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ISyE 6644 — Spring 2018 — Practice Test #3
Hi One Last Time, Gentle Class!
Our Test 3 will cover every freaking thing in the course! However, I’ll place more weight
on the new stuff since Test 2.
Practice Test 3 below is actually an augmentation of the Test 3 originally given in my
“live” 6644 class during the Fall 2013 semester. Your Test 3 won’t be as long. You’ll get
to have 3 cheat sheets (6 sides total) and plenty of time. ,
Also, all of our test questions will be multiple choice, in spite of what you see below.
Dave
1. If X and Y have joint p.d.f. f(x, y) = c(1 − x)y, 0 ≤ y ≤ x ≤ 1, for some
appropriate constant c, find E[X].
Solution: First of all,
1 =
∫ 1
0
∫ x
0
c(1− x)y dy dx = c
2
∫ 1
0
(x2 − x3) dx = c
24
,
so that c = 24. Then
fX(x) =
∫ ∞
−∞
f(x, y) dy
=
∫ x
0
24(1− x)y dy
= 12(x2 − x3), 0 < x < 1.
Thus,
E[X] =
∫ 1
0
12(x3 − x4) dx = 0.6. 2
22. If X and Y are i.i.d. Pois(3) RV’s, find Var(XY ).
Solution:
Var(XY ) = E[(XY )2]− (E[XY ])2
= E[X2]E[Y 2]− (E[X]E[Y ])2 X, Y indep
= (E[X2])2 − (E[X])4 X, Y i.i.d.
= (Var(X) + (E[X])2)2 − (E[X])4 X, Y i.i.d.
= (3 + 9)2 − 34 = 63. 2
3. TRUE or FALSE? The runs “up and down” test is most often used to test for
goodness-of-fit of observations.
Solution: FALSE. It’s a test for independence. 2
4. Draw an Arena DECIDE module.
Solution: It’s a sideways diamond. 2
5. In what Arena template would you find the Sequence spreadsheet?
Solution: Advanced Transfer. 2
6. In Arena, what module can you use to merge multiple entities into one?
Solution: Batch. 2
7. Arena Program (this will take a little work): Explicitly describe and show how to
implement in Arena the following inventory system. Customers arrive according
to a Poisson process at the rate of 10/day. Each customer demands a discrete
uniform number of items from 1 to 8 (like an 8-sided die toss). When inventory
goes below 50, we order 200 items. We allow for backlogs and lead-times.
I want appropriate English verbiage and Arena block diagrams with adequate
explanation of the model. Also discuss what stats you’d collect and how you’d do it.
38. TRUE or FALSE? Suppose that U1, U2, . . . are truly i.i.d. U(0,1) random variables.
Then a 95% chi-square goodness-of-fit test for uniformity will incorrectly reject
uniformity of the observations about 5% of the time.
Solution: TRUE. (That’s what Type I error is.) 2
9. TRUE or FALSE? Suppose Z1, . . . , Z6 are i.i.d. standard normal random variables
obtained by the Box–Muller method. Then
∑6
i=1 Z
2
i ∼ Erlang3(1/2).
Solution: TRUE. The sum of 6 i.i.d. Z2i random variables is a χ
2(6), which is
itself an Erlang3(1/2). 2
10. Suppose X1, X2, . . . , X100 are i.i.d. with Pr(X = 0) = Pr(X = 2) = 0.5. Define
the sample mean X¯ ≡ ∑100i=1Xi/100. Use the Central Limit Theorem to find an
approximate expression for Pr(X¯ < 1.1).
Solution: Note that E[X] =

x xPr(X = x) = 1 and E[X
2] =

x x
2 Pr(X =
x) = 2, so that Var(X) = E[X2] − (E[X])2 = 1. These results makes sense since
X ∼ 2Bern(0.5).
Anyway, the CLT now implies that X¯ ≈ Nor(µ, σ2/n) = Nor(1, 0.01), and so
Pr(X¯ < 1.1) = Pr
(
X¯ − 1√
0.01
<
1.1− 1√
0.01
)
≈ Pr(Nor(0, 1) < 1) = 0.8413. 2
11. If U, V are i.i.d. U(0,1), what’s the distribution of −`n(√U)− `n(√V )?
Solution:
−`n(

U)− `n(

V ) = −1
2
`n(U)− 1
2
`n(V ) ∼ Erlang2(2). 2
12. TRUE or FALSE? If we want to generate X ∼ Pois(4.5), then it’s better to use a
normal approximation than acceptance-rejection.
4Solution: FALSE. (The normal approximation requires larger λ for the asymp-
totics to work and the efficiency to kick in.) 2
13. TRUE or FALSE? The acceptance-rejection’s majorizing function t(x) is usually
a p.d.f.
Solution: FALSE. (It integrates to something > 1.) 2
14. Consider the stationary first-order exponential autoregressive process (EAR(1)),
Xi =
{
αXi−1, w.p. α
αXi−1 + i, w.p. 1− α,
where X0 and the i’s are i.i.d. Exp(λ), and 0 < α < 1. Find Cov(X0, X1).
Solution:
Cov(X0, X1) = Cov
(
X0, αX0
)
α + Cov
(
X0, αX0 + 1
)
(1− α)
= α2Var(X0) + α(1− α)Var(X0) + 0
= α/λ2. 2
15. If X1, . . . , Xn are i.i.d. Exp(1/9), what is the expected value of the sample variance
S2?
Solution: E[S2] = σ2 = 81. 2
16. If X1, X2, X3 are i.i.d. normal, with X1 = 3, X2 = 2, and X3 = 7, what is the MLE
for E[X2i ]?
Solution: By invariance, the MLE is
Ê[X2i ] = µˆ
2 + σˆ2 = X¯2 +
n− 1
n
S2 = 42 +
14
3
= 20.67. 2
517. Find x such that e2x = 1/x. (Get within two decimals.)
Solution: Bisection search quickly reveals that x is about 0.4265. 2
18. TRUE or FALSE? The square root of the sample variance is unbiased for the
standard deviation.
Solution: FALSE. (E[S2] = σ2 ; E[S] = σ.) 2
19. Suppose we’re conducting a χ2 goodness-of-fit test to determine whether or not
200 i.i.d. observations are from a Johnson distribution, which has 4 parameters
that must be estimated. If we divide the observations into 10 equal-probability
intervals, how many degrees of freedom will our test have?
Solution: 10− 4− 1 = 5. 2
20. We are interested in seeing if the number of emergency department visits occurring
each day at the Georgia Tech clinic is Binomial(5,0.5). Below are the results for a
200-day period. We’ll assume that the numbers from day to day are i.i.d.
# of visits # of days
0 6
1 30
2 61
3 62
4 34
5 7
Thus, for example, there were 30 days during when the ED had exactly 1 visit.
We’ll perform a 95% χ2 goodness-of-fit test to see if the number of accidents each
day is Binomial(5,0.5).
(a) How many intervals will you use for your test?
Solution: Let X denote the number of visits on a particular day. Under
the null hypothesis, the expected number of occurrences of i visits is Ei =
nPr(X = i) = 200
(
5
i
)
(0.5)5. So we have the following table.
6i Oi Ei
0 6 6.25
1 30 31.25
2 61 62.5
3 62 62.5
4 34 31.25
5 7 6.25
Since all of the Ei’s are ≥ 5, we can use all of the cells. Thus, the answer is 6
intervals. 2
(b) What is your statistic value?
Solution: χ20 =
∑5
i=0(Oi − Ei)2/Ei = 0.432. 2
(c) What is your conclusion? Binomial(5,0.5) or not?
Solution: Since the previous answer is so small, we don’t really need to look
up the quantile, but I’ll do it anyway. The critical value is χ2α,k−1 = χ
2
0.05,5 =
11.07, indicating a fail to reject. (So, yes, we’ll assume it’s Binomial.) 2
21. Consider the following PRN’s: 0.18, 0.92, 0.61, 0.33. If we use the Kolmorogov-
Smirnov goodness-of-fit test to see if these numbers are U(0,1), what is the value
of the test statistic?
Solution: Let’s make the usual table with the ordered PRN’s.
i 1 2 3 4
R(i) 0.18 0.33 0.61 0.92
i
n
−R(i) 0.07 0.17 0.14 0.07
R(i) − i−1n 0.18 0.08 0.11 0.18
This indicates that D+n = maxi[
i
n
−R(i)] = 0.17 and D−n = maxi[R(i)− i−1n ] = 0.18,
so that Dn = max(D
+, D−) = 0.18. 2
22. Do the PRN’s in Question 21 pass the Kolmogorov–Smirnov goodness-of-fit test
for uniformity at level α = 0.05?
Solution: From the one-sided table, we have Dα,n = D0.05,4 = 0.565. Since
Dn < Dα,n, we fail to reject uniformity. 2
723. Consider a stationary stochastic process X1, X2, . . ., with covariance function
Rk = Cov(X1, X1+k) = 3− k for k = 0, 1, 2, 3, and Rk = 0 for k ≥ 4. Find Var(X¯4).
Solution: From class notes, we have
Var(X¯n) =
1
n
[
R0 + 2
n−1∑
k=1
(
1− k
n
)
Rk
]
=
1
4
[
R0 + 2(
3
4
)R1 + 2(
2
4
)R2 + +2(
1
4
)R3
]
=
1
4
[
3 + 2(
3
4
)2 + 2(
2
4
)1 + +2(
1
4
)0
]
= 1.75. 2
24. Consider the following (approximately normal) average waiting times from 4 inde-
pendent replications of a complicated queueing network. Suppose that each output
is based on the average of 500 waiting times:
30 40 10 50
Use the method of independent replications to calculate a two-sided 90% confidence
interval for the mean µ.
Solution: We use b = 4 reps here. Z¯b =
∑b
i=1 Zi/b = 32.5 and S
2
Z =
1
b−1
∑b
i=1(Zi−
Z¯b)
2 = 291.67. Then the desired CI is
µ ∈ Z¯b ± tα/2,b−1

S2Z/b
= 32.5± t0.05,3

291.67/4
= 32.5± 2.353(8.539)
= 32.5± 20.09 = [12.4, 52.6]. 2
25. Which is the method of batch means more appropriate for: terminating or
steady-state simulations?
Solution: Steady-State. 2
826. Which is usually a better way to deal with initialization bias in steady-state
simulation analysis: (i) make an extremely long run to overwhelm the bias, or (ii)
perform truncation?
Solution: Truncate. 2
27. Consider the following 10 snowfall totals in Buffoonalo, NY over consecutive years:
130 94 125 112 150 123 141 133 128 152
Use the method of batch means to calculate a two-sided 90% confidence interval
for the mean µ. In particular, use two batches of size 5.
Solution: We use b = 2 batches here here. X¯n = 128.8, the batch means are
X¯1,5 = 122.2 and X¯2,5 = 135.4, and the batch means estimator for the variance
parameter is
V̂B =
m
b− 1
b∑
i=1
(X¯i,m − X¯n)2 = 5
2∑
i=1
(X¯i,5 − X¯10)2 = 435.6.
Then the desired CI is
µ ∈ X¯n ± tα/2,b−1

V̂B/n
= 128.8± t0.05,1

435.6/10
= 128.8± 6.314(6.6)
= 128.8± 41.67 = [87.1, 170.5]. 2
28. Suppose [0, 1] is a 90% confidence interval for the mean µ based on 10 independent
replications of size 1000. Now the boss has decided that she wants a 99% CI for
2µ based on those same 10 replications of size 1000. What is it?
Solution: Let’s get a 99% CI for µ first. To begin with, the original 90% CI for µ
is
µ ∈ Z¯b ± t0.05, b−1

S2Z
b
= 0.5± 0.5.
9This implies that the 99% CI for µ will have half-length
t0.005, b−1

S2Z
b
=
t0.005, b−1
t0.05, b−1
t0.05, b−1

S2Z
b
=
t0.005, 9
t0.05, 9
(0.5) =
3.250
1.833
(0.5) = 0.887.
Now we can get the 99% CI for µ:
µ ∈ Z¯b ± t0.005, b−1

S2Z
b
= 0.5± 0.887.
This immediately implies that the 99% CI for 2µ is
2µ ∈ 1± 1.77 = [−0.77, 2.77]. 2
29. Suppose I use the method of overlapping batch means with sample size n = 10000
and batch size m = 500. Approximately how many degrees of freedom will the
resulting variance estimator have?
Solution: Denote b = n/m = 20. You get approximately 3
2
(b − 1) = 28.5 d.f.
(Will also accept 3b/2 = 30 or anything reasonably close.) 2
30. If W(t) is a standard Brownian motion process and a < b, find Pr(W(a) Solution: W(a) −W(b) is normally distributed with mean 0. This immediately
yields a probability of 1/2. 2
31. Suppose that A =
∫ 1
0
B(t) dt is the area under a Brownian bridge process. Find
Pr(A > 1/

12).
Solution: From class notes, we have A ∼ Nor(0, 1
12
). Then
Pr(A > 1/

12) = Pr(Nor(0, 1) > 1) = 0.1587. 2
32. We are studying the waiting times arising from two queueing systems. Suppose
we make 4 independent replications of both systems, where the systems are simu-
lated independently of each other. Assuming that the average waiting time results
from each replication are approximately normal, find a two-sided 95% CI for the
difference in the means of the two systems.
10
replication system 1 system 2
1 10 25
2 20 10
3 5 40
4 30 30
Solution: This is a two-sample CI problem assuming unknown and unequal vari-
ances. We have X¯ = 16.25, Y¯ = 26.25, S2X = 122.917, and S
2
Y = 156.25. The
estimated d.f. is
ν ≡
(
S2X
n
+
S2Y
m
)2
(S2X/n)
2
n+1
+
(S2Y /m)
2
m+1
− 2 = 7.86 = 7.
Then the appropriate CI is
µX − µY ∈ X¯ − Y¯ ± tα/2,ν

S2X
n
+
S2Y
m
= −10± t0.025,7

122.9
4
+
156.3
4
= −10± 2.365(8.355)
= −10± 19.76 = [−29.76, 9.76]. 2
33. This is sort of the same as Question 32, except we have now used common random
numbers to induce positive correlation between the results of the two systems.
Again find a two-sided 95% CI for the difference in the means of the two systems.
replication system 1 system 2
1 10 25
2 20 30
3 5 10
4 30 40
Solution: This is a paired-t CI problem assuming unknown variance of the differ-
ences.
replication Xi Yi Di
1 10 25 −15
2 20 30 −10
3 5 10 −5
4 30 40 −10
11
Now,
µX − µY ∈ D¯ ± tα/2,n−1

S2D
n
= −10± t0.025,3

16.67
4
= −10± 3.182(2.041)
= −10± 6.50 = [−16.50, −3.5]. 2
34. Let’s use the basic Monte Carlo technique from class to integrate I =
∫ 1
0
ex dx.
(a) First of all, what is the exact value of I?
Solution: I = e− 1 = 1.718. 2
(b) Use the PRN’s 0.95, 0.63, 0.15, and 0.42 to estimate I.
Solution: Iˆn =
1
n
∑n
i=1 e
Ui = 1.787. 2
(c) Now use antithetics to estimate I.
Solution: I˜n =
1
n
∑n
i=1 e
1−Ui = 1.656. 2
(d) Combine your last two answers.
Solution: I¯n =
1
2
(Iˆn + I˜n) = 1.721, which is a great answer. 2
35. Suppose that I’m interested in selecting the most popular television show during a
particular time period. What kind of selection problem is this — (a) normal, (b)
multinomial, or (c) Bernoulli?
Solution: (b). 2
36. There may also be a couple additional easy ranking and selection questions from
Module 10.

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