Key
The 1st Hourly
Math 1107
Summer Term 2008
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 | Triplet of Dice |Random Variable
We have a triplet of dice – a fair d2 {faces 1,
2}, a fair d3 {faces 1, 2, 3} and a loaded d4 {faces 0, 1, 2, 3}
– note the probability model for the d4 below.
|
Face |
Probability |
|
0 |
0.40 |
|
1 |
0.25 |
|
2 |
0.25 |
|
3 |
0.10 |
|
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-triplet.
1. List the possible triplets of face values, and compute
a probability for each triplet of face values.
Write the triplets as (d2,d3,d4).
There are 2*3*4=24 triplets.
Pr{(1,1,0)} =
Pr{1 from d2}*Pr{1 from d3}*Pr{0 from d4} = (1/2)*(1/3)*(4/10) = 4/60 = 16/240
Pr{(1,2,0)} =
Pr{1 from d2}*Pr{2 from d3}*Pr{0 from d4} = (1/2)*(1/3)*(4/10) = 4/60 = 16/240
Pr{(1,3,0)} = Pr{1
from d2}*Pr{3 from d3}*Pr{0 from d4} = (1/2)*(1/3)*(4/10) = 4/60 = 16/240
Pr{(2,1,0)} =
Pr{2 from d2}*Pr{1 from d3}*Pr{0 from d4} = (1/2)*(1/3)*(4/10) = 4/60 = 16/240
Pr{(2,2,0)} =
Pr{2 from d2}*Pr{2 from d3}*Pr{0 from d4} = (1/2)*(1/3)*(4/10) = 4/60 = 16/240
Pr{(2,3,0)} =
Pr{2 from d2}*Pr{3 from d3}*Pr{0 from d4} = (1/2)*(1/3)*(4/10) = 4/60 = 16/240
Pr{(1,1,1)} =
Pr{1 from d2}*Pr{1 from d3}*Pr{1 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(1,2,1)} =
Pr{1 from d2}*Pr{2 from d3}*Pr{1 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(1,3,1)} =
Pr{1 from d2}*Pr{3 from d3}*Pr{1 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(2,1,1)} =
Pr{2 from d2}*Pr{1 from d3}*Pr{1 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(2,2,1)} =
Pr{2 from d2}*Pr{2 from d3}*Pr{1 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(2,3,1)} =
Pr{2 from d2}*Pr{3 from d3}*Pr{1 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(1,1,2)} =
Pr{1 from d2}*Pr{1 from d3}*Pr{2 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(1,2,2)} =
Pr{1 from d2}*Pr{2 from d3}*Pr{2 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(1,3,2)} =
Pr{1 from d2}*Pr{3 from d3}*Pr{2 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(2,1,2)} =
Pr{2 from d2}*Pr{1 from d3}*Pr{2 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(2,2,2)} =
Pr{2 from d2}*Pr{2 from d3}*Pr{2 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(2,3,2)} =
Pr{2 from d2}*Pr{3 from d3}*Pr{2 from d4} = (1/2)*(1/3)*(1/4) = 1/24 = 10/240
Pr{(1,1,3)} =
Pr{1 from d2}*Pr{1 from d3}*Pr{3 from d4} = (1/2)*(1/3)*(1/10) = 1/60 = 4/240
Pr{(1,2,3)} =
Pr{1 from d2}*Pr{2 from d3}*Pr{3 from d4} = (1/2)*(1/3)*(1/10) = 1/60 = 4/240
Pr{(1,3,3)} =
Pr{1 from d2}*Pr{3 from d3}*Pr{3 from d4} = (1/2)*(1/3)*(1/10) = 1/60 = 4/240
Pr{(2,1,3)} =
Pr{2 from d2}*Pr{1 from d3}*Pr{3 from d4} = (1/2)*(1/3)*(1/10) = 1/60 = 4/240
Pr{(2,2,3)} =
Pr{2 from d2}*Pr{2 from d3}*Pr{3 from d4} = (1/2)*(1/3)*(1/10) = 1/60 = 4/240
Pr{(2,3,3)} =
Pr{2 from d2}*Pr{3 from d3}*Pr{3 from d4} = (1/2)*(1/3)*(1/10) = 1/60 = 4/240
2. Define SUM as the sum of the face values in the
triplet. Compute and list the possible values for the variable SUM, and compute
a probability for each value of SUM.
SUM[(1,1,0)}] =
1+1+0 = 2
SUM[(1,2,0)}] =
1+2+0 = 3
SUM[(1,3,0)}] =
1+3+0 = 4
SUM[(2,1,0)}] =
2+1+0 = 3
SUM[(2,2,0)}] =
2+2+0 = 4
SUM[(2,3,0)}] =
2+3+0 = 5
SUM[(1,1,1)}] =
1+1+1 = 3
SUM[(1,2,1)}] =
1+2+1 = 4
SUM[(1,3,1)}] =
1+3+1 = 5
SUM[(2,1,1)}] =
2+1+1 = 4
SUM[(2,2,1)}] =
2+2+1 = 5
SUM[(2,3,1)}] =
2+3+1 = 6
SUM[(1,1,2)}] =
1+1+2 = 4
SUM[(1,2,2)}] =
1+2+2 = 5
SUM[(1,3,2)}] =
1+3+2 = 6
SUM[(2,1,2)}] =
2+1+2 = 5
SUM[(2,2,2)}] =
2+2+2 = 6
SUM[(2,3,2)}] =
2+3+2 = 7
SUM[(1,1,3)}] =
1+1+3 = 5
SUM[(1,2,3)}] =
1+2+3 = 6
SUM[(1,3,3)}] =
1+3+3 = 7
SUM[(2,1,3)}] =
2+1+3 = 6
SUM[(2,2,3)}] =
2+2+3 = 7
SUM[(2,3,3)}] =
2+3+3 = 8
Pr{SUM=2} = Pr{(1,1,0)} = 16/240
Pr{SUM=3} = Pr{One of (1,2,0),
(2,1,0), (1,1,1) Shows} = Pr{(1,2,0)}+
Pr{(2,1,0)}+ Pr{(1,1,1)} = (16/240)+(16/240)+(10/240)
= 42/240
Pr{SUM=4} = Pr{One of (1,3,0), (2,2,0) , (1,2,1), (2,1,1),
(1,1,2) Shows} = Pr{(1,3,0)}+Pr{(2,2,0)}+Pr{(1,2,1)}+Pr{(2,1,1)}+Pr{(1,1,2)} =
(16/240)+(16/240)+(10/240)+(10/240)+(10/240) = 62/240
Pr{SUM=5} = Pr{One of (2,3,0), (1,3,1), (2,2,1), (1,2,2), (2,1,2),
(1,1,3) Shows} = Pr{(2,3,0)}+Pr{(1,3,1)}+Pr{(2,2,1)}+Pr{(1,2,2)}+Pr{(2,1,2)}
+ Pr{(1,1,3))=
(16/240)+(10/240)+ (10/240)+ (10/240)+ (10/240) + (4/240) = 60/240
Pr{SUM=6} = Pr{One of (2,3,1), (1,3,2), (2,2,2), (1,2,3), (2,1,3) Shows}
= Pr{(2,3,1)}+Pr{(1,3,2)}+Pr{(2,2,2)}+ Pr{(1,2,3)}+Pr{(2,1,3)}
= (10/240)+ (10/240)+ (10/240)+(4/240)+(4/240) = 38/240
Pr{SUM=7} = Pr{One of (2,3,2), (1,3,3), (2,2,3) Shows} = Pr{(2,3,2)}+ Pr{(1,3,3)}+Pr{(2,2,3)} = (10/240)+(4/240) +(4/240) = 18/240
Pr{SUM=8} = Pr{(2,3,3)} = 4/240
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 sampling with replacement,
approximately 96.6% of US resident pregnancies yielding one or more live births
are a singleton pregnancies.
In long runs of sampling with replacement, approximately
3.3% of US resident pregnancies yielding one or more live births are twin
pregnancies.
In long runs of sampling with replacement,
approximately 0.1% of US resident pregnancies yielding one or more live births
are triplet or more pregnancies.
2. Compute and
discuss Perfect Samples for n=2500.
Esingleton = n*Psingleton
= 2500*0.9660 = 2415
Etwin = n*Ptwin =
2500*0.033 = 82.5
Etriplet+ = n*Ptriplet+
= 2500*0.001 = 2.5
In random samples of 2500 US resident pregnancies
yielding one or more live births, approximately 2415 sampled pregnancies are
singleton pregnancies.
In random samples of 2500 US resident
pregnancies yielding one or more live births, approximately 82.5 sampled
pregnancies are twin pregnancies.
In random samples of 2500 US resident
pregnancies yielding one or more live births, approximately 2.5 sampled
pregnancies are triplet or more pregnancies.
Case Three | Color Slot Machine | Conditional
Probabilities
Here is our slot machine – on
each trial, it produces a 10-color sequence, using the table below:
|
Sequence* |
Probability |
|
RRBBRRYRRR |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered
from left to right: (1st 2nd 3rd 4th
5th6th7th 8th 9th 10th
)
Compute the following conditional probabilities:
Pr{ Yellow Shows Exactly Twice
| Blue Shows}
|
Sequence* |
Probability |
|
RRBBRRYRRR |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
Pr{Blue Shows} =
Pr{One of RRBBRRYRRR, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY,
RRYYGRRBBY, YYGBYYBGRR Shows} =
Pr{RRBBRRYRRR} +
Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} +
Pr{RRYYGRRBBY} + Pr{YYGBYYBGRR} = .1+.1+.15+.1+.25+.1+.2 = 1.00
|
Sequence* |
Probability |
|
|
|
|
Total |
0 |
Pr{ Yellow Shows
Exactly Twice and Blue Shows} = 0
Pr{ Yellow Shows Exactly
Twice | Blue Shows} = Pr{ Yellow Shows Exactly Twice and Blue Shows}/Pr{Blue
Shows} = 0/1 =0
Pr{ Green Shows | “BR” Shows }
|
Sequence* |
Probability |
|
RRBBRRYRRR |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
Total |
.45 |
Pr{ “BR” Shows } =
Pr{One of RRBBRRYRRR, RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG Shows} = Pr{RRBBRRYRRR}+
Pr{ RRGGRGBRRB}+
Pr{BBYYGGYGBR}+
Pr{GRRGGYBRGG} = .1+.1+.15+.1 = .45
Pr{Green Shows and
“BR” Shows}
|
Sequence* |
Probability |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
Total |
.35 |
Pr{ Green Shows and
“BR” Shows } = Pr{One of RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG Shows} = Pr{ RRGGRGBRRB}+
Pr{BBYYGGYGBR}+
Pr{GRRGGYBRGG} =.1+.15+.1 = .35
Pr{ Green Shows | “BR”
Shows } = Pr{ Green Shows and “BR” Shows }/Pr{ “BR” Shows } = .35/.45 = 7/9
Pr{ Red Shows | Green Shows}
|
Sequence* |
Probability |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
.90 |
Pr{Green Shows} =
Pr{One of RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY, RRYYGRRBBY,
YYGBYYBGRR Shows} =
Pr{RRBBRRYRRR} +
Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} + Pr{RRYYGRRBBY}
+ Pr{YYGBYYBGRR} = .1+.15+.1+.25+.1+.2 = .90
Pr{ Red Shows and Green Shows}
|
Sequence* |
Probability |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
.90 |
Pr{ Red and Green
Show } = Pr{One of RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY, RRYYGRRBBY,
YYGBYYBGRR Shows} =
Pr{RRBBRRYRRR} +
Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} + Pr{RRYYGRRBBY}
+ Pr{YYGBYYBGRR} = .1+.15+.1+.25+.1+.2 = .90
Pr{ Red Shows |
Green Shows} = Pr{ Red Shows and Green Shows}/Pr{ Green Shows} = .9/.9 = 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{“YR”
Shows }
|
Sequence* |
Probability |
|
RRBBRRYRRR |
.10 |
|
BGYGYRYGYY |
.25 |
|
Total |
.35 |
Pr{Blue Shows} =
Pr{One of RRBBRRYRRR, BGYGYRYGYY Shows} =
Pr{RRBBRRYRRR}
+ Pr{BGYGYRYGYY} = .1+.25 = .35
2. Pr{ Green
Shows 2nd or 3rd }
|
Sequence* |
Probability |
|
RRGGRGBRRB |
.10 |
|
BGYGYRYGYY |
.25 |
|
YYGBYYBGRR |
.20 |
|
Total |
.55 |
Pr{{ Green Shows 2nd
or 3rd } = Pr{One of RRGGRGBRRB, BGYGYRYGYY, YYGBYYBGRR
Shows} =
Pr{RRGGRGBRRB}
+ Pr{BGYGYRYGYY} + Pr{YYGBYYBGRR} = .1+.25+.1+.2 = .55
3. Pr{ Green
Shows } – Use the Complementary Rule
|
Sequence* |
Probability |
|
RRBBRRYRRR |
.10 |
|
Total |
.10 |
Pr{No Green Shows} =
Pr{RRBBRRYRRR} = .10
Pr{Green Shows } = 1 – Pr{No Green Shows } =
1 –.1 = .90
Optional Check
|
Sequence* |
Probability |
|
RRGGRGBRRB |
.10 |
|
BBYYGGYGBR |
.15 |
|
GRRGGYBRGG |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRYYGRRBBY |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
.90 |
Pr{Blue Shows} =
Pr{One of RRGGRGBRRB, BBYYGGYGBR, GRRGGYBRGG, BGYGYRYGYY, RRYYGRRBBY,
YYGBYYBGRR Shows} =
Pr{RRGGRGBRRB} +
Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} + Pr{BGYGYRYGYY} + Pr{RRYYGRRBBY} +
Pr{YYGBYYBGRR} =.1+.15+.1+.25+.1+.2 = .90
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. A sample of
college students is needed for a sample survey. The people running the study
decide on the following: they divide the population of colleges and
universities into groups based upon enrollment size and whether the college or
university is private or public; next, they used judgment to select one school
from each group. Then, a random sample of students was selected from each
selected school.
The first stage of
sampling is not random – every stage of sampling should be random.
2. 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. Suppose we have a new candidate
treatment, and further suppose that a standard treatment for a similar (but different)
disease is available. A comparative clinical trial is proposed that would
compare these treatments in patients with condition Z.
The “standard”
treatment isn’t standard for Condition Z.
3.
A sample survey design employs a random sample of the
United Kingdom (England, Wales, Scotland, and Northern Ireland) adult subjects
(18 years or older, non-institutionalized). The research objective of the
survey is to determine attitudes of Irish* subjects regarding Birth Control.
Assume that the Survey Instrument is written properly and delivered in an
unbiased way.
*
Ireland is split into two countries: Northern Ireland and Ireland
The sampling
includes non-Irish subjects, and excludes Irish subjects in N. Ireland.
4. 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 sampling
excludes private high schools, and the response rate is too low.
Case Six | Clinical Trial Sketch | Study of
Tamoxifen and Raloxifene (STAR) for the Prevention of Breast Cancer
The purpose of this study is
to examine the performance of the drug Raloxifene
(relative to the drug Tamoxifen) in
reducing the incidence of breast cancer in postmenopausal women1 who are at increased risk
of the disease2.
1. Postmenopausal women at increased risk for developing
invasive breast cancer, who meet one of the following criteria: At least 12 months since spontaneous menstrual
bleeding; Prior documented
hysterectomy and the surgical removal of fallopian tubes and ovaries; At least 55 years of age with prior hysterectomy
with or without surgical removal of the ovaries; Aged 35 to 54 years with a prior hysterectomy without surgical
removal of the ovaries or with a status of ovaries unknown with documented
follicle-stimulating hormone level demonstrating elevation in postmenopausal
range.
2. Women without prior breast cancer, but who are at
elevated risk for breast cancer: Histologically confirmed lobular
carcinoma in situ treated by local excision only or at least 1.66% probability
of invasive breast cancer within 5 years using Breast Cancer Risk Assessment
Profile; No clinical evidence of
malignancy on physical exam within the past 180 days; No evidence of suspicious or malignant disease on bilateral
mammogram within the past year; No
bilateral or unilateral prophylactic mastectomy and No prior invasive breast cancer or intraductal carcinoma in situ
Objectives: Determine whether Raloxifene is more or less effective than Tamoxifen in significantly reducing the incidence rate of invasive breast cancer in postmenopausal women; Evaluate the effects of Tamoxifen and Raloxifene on the incidence of intraductal carcinoma in situ, lobular carcinoma in situ, endometrial cancer, ischemic heart disease, fractures of the hip and spine, or Colles' fractures of the wrist in these participants; Evaluate the toxic effects of these regimens in these participants and Determine the effect of these regimens on the quality of life of these participants.
Sketch a comparative
clinical trial to evaluate the drug Raloxifene (relative to the drug Tamoxifen) in reducing the incidence of
breast cancer in postmenopausal women1
who are at increased risk of the disease2.
Solution
Purpose
of Treatment: The purpose
of this study is to examine the performance of the drug Raloxifene (relative to the drug Tamoxifen) in reducing the incidence of breast cancer in
postmenopausal women1 who
are at increased risk of the disease2.
Eligible subjects are:
1. postmenopausal women at increased risk for developing invasive breast
cancer, who meet one of the following criteria: At least 12 months since spontaneous
menstrual bleeding; Prior documented
hysterectomy and the surgical removal of fallopian tubes and ovaries; At least 55 years of age with prior
hysterectomy with or without surgical removal of the ovaries; Aged 35 to 54 years with a prior
hysterectomy without surgical removal of the ovaries or with a status of
ovaries unknown with documented follicle-stimulating hormone level
demonstrating elevation in postmenopausal range.
2. Women without prior breast cancer, but
who are at elevated risk for breast cancer: Histologically
confirmed lobular carcinoma in situ treated by local excision only or at least
1.66% probability of invasive breast cancer within 5 years using Breast Cancer
Risk Assessment Profile; No clinical
evidence of malignancy on physical exam within the past 180 days; No evidence of suspicious or malignant
disease on bilateral mammogram within the past year; No bilateral or unilateral prophylactic mastectomy and No prior invasive breast cancer or
intraductal carcinoma in situ. The eligible patients are briefed as to
the details and potential risks and benefits of study participation, and those
who give informed consent and who meet all inclusion and exclusion
requirements are enrolled in the trial.
Study
treatments include Raloxifene and Tamoxifen. Enrolled subjects are randomly
assigned either to Raloxifene with PlaceboTamoxifen or to
Tamoxifen with PlaceboRalixifene. Double-blinding is employed, so that neither the subjects
nor the clinical workers know the actual individual treatment
assignments.
Subjects are then followed for: Incidence of invasive breast cancer in postmenopausal women; Incidence of intraductal carcinoma in situ, Incidence of lobular carcinoma in situ, Incidence of endometrial cancer, Incidence of ischemic heart disease, Incidence of fractures of the hip and spine, and Incidence of Colles' fractures of the wrist, Toxic effects of the medications, and Quality of Life.
Work all six (6) cases.