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Unformatted text preview: CEE 304  Uncertainty Analysis In Engineering
W 1996 Final Exam W Exam is open notes and openbook.
The exam lasts 150 minutes,3:00 pm  5:30 pm, and there are 150 points.
Some students will be taking the exam from 5 pm to 7:30 pm.
SHOW WORK! 1. (25 pts) Have you seen all of the Christmas lights in some large displays this
year? Bulbs seem to burn out at random times. Suppose that on a given evening
there are 100 lights decorating a home and that each light fails during the evening
with probability 0.002. When lights fail they are replaced the next day. (a) What is the probability that 2 or more lights fail on the same evening? (b) What is the mean and variance of the number of lights that fail in one
evening? (c) What is the mean and variance of the number of lights that fail
over a week (seven evenings)? Suppose that one considered a large building covered with lights and recorded
the actual times during the evening when bulbs failed. (d) Why might that system be well described by a Poisson process? (e) Suppose that when the lights are on, the arrival rate of failures is 0.2 per
hour. During a week (100 hours of operation), what is the mean and variance of
the number of failures that will occur? (f) If the height above the sidewalk of bulbs that fail is uniformly distributed
between zero and 300 feet, what is the probability that during a week there are no failures that occur above 275 feet?
(g) On average, how many hours pass before some bulb above 275 feet fails? 2. (10 pts) The maximum recorded windspeeds in each year of record can be
modeled by a lognormal distribution. Suppose that such maximum annual
windspeeds have a mean of 48 miles / hour with a standard deviation of 12
miles/ hour. If you wanted to design for the windspeed which is exceeded with
only a 0.5% (p = 0.005) probability, what should be your design value? 3. (15 pts) An engineering student got a summer job setting off fireworks on the 4th of July and other summer events. For a particular display, the bomb is shot
from the tube so that after t seconds it is at a height H = (100 m/sec) t—0.5 (10 m/sec2) t2 = 500—5(lO—t)2 m 0 s t s 20 sec CEE 304  Uncertainty Analysis In Engineering
*** 1996 Final Exam *** Unfortunately timers on bombs are not accurate. The student estimates that
timers will yield a time to detonation after launch that is normally distributed with u = 10 secs and o = 1.2 secs. a) What is the probability that the bomb explodes more than 495 meters
above the ground? ‘ '
b) What is the mean of the height at which the bomb will explode?
c) OUTLINE how to obtain the pdf of height H at which bomb explodes.
(You need NOT do the actual derivation—describe the steps.)
What would make the derivation difficult?
What might be unusual about pdf of H? 4. (5 pts) A charcoal filter can remove chlorinated hydrocarbons from drinking
water. An environmental engineer collected 15 independent samples of water
that passed through such a filter. The average of the 15 observed removal rates
was i = 99.37% with sample standard deviation 5 = 0.21%. Assuming that the
measurements are normally distributed about the true removal rate it, what is a
90% confidence interval for u? 5. (20 pts) For the filter described in problem 4, the manufacturer claims that the
product should remove at least 99% of the chlorinated hydrocarbons. Use the
fact that in 12 of the 15 samples the calculated removal rate exceeded 99% to
evaluate the manufacturer's claim. a) What are the appropriate hypotheses? b) What is the rejection region for a 5% test? c) What is the p—value for the observed sample? (1) If the probability that the measured removal rate exceeds 99% is 0.70,
what is the probability your test accepts the null hypothesis? e) What is the probability you accept the null hypothesis when it is true? 6. (10 pts) Suppose that a group of students decides to analyze the actual value of
sleeping versus cramming before exams. 12 students agree to participate in the
experiment. In one of their classes they agree to sleep the night before one
prelim, and before the other prelim they stay up all night cramming. The
normalized scores appear below for each student and each prelim. a) What are the appropriate hypotheses for a Wilcoxon signedrank test? b) For a = 5%, what is the rejection region for the Wilcoxon signedrank test?
c) What is the pvalue for this sample using that test? CEE 304  Uncertainty Analysis In Engineering
*“ 1996 Final Exam *** Student With sleep_ Cramming Difference
1 95 71 24
2 84 69 15
3 91 99 8
4 69 55 14
5 64 42 22
6 72 93 21
7 55 22 33
8 71.5 75 3.5
9 57 48 9
10 81.5 59 22.5
11 78 61.5 16.5
12 64.5 16.5 48
Average 73.5 59.3 14.3
St. Dev. 12.7 25.0 18.6 7. (10 pts) For data in problem 6, what pvalue do you obtain with a onesample
t test? If the true mean of the differences were 12 points and the true standard deviation were 15 points, what would the type 11 error be for a test with on = 5%? 8. (15 pts) I took a sample of the weights of 7 men and 5 women in the class.
View these 12 values as a random sample of men and women. a) Using the most powerful nonparametric method you can, set up a test of whether men weigh more than women. What is rejection region for a=5%?
b) What is the pvalue for these results with that nonparametric test? c) What is the pvalue for a pooled ttest?
d) Without looking at the results you obtained, which of the 2 tests would you
recommend to a client ? Why ? Men Women
Weight Weight
130 114
195 90
207 147
202 193
140 139
152
180
Average 172.29 136.60
St. Dev. 31.37 38.66 CEE 304  Uncertainty Analysis In Engineering
*“ 1996 Final Exam *“ 9. (30 pts) I took the first 9 students from the data the class provided on the
weight (pounds) and height (inches) of students in the class, which yielded: height = 67.0 inches sheight = 4.95 inches
weight = 145.6 lbs sweight = 36.6 lbs 2 (weighti  weight )(heighti  h—eight ) = 1148.6
a) What are the leastsquares estimates of a and B for the model weight = a + [3 (height) + (error) ? b) Compute an unbiased estimate of the variance of the model's residual error?
c) What is the standard deviation of the estimator of B? d) What is the rejection region for a 10% test of whether or not [3 =0?
Why have you selected either a onesided or twosided test?
What is the pvalue for this case? e) Using your fitted line, what is your estimate of the weight of a person who is
5 feet 4 inches tall? Given the data provided, what is a 95% confidence interval
for the weight of a randomly selected person who is 5 feet 4 inches tall? 10. (10 pts) Student comments encouraged me to try to formulate a problem that
addresses important ideas what I really hope students learned in the course. Here goes. (a) What fundamental problem does the collection of knowledge and
mathematical relationships called "probability theory” address?
What key concepts does probability theory employ to address those issues? (b) Why might statistics be considered the science of decision making (under
uncertainty)? What key concepts does statistics employ to address such issues? Best of success in the rest of your exams. Iery Stedinger CEE 304  UNCERTAIN'IY ANALYSIS IN ENGINEERING
Final Exam Solution ~ 1996 1a) Pr[ N 2 2 I = 1 — Pr[N=0]  Pr[N=1] = 1*(1p)100 + 100(1p)99p1 = 1 — 0.819 — 0.164: 0.0174 4 pts‘
b) Mean one evening = np = 100(0.002) = 0.2; Var = np(1p) = 0.2 3 pts
c) Mean 7 evenings = 7np = 100(0.002) = 1.4; Var = 7np(1p) = 1.4 3 pts The numbers each evening Ni are independent so variances add. d) Poisson processes requires events to occur one at a time, arrive with a
constant rate, and occur independently. All three of these conditions are met for
light bulbs burning out during the evening hours. It is not met if one considers
bulbs that fail due to a power surge, or when they are first turned on; in those cases multiple failures would occur at the same time. 5 pts e) Number of failures is Poisson, 7» = 0.2 per hour, t= 100 hours,
E[X] = Var[X] = kt = 20 4 pts
f) Uniform distribution implies that F(H) = H/ 300 ft for 0 s H S 300.
Probability that all failures are below 275 feet, when A = 0.2 per hr and t= 100 hr:
Pr[ H s 275 ft] = exp{ — kt[1 F(275) ] } = exp{ — 20 [25/300 ] } = 0.189. 4 pts
g) Consider a new Poisson process with N“ = 0.2 (25 / 300) per hr = 0.0167 per hr. The mean wait (is exponential with) mean 1/ 1* = 60 hours. 2 pts 2. Recall the relationship between the realspace and logspace moments of a
lognormal distribution. clogs = sqrt{ln[1+(orea1/ureal)2]} = sqrt{ln[1+(12/48)2]} = 0.246 (2 pts)
ulog = 1n[urea1] — 0.5 clogsz = 3.84. (2 pts)
ulog + 20995 Clogs = “log + 2.576 Clogs = 4.47 (4 pts) . >995 %tile windspeed is exp(4.445) = 87.6 (2 pts) 3a) Pr[ H 2 495] = Pr[ 500  5(10 — T)2 2 495] = Pr[ 5 2 5(10 —T)2] = Pr[12 I T  10 I] = Pr[ I T— 10 I /1.2 g 0.833] = 1 — 2 (Ix0.833) = 1  2*0203 = 60%. 5 pts b) E{ 500 — 5(10 — T)2 I = 500  5c2 = 500 — 5(1.2)2 = 492.8. Almost at top!. 5 pts c) The safe way is to first compute FH(h) = Pr[ 500 — 5(10 — T)2 S h ] = Pr[ (T10)2 2 (500h)/5 I = Pr[ I T10 I/1.2 2 Jioo—h/s /1.2] =Pr[ IZI 2 J1oo—h/5 /1.2] This can be expressed in terms of standard normal cdf CD, and by taking the
derivative of that expression the pdf for H is obtained. 3 pts {{{ The actual expression is: FH(h) = 2 (D[ — v 100—h/5 /1.2] for 0 s h s 500;
so fH(h) = 2¢[ —J100—h/5 /1.2 ]/[2*5*1.2(/100—h/5 ] ; ¢[z] = exp(zz/2)/ 75 II} CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Final Exam Solution ~ 1996 Problems occur because original function is not monotonic and thus simple
formulas do not apply: HS h results in two unconnected Tregions. 1 pt
Because of square root, density of H will goes to INFINITY at h=500 m. 1 pt 4. 90% confidence interval for the true p. given it = 99.37%, s = 0.21%, n = 15
is i i S/ [if = ' 1.15ng = 5 PIS 5. SIGN TEST on differences di = (xi  0.99)! If in 12 of the 15 samples value exceeded 99%. Then in 3 of 15 it did not. a) HO: p(+) = p() = 0.5; Ha: p(+) > p(—) because we wish to demonstrate that data proves claim. 4 pts
b) Count negative signs 5. Expect few. Reject S s 4 Yields a = 5.9%. 4 pts To be strictly correct should reject S s 3 => or = 1.8%. Accepted either answer.
[On can count positive signs, then rejecting if S... 2 11 corresponds to S s 4] c) pvalue for S = 3 is 1.8% = Pr[ S s 3 if p(+) = p() = 0.5 ]. 4 pts
d) If probability of minus sign is 0.30 = 1 — 0.70, then observe S s 4 about 52% of
time. Pr[accept null hypothesis when S 2 5] = 1 — 0.52 = 48%. 4 pts e) If use 5 s 4, Pr[accept null hypothesis when it is true] = 1 — 0.059 = 94.1%. 4 pts 6. a) PAIRED data => Use the differences (xi — yi)! Participants do not seem to know which will be more effective: sleeping or
craming. No preference is specified. Therefore use Ho: median = 0 (also that any + and  observations equally like to assume any rank); Ha: median ¢ 0. 3 pts
b) With n=12 and a twosided test, reject if W 2 64
orWs12(12+1)/264=7864 =14. 3pts Sum of all ranks is 78=12(12+1) / 2, so the complement of 64 is 14. OR because Wdistribution symmetric: 14 = Zuw  64 = 2*12(12+1) / 4  64. c) Observe sum of positive ranks is 68 (sum of negative ranks is R(3.5) + R(8) +
R(21) = 1 + 2 + 7 = 10; where 10+68 = 78); pvalue is 2*0.01 = 2%. 4 pts TEST OF MEDIAN = 0.000000 VERSUS MEDIAN N.E. 0.000000
N FOR WILCOXON ESTIMATED
N TEST STATISTIC PVALUE MEDIAN Diffs 12 12 68.0 0.025 15.37
(Minitab use normal approx. for pvalue.) 7. a) Onesample two—sided t test. t = M *14.3/ 18.6 = 2.66 3 pts
for t with doff = 11; pvalue = 2.2%  a little bigger than 2%. 3 pts b) Cl: 12/ 15 = 0.8, n=12, twosided test a=5% yields [3 ~ 0.40 4 pts CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Final Exam Solution ~ 1996 TEST OF MU = 0.00 VS MU N.E. 0.00
N MEAN STDEV SE MEAN T P VALUE
C3 12 14.29 18.59 5.37 2.66 0.022 8. a) Twosample WilcoxonMannWhitney test: Ha: medianm = medianw
Onesided test of whether men weight more: Ha: medianm > medianw
Summing the ranks of the women yields W = 43 yields a = 5.3% when smallest observations have largest ranks. If assign small values small ranks then need to find corresponding lowertail value:
Use W1 = 2”“,  wu = 2[5*(12+1)/2]  43 = 65  43 = 22. 4 pts b) If assign small values small ranks then w = 22 = 1+2+4+6+9.
For n = 5, m = 7, then w= 22 or 43 is 5.3% critical point: pvalue = 5.3%. 4 pts MINITAB> MannWhitney Confidence Test women N = 5 Median = 139.00
men N = 7 Median = 180.00
W = 22.0 Test of ETAl ETAZ vs. ETAl 1.1:. ETAZ is significant at 0.0522
Cannot reject at alpha = 0.05 (Minitab use normal approx. for pvalue.) c) Compute first the pooled st. dev sp = 34.5 with 10 degrees of freeedom. Pooled t value = (172.29136.60)/ [Sp A/ 1 / 5 +1 / 7 ]= —1.77,
pvalue for one—sided test with 10 doff is about 6% 4 pts MTB > TwoSample 95.0 'women' 'men';
SUBC> Alternative —1; Pooled.
TWOSAMPLE T FOR women VS men N MEAN STDEV SE MEAN
women 5 136.6 38.7 17
men 7 172.3 31.4 12
TTEST MU women = MU men (VS LT): = l.77 P=0.054 DF= lO
POOLED STDEV = 34.5 d) There is no assurrance or solid evidence that data is normal; the sample sizes
are small so that assumption is important for computation of critical value of t
statistic. Thus assumptions of the t analysis are unsupported. Recommend
Wilcoxon analysis that makes no such unsupported assumptions and provides a
powerful and honest test of the hypotheses! 3 pts CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Final Exam Solution ~ 1996 9. REGRESSION: Minitab gives
The regression equation is C4 =  247 + 5.86 C3 Predictor Coef Stdev t—ratio p Constant 247.0 114.3 2.16 0.067 C3 5.860 1.702 3.44 0.011 s = 23.83
a) a = 247.0; b = 5.860 3+3 pts
b) 5,.2 = (23.83)2 4 pts
c) StDev(b) = 1.702 4 pts
d) Expect weight to increase with height
> HO: [3 = 0 reject if T > t0.10,7 = 1.415 4 pts For this data obtain, t = 3.44 so for df = 7, obtain pvalue = 0.05 (right on). 4 pts
e) For 5 feet 4 inches have height of 64 inches. Prediction for y is 128 lbs. 2 pts
For a single randomly person, corresponding to one future observation: 1 (x—i)2 a+bxi t0.025,7 se 1+H+ n
2 (Xii)2
i=1 where a + b x = 247.0 + 5.860 * 64 = 128 lbs.
SD of prediction is 25.7, so for t0.025,7 = 2.365 obtain a 95% prediction interval for a future observation of 67.3 to 189 lbs. WHOW! 6 pts 10a. The fundamental problem probability theory addresses is how to determine
the relative likelihood or "probability" that various combinations of events
occur. Thus given the rules of an “experiment” and all possible outcomes,
probability theory should determine the probability of every possible outcomes. Key concepts are the idea of a random experiment, independence, and a random
variable, which is a numerical value generated by such an experiment. 10b. Statistics is the science of decision making under uncertainty because
hypothesis testing codifies a systematic procedure for choosing between two (or
more) hypothesis when data does not precisely determine what is true. Key concepts are that of specific hypotheses, a decision rule based upon a test
statistic that summarizes the data, and the type I and II errors. ...
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 Stedinger

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