Key | The
Comprehensive Final Examination | Math 1107 | Spring 2011 | 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
youre done: Print your name on a blank sheet of paper. Place your toolsheets,
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.
Initial:___
_____________________________________________________________________________________
Name
(PRINTED)
Signature
Date
Case One | Color Slot Machine |
Probability Computation Rules
Here is our slot machine on each trial,
it produces a 6-color sequence, using the table below:
Sequence* |
Probability |
BBBBBB |
.10 |
BBGGBG |
.10 |
RRYYGG |
.15 |
GBBGGY |
.10 |
RGBYYR |
.25 |
BBYYGR |
.10 |
YBGRYY |
.20 |
Total |
1.00 |
* B-Blue, G-Green, R-Red,
Y-Yellow, Sequence is numbered as 1st to 6th
, from left to right: (1st 2nd 3rd 4th
5th6th )
Compute the following probabilities
if a rule is specified, you must use that rule:
Pr{ Blue Shows | Green Shows } This
is a conditional probability.
Sequence* |
Probability |
BBGGBG |
.10 |
RRYYGG |
.15 |
GBBGGY |
.10 |
RGBYYR |
.25 |
BBYYGR |
.10 |
YBGRYY |
.20 |
Total |
0.90 |
Pr{ Green Shows } =
Pr{One of BBGGBG, RRYYGG, GBBGGY, RGBYYR, BBYYGR
or YBGRYY Shows} =
Pr{BBGGBG}+Pr{RRYYGG}+Pr{GBBGGY}+Pr{RGBYYR}+Pr{BBYYGR}+Pr{YBGRYY} =
.10 + .15 + .10 + .25 + .10 + .20 = 0.90
Sequence* |
Probability |
BBGGBG |
.10 |
GBBGGY |
.10 |
RGBYYR |
.25 |
BBYYGR |
.10 |
YBGRYY |
.20 |
Total |
0.75 |
Pr{ Blue and Green Show } =
Pr{One of BBGGBG, GBBGGY, RGBYYR, BBYYGR
or YBGRYY Shows} =
Pr{BBGGBG}+ Pr{GBBGGY}+Pr{RGBYYR}+Pr{BBYYGR}+Pr{YBGRYY} =
.10 + .10 + .25 + .10 + .20 = 0.75
Pr{ Blue Shows | Green Shows } = Pr{ Blue and Green Show }/ Pr{ Green
Shows } = .75/.90
Pr{ Red and Blue Show } Use the
additive rule.
Sequence* |
Probability |
RGBYYR |
.25 |
BBYYGR |
.10 |
YBGRYY |
.20 |
Total |
.55 |
Pr{ Red and Blue Show } =
Pr{One of RGBYYR, BBYYGR or YBGRYY Shows} =
Pr{RGBYYR}+Pr{BBYYGR}+Pr{YBGRYY} =
.25 + .10 + .20 = 0.55
Pr{ Yellow Shows } Use the
complementary rule.
Other Event = Yellow
Does Not Show
Sequence* |
Probability |
BBBBBB |
.10 |
BBGGBG |
.10 |
Total |
.20 |
Pr{Yellow does not Show} = Pr{ BBBBBB } + Pr{ BBGGBG } = .10 + .10 =
0.20
Pr{Yellow Shows} = 1 Pr{Yellow does not Show} = 1 0.20 = 0.80
Optional Check:
Sequence* |
Probability |
RRYYGG |
.15 |
GBBGGY |
.10 |
RGBYYR |
.25 |
BBYYGR |
.10 |
YBGRYY |
.20 |
Total |
.80 |
Pr{Yellow Shows} = Pr{One of RRYYGG, GBBGGY, RGBYYR, BBYYGR or YBGRYY
Shows} = Pr{RRYYGG} + Pr{GBBGGY} + Pr{RGBYYR} + Pr{BBYYGR} + Pr{YBGRYY } =
.15 + .10 + .25+ .10 + .20 = .80
Show full work and detail for full
credit.
Case Two | Descriptive Statistics | C-reactive
Protein
Inflammation is the first response of the
immune system to infection or irritation. Inflammation is characterized by
redness, heat, swelling, pain and dysfunction of the organs involved. The C-reactive protein (CRP) owes its
name to the ability of this protein to precipitate pneumococcal
C-polysaccharide in the presence of calcium. It was first discovered in 1930 by
Tillet and Frances. CRP is a direct measure of the level of inflammation in the
body. CRP Elevations in CRP are seen
in infectious, inflammatory, and malignant diseases but also with pregnancy,
and trauma. Consider a random sample of
adult US residents. From each sample subject we obtain serum, and determine
the serum CRP level as milligrams CRP per deciliter serum (mg/dL). Here is the
CRP data:
0.1, 0.15, 0.20, 0.3, 0.35, 0.40,
0.47, 0.49, 0.52, 0.57, 0.63, 0.65, 0.70, 0.72, 0.73, 0.75, 0.77, 0.80, 0.83,
0.90, 0.92, 0.93, 0.94, 0.95, 0.98, 1.03, 1.10, 1.15, 1.20, 1.25, 2.5, 3.4,
5.6, 5.4, 5.5, 5.8, 6.0, 7.5, 8.9, 9.5, 11.1, 11.5, 12, 13, 15, 20, 25, 27, 30,
35
Compute and interpret the following statistics:
sample size, p00, p25, p50, p75, p100, (p100-p50), (p75-p25) and (p50-p00). Show your work, and discuss completely for
full credit.
Numbers
n p00
p25 p50 p75
p100 R42 R31
R20
50 0.1
0.7 1.005 7.5
35 33.995 6.8
0.905
R42 = p100 p50 = 35 1.005 =
33.995
R31 = p75 p25 = 7.5 0.7 =
6.8
R20 = p50 p00 = 1.005 0.1 = 0.905
Interpretation
There are 50 US Resident
Adults in our sample.
The adult in our sample with
the lowest CRP level has CRP at 0.1 mg/dL.
Approximately 25% of the
adults in our sample have CRP levels at 0.7 mg/dL or less.
Approximately 50% of the
adults in our sample have CRP levels at 1.005 mg/dL or less.
Approximately 75% of the
adults in our sample have CRP levels at 7.5 mg/dL or less.
The adult in our sample with
the highest CRP level has CRP at 35 mg/dL.
Approximately 50% of the adults in our sample have CRP levels
between 1.005 and 35 mg/dL. The largest difference in CRP between any pair of
adults in this upper half sample is 33.995 mg/dL.
Approximately 50% of the adults in our sample have CRP levels
between 0.7 and 7.5 mg/dL. The largest difference in CRP between any pair of
adults in this middle half sample is 6.8 mg/dL.
Approximately 50% of the adults in our sample have CRP levels
between 0.1 and 1.005 mg/dL. The largest difference in CRP between any pair of
adults in this lower half sample is 0.905 mg/dL.
Case
Three | Confidence Interval Proportion | C-reactive Protein
Using the context and
data from case two, consider the proportion of adult US residents present CRP
levels at 10 or higher with 95% confidence interval for this population
proportion. Show your work. Completely discuss and interpret your test
results, as indicated in class and case study summaries.
Event = CRP at 10 mg/dL
or more
0.1, 0.15, 0.20, 0.3,
0.35, 0.40, 0.47, 0.49, 0.52, 0.57, 0.63, 0.65, 0.70, 0.72, 0.73, 0.75, 0.77,
0.80, 0.83, 0.90, 0.92, 0.93, 0.94, 0.95, 0.98, 1.03, 1.10, 1.15, 1.20, 1.25,
2.5, 3.4, 5.6, 5.4, 5.5, 5.8, 6.0, 7.5, 8.9, 9.5, 11.1, 11.5, 12, 13, 15 | 20, 25, 27, 30, 35
Numbers
n = 50
e = sample event count =
10
p = sample event
proportion = e/n = 10/50 = .20
1 p = 1 .20 = .80
standard error proportion = sdp =
sqrt(p*(1 p)/n) = sqrt(.2*.8/50) ≈ sqrt(0.0032) ≈ 0.056569
From 2.00 0.022750 0.95450, Z≈2.00
lower95 = p (Z*sdp)
≈ .20 (2*0.056569) ≈ 0.086862915
upper95 = p + (Z*sdp)
≈ .20 + (2*0.056569) ≈ 0.313137085
Interpretation
We estimate the population
proportion of US resident adults whose serum C Reactive Protein (CRP) level is
at 10 mg/dL or greater.
Each member of the family
of samples is a single random sample of 50 adults. The family of samples
consists of possible samples of this type.
From each member sample,
compute:
e = sample event count
(serum CRP is 10 mg/dL or greater), p = sample event proportion = e/n, standard error
proportion = sdp = sqrt(p*(1 p)/n), and the interval as
[lower95 = p (Z*sdp), upper95 = p + (Z*sdp)]. Doing this for each member of
the family of samples yields a family of intervals, approximately 95% of which
capture the population proportion of US resident adults whose serum C Reactive
Protein (CRP) level is at 10 mg/dL or greater. If our interval is in this 95%
supermajority, then between 8.68% and 31.31% of US Resident adults have serum
CRP levels of 10 mg/dL or greater.
Case Four | Median Test | Green Lynx Spiders
The green lynx spider, Peucetia
viridans, is a conspicuous, large, bright green spider found on many
kinds of shrub-like plants throughout the southern United States and is the
largest North American lynx spider. Although it is common throughout Florida
and aggressively attacks its insect prey, it very seldom bites humans. Lynx spiders get their name from the way
that they sometimes pounce on their prey in a catlike fashion. |
A random sample of green lynx spiders (male) yields the following
body lengths (excluding legs), in millimeters per spider:
4.2, 4.7, 5.5, 5.7, 7.8 |
8.5, 9.8, 10.2, 10.9, 11.2 | 12.0, 12.1, 12.2, 12.2, 12.3 | 12.5, 12.5, 12.6,
12.6, 12.7
12.8, 12.9, 13.1, 13.2, 13.5
| 13.6, 14.2, 15.8, 16.4, 20.2
Test the following:
null (H0): The median body length for male Green Lynx spiders is 14
mm (h = 14) against the alternative (H1): h ≠ 14. Completely discuss and interpret
your test results, as indicated in class and case study summaries. Show
all work and detail for full credit.
4.2, 4.7, 5.5, 5.7, 7.8 | 8.5,
9.8, 10.2, 10.9, 11.2 | 12.0, 12.1, 12.2, 12.2, 12.3 | 12.5, 12.5, 12.6, 12.6,
12.7
12.8, 12.9, 13.1, 13.2, 13.5 |
13.6, 14.2, 15.8, 16.4, 20.2
n = 30
Sample Count Below 14: 26
Sample Count Above 14: 4
Sample Error for Two-Sided Test = Max(26, 4) =
26
From the row 30 26 0.00003, base p-value = 0.00003
p-value for two-sided test = 2*0.00003 =
0.00006
Interpretation
Our test concerns the median Green Lynx body length (male).
Each member of the family of samples is a single random sample of
30 Green Lynx spiders (male). The family of samples consists of all possible
samples of this type.
From each member of the family of samples, compute the sample error
as the maximum of the number of spiders in the sample whose body lengths are strictly
less than 14 millimeters, and the number of spiders in the sample whose spider
legths are strictly larger than 14 millimeters. Computing this error for each
member sample yields a family of errors.
If the true population Green Lynx spider body length (male) is 14
millimeters, then approximately 0.0006% of the samples yield errors equal to or
more extreme than our sample. The sample presents highly significant evidence
against the null hypothesis.
Table: 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
0.146860 0.70628 1.10
0.135670 0.72867 1.15 0.125070 0.74986 1.20 0.115070 0.76986 1.25 0.105650 0.78870 1.30 0.096800 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 0.0034670 0.99307 2.75 0.0029798 0.99404 2.80 0.0025551 0.99489 2.85 0.0021860 0.99563 2.90 0.0018658 0.99627 2.95 0.0015889 0.99682 3.00 0.0013499 0.99730 |
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n error base p-value 25 1 1.00000 25 2 1.00000 25 3 0.99999 25 4 0.99992 25 5 0.99954 25 6 0.99796 25 7 0.99268 25 8 0.97836 25 9 0.94612 25 10 0.88524 25 11 0.78782 25 12 0.65498 25 13 0.50000 25 14 0.34502 25 15 0.21218 25 16 0.11476 25 17 0.05388 25 18 0.02164 |
Table: Medians n error base p-value 25 19 0.00732 25 20 0.00204 25 21 0.00046 25 22 0.00008 25 23 0.00001 25 24
0.00000 25 25
0.00000 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 30 11 0.95063 |
n error base p-value 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 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
<0.00001 30 28
<0.00001 30 29
<0.00001 30 30
<0.00001 |
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