Instructor Key
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 );
This
copy of the hourly.
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 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
Probability
Computational Rules and Random Variable
Pair of Dice
We have a pair of dice – a fair d2 {faces 0, 8} and a d4
{faces 1,3,5,7} – note the probability model for the d4 below.
|
Face |
Probability |
|
1 |
0.15 |
|
3 |
0.35 |
|
5 |
0.25 |
|
7 |
0.25 |
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-pair.
Show full detail for full credit.
1.a) List the possible pairs of face values,
and compute a probability for each pair of face values.
1.a – Pairs with
Probabilities
|
Face
Value Probability |
1 .15 |
3 .35 |
5 .25 |
7 .25 |
|
0 .50 |
(1,0) |
(3,0) |
(5,0) |
(7,0) |
|
8 .50 |
(1,8) |
(3,8) |
(3,8) |
(3,8) |
Pr{(1,0)}=Pr{1
from d4}*Pr{0from d2}= .15*.50 = .075
Pr{(1,8)}=Pr{1
from d4}*Pr{8 from d2}= .15*.50 = .075
Pr{(3,0)}=Pr{3
from d4}*Pr{0 from d2}= .35*.50 = .175
Pr{(3,8)}=Pr{3
from d4}*Pr{8 from d2}= .35*.50 = .175
Pr{(5,0)}=Pr{5
from d4}*Pr{0 from d2}= .25*.50 = .125
Pr{(5,8)}=Pr{5
from d4}*Pr{8 from d2}= .25*.50 = .125
Pr{(7,0)}=Pr{7
from d4}*Pr{0 from d2}= .25*.50 = .125
Pr{(7,8)}=Pr{7
from d4}*Pr{8 from d2}= .25*.50 = .125
1.b) Define THING as
follows. When the higher face value in the pair is less than or equal to 4,
define THING as the product of the
face values in the pair. Otherwise, define THING as the sum of the face values
in the pair. Compute and list the possible values for the variable THING, and
compute a probability for each value of THING.
1.b – Thing with
Probabilities
|
Face Value Probability |
1 .15 |
3 .35 |
5 .25 |
7 .25 |
|
0 .50 |
(1,0) High £ 4 1*0 = 0 |
(3,0) High £ 4 3*0=0 |
(5,0) High > 4 5+0=5 |
(7,0) High > 4 7+0=7 |
|
8 .50 |
(1,8) High > 4 8+1=9 |
(3,8) High > 4 3+8=11 |
(5,8) High > 4 5+8=13 |
(7,8) High > 4 7+8=15 |
Pr{THING=0} =
Pr{(1,0)}+Pr{(3,0)} = .075+.175 = .25
Pr{THING=5} =
Pr{(5,0)} = .125
Pr{THING=7} =
Pr{(7,0)} = .125
Pr{THING=9} =
Pr{(1,8)} = .075
Pr{THING=11} =
Pr{(3,8)} = .175
Pr{THING=13} =
Pr{(5,8)} = .125
Pr{THING=15} =
Pr{(7,8)} = .125
Case Two
Clinical Trial
Sketch
Post-Lumpectomy
Radiotherapy – Breast Cancer
Lumpectomy:
A
lumpectomy is one type of surgery for breast cancer. The malignant tumor and a
surrounding margin of normal breast tissue are removed. Lymph nodes in the
armpit may also be removed.
Sketch a clinical trial for intra-operative radiotherapy versus standard
radiotherapy in post-lumpectomy breast cancer patients.
For full credit, discuss completely, per examples discussed in class
and on the web page.
Condition: Breast
Cancer with Lumpectomy Indicated
Subjects have
breast cancer that is indicated for treatment via lumpectomy plus radiotherapy.
Those potential subjects who are briefed as to the potential risks and benefits
of study participation and who give informed consent are checked for inclusion
and exclusion criteria. Those who qualify are then enrolled in the study.
Treatments:
Lumpectomy plus intra-operative single-dose radiotherapy (L1); Lumpectomy plus
standard six week post-operative radiotherapy (L2).
Enrolled subjects
are randomly assigned to either L1 or
L2. The usual standard is that we employ double-blinding, and in this case
would require L1 to include a placebo post-operative radiotherapy cycle, and
would require L2 to include a placebo intra-operative sham probe. Most likely
this would not be done in a real clinical trial.
We track our
subjects for: Breast Cancer Status, Survival Status and Survival Time, Adverse
Events, Toxicity, and Quality of Life.
Case Three
Probability
Computational Rules
Color Slot Machine
Here is our slot machine – on each trial, it produces a 4-color
sequence, using the table below:
|
Sequence* |
Probability |
|
GBRB |
.25 |
|
BBGG |
.10 |
|
GBBR |
.25 |
|
RGYB |
.10 |
|
BBYY |
.10 |
|
RRYY |
.10 |
|
YBGR |
.10 |
|
Total |
1.00 |
* B-Blue, G-Green,
R-Red, Y-Yellow, Sequence is numbered as
1st to
4th , from left to right: (1st 2nd 3rd
4th)
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.a) Pr{ Red Shows 1st
or 3rd }
Pr{R 1st
or 3rd} = Pr{GBRB} + Pr{RGYB} + Pr{RRYY} = .25+.10 + .10 = .45
3.b) Pr{ Red Shows and
Blue Does Not Show }
Pr{R Shows and B
doesn’t} = Pr{RRYY} =.10
3.c) Pr{ Yellow Shows} - Use the
Complementary Rule.
Pr{Y does not
show} = Pr{GBRB} + Pr{BBGG} + Pr{GBBR} = .25 + .10 + .25 = .60
Pr{Y shows} = 1 -
.60 = .40
Check: Pr{Y
shows} = Pr{RGYB}+ Pr{BBYY}+ Pr{RRYY}+ Pr{YBGR} = .1+.1+.1+.1 = .40
Show full detail for full credit.
Case Four
Conditional
Probability
Color
Bowl/Draws without Replacement
We have a bowl containing the following colors and counts of balls
(color@count):
White @ 1, Black
@ 2, Blue @ 4, Green @ 4, Red @ 3, Yellow @ 1
Each trial of our experiment consists of five draws without
replacement from the bowl. Compute the following conditional probabilities.
Compute these
directly.
|
Color |
Before 1st
Draw |
|
White |
1 |
|
Black |
2 |
|
Blue |
4 |
|
Green |
4 |
|
Red |
3 |
|
Yellow |
1 |
|
Total |
15 |
4.a) Pr{ Yellow shows 2nd | Yellow shows 1st}
|
Color |
Before 1st
Draw |
Before 2nd
Draw |
|
White |
1 |
1 |
|
Black |
2 |
2 |
|
Blue |
4 |
4 |
|
Green |
4 |
4 |
|
Red |
3 |
3 |
|
Yellow |
1 |
0 |
|
Total |
15 |
14 |
Pr{Y 2nd
| Y 1st} = 0/14 = 0
4.b) Pr{ Black shows 5th
| Black shows 1st, Red shows 2nd, Blue shows 3rd,
and Green shows 4th }
|
Color |
Before 1st
Draw |
Before 2nd
Draw |
Before 3rd
Draw |
Before 4th
Draw |
Before 5th
Draw |
|
White |
1 |
1 |
1 |
1 |
1 |
|
Black |
2 |
1 |
1 |
1 |
1 |
|
Blue |
4 |
4 |
4 |
3 |
3 |
|
Green |
4 |
4 |
4 |
4 |
3 |
|
Red |
3 |
3 |
2 |
2 |
2 |
|
Yellow |
1 |
1 |
1 |
1 |
1 |
|
Total |
15 |
14 |
13 |
12 |
11 |
Pr{ Black 5th
| Black 1st, Red 2nd, Blue 3rd, Green 4th
} = 1/11
4.c) Pr{ Blue shows 3rd
| Blue shows 1st, Blue shows 2nd }
|
Color |
Before 1st
Draw |
Before 2nd
Draw |
Before 3rd
Draw |
|
White |
1 |
1 |
1 |
|
Black |
2 |
2 |
2 |
|
Blue |
4 |
3 |
2 |
|
Green |
4 |
4 |
4 |
|
Red |
3 |
3 |
3 |
|
Yellow |
1 |
1 |
1 |
|
Total |
15 |
14 |
13 |
Pr{ Blue 3rd|Blue
1st, Blue 2nd} = 2/13
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.
5.a) 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 is not an appropriate treatment for the design. Use the appropriate
standard treatment for the disease at hand.
5.b) 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.
Placebo alone is
not an appropriate treatment option for people with Hepatitis C. Informed
consent may also be lacking in this design.
5.c) A sample survey design targets a random
sample of residents of metro Atlanta with a well- designed questionnaire concerning driving/automotive safety
practices. The people running this survey sample design want to say that their
results will describe the driving/automotive safety practices of all Georgia
drivers.
Use a
representative sample covering all of Georgia, not just metro Atlanta.
5.d) A random sample of parents of
college/university first-year undergraduate students is surveyed about the
study practices of their children. The survey questionnaire was properly
written, and the sample of parents reasonably selected. The parents responded
to questions about their children's study habits.
Parents’ reports
may not be reliable in describing their children’s study behavior.
Case Six
Perfect Samples
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. So the blood types are O, A, B, AB.
Suppose that the probabilities for blood types for US
residents are noted below:
|
Blood Type |
Probability |
|
O |
.40 |
|
A |
.30 |
|
B |
.10 |
|
AB |
.20 |
|
Total |
1.00 |
Show all work and detail for full credit.
6.a) Interpret each
probability using the Long Run Argument. Be specific and complete for full
credit.
In long runs of
draws with replacement from this human population, approximately 40% of sampled
subjects will show type O; approximately 30% of sampled subjects will show type
A; approximately 10% of sampled subjects will show type B and approximately 20%
of sampled subjects will show type AB.
6.b) Compute the perfect
sample of n=150 US residents, and describe the relationship of this perfect
sample to real random samples of US residents.
|
Blood Type |
Probability |
Expected Count
(n=150) |
|
O |
.4 |
.4*150=60 |
|
A |
.3 |
.3*150=45 |
|
B |
.1 |
.1*150=15 |
|
AB |
.2 |
.2*150=30 |
|
Total |
1.0 |
1.0*150=150 |
EO =
150*PO = 150*.4 = 60
EA =
150*PA = 150*.3 = 45
EB =
150*PB = 150*.1 = 15
EAB =
150*PAB = 150*.2 = 30
Real random
samples of 150 subjects will present approximately 60 type O, approximately 45
type A, approximately 15 type B and approximately type AB subjects.
Be certain that you have worked all six (6) cases.
Performance Summaries
|
n |
m |
p00 |
p10 |
p25 |
p50 |
p75 |
p90 |
p100 |
|
28 |
78.2825 |
35 |
51.67 |
64.585 |
84.92 |
91.67 |
97.5 |
100 |
There were n=28
scores for the first hourly.
The sample mean
score was approximately 78.3%.
The minimum score
was 35%.
Approximately 90%
of the scores were at or above 51.67%.
Approximately 75%
of the scores were at or above 64.59%.
Approximately 50%
of the scores were at or above 84.92%.
Approximately 25%
of the scores were at or above 91.67%.
Approximately 10%
of the scores were at or above 97.50%.
The maximum score
was 100%.
|
percent score |
count |
percent |
|
<60 |
6 |
21.43 |
|
[60,70) |
2 |
7.14 |
|
[70,80) |
4 |
14.29 |
|
[80,90) |
5 |
17.86 |
|
[90,100) |
9 |
32.14 |
|
100 |
2 |
7.14 |