Key | The Comprehensive
Final Examination | Math 1107 | Fall Semester 2010 | CJ Alverson
Protocol
You will use only the
following resources: Your
individual calculator; Your individual tool-sheets
(two (2) 8.5 by 11 inch sheets); 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 tool-sheets, test and
work under this sheet, and turn it all in to me. Do not share information
with any other students during this test.
Sign and Acknowledge: I
agree to follow this protocol. Initial: ______
______________________________________________________________________________
Name
(PRINTED)
Signature
Date
Case One | Probability and Random Variables | Color Slot Machine
Here is our slot machine
– on each trial, it produces an 8-color sequence, using the table below:
|
Sequence* |
Probability |
|
RRBBRRYR |
.10 |
|
RRRGBRRB |
.11 |
|
BYGGYGBR |
.14 |
|
GRGYBRGG |
.10 |
|
YYYRYGYY |
.20 |
|
RYGRRBBY |
.15 |
|
YYYYBGRR |
.20 |
|
Total |
1.00 |
Define ThingYC as
the number of times that yellow shows on a sequence. Compute the values and
probabilities for ThingYC.
|
Sequence* |
Probability |
ThingYC(Yellow Count) |
|
RRBBRRYR |
.10 |
1 |
|
RRRGBRRB |
.11 |
0 |
|
BYGGYGBR |
.14 |
2 |
|
GRGYBRGG |
.10 |
1 |
|
YYYRYGYY |
.20 |
6 |
|
RYGRRBBY |
.15 |
2 |
|
YYYYBGRR |
.20 |
4 |
|
Total |
1.00 |
|
Pr{ ThingYC
= 0 } = Pr{ RRRGBRRB } = 0.11
Pr{ ThingYC
= 1 } = Pr{ RRBBRRYR } + Pr{ GRGYBRGG } = 0.10 + 0.10 = 0.20
Pr{ ThingYC
= 2 } = Pr{ BYGGYGBR } + Pr{ RYGRRBBY } = 0.14 + 0.15 = 0.29
Pr{ ThingYC
= 4 } = Pr{ YYYYBGRR } = 0.20
Pr{ ThingYC
= 6 } = Pr{ YYYRYGYY } = 0.20
Define RVGR as 1 if “GR” shows on a sequence, and as 0 if “GR” does not show on a sequence. Compute
the values and probabilities for RVGR.
Show all work in full detail for full credit.
|
Sequence* |
Probability |
RVGR(“GR” Shows) |
|
RRBBRRYR |
.10 |
0/No |
|
RRRGBRRB |
.11 |
0/No |
|
BYGGYGBR |
.14 |
0/No |
|
GRGYBRGG |
.10 |
1/Yes |
|
YYYRYGYY |
.20 |
0/No |
|
RYGRRBBY |
.15 |
1/Yes |
|
YYYYBGRR |
.20 |
1/Yes |
|
Total |
1.00 |
|
Pr{ RVGR = 1 (“GR” Shows) } =
Pr{ One of GRGYBRGG, RYGRRBBY
or YYYYBGRR Shows } =
Pr{ GRGYBRGG } + Pr{ RYGRRBBY
} + Pr{ YYYYBGRR } = 0.10 + 0.15 + 0.20 = 0.45
Pr{ RVGR = 0 (“GR” Does Not Show) } =
1 – Pr{ RVGR = 1 (“GR” Shows) } = 1 – 0.45 = 0.55
or do it the long way …
Pr{ RVGR = 0 (“GR” Does Not Show) } =
Pr{ One of RRBBRRYR, RRRGBRRB,
BYGGYGBR or YYYRYGYY Shows } =
Pr{ RRBBRRYR } + Pr{ RRRGBRRB
} + Pr{ BYGGYGBR } + Pr{ YYYRYGYY } =
0.10 + 0.11 + 0.14 + 0.20
= 0.35 + 0.20 = 0.55
Case Two | Summary Interval | 2009
Fictitious City Iron Man
An Iron Man
event comprises a 2.4 mile swim, 112 mile
bike course and a full marathon (26.2 mile run). Suppose that we have a random sample of finishers of the
2009 Fictitious City Iron Man, whose finishing times (in hours) are listed
below:
8.25, 8.56, 8.75, 9.10, 9.12, 9.20, 9.22, 9.25,
9.45, 9.85, 10.15, 10.18, 10.20, 10.25, 10.30, 10.46, 10.56, 10.80, 11.05 11.23,
11.30, 11.52, 11.56, 11.60, 11.75, 11.80, 11.95, 12.05, 12.15, 12.18, 12.20, 12.25,
12.28, 12.30, 12.35, 12.45 12.50 12.59, 12.60, 12.70, 12.73, 12.85, 12.90,
13.10, 13.25, 14.15, 15.25, 16.75, 17.30, 18.10
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 your work, and discuss completely for full
credit.
lower2 = m – (2*sd) ≈ 11.6878
– (2*2.10451)
≈ 7.47878
upper2 = m + (2*sd) ≈ 11.6878 + (2*2.10451) ≈
15.8968
lower3 = m – (3*sd) ≈ 11.6878
– (3*2.10451)
≈ 5.37427
upper3 = m + (3*sd) ≈ 11.6878 + (3*2.10451) ≈
18.0013
There are 50 people who
finished the 2009 Fictitious City Iron Man in our sample. At least 75% of the
people in our sample finished the 2009 Fictitious City Iron Man in between 7.48
and 15.90 hours. At least 89% of the people in our sample finished the 2009
Fictitious City Iron Man in between 5.37 and 18.00 hours.
If the finish times for
the 2009 Fictitious City Iron Man cluster symmetrically around a central value,
becoming rare as distance from the center increases, then: Approximately 95% of
the people in our sample finished the 2009 Fictitious City Iron Man in between
7.48 and 15.90 hours and approximately 100% of the people in our sample
finished the 2009 Fictitious City Iron Man in between 5.37 and 18.00 hours.
Case Three | Confidence Interval, Population Mean | 2009 Fictitious City Iron
Man
Using the context and data of
Case Two, consider the population mean finish time for the 2009
Fictitious City Iron Man.
Compute and interpret a 95% confidence interval for this population mean. Show
your work. Completely discuss and interpret your results, as indicated in class
and case study summaries.
lower95 = m – ( Z*sd / sqrt(n) ) ≈ 11.6878 – (2*2.10451/sqrt(50)) ≈ 11.0926
upper95 = m + ( Z*sd / sqrt(n) ) ≈ ≈ 11.6878 – (2*2.10451/sqrt(50))
≈ 12.2830
Our population is the
population of people who finished the 2009 Fictitious City Iron Man, and our
population mean is the population mean finish time.
Each member of the Family
of Samples is a single random sample of 50 people who finished the 2009
Fictitious City Iron Man.
From
each member sample, compute: sample mean finish time m,
sample standard deviation sd, then from the means
table row 2.00 0.02275 0.95450, use Z = 2. Then compute
the interval
[lower95
= m – ( Z*sd /
sqrt(n) ), upper95 = m + ( Z*sd / sqrt(n) )].
Doing this for each member of the family of
samples yields a family of intervals.
Approximately 95% of the intervals in the family
cover the unknown population mean. If our interval resides in this 95%
supermajority of member intervals, then the true population
mean finish time for the 2009
Fictitious City Iron Man is between 11.1 and 12.3 hours.
Case Four | Median Test | Pick’s Disease and Survival Time
Pick's disease (Frontotemporal
Dementia) is a relatively rare, degenerative
brain illness that causes dementia. The first description of the disease was
published in 1892 by Arnold Pick. Pick's disease is marked by "Pick
bodies", rounded, microscopic structures found within affected cells.
Neurons swell, taking on a "ballooned" appearance. Pick's disease is
usually sharply confined to the front parts of the brain, particularly the
frontal and anterior temporal lobes. The first symptoms of Pick's disease are
often personality change, and a decline in function at work and home.
Eventually, they enter a terminal vegetative state. Suppose
that we identify a random sample of deceased cases of Pick’s Disease – time
from initial diagnosis to death is given in years:
0, 0, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7,
8, 8, 8, 8, 11, 12, 13, 15.
Test
the following: null (H0): The median time to death from diagnosis is
9 years (h = 9) against
the alternative (H1): h
< 9. Show
your work. Completely discuss and interpret your results, as indicated in class
and case study summaries.
0, 0, 1, 1, 1 | 2, 2, 3, 3, 4 | 4, 5, 5, 6, 6 | 6,
7, 7, 7, 7 | 7, 7, 8, 8, 8 | 8, 11, 12, 13, 15.
Alternative Hypothesis:
“Guess is too Large”
Error: Count Below Guess
Error = #{sample
cases surviving strictly less than 9 years past diagnosis } = 26
n = 30
From 30 26 0.00003, p ≈ 0.00003 (0.003%)
Null Hypothesis: The median time to death from
diagnosis is 9 years
Alternative
Hypothesis: The median time to death from
diagnosis < 9 years (Guess is too Large)
Error Function: Number of Cases Surviving Strictly Less Than 9
Years
Our population consists of people with Pick’s disease. Our null
hypothesis is that the population median time to death from diagnosis is 9 years.
Each member of the family of samples (FoS) is a single random sample of 30 people who died
with Pick’s disease. The FoS
consists of all possible samples of this type.
From each member of the (FoS),
compute an error as the number of sample cases who survived strictly less than
9 years after diagnosis with Pick’s disease. Computing this error for each
member of the FoS forms a
family of errors (FoE).
If the true population median time to death after diagnosis with
Pick’s disease is 9 years, then approximately 0.003% member samples from the FoS yield errors as bad as or worse than our error. The
sample presents highly significant evidence against the null hypothesis.
Table 1: Means and Proportions
|
Z(k) PROBRT ROBCENT 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 0.08850
0.82298 1.40 0.08075
0.83849 1.45 0.07352
0.85294 1.50 0.06680
0.86639 1.55 0.06057
0.87886 1.60 0.05479
0.89040 1.65 0.04947
0.90106 1.70 0.04456
0.91087 1.75 0.04005
0.91988 1.80 0.03593
0.92814 1.85 0.03215
0.93569 1.90 0.02871
0.94257 1.95 0.02558
0.94882 2.00 0.02275
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 |
Table 2. Medians
|
n error base p-value 30 1 1.00000 30 2 1.00000 30 3 1.00000 30 4 1.00000 30 5 0.99997 30 6 0.99984 30 7 0.99928 30 8 0.99739 30 9 0.99194 30 10 0.97861 |
n error base p-value 30 11 0.95063 30 12 0.89976 30 13 0.81920 30 14 0.70767 30 15 0.57223 30 16 0.42777 30 17 0.29233 30 18 0.18080 30 19 0.10024 30 20 0.04937 |
n error base p-value 30 21 0.02139 30 22 0.00806 30 23 0.00261 30 24 0.00072 30 25 0.00016 30 26 0.00003 30 27 0.00000 30 28 0.00000 30 29 0.00000 30 30
0.00000 |