Key
The Comprehensive Final Examination
Math 1107
Spring 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
hourly.
Sign and
Acknowledge: I agree to follow this
protocol. Initial:___
______________________________________________________________________________________
Name
(PRINTED)
Signature Date
Case One |
Random Variables | Color Slot Machine
Here is our slot
machine – on each trial, it produces a color sequence, using the table below:
|
Sequence* |
Probability |
|
BBRRYRRR |
.10 |
|
RRRGBRRB |
.10 |
|
BBYYYGBR |
.15 |
|
GRRGGYGG |
.10 |
|
BGYGYRYG |
.25 |
|
YYGRRBBY |
.10 |
|
YYYYBGRR |
.20 |
|
Total |
1.00 |
*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered from left to right
Consider the random variable GC, defined as the number of times that
green shows in the color sequence. List the values of GC, and compute the
probabilities for those values.
|
Sequence* |
Probability |
Green
Count |
|
BBRRYRRR |
.10 |
0 |
|
RRRGBRRB |
.10 |
1 |
|
BBYYYGBR |
.15 |
1 |
|
GRRGGYGG |
.10 |
5 |
|
BGYGYRYG |
.25 |
3 |
|
YYGRRBBY |
.10 |
1 |
|
YYYYBGRR |
.20 |
1 |
|
Total |
1.00 |
|
Pr
Pr
Pr
Pr
Consider the random variable RG, defined as 1 if “RG” shows in the color
sequence and as 0 if “RG” does not show in the color sequence..
List the values of RG, and compute the probabilities for those values.
|
Sequence* |
Probability |
RG |
|
BBRRYRRR |
.10 |
0 |
|
RRRGBRRB |
.10 |
1 |
|
BBYYYGBR |
.15 |
0 |
|
GRRGGYGG |
.10 |
1 |
|
BGYGYRYG |
.25 |
0 |
|
YYGRRBBY |
.10 |
0 |
|
YYYYBGRR |
.20 |
0 |
|
Total |
1.00 |
|
Pr
Pr
.10 + .15 + .25 + .10 + .20 = .80
Pr
Consider the random variable B2, defined as 1 if blue shows exactly twice in the color
sequence and as 0 otherwise. List the values of B2, and compute the
probabilities for those values.
|
Sequence* |
Probability |
B2 |
|
BBRRYRRR |
.10 |
1 |
|
RRRGBRRB |
.10 |
1 |
|
BBYYYGBR |
.15 |
0 |
|
GRRGGYGG |
.10 |
0 |
|
BGYGYRYG |
.25 |
0 |
|
YYGRRBBY |
.10 |
1 |
|
YYYYBGRR |
.20 |
0 |
|
Total |
1.00 |
|
Pr
Pr
Pr
Pr
Show all work for full credit.
Case Two | Summary Intervals | Sea Weaselsä
Sea Weaselsä are instant
pets, distributed as a kit - Sea Weasel Eggsä , Sea Weasel Water
Conditionerä and Sea Weasel Foodä . A Sea
Weaselä Kit is started by placing the Sea Weasel Eggsä , Sea
Weasel Water Conditionerä and some water in a container. The eggs will then
hatch, producing Sea Weaselsä , and some of them will survive. A random sample of Sea
Weaselä Kits is selected, each selected kit is started, and
the number of surviving Sea Weaselsä is counted one week
after start. Here are the Sea Weaselä counts per kit
20 23 25 26 27 32 35 41 42 42
45 47 48 51 56 60 61 62 65 70 73 76 84 89 90 91 91 92 92
96 98 104 109 117 124 130 132 135 138 140 142 145 151 160 162 165 166 167 171
174 174 176 178 179 181 182 183 185 187 275
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.
Numbers
n m sd lower2 upper2
lower3 upper3
60 108.033
58.1273 -8.22119 224.288
-66.3485 282.415
lower2 = m –
(2*sd) = 108.033 –
(2*58.1273) » -8.22119 [0]
upper2 = m +
(2*sd) = 108.033 + (2*58.1273) » 224.288
lower3 = m – (3*sd) = 108.033 – (3*58.1273) » -66.3485 [0]
lower3 = m +
(3*sd) = 108.033 + (3*58.1273) » 282.415
Interpretation
At least 75% of
the Sea Weasel kits in the sample yield between 0 and 224 Sea Weasels.
At least 89% of
the Sea Weasel kits in the sample yield between 0 and 282 Sea Weasels.
If the Sea
Weasel kit yields cluster symmetrically around a central value, becoming more
rare as the kit yields fall farther out from the central value, then:
approximately
95% of the Sea Weasel kits in the sample yield between 0 and 224 Sea Weasels.
and
approximately 100% of the Sea Weasel kits in the sample yield between 0 and 282
Sea Weasels.
Case Three | Hypothesis Test,
Median | Sea Weaselsä
Using the data
and context from Case Two, test the following: null (H0): The median
Sea
Weaselä Kits yield is 100 weasels (h = 100) against the alternative (H1):
h < 100.
Show your work. Fully discuss the results. This discussion must include a clear discussion of
the population and the population median, the family of samples, the family of
errors and the interpretation of the p-value.
Numbers
20 23 25 26 27 32 35 41 42 42 45 47 48 51 56 60 61 62 65 70 73 76 84 89 90 91 91 92 92 96 98 104 109 117 124 130 132 135 138 140 142 145 151 160 162 165 166
167 171 174 174 176 178 179 181 182 183 185 187 275
Null: “Median Kit Yield = 100 Weasels”
Alternative: “Guess is too Large”
Error Form: Count Strictly Below 100
n=60
error = number of kits in the sample
yielding strictly fewer than 100 Sea Weasels = 31
From the row: 60 31 0.34944, p»0.34944
Interpretation
Each member of the Family of Samples (FoS) is a random sample of 60 Sea Weasel kits. The FoS consists of all possible samples of this type. From
each member sample of the FoS, compute the sample
error as the number of kits in the sample yielding strictly fewer than 100 Sea
Weasels. Computing
this error for each member of the FoS yields the Family of Errors. If the median Sea Weasel
kit yield is 100 Sea Weasels, then approximately 34.9% of the Family of Samples
yield errors as extreme as or more extreme than our single computed error. Our
sample does not seem to present significant evidence against the null
hypothesis.
Case Four | Confidence Interval: Population Mean | BarrelCorpÔ Survival Times
BarrelCorpÔ manufactures barrels and wishes to ensure the
strength and quality of its barrels. Chimpanzees traumatized the company owner
as a youth; so the company uses the following test (Angry_Barrel_of_Monkeys_Test)
of its barrels: Ten (10) chimpanzees are loaded into the barrel. The
chimpanzees are exposed to Angry!Monkey!Gas!Ô, an agent guaranteed to drive the chimpanzees to a
psychotic rage. The angry, raging, psychotic chimpanzees then destroy the
barrel from the inside in an angry, raging, psychotic fashion. The survival
time, in minutes, of the barrel is noted. A random sample of 50 BarrelCorpÔ barrels is evaluated using the Angry_Barrel_of_Monkeys_Test,
and the survival time in minutes of each barrel is noted. The survival time of
each barrel is listed below:
12, 12, 13, 14, 15, 16,
16, 17, 18, 18, 22, 23, 25, 26, 27, 29, 30, 32, 32, 33, 34, 35, 36, 37, 35, 35,
36, 38, 40, 40, 41, 42, 42, 42, 43, 43, 44, 45, 45, 47, 48, 48, 49, 50, 50, 54,
72, 77, 84, 86, 88, 93, 95, 97, 99, 118
Estimate the population mean BarrelCorpÔ barrel survival time under
the Angry Barrel of Monkeys Test with 95% confidence. That is,
compute and discuss a 95% confidence interval for this population mean. Show your work. Fully
discuss the results. This
discussion must include a clear discussion of the population and the population
mean, the family of samples, the family of intervals and the interpretation of
the interval.
Work all four (4)
cases.
Numbers
n m sd se
lower95 upper95
56 43.5357 25.3886
3.39269 36.7503 50.3211
From row: 2.00 0.022750 0.95450, Z»2
lower95 = m – (z*(sd/sqrt(56)) = 43.5357 – (2* 25.3886/sqrt(56)) » 36.7503
upper95 = m + (z*(sd/sqrt(56)) = 43.5357 + (2* 25.3886/sqrt(56)) » 50.3211
Report the interval as [36.8, 50.3].
Interpretation
Each member of
the Family of Samples(FoS) is a random sample of 56 BarrelCorp barrels – the FoS
consists of every possible sample of this type. From each member of the FoS, compute the sample mean survival time (in minutes) of
the barrels in the sample, the sample standard deviation, and then the interval
[lower95 = m – (z*(sd/sqrt(56)), upper95 = m + (z*(sd/sqrt(56))]. Computing an
interval in this way from each member of the FoS
yields a Family of Intervals(FoI), approximately 95%
of which cover the population mean survival time (in minutes) for BarrelCorps barrels. If our single computed interval is in
this 95% supermajority, then the population mean survival time for BarrelCorps barrels undergoing the AngryMonkey
test is between 36.8 and 50.3 minutes.
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 |