ISYE 6644 Exam 3 Practice Questions – Spring 2026 Complete Study Guide with Verified Questions, Answers & Rationales. Georgia Institute Of Technology.

ISYE 6644 Exam 3 Practice Questions – Spring
2026 Complete Study Guide with Verified
Questions, Answers & Rationales. Georgia
Institute Of Technology.



Section 1: Input Modeling and Goodness-of-Fit

Question 1
You have collected 100 data points and want to fit a distribution. The Kolmogorov-
Smirnov test statistic DnDn is calculated as 0.15. The critical value
at α=0.05α=0.05 for n=100n=100 is approximately 1.36/n=0.1361.36/n=0.136.
What is the correct interpretation?


 A. Fail to reject H0H0; the fitted distribution is adequate
 B. Reject H0H0; the fitted distribution is not a good fit
 C. The test is inconclusive
 D. Accept H0H0 with 95% confidence

Correct Answer: B

Explanation: Since the test statistic Dn=0.15Dn=0.15 exceeds the critical value 0.136, we
reject the null hypothesis that the data come from the fitted distribution. This indicates a
poor fit at the 5% significance level.

, Question 2
Which goodness-of-fit test is most appropriate for testing uniformity of pseudo-random
numbers when the parameters are estimated from the data?

 A. Kolmogorov-Smirnov test
 B. Chi-square test
 C. Anderson-Darling test
 D. Shapiro-Wilk test

Correct Answer: B

Explanation: The chi-square goodness-of-fit test is preferred when parameters are
estimated because it naturally accounts for the reduction in degrees of freedom. The
Kolmogorov-Smirnov test is designed for fully specified distributions with known
parameters.




Question 3
TRUE or FALSE? In a chi-square goodness-of-fit test with 5 intervals and 2 estimated
parameters, the degrees of freedom are 5−1−2=25−1−2=2.

 A. True
 B. False

Correct Answer: A

Explanation: For the chi-square goodness-of-fit test, degrees of freedom = k−1−sk−1−s,
where kk is the number of intervals and ss is the number of estimated parameters.
With k=5k=5 and s=2s=2, df = 2.

, Question 4
Suppose you perform a chi-square goodness-of-fit test with 8 equal-probability
intervals and 3 estimated parameters, obtaining a test statistic of χ02=12.5χ02=12.5.
At α=0.05α=0.05, what is your conclusion? (Critical value χ0.05,42=9.49χ0.05,42=9.49)


 A. Fail to reject H0H0
 B. Reject H0H0
 C. The test is inconclusive
 D. Accept H0H0

Correct Answer: B

Explanation: Degrees of freedom = 8−1−3=48−1−3=4. Since 12.5 > 9.49, we reject the
null hypothesis. The fitted distribution is not a good fit at the 5% significance level.




Question 5
TRUE or FALSE? The Kolmogorov-Smirnov test is generally more powerful than the chi-
square test for small sample sizes.

 A. True
 B. False

Correct Answer: A

Explanation: The Kolmogorov-Smirnov test is more powerful than the chi-square test for
small samples because it uses all individual data points rather than binned data. The chi-
square test loses information through binning.

, Question 6
What is the correct degrees of freedom for a chi-square goodness-of-fit test with 10
intervals, testing whether data follow a normal distribution with both mean and variance
estimated from the sample?

 A. 9
 B. 8
 C. 7
 D. 6

Correct Answer: C

Explanation: Degrees of freedom = k−1−s=10−1−2=7k−1−s=10−1−2=7,
where s=2s=2 for the estimated mean and variance.




Question 7
TRUE or FALSE? The Anderson-Darling test gives more weight to the tails of the
distribution than the Kolmogorov-Smirnov test.

 A. True
 B. False

Correct Answer: A

Explanation: The Anderson-Darling test is more sensitive to discrepancies in the tails,
making it particularly useful when tail behavior is important for the simulation model.