr{BBYYG}=

.10 + .10 = .20

Pr{REDCOUNT=1} =

Pr{One of GBRBY, GBBRG, RGYBG or YBGRY Shows } =

Pr{GBRBY}+ Pr{GBBRG}+ Pr{RGYBG}+ Pr{YBGRY} =

.25 + .25 + .10 + .10 = .70

Pr{REDCOUNT=2} =

Pr{ RRYYB} = .10

Define BGCOUNT as the number of times that “BG” shows in the sequence. Compute the probability model for BGCOUNT – Compute the values of BGCOUNT, showing in detail how these values are computed from each color sequence. Compute the probability for each value of BGCOUNT, showing full detail.

Sequence*	Probability	BGCOUNT
GBRBY	.25	0
BBGGB	.10	1
GBBRG	.25	0
RGYBG	.10	1
BBYYG	.10	0
RRYYB	.10	0
YBGRY	.10	1
Total	1.00

Pr(BGCOUNT=1} = Pr{One of BBGGB, RGYBG or YBGRY Shows} =

Pr{BBGGB}+ Pr{RGYBG}+ Pr{YBGRY} = .10 + .10 + .10 = .30

Pr(BGCOUNT=0} =

Pr{One of GBRBY, GBBRG, BBYYG, RRYYB Shows} =

Pr{GBRBY}+ Pr{GBBRG}+ Pr{BBYYG}+ Pr{RRYYB} = .25 + .25 + .10 + .10 = .70

Define YELLOW as 1 if yellow shows in the sequence and 0 if yellow does not show in the sequence.

Compute the probability model for YELLOW. Compute the values of YELLOW, showing in detail how

these values are computed from each color sequence. Compute the probability for each value of, showing

full detail.

Sequence*	Probability	YELLOW
GBRBY	.25	1
BBGGB	.10	0
GBBRG	.25	0
RGYBG	.10	1
BBYYG	.10	1
RRYYB	.10	1
YBGRY	.10	1
Total	1.00

Pr(YELLOW=1} = Pr{One of GBRBY, RGYBG, BBYYG, RRYYB, YBGRY } =

Pr{GBRBY}+ Pr{RGYBG}+ Pr{BBYYG}+ Pr{RRYYB}+ Pr{YBGRY } =

.25 + .10 + .10 + .10 + .10 = .65

Pr(YELLOW=0} =

Pr{One of BBGGB or GBBRG Shows} =

Pr{BBGGB}+ Pr{GBBRG} = .10 + .25 = .35

Show all work in full detail for full credit.

Case Two | Clinical Trial Sketch | Prevention Cerebral Toxoplasmosis in HIV-Infected Patients

HIV and Human Immune response: HIV infects cells in the immune system and the central nervous system. The main type of cell that HIV infects is the T helper lymphocyte. These cells play a crucial role in the immune system, by coordinating the actions of other immune system cells. A large reduction in the number of T helper cells seriously weakens the immune system. Once it has found its way into a cell, HIV produces new copies of itself, which can then go on to infect other cells. Over time, HIV infection leads to a severe reduction in the number of T helper cells available to help fight disease. Opportunistic infections and cancers are those that a healthy immune system would normally prevent. Treatment for the specific infection or cancer is often carried out, but the underlying cause is the action of HIV as it erodes the immune system.

Cerebral toxoplasmosis is one of the most frequently encountered opportunistic infections in the course of AIDS. The mortality (death) rate is estimated to be greater than 50 percent. Toxoplasmosis is a disease caused by an obligate intracellular protozoal parasite, Toxoplasma gondii, whose name was derived from the crescent shape of the parasite (toxon is Greek for "arc"), as well as the name of the North African rodent in which it was first observed, Ctenodactylus gundi. T gondii is one of the most successful protozoal parasites; it infects the nucleated cells of virtually all warm-blooded animals. Some species of felines are the definitive host for sexual reproduction of the parasite; however, asexual reproduction occurs in secondary hosts, such as rodents, livestock, birds, and humans, culminating in the formation of tissue cysts, which persist for the lifespan of the secondary host.Pyrimethamine is an antiparasitic drug. It prevents the growth and reproduction of parasites. Pyrimethamine is used to treat and prevent malaria. Pyrimethamine is also used in the treatment of toxoplasmosis.

Given the likely adverse consequences of cerebral toxoplasmosis in HIV-Infected patients, successful prevention of this parasitic infection is highly desirable. Pyrimethamine is an established treatment for existing infections, and might be effective in the prevention of cerebral toxoplasmosis in HIV-infected patients.

Sketch a basic clinical trial for the use of Pyrimethamine in the prevention of cerebral toxoplasmosis in HIV-Infected patients. Make your sketch concise and complete, following the style demonstrated in class, in the second hourly and in case study summaries.

Our objective is to evaluate the value of preventing cerebral toxoplasmosis in patients with HIV.

We recruit volunteers who are infected with HIV, but are free of cerebral toxoplasmosisOnce informed of the possible risks, benefits and details of study participation, the patients who have given informed consent, those patients meeting the necessary inclusion and exclusion criterion are enrolled in the trial.

Once enrolled, subjects are randomly assigned to either Placebo-only group, employing a placebo version of Pyrimethamine or to the Pyrimethamine group. Double-blinding is employed, in which neither the subjects (nor their proxies) nor the clinical personnel know the individual assignments to treatment.

Treated subjects are then followed for safety and toxicity, for quality-of-life, for cerebral toxoplasmosis infection status and for cerbreal toxoplasmosis-related mortality.

Case Three | Summary Intervals | Duchenne Muscular Dystrophy Survival Time

Duchenne muscular dystrophy (DMD) is an inherited disorder characterized by rapidly progressive muscle weakness which starts in the legs and pelvis and later affects the whole body. Suppose that we follow individuals diagnosed with DMD from diagnosis until death, noting age at death in months. Consider a random sample of individuals who were diagnosed with, and died with DMD. Age at death in months follows below:

17 22 37 45 50 65 73 87 93 102 112 123 127 130 137 138 143 150 156 161 169 173 177 179 180 181 181 182 184 185 186 186 188 189 190 190 190 192 193 193 194 194 195 196 196 197 197 199 199 200 200 201 203 204 207 210 213 215 217 219 220 223 227 228 230 231 233 234 235 237 239 240 240 241

Let m denote the sample mean, and sd the sample standard deviation. Compute and interpret the intervals  m ± 2sd and m ± 3sd, using Tchebysheff’s Inequalities and the Empirical Rule. Be specific and complete. Show complete detail and work for full credit. Follow case study solutions and sample hourly keys in presenting your solutions.

Numbers

n = 74

m = 175.27027

sd = 55.4837499

Lower bound, short interval = m - 2*sd = 175.27027 - 2*55.4837499 = 64.30

Upper bound, short interval = m + 2*sd = 175.27027 + 2*55.4837499 = 286.24

[64.30, 286.24]

Lower bound, long interval = m - 3*sd = 175.27027 - 3*55.4837499 = 8.82

Upper bound, long interval = m + 3*sd = 175.27027 + 3*55.4837499 = 341.72

[8.82, 341.72]

Discussion

At least 75% of the Duchenne Muscular Dystrophy patients in the sample survive between 64.30 and 286.24 months. At least 89% of the Duchenne Muscular Dystrophy patients in the sample survive between 8.82 and 341.72 months.

If the Duchenne Muscular Dystrophy patient survival times cluster symmetrically around a central value, with extreme values on either side of the central value becoming increasingly rare, then:

Approximately 95% of the Duchenne Muscular Dystrophy patients in the sample survive between 64.30 and 286.24 months. Approximately 100% of the Duchenne Muscular Dystrophy patients in the sample survive between 8.82 and 341.72 months.

Case Four | Confidence Interval for Mean | Duchenne Muscular Dystrophy Survival Time

Using the sample from Case Three, estimate the population mean survival time in months for patients with DMD with 97% confidence. That is, compute and discuss a 97% confidence interval for this population mean. Provide concise and complete details and discussion as demonstrated in the case study summaries.

Numbers

n = 74

m = 175.27027

sd = 55.4837499

From 2.20 0.013903 0.97219, Z = 2.20

Lower bound = m - Z*(sd/sqrt(n)) = 175.27027 – 2.2*(55.4837499/sqrt(74)) = 161.08

Upper bound = m + Z*(sd/sqrt(n)) = 175.27027 + 2.2*(55.4837499/sqrt(74)) = 189.46

[161.08, 189.46]

Interpretation

Our population is the population of patients with Duchenne Multiple Dystrophy(DMD) and our population mean is the population mean survival time in months.

Our Family of Samples (FoS) consists of every possible random sample of 74 DMD patients.

From each member sample of the FoS, we compute the sample mean (m) and standard deviation(sd) and then compute the interval

[m – 2.2*(sd/sqrt(n)), m + 2.2*(sd/sqrt(n))].

Computing this interval for each member sample of the FoS, we obtain a Family of Intervals (FoI), approximately 97% of which cover the true population mean survival time in months for patients with Duchenne Muscular Dystrophy.

If our interval, [161.08, 189.46] is among the approximate 97% super-majority of intervals that cover the population mean, then the true population mean survival time in months for patients with Duchenne Muscular Dystrophy is between 161.08 and 189.46 months.

Table 1. Means and Proportions

Z(k) PROBRT PROBCENT 0.05 0.48006 0.03988 0.10 0.46017 0.07966 0.15 0.44038 0.11924 0.20 0.42074 0.15852 0.25 0.40129 0.19741 0.30 0.38209 0.23582 0.35 0.36317 0.27366 0.40 0.34458 0.31084 0.45 0.32636 0.34729 0.50 0.30854 0.38292 0.55 0.29116 0.41768 0.60 0.27425 0.45149 0.65 0.25785 0.48431 0.70 0.24196 0.51607 0.75 0.22663 0.54675 0.80 0.21186 0.57629 0.85 0.19766 0.60467 0.90 0.18406 0.63188 0.95 0.17106 0.65789 1.00 0.15866 0.68269

Z(k) PROBRT PROBCENT 1.05 0.14686 0.70628 1.10 0.13567 0.72867 1.15 0.12507 0.74986 1.20 0.11507 0.76986 1.25 0.10565 0.78870 1.30 0.09680 0.80640 1.35 0.088508 0.82298 1.40 0.080757 0.83849 1.45 0.073529 0.85294 1.50 0.066807 0.86639 1.55 0.060571 0.87886 1.60 0.054799 0.89040 1.65 0.049471 0.90106 1.70 0.044565 0.91087 1.75 0.040059 0.91988 1.80 0.035930 0.92814 1.85 0.032157 0.93569 1.90 0.028717 0.94257 1.95 0.025588 0.94882 2.00 0.022750 0.95450

Z(k) PROBRT PROBCENT 2.05 0.020182 0.95964 2.10 0.017864 0.96427 2.15 0.015778 0.96844 2.20 0.013903 0.97219 2.25 0.012224 0.97555 2.30 0.010724 0.97855 2.35 0.009387 0.98123 2.40 0.008198 0.98360 2.45 0.007143 0.98571 2.50 0.006210 0.98758 2.55 0.005386 0.98923 2.60 0.004661 0.99068 2.65 0.004025 0.99195 2.70 .0034670 0.99307 2.75 .0029798 0.99404 2.80 .0025551 0.99489 2.85 .0021860 0.99563 2.90 .0018658 0.99627 2.95 .0015889 0.99682 3.00 .0013499 0.99730