INSTRUCTOR ANSWER GUIDE
OpenStax Introductory Business Statistics 2e
Instructor Answer Guide
,OpenStax Introductory Business Statistics 2e
Instructor Answer Guide
CHAPTER 11: FACTS ABOUT THE CHI-SQUARE DISTRIBUTION
Exercise 1. If the number of degrees of freedom for a chi-square distribution is 25, what is the
population mean and standard deviation?
Solution mean = 25 and standard deviation = 7.0711
Exercise 2. If df > 90, the distribution is _____________. If df = 15, the distribution is
________________.
Solution approximately normal, skewed right.
Exercise 3. When does the chi-square curve approximate a normal distribution?
Solution when the number of degrees of freedom is greater than 90
Exercise 4. Where is μ located on a chi-square curve?
Solution just to the right of the peak
Exercise 5. Is it more likely the df is 90, 20 or two in the graph?
Figure 11.10
Solution: df=2
Exercise 6. An archer’s standard deviation for his hits is six (data is measured in distance from the
center of the target). An observer claims the standard deviation is less.
What type of test should be used?
Solution a test of a single variance
Exercise 7. An archer’s standard deviation for his hits is six (data is measured in distance from the
center of the target). An observer claims the standard deviation is less.
State the null and alternative hypotheses.
Solution H0: σ2 = 36; Ha: σ2 < 36
Exercise 8. An archer’s standard deviation for his hits is six (data is measured in distance from the
2
May 28, 2026
,OpenStax Introductory Business Statistics 2e
Instructor Answer and Solution Guide
Chapter 11: Facts about the Chi-Square Distribution
center of the target). An observer claims the standard deviation is less.
Is this a right-tailed, left-tailed, or two-tailed test?
Solution a left-tailed test
Exercise 9. The standard deviation of heights for students in a school is 0.81. A random sample of
50 students is taken, and the standard deviation of heights of the sample is 0.96. A
researcher in charge of the study believes the standard deviation of heights for the
school is greater than 0.81.
What type of test should be used?
Solution a test of a single variance
Exercise 10. The standard deviation of heights for students in a school is 0.81. A random sample of
50 students is taken, and the standard deviation of heights of the sample is 0.96. A
researcher in charge of the study believes the standard deviation of heights for the
school is greater than 0.81.
State the null and alternative hypotheses.
Solution H0: σ2 = 0.812; Ha: σ2 > 0.812
Exercise 11. The standard deviation of heights for students in a school is 0.81. A random sample of
50 students is taken, and the standard deviation of heights of the sample is 0.96. A
researcher in charge of the study believes the standard deviation of heights for the
school is greater than 0.81.
df = _______
Solution 49
Exercise 12. The average waiting time in a doctor’s office varies. The standard deviation of waiting
times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s
office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the
variance of waiting times is greater than originally thought.
What type of test should be used?
Solution a test of a single variance
Exercise 13. The average waiting time in a doctor’s office varies. The standard deviation of waiting
times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s
office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the
variance of waiting times is greater than originally thought.
What is the test statistic?
Solution 42.17
Exercise 14. The average waiting time in a doctor’s office varies. The standard deviation of waiting
3
May 28, 2026
, OpenStax Introductory Business Statistics 2e
Instructor Answer Guide
times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s
office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the
variance of waiting times is greater than originally thought.
What can you conclude at the 5% significance level?
Solution We decline to reject the null hypothesis. There is not sufficient evidence to conclude
that the standard deviation of waiting times is greater than 3.4.
Exercise 15. Determine the appropriate test to be used. An archeologist is calculating the
distribution of the frequency of the number of artifacts they find in a dig site. Based
on previous digs, the archeologist creates an expected distribution broken down by
grid sections in the dig site. Once the site has been fully excavated, the archeologist
compares the actual number of artifacts found in each grid section to see if their
expectation was accurate.
Solution a goodness-of-fit test
Exercise 16. Determine the appropriate test to be used. An economist is deriving a model to
predict outcomes on the stock market. They create a list of expected points on the
stock market index for the next two weeks. At the close of each day’s trading, the
economist records the actual points on the index. They want to see how well the
model matched what actually happened.
Solution a goodness-of-fit test
Exercise 17. Determine the appropriate test to be used. A personal trainer is putting together a
weight-lifting program for their clients. For a 90-day program, they expect each
client to lift a specific maximum weight each week. As the program goes along, the
trainer records the actual maximum weights their clients lifted. They want to know
how well their expectations met with what was observed.
Solution a goodness-of-fit test
Exercise 18. A teacher predicts that the distribution of grades on the final exam will be as in Table
11.21.
Grade Proportion
A 0.25
B 0.30
C 0.35
D 0.10
Table 11.21
The actual distribution for a class of 20 is in Table 11.22.
Grade Frequency
A 7
4
May 28, 2026
OpenStax Introductory Business Statistics 2e
Instructor Answer Guide
,OpenStax Introductory Business Statistics 2e
Instructor Answer Guide
CHAPTER 11: FACTS ABOUT THE CHI-SQUARE DISTRIBUTION
Exercise 1. If the number of degrees of freedom for a chi-square distribution is 25, what is the
population mean and standard deviation?
Solution mean = 25 and standard deviation = 7.0711
Exercise 2. If df > 90, the distribution is _____________. If df = 15, the distribution is
________________.
Solution approximately normal, skewed right.
Exercise 3. When does the chi-square curve approximate a normal distribution?
Solution when the number of degrees of freedom is greater than 90
Exercise 4. Where is μ located on a chi-square curve?
Solution just to the right of the peak
Exercise 5. Is it more likely the df is 90, 20 or two in the graph?
Figure 11.10
Solution: df=2
Exercise 6. An archer’s standard deviation for his hits is six (data is measured in distance from the
center of the target). An observer claims the standard deviation is less.
What type of test should be used?
Solution a test of a single variance
Exercise 7. An archer’s standard deviation for his hits is six (data is measured in distance from the
center of the target). An observer claims the standard deviation is less.
State the null and alternative hypotheses.
Solution H0: σ2 = 36; Ha: σ2 < 36
Exercise 8. An archer’s standard deviation for his hits is six (data is measured in distance from the
2
May 28, 2026
,OpenStax Introductory Business Statistics 2e
Instructor Answer and Solution Guide
Chapter 11: Facts about the Chi-Square Distribution
center of the target). An observer claims the standard deviation is less.
Is this a right-tailed, left-tailed, or two-tailed test?
Solution a left-tailed test
Exercise 9. The standard deviation of heights for students in a school is 0.81. A random sample of
50 students is taken, and the standard deviation of heights of the sample is 0.96. A
researcher in charge of the study believes the standard deviation of heights for the
school is greater than 0.81.
What type of test should be used?
Solution a test of a single variance
Exercise 10. The standard deviation of heights for students in a school is 0.81. A random sample of
50 students is taken, and the standard deviation of heights of the sample is 0.96. A
researcher in charge of the study believes the standard deviation of heights for the
school is greater than 0.81.
State the null and alternative hypotheses.
Solution H0: σ2 = 0.812; Ha: σ2 > 0.812
Exercise 11. The standard deviation of heights for students in a school is 0.81. A random sample of
50 students is taken, and the standard deviation of heights of the sample is 0.96. A
researcher in charge of the study believes the standard deviation of heights for the
school is greater than 0.81.
df = _______
Solution 49
Exercise 12. The average waiting time in a doctor’s office varies. The standard deviation of waiting
times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s
office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the
variance of waiting times is greater than originally thought.
What type of test should be used?
Solution a test of a single variance
Exercise 13. The average waiting time in a doctor’s office varies. The standard deviation of waiting
times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s
office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the
variance of waiting times is greater than originally thought.
What is the test statistic?
Solution 42.17
Exercise 14. The average waiting time in a doctor’s office varies. The standard deviation of waiting
3
May 28, 2026
, OpenStax Introductory Business Statistics 2e
Instructor Answer Guide
times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s
office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the
variance of waiting times is greater than originally thought.
What can you conclude at the 5% significance level?
Solution We decline to reject the null hypothesis. There is not sufficient evidence to conclude
that the standard deviation of waiting times is greater than 3.4.
Exercise 15. Determine the appropriate test to be used. An archeologist is calculating the
distribution of the frequency of the number of artifacts they find in a dig site. Based
on previous digs, the archeologist creates an expected distribution broken down by
grid sections in the dig site. Once the site has been fully excavated, the archeologist
compares the actual number of artifacts found in each grid section to see if their
expectation was accurate.
Solution a goodness-of-fit test
Exercise 16. Determine the appropriate test to be used. An economist is deriving a model to
predict outcomes on the stock market. They create a list of expected points on the
stock market index for the next two weeks. At the close of each day’s trading, the
economist records the actual points on the index. They want to see how well the
model matched what actually happened.
Solution a goodness-of-fit test
Exercise 17. Determine the appropriate test to be used. A personal trainer is putting together a
weight-lifting program for their clients. For a 90-day program, they expect each
client to lift a specific maximum weight each week. As the program goes along, the
trainer records the actual maximum weights their clients lifted. They want to know
how well their expectations met with what was observed.
Solution a goodness-of-fit test
Exercise 18. A teacher predicts that the distribution of grades on the final exam will be as in Table
11.21.
Grade Proportion
A 0.25
B 0.30
C 0.35
D 0.10
Table 11.21
The actual distribution for a class of 20 is in Table 11.22.
Grade Frequency
A 7
4
May 28, 2026
