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Unformatted text preview: CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
finafflxam
Monday, December 17, 1990 Exam is open notes and ripenbook.
The exam lasts 150 minutes and there are 150 points.
SHOW WORK! 1. (10 points) An engineer wishes to employ the distribution Fx(x)=1— e‘7x{1+yx+(yx)2/2} for x20 fx(x) = y3x2e‘Yx /2 . for x _>_ o
to model the strength of concrete columns. Given a random sample of n
independent observations {X1, X2, , Xn}, what is the maximum likelihood
estimator of the unknown shape parameter Y? 2. (6 points) What is one common advantage of
(a) methodof—moment estimators over maximumlikelihood estimators,
and (b) maximumlikelihood estimators over methodof—moment estimators? 3. (10 points) Consider the distributionlwith pdf. f(x) = a(oc+1)x°“1(1 — x) for o s x s 1
0 otherwise. What is the method of moments estimator of a 2 0? 4. (10 points) Experience has shown that the strength of a certain type of column
can be described by a Weibull distribution with a shape parameter k equal to 5. If
you know that for a particular geometry and size of column the median strength
is 50,000 psi, what would be the 1 percentile of the distribution of strengths? 5. (10 points) Consider the 9 observations
75 6 138 53 86 42 50 278 13 which may have been drawn from a normal distribution. Use the attached
probability paper to construct a visual test of normality. Indicate the numerical
values of the quantities you are plotting. What is your impression? CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
,‘Finalﬁxam
Monday, December 17, 1990 6. (10 points) An environmental engineer is concerned with the concentration
of zebra mussel larvae in lake water. Last week, 10 measurements of mussel concentrations were made; the observations had a sample average of 11,530 with a sample standard deviation of 4,320. What is a 80% confidence interval for the
actual concentration of larvae that week? (Assume that the concentrations are normally distributed about the true weekly mean concentration.) What is the probability the particular confidence interval just constructed with the given data contains the true larval concentration for last week? 7. (16 points) At a Boy Scout campout two patrols got into an argument about
whose members were better at orienteering: using a compass and one's pace to
follow a series of bearing and distance directions to arrive at a specified location.
To resolve the dispute the senior patrol leader challenged them to a test. All of
the boys were given a different compass course which they were to complete.
The distances the individuals ended up from their targets were measured and the results are presented below. (i) How would it be reasonable to determine who won? [Careful] (ii) What are the appropriate hypotheses to test whether
either patrol is significantly better than the other? (iii) At the 10% level, using the most powerful nonparametric test you can,
what do you conclude? (iv) Approximately what. is the pvalue for the results in (iii)? (v) At the 10% level, what do you conclude using a twosample t test? (vi) Approximately what is the pvalue of the results in (v)? Ravens Eagles
18 20
13 24
120 66
19 26
22 42
8 30
_ Zﬁ
# of boys in patrol 6 7
Average 33.33 33.00 Stand. Deviation 42.74 16.22 CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
{FinalﬂSxam
Monday, December 17, 1990 8. (25 points) Windspeeds over the ocean can be estimated from ground radar or
by Seasat satellite. Simultaneous measurements for the same points were made
by the two methods to see if they yielded consistent results, yielding the data set: Te_st M Satellite Difference
1 4.15 4.18 0.03
2 3.68 3.94 0.26
3 3.42 5.98 2.56
4 4.85 6.30 1 .45
5 4.51 3.92 0.59
6 6.40 6.81 —0.41
7 4.30 5.85 1.55
8 3.56 3.02 0.54
9 3.65 5.26 —1.61
10 4.11 3.25 0.86
.11 115 & i114
Average 4.19 4.85 0.67
Standard Deviation 0.87 1.28 1.12 To test the indicated hypothesis of consistency at the 5% level, what are the
rejection regions of each of the following three tests, and the pvalues for each
data set with each test?
(a) a paired ttest,
(b) a Wilcoxon Sign Rank test, and (c) the Sign test (you can pretend that n=10 for use of tables).
(d) Let Di equal the difference between the paired observations. For a ttest with a = 5%, when
E[Di] = 4.7; Var[Di] = (1.0)2,
what sample size would one need to achieve a type H error less than 20%? (e) For E[Di] = —0.7; Var[Di] = (1.0)2, and n = 10, what are the type I and II mm
for a ttest? 9. (6 points) Eastern Airlines would send a shuttle airplane from Boston to New
York during afternoon hours as soon as they filled a plane with 80 passengers.
At that hour passengers arrived at the gate independently at a rate of 3 per
minute. (a) If a plane just left so that the gate is empty, what is the mean and variance of
the time until the next plane is full with its 80 passengers? ' (b) What is the probability the gap between two passengers is > 1 minute? CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
finaIZSxam
Monday, December 17, 1990 10. (47 points) A geotechnical engineer is interested in the shear strength of soil
in a clay stratum at various depths. She obtained 9 measurements of shear
strengths (in kips per square foot  ksf ) at 9 depths depth x: 1.9 3.1 4.2 9.3 10 meters
strength y. 0.3 0.6 0.5 1.3 1.6 ksf
n = 9 )7 = 6.044 f: 0.911
SX = SY = and 2 (xi §_)(yi Y) = 8.196 a) What are the leastsquares estimates of a and B for the simple linear model
Y=a+Bx+8 relating depth x with shear strength .Y, with errors 8 due to soil variability and laboratory measurement percision? b) What is an unbiased estimate of the standard deviation of the model errors 8? c) Estimate the variances of the least squares estimators of 0c and B? d) What is R2 and adjusted R2 for this data set? e) Before gathering this data the geotechnical engineer believed that shear
strength should increase with depth due to the structure and clay content of
the stratum. With this presupposition, what are the appropriate null and
alternative hypotheses on [3? What is the rejection region for the 5% test of
whether B = 0? What is the approximate pvalue for this data set? ' f) Using your fitted line, what is the best estimate of the value of y for a depth of
x = 7.5 meters? What is a 95% confidence interval for the shear strength that
would be obtained if another measurement was made at that depth? CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Solutions Final Exam 1990 1. The density function is fx(x) = 73x2e—Yx /2 for x Z 0
Hence the likelihood function is
L(k) = n 54in k) = H 73x12e_7xi/2
It's easier to work with the loglikelihood function:
ln{ L(l<) } = 2 { 3 ln(y) —1n(2) + 2 ln(xi) — 'yxi ]
= 3n ln(y) — n ln(2) + 2 E ln(xi) — 72 xi
The maximum of this function can be found by setting its derivative to zero:
0 = d ln{ L(Y) l/dy = 3n/y ~—§.‘.xi yielding the MLE " y: 3n/2‘. xi 2. 0 Method of moment estimators are often easier. 0 Statisticians prefer maximum likelihood estimators over methodof
moments estimators because MLEs have a sounder theoretical basis, are more efficient (i.e. have a smaller variance) asymptotically, and are asymptotically
unbiased. Also, MLEs can be used with catagorical data. 3. Given the pdf
f(x) = a(a+1)x°‘_1(1 — x) for 0 S x S 1 E[X] = I x f(x) dx = I x a(a+1) xala  x) dx = a(a+1) {1/(oc+1) — 1/(a+2) } =
(X/(a+2). Method of moments estimator of on found by solving »x = on /((1+2) to
obtain A(X = 2»X /(1 — »X) > 0. 4. For the Weibull distribution F(x) = 1  exp{ bxl< }
Given k = 5 and 0.5 = F(50,000), one solves 0.5 = 1 — exp{ b(50,000)5}
to obtain b = 2.22xlO'24. The 1 percentile is then obtained by solving
0.01 = 1 — exp{2.22x10'24(x0_01)5 } yielding x001 = 21,440. 5. A table is presented below. A probability plot can be obtained by plotting
xa) versus the plotting positions pi using normal probability paper. Rank pi = (i3/8)/(n+1/4) 01(1)? X0)
1 0.068 154 6
2 0.176 0.96 13
3 0.284 0.59 42
4 0.392 0.28 50
5 0.500 0.00 53
6 0.608 0.28 75 CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Solutions Final Exam 1990 7 0.716 0.59 86
8 0.824 0.96 138
9 0.932 1.54 273 300 250
200
150 _
100 / n’I
.’"5'°"/ I.F=./I aI I I ,. ~2.001.50 1.000.50 0.00 0.50 1.00 1.50 2.00 Line doesn't look very straight. Observations are pilling up around zero.
Try a lognormal or Weibull distribution. 6. An 80% confidence interval for the true mean larval concentration when a?
= 11,530 and s = 4,320 based upon a sample of size 10 is ‘ 3x it0.10195/sqrt(10)=9641 to 13419 for tum: 1.383. The true larval concentration either is, or is not, contained in this particular
interval for last week, i.e. p = 0 or 1! 7. BSA orienteering competition (i) A common way to determine who won would be to compare the two
group's mean distance. However, the single outlier in the Ravan patrol
dominates the average score for the whole patrol. One could avoid such
problems by reporting the median of the two patrols, dropping the largest one or
two, or comparing the average ranks; comparing the average ranks is
equivalent to use of the Wilcoxon rank sum test. (ii) H0 is that there is no difference in the populations Ha is that one or the other patrol is better (twosided test) (iii) Ranking the data and assigning rank 1 to 8 and thirteen to 120, the
sum of the ranks for the Raven patrol (which has fewer members) is w = 29.
Unfortunately, the tables are for upper tail values; 29 corresponds to an upper
tail value of w‘= m(m+n+1) — 29 = 55. Table A10 shows that c = 54 has an
exceedance probability of 5.1%. Thus a rejection region for a 10.2% test is W S 30
and W 2 54. w = 55 is significant. (iv) Interpolating, W = 29, corresponding to W = 55, yields a pvalue of
2(0.04) = 8% for a twosided test. (v) Using a pooled ttest with 11 degrees of freedom yields t = 0.019 . (vi) Note: t = 0.019 = 0.02 = 0.0; the pvalue for t=0 for a two sided test is
100%. Actually, using ztables as an approximation, one obtains p = 98%. CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Solutions Final Exam 1990 8. Windspeed data
(a) the value of the paired t is sqrt(11)[0.67/1.12] = 1.98 with 11—1 = 10
degrees of freedom. 0104,11 = 1.98 so for a twosided test this result is significant at about the 2(4%) = 8% level.
A rejection region for a 5% level test would be IT  > “1025,10 = 2.228. (b) Consider the Wilcoxzon signed—rank test. The sum of all the ranks is
n(n+1)/2 = 66. The sum of the ranks of the positive differences is 8+ = 15, corresponding to S_ = 51. S_ = 51 corresponds to a onesided pvalue of 4%; the twosided p—value is again 8%, as with the ttest. Rejection region for a 5% test is
8+ or S. greater than 55 [actual or = 2(2.7%) = 5.4%]. (c) Only three differences are positive. The probability of zero, one, two or three successes in eleven trials were p = 0.5 is (0.5)“[ 1 + 11 + 10*11/2 +
9*10*11/ 3] = 0.00049{ 1 + 10 + 55 + 165} = 0.113. Thus the pvalue for a twocsided
sign test is 22%. A symmetric 2sided test would reject only if there were 0 or 1, 10 or 11 negative signs, corresponding to an a of 1.2%; this doesn't seem very
satisfactory. (d) d = Au/ 0' = 0.7/1.0 = 0.7; using Table A13 for a twosided 5% test yields
df = 19 so n = 20. One could use as an (poor) approximation n = [ (zB + zw2)/ d ]2
= [(0.84 + 1.96)/0.7]2 = 16. (e) Again d = 0.7; Type I error on = 5%; Type H error = 50% from Table A13.
The Type I error is generally given. 9. Here the arrival rate for this Poisson processes is 2» = 3 per minute.
(a) The waiting time until the kth arrival is gamma distributed with on = k and [3 = 1/1/ . Thus the mean is 043 = 80/3 ; whereas the variance is 0432 = 80/32 =
8.89. (b) This can be calculated by noting the time between arrivals is
exponential with parameter A = 3 per minute; the probability of waiting more than 1 minute is P[ X > 1 ] = 1 — F(x) = exp(—k*1) = exp(—3) = 0.05. Alternatively,
using the Poisson distribution, the probability of waiting more than a minute is P[no arrivals in one minute] = (kt)0exp(—lt)/0! =exp(—l*1) = 0.05. CEE 304  UNCERTAINTY ANALYSIS IN ENGINEERING
Solutions Final Exam 1990 10. (47 points) Here is the big regression problem.
a) The least—squares estimates of on and B are a = 0.104 and b = 0.133. b) The unbiased estimate of the variance of e is se2 = 0.054 = (.231)2. 62 4 2
c) Var(b) = n—‘—— =8.72x10 =(0.0295) .
Z (XiE2
i=1
Var(a) = 63 % + n—L = 0.038 = (0.194)?
2 (xi702 l=1 2 = _(Residual sumofsquares) _ _ _ 2 _ 2 _ _
d) R [1 mam a1 sum_of_squms) 1 (n k)se /[(n my 10.745, Adjusted R? = 1  seZ/sy2 = 0.71 e) Ho: [3 = 0; Ha: [3 > 0. Requires a onesided test; reject Ho if T > t0_05l7 = 1.860.
T = (b  0)/sd(b) = 4.52; clearly' significant! pvalue near. 0.001.
The rejection region for the 10% test of whether or not on = 0 is t > t0.05’6 = 1.943; Yes “a is signiﬁcantly different from zero. f) Using your fitted line, the natural estimate of the mean value of y for x = 7.5
meters is a + bx = 1.105 ksf. A 95 percent confidence interval for a future observation is
_— 2
a+bxi t0.025,7S 1+}11+n(x—x)
‘ V Z (xi—i )2
' 1 which for x = 7.5 and toms; = 2.365 yields 0.52 ksf 1t; 1.69 ksf. What assumptions justify the use of leastsquares regression for the estimation
of the model parameters? With what kinds of data might these estimators
perform poorly? ...
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This note was uploaded on 02/02/2008 for the course CEE 3040 taught by Professor Stedinger during the Fall '08 term at Cornell.
 Fall '08
 Stedinger

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