ISYE 6644 Week 13 Homework
Score for this quiz: 5 out of 5 Submitted Apr 18
at 2:04pm
This attempt took 181 minutes.
Question 1
pts
(Lesson 10.1: Introduction to Comparison of Systems.) Suppose we are dealing with i.i.d.
normal observations with unknown variance. Which of the following is true about a 95%
confidence interval for the mean \mu?
a. If you take more observations, the CI will always shrink.
Correct!
b. We are 95% sure that our CI will actually contain the unknown value of \ mu.
c. If we calculate 100 of these CI's, exactly 95 will actually contain \mu.
d. A 99% CI based on the same data will be shorter than the corresponding 95% CI.
(b). [(a) would be correct if you said usually'' instead of "always''. (c) would be OK if you said
"about'' instead of "exactly''. (d) is simply 180^{\rm o} wrong.]
Question 2
pts
(Lesson 10.3: Confidence Intervals for the Difference in Two Means.) We are studying the
waiting times arising from two queueing systems. Suppose we make 4 independent replications
of both systems, where the systems are simulated independently of each other.
\begin{array}{c|c|c} \text{replication} &\text{system 1} & \text{system 2}\\ \hline 1 & 10 & 25
\\ 2 & 20 & 10 \\ 3 & 5 & 40 \\ 4 & 30 & 30 \\ \ end{array}
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.
a. [-30, -5]
, b. [-15, 5]
Correct!
c. [-29.76, 9.76]
d. [-35, 10]
e. [5, 30]
This is a two-sample CI problem assuming unknown and unequal variances.
We have
\bar{X} = 16.25 and \bar{Y} = 26.25 with n=4 reps, and S_X^2 = 122.917 and S_Y^2 = 156.25
with m=4 reps. The estimated degrees of freedom is
\nu \; \equiv \; \frac{ \Big( \frac{S_X^2}{n} + \frac{S_Y^2}{m} \Big)^2} {\ frac{(S_X^2/n)^2}{n+1}
+ \frac{(S_Y^2/m)^2}{m+1}} - 2 \; = \; 7.86 \; =
\; 7.
Then the appropriate CI is
\begin{align} \mu_X - \mu_Y \in & \bar{X} - \bar{Y} \pm t_{\alpha/2,\nu} \ sqrt{\frac{S_X^2}{n}
+ \frac{S_Y^2}{m}} \\= & -10 \pm t_{0.025,7} \ sqrt{\frac{122.9}{4} + \frac{156.3}{4}} \\= & -10
\pm 2.365 (8.355) \\ = & -10 \pm 19.76 \; = \; [-29.76,\, 9.76]. \end{align}
Thus, the answer is (c).
Question 3
pts
(Lesson 10.6: Common Random Numbers.) This is sort of the same as Question 2, except we
have now used common random numbers to induce positive correlation between the results of
the two systems. \begin{array}{c|c|c} \text{replication} &\text{system 1} & \text{system 2}\\
\hline 1 & 10 & 25 \\ 2 & 20 & 30 \\ 3 & 5 & 10 \\ 4 & 30 & 40 \\ \ end{array}
Again find a two-sided 95% CI for the difference in the means of the two systems.
a. [-30, -5]
b. [-15, 5]
Correct!
c. [-16.5, -3.5]
Score for this quiz: 5 out of 5 Submitted Apr 18
at 2:04pm
This attempt took 181 minutes.
Question 1
pts
(Lesson 10.1: Introduction to Comparison of Systems.) Suppose we are dealing with i.i.d.
normal observations with unknown variance. Which of the following is true about a 95%
confidence interval for the mean \mu?
a. If you take more observations, the CI will always shrink.
Correct!
b. We are 95% sure that our CI will actually contain the unknown value of \ mu.
c. If we calculate 100 of these CI's, exactly 95 will actually contain \mu.
d. A 99% CI based on the same data will be shorter than the corresponding 95% CI.
(b). [(a) would be correct if you said usually'' instead of "always''. (c) would be OK if you said
"about'' instead of "exactly''. (d) is simply 180^{\rm o} wrong.]
Question 2
pts
(Lesson 10.3: Confidence Intervals for the Difference in Two Means.) We are studying the
waiting times arising from two queueing systems. Suppose we make 4 independent replications
of both systems, where the systems are simulated independently of each other.
\begin{array}{c|c|c} \text{replication} &\text{system 1} & \text{system 2}\\ \hline 1 & 10 & 25
\\ 2 & 20 & 10 \\ 3 & 5 & 40 \\ 4 & 30 & 30 \\ \ end{array}
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.
a. [-30, -5]
, b. [-15, 5]
Correct!
c. [-29.76, 9.76]
d. [-35, 10]
e. [5, 30]
This is a two-sample CI problem assuming unknown and unequal variances.
We have
\bar{X} = 16.25 and \bar{Y} = 26.25 with n=4 reps, and S_X^2 = 122.917 and S_Y^2 = 156.25
with m=4 reps. The estimated degrees of freedom is
\nu \; \equiv \; \frac{ \Big( \frac{S_X^2}{n} + \frac{S_Y^2}{m} \Big)^2} {\ frac{(S_X^2/n)^2}{n+1}
+ \frac{(S_Y^2/m)^2}{m+1}} - 2 \; = \; 7.86 \; =
\; 7.
Then the appropriate CI is
\begin{align} \mu_X - \mu_Y \in & \bar{X} - \bar{Y} \pm t_{\alpha/2,\nu} \ sqrt{\frac{S_X^2}{n}
+ \frac{S_Y^2}{m}} \\= & -10 \pm t_{0.025,7} \ sqrt{\frac{122.9}{4} + \frac{156.3}{4}} \\= & -10
\pm 2.365 (8.355) \\ = & -10 \pm 19.76 \; = \; [-29.76,\, 9.76]. \end{align}
Thus, the answer is (c).
Question 3
pts
(Lesson 10.6: Common Random Numbers.) This is sort of the same as Question 2, except we
have now used common random numbers to induce positive correlation between the results of
the two systems. \begin{array}{c|c|c} \text{replication} &\text{system 1} & \text{system 2}\\
\hline 1 & 10 & 25 \\ 2 & 20 & 30 \\ 3 & 5 & 10 \\ 4 & 30 & 40 \\ \ end{array}
Again find a two-sided 95% CI for the difference in the means of the two systems.
a. [-30, -5]
b. [-15, 5]
Correct!
c. [-16.5, -3.5]
