The Comprehensive Final Examination | Math 1107 | Fall
Semester 2009 | 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 toolsheet,
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:
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follow this protocol. Initial:___
______________________________________________________________________________________
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(PRINTED)
Signature Date
Case One – Hypothesis Test: Population
Median | Glioblastoma Multiforme
Glioblastoma multiforme
(GBM) is the highest grade glioma
tumor and is the most malignant form of astrocytomas.
These tumors originate in the brain. GBM tumors grow rapidly, invade nearby
tissue and contain cells that are very malignant. GBM are among the most common
and devastating primary brain tumors in adults. Suppose that we have a random
sample of GBM patients, with survival time (in months) listed below:
0, 1, 2, 2, 3 | 3, 3, 4, 4, 4| 4, 5, 5, 5, 5| 5, 5, 6,
6, 6 | 7, 7, 8, 8, 8| 9, 10, 10, 10, 11| 12, 13, 13, 13, 14
14, 14, 15, 16, 16 | 17, 17, 18, 18, 19 | 19, 20, 23,
23, 24 | 27, 28, 30, 30, 31 | 36, 38, 40, 58, 60
Test the following: null (H0):
The median survival time (in months after diagnosis) for patients who have died
with glioblastoma multiforme is 25 (h = 25) against the alternative (H1):
h < 25. Show your work.
Completely discuss and interpret your test results, as indicated in class and
case study summaries. Fully discuss the testing
procedure and results.
Null Hypothesis: Median Survival Time = 25 Months
Alternative Hypothesis: Median Survival Time < 25 Months (Guess is too Large)
Error Function: Number of Sample Patients Surviving Strictly Less Than 25 Months
0, 1, 2, 2, 3 | 3, 3, 4, 4, 4| 4, 5, 5, 5,
5| 5, 5, 6, 6, 6 | 7, 7, 8, 8, 8| 9, 10, 10, 10, 11| 12, 13, 13, 13, 14
14, 14, 15, 16, 16 | 17, 17, 18, 18, 19 | 19, 20, 23, 23, 24 | 27, 28, 30, 30, 31 | 36, 38, 40, 58, 60
sample error = Number of Sample Patients Surviving Strictly Less Than 25 Months = 50
n = sample size = 60
from 60 50 <0.00001, p < 0.00001 < .01 (p is strictly less than 1/100,000)
Our population consists of patients who have been diagnosed with and who have died with glioblastoma multiforme (GBM). Our null hypothesis is that the population median survival time for this population is 25 months.
Each member of the family of samples (FoS) is a single random sample of 60 patients who have been diagnosed with and who have died with glioblastoma multiforme (GBM). The FoS consists of all possible samples of this type.
From each member of the (FoS), compute an error as the number of sample GBM patients surviving strictly less than 25 months. Computing this error for each member of the FoS forms a family of errors (FoE).
If the true population median survival time for GBM patients is 25 months, then fewer than .001% of member samples from the FoS yield errors as bas as or worse than our error. The sample presents highly significant evidence against the null hypothesis.
Case Two | Confidence
Interval: Population Proportion |
The objective of the
Framingham Heart Study was to identify the common factors or characteristics
that contribute to Cardiovascular disease (CVD) by following its development
over a long period of time (since 1948) in
a large group of participants who had not yet developed overt symptoms of CVD
or suffered a heart attack or stroke. Blood pressure is a measurement of the
force applied to the walls of the arteries as the heart pumps blood through the
body. Blood pressure readings are measured in millimeters of mercury (mm Hg)
and usually given as two numbers: the systolic blood pressure (SBP) reading,
representing the maximum pressure exerted when the heart contracts and the diastolic blood pressure (DBP) reading, representing the
pressure in the arteries when the heart is at rest. Consider the systolic to
diastolic blood pressure ratio R = SBP/DBP. A sample of FHS adult subjects yields the following ratios:
1.86 1.71 1.43 1.82 1.62 2.57 1.53 1.42 1.55 1.75 1.95 1.58
2.33 1.69 1.56 1.85 1.51 1.50 1.72 1.50 2.06 1.36 1.78 1.56
1.62 1.75 1.78 1.52 1.83 1.60 1.54 1.80 2.13 1.67 1.86 1.60
1.48 1.47 1.45 2.44 2.50 2.05 1.50 1.79 1.82 1.59 1.74 1.86
1.64 1.54 2.03 1.91 1.92 1.97 1.54 1.67 2.00 1.63 1.82 1.49
Consider the proportion of
2.57, 2.33, 2.06, 2.13, 2.44 | 2.50, 2.05, 2.03, 2.00
sample size n=60
event = “diastolic
blood pressure ratios of 2 or greater (R
³ 2)”
sample event
count = e = 9
p = e/n = 9/60 = .15
1 – p
= 1 – .15 = .85
sdp = sqrt( (p*(1 – p) ) = sqrt(
(.15*.85) ) = 0.046098
from 2.10 0.017864 0.96427, z=2.10
lower96 = p – (z*sdp) = .15 –
(2.1*0.046098) = 0.053195
upper96 = p + (z*sdp) = .15 +
(2.1*0.046098) = 0.24681
Report the interval as [0.053,0.246].
Interpretation
Our population is the population of
Framingham Heart Study subjects and our population proportion is the proportion
of Framingham Heart Study subjects with systolic to
diastolic blood pressure ratios of 2 or
greater (R ³ 2).
Our Family of Samples (FoS) consists of every possible random sample of 60
Framingham Heart Study subjects.
From each member
sample of the FoS, we compute the sample proportion p
of Framingham Heart Study subjects with systolic to diastolic blood
pressure ratios of 2 or greater (R ³ 2), sdp = sqrt( (p*(1 – p)/60 ) and then compute the interval
[p – (2.1*sdp),
p + (2.1*sdp)].
Computing this interval for each member
sample of the FoS, we obtain a Family of Intervals (FoI), approximately 96% of which cover the true population proportion
of Framingham Heart Study subjects with systolic to
diastolic blood pressure ratios of 2 or
greater (R ³ 2).
If our interval is among the approximate
96% super-majority of intervals that cover the population proportion, then
between 5.3% and 24.6% of Framingham
Heart Study subjects
with systolic to diastolic blood pressure ratios of 2 or greater (R ³ 2).
Case Three – Probability
Rules | Pair of Color Slot Machines
Consider a pair of color slot machines, described
by the tables below. Assume that the probabilities are correct, and that the
machines operate in a mutually independent fashion. Our experiment consists of
observing pairs of sequences from the machines.
Machine
1 |
Machine
2 |
||
Sequence* |
Probability |
Sequence* |
Probability |
BRRYR |
.10 |
BGGRY |
.10 |
GGYBR |
.20 |
YBBYG |
.25 |
GYRYG |
.25 |
GGGYY |
.30 |
BYYBG |
.45 |
RRBBB |
.35 |
Total |
1 |
Total |
1 |
*B-Blue, G-Green, R-Red, Y-Yellow, sequences are numbered from
left to right.
Compute Pr
Machine
1 |
Machine
2 |
||
Sequence* |
Probability |
Sequence* |
Probability |
BRRYR |
.10 |
BGGRY |
.10 |
GGYBR |
.20 |
YBBYG |
.25 |
GYRYG |
.25 |
GGGYY |
.30 |
BYYBG |
.45 |
RRBBB |
.35 |
Total |
1 |
Total |
1 |
*B-Blue, G-Green, R-Red, Y-Yellow, sequences are numbered from left
to right.
Identify the qualifying pairs: Both Blue and Red Show in the Pair: (BRRYR
, BGGRY ),( BRRYR, YBBYG ),( BRRYR , RRBBB ),
(GGYBR , BGGRY ),( GGYBR , YBBYG ),( GGYBR ,
RRBBB ), (GYRYG , BGGRY ),( GYRYG , YBBYG ),(
GYRYG , RRBBB ), (BYYBG , BGGRY ),( BYYBG ,
RRBBB ).
Pr
Pr
Pr
Pr
Pr
(.10*.10) +
(.10*.25) + (.10*.35) + (.20*.10) + (.20*.25) + (.20*.35) + (.25*.10) +
(.25*.25) + (.25*.35) +
(.45*.10) +
(.45*.35) = (.1*.70)
+ (.2*1.4) + (.45*.45) = .07 + .028 + .2025 = .3005
Compute Pr
Machine
1 |
Machine
2 |
||
Sequence* |
Probability |
Sequence* |
Probability |
BRRYR |
.10 |
BGGRY |
.10 |
GGYBR |
.20 |
YBBYG |
.25 |
GYRYG |
.25 |
GGGYY |
.30 |
BYYBG |
.45 |
RRBBB |
.35 |
Total |
1 |
Total |
1 |
*B-Blue, G-Green, R-Red, Y-Yellow, sequences are numbered from left
to right.
Other Event = “Red Does Not Show in the
Pair” = “Red Shows in Neither Slot”
Qualifying Pairs: (BYYBG , YBBYG), ( BYYBG, GGGYY)
Pr
Pr
So then
Pr
Case Four – Descriptive
Statistics | Fictitious
Spotted Toad
The Fictitious Spotted Toad is a
native species of
9, 10, 10, 10, 11 | 12, 12, 12, 12, 13 |
15, 16, 17, 18, 18 | 18, 18, 19, 19, 20 | 20, 21, 22, 22, 23
24, 24, 25, 25, 26 | 29, 29, 29, 31, 31|
33, 33, 33, 33, 33 | 34, 39, 40, 40, 40 | 43, 43, 45, 47, 47
50, 56, 56, 56, 60 | 63, 65, 66, 67, 98
Compute and interpret the following
statistics: sample size, p00, p25, p50, p75, p100, (p100
– p50), (p75 – p25), (p50 –
p00).
n=60
min = p00 = 9
p25 = 18
p50 = 27.5
p75 = 41.5
max = p100 = 98
range42 = (p100 – p50) = 98 – 27.5 = 70.5
range31 = (p75 – p25) = 41.5 – 18 = 23.5
range20 = (p50 – p00) = 27.5 – 9 = 18.5
There are 60 Fictitious Spotted Toads in
the sample.
The toad in the sample with the fewest
spots has 9 spots.
Approximately 25% of the toads in the
sample have 18 or fewer spots.
Approximately 50% of the toads in the
sample have 27.5 or fewer spots.
Approximately 75% of the toads in the
sample have 41.5 or fewer spots.
The toad in the sample with the most spots
has 98 spots.
Approximately 50% of the toads in the
sample have between 9 and 27.5 spots. The largest possible difference in spot
count between any pair of toads in the lower half sample is 18.5 spots.
Approximately 50% of the toads in the
sample have between 18 and 41.5 spots. The largest possible difference in spot
count between any pair of toads in the middle half sample is 23.5 spots.
Approximately 50% of the toads in the
sample have between 27.5 and 98 spots. The largest possible difference in spot
count between any pair of toads in the upper half sample is 70.5 spots.
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 |
Table 2. Medians
n error base p-value 60
0 1.00000 60 1
1.00000 60 2
1.00000
60 3
1.00000 60 4
1.00000 60 5
1.00000 60 6
1.00000 60 7
1.00000 60 8
1.00000 60 9
1.00000 60 10
1.00000 60 11
1.00000 60 12
1.00000 60 13
0.99999 60 14
0.99998 60 15
0.99993
60 16
0.99980 60 17
0.99947 60 18
0.99866 60 19
0.99689 60 20 0.99326 |
n error base p-value 60
21 0.98633 60 22
0.97405 60 23
0.95377
60 24
0.92250 60 25
0.87747 60 26
0.81685 60 27
0.74052 60 28
0.65056 60 29
0.55129
60 30 0.44871 60 31
0.34944 60 32
0.25948 60 33
0.18315 60 34
0.12253 60 35
0.07750 60 36
0.04623
60 37
0.02595 60 38
0.01367 60 39
0.00674 60 40
0.00311 |
n error base p-value 60
41 0.00134 60 42
0.00053 60 43
0.00020 60 44
0.00007
60 45
0.00002 60 46
0.00001 60 47
<0.00001 60 48
<0.00001 60 49 <0.00001 |