Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 11: The Chi-Square Distribution

65 problems · 2 figures

11-1 §11.1 source ↗

If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation?

Answer

mean = 25 and standard deviation = 7.0711

11-3 §11.1 source ↗

When does the chi-square curve approximate a normal distribution?

Answer

when the number of degrees of freedom is greater than 90

11-5 §11.1 source ↗

Is it more likely the df is 90, 20, or two in the graph?
Figure 11.13 in the book ↗

Answer

df = 2

11-7 §11.2 source ↗

Determine the appropriate test to be used in the next three exercises.

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.

Answer

a goodness-of-fit test

11-9 §11.2 source ↗

Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in Table 11.27.

The actual distribution for a class of 20 is in Table 11.28.

df= ______

Table 11.27 in the book ↗
Grade Proportion
A 0.25
B 0.30
C 0.35
D 0.10
Table 11.28 in the book ↗
Grade Frequency
A 7
B 7
C 5
D 1

Answer

3

11-11 §11.2 source ↗

Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in Table 11.27.

The actual distribution for a class of 20 is in Table 11.28.

χ2 test statistic = ______

Table 11.27 in the book ↗
Grade Proportion
A 0.25
B 0.30
C 0.35
D 0.10
Table 11.28 in the book ↗
Grade Frequency
A 7
B 7
C 5
D 1

Answer

2.04

11-13 §11.2 source ↗

Use the following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in Table 11.27.

The actual distribution for a class of 20 is in Table 11.28.

At the 5% significance level, what can you conclude?

Table 11.27 in the book ↗
Grade Proportion
A 0.25
B 0.30
C 0.35
D 0.10
Table 11.28 in the book ↗
Grade Frequency
A 7
B 7
C 5
D 1

Answer

We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.

11-15 §11.2 source ↗


Use the following information to answer the next nine exercises: The cumulative number of COVID-19 related cases reported for Santa Clara County for a certain time period is broken down by ethnicity as in Table 11.29.

The percentage of each ethnic group in Santa Clara County is as in Table 11.30.

H0: _______

Table 11.29 in the book ↗
Ethnicity Number of Cases
White 2,229
Hispanic/Latino 1,157
Black/African-American 457
Asian, Pacific Islander 232
Total = 4,075
Table 11.30 in the book ↗
Ethnicity Percentage of total county population Number expected (round to two decimal places)
White 42.9% 1748.18
Hispanic/Latino 26.7%
Black/African-American 2.6%
Asian, Pacific Islander 27.8%
Total = 100%

Answer

H0: the distribution of COVID-19 cases follows the ethnicities of the general population of Santa Clara County.

11-17 §11.2 source ↗


Use the following information to answer the next nine exercises: The cumulative number of COVID-19 related cases reported for Santa Clara County for a certain time period is broken down by ethnicity as in Table 11.29.

The percentage of each ethnic group in Santa Clara County is as in Table 11.30.

Is this a right-tailed, left-tailed, or two-tailed test?

Table 11.29 in the book ↗
Ethnicity Number of Cases
White 2,229
Hispanic/Latino 1,157
Black/African-American 457
Asian, Pacific Islander 232
Total = 4,075
Table 11.30 in the book ↗
Ethnicity Percentage of total county population Number expected (round to two decimal places)
White 42.9% 1748.18
Hispanic/Latino 26.7%
Black/African-American 2.6%
Asian, Pacific Islander 27.8%
Total = 100%

Answer

right-tailed

11-19 §11.2 source ↗


Use the following information to answer the next nine exercises: The cumulative number of COVID-19 related cases reported for Santa Clara County for a certain time period is broken down by ethnicity as in Table 11.29.

The percentage of each ethnic group in Santa Clara County is as in Table 11.30.

χ2 test statistic = _______

Table 11.29 in the book ↗
Ethnicity Number of Cases
White 2,229
Hispanic/Latino 1,157
Black/African-American 457
Asian, Pacific Islander 232
Total = 4,075
Table 11.30 in the book ↗
Ethnicity Percentage of total county population Number expected (round to two decimal places)
White 42.9% 1748.18
Hispanic/Latino 26.7%
Black/African-American 2.6%
Asian, Pacific Islander 27.8%
Total = 100%

Answer

2016.136

11-21 §11.2 source ↗


Use the following information to answer the next nine exercises: The cumulative number of COVID-19 related cases reported for Santa Clara County for a certain time period is broken down by ethnicity as in Table 11.29.

The percentage of each ethnic group in Santa Clara County is as in Table 11.30.

Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value.

Let α = 0.05

Decision: ________________

Reason for the Decision: ________________

Conclusion (write out in complete sentences): ________________

Table 11.29 in the book ↗
Ethnicity Number of Cases
White 2,229
Hispanic/Latino 1,157
Black/African-American 457
Asian, Pacific Islander 232
Total = 4,075
Table 11.30 in the book ↗
Ethnicity Percentage of total county population Number expected (round to two decimal places)
White 42.9% 1748.18
Hispanic/Latino 26.7%
Black/African-American 2.6%
Asian, Pacific Islander 27.8%
Total = 100%
Figure 11.14 in the book ↗

Answer

Graph: Answers may vary.

Decision: Reject the null hypothesis.

Reason for the Decision: p-value < alpha

Conclusion (write out in complete sentences): The make-up of COVID-19 cases does not fit the ethnicities of the general population of Santa Clara County.

11-23 §11.3 source ↗

Determine the appropriate test to be used in the next three exercises.

A pharmaceutical company is interested in the relationship between age and presentation of symptoms for a common viral infection. A random sample is taken of 500 people with the infection across different age groups.

Answer

a test of independence

11-25 §11.3 source ↗

Determine the appropriate test to be used in the next three exercises.

A marathon runner is interested in the relationship between the brand of shoes runners wear and their run times. They take a random sample of 50 runners and records their run times as well as the brand of shoes they were wearing.

Answer

a test of independence

11-27 §11.3 source ↗


Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. Table 11.31 shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance they must travel.

df = _______

Table 11.31 in the book ↗
Traveling Distance Third class Second class First class Total
1–100 miles 21 14 6 41
101–200 miles 18 16 8 42
201–300 miles 16 17 15 48
301–400 miles 12 14 21 47
401–500 miles 6 6 10 22
Total 73 67 60 200

Answer

8

11-29 §11.3 source ↗


Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. Table 11.31 shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance they must travel.

How many passengers are expected to travel between 401 and 500 miles and purchase first-class tickets?

Table 11.31 in the book ↗
Traveling Distance Third class Second class First class Total
1–100 miles 21 14 6 41
101–200 miles 18 16 8 42
201–300 miles 16 17 15 48
301–400 miles 12 14 21 47
401–500 miles 6 6 10 22
Total 73 67 60 200

Answer

6.6

11-31 §11.3 source ↗


Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. Table 11.31 shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance they must travel.

What is the p-value?

Table 11.31 in the book ↗
Traveling Distance Third class Second class First class Total
1–100 miles 21 14 6 41
101–200 miles 18 16 8 42
201–300 miles 16 17 15 48
301–400 miles 12 14 21 47
401–500 miles 6 6 10 22
Total 73 67 60 200

Answer

0.0435

11-33 §11.3 source ↗

Table 11.32 Smoking Levels by Ethnicity (Observed) in the book ↗
Smoking Level Per Day Black Native Hawaiian Hispanic/Latino Japanese American White TOTALS
1-10
11-20
21-30
31+
TOTALS


Use the following information to answer the next eight exercises: An article in the New England Journal of Medicine, discussed a study on smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 Black people, 2,745 Native Hawaiian people, 12,831 Hispanic/Latino people, 8,378 Japanese American people and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 Black people, 3,062 Native Hawaiian people, 4,932 Hispanic/Latino people, 10,680 Japanese American people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 Black people, 1,419 Native Hawaiian people, 1,406 Hispanic/Latino people, 4,715 Japanese American people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 Black people, 788 Native Hawaiian people, 800 Latino people, 2,305 Japanese American people, and 3,970 White people.

Complete the table.

Answer

Smoking Level Per Day Black Native Hawaiian Hispanic/Latino Japanese White Totals
1-10 9,886 2,745 12,831 8,378 7,650 41,490
11-20 6,514 3,062 4,932 10,680 9,877 35,065
21-30 1,671 1,419 1,406 4,715 6,062 15,273
31+ 759 788 800 2,305 3,970 8,622
Totals 18,830 8,014 19,969 26,078 27,559 10,0450
Table 11.60
Table 11.60 in the book ↗
Smoking Level Per Day Black Native Hawaiian Hispanic/Latino Japanese White Totals
1-10 9,886 2,745 12,831 8,378 7,650 41,490
11-20 6,514 3,062 4,932 10,680 9,877 35,065
21-30 1,671 1,419 1,406 4,715 6,062 15,273
31+ 759 788 800 2,305 3,970 8,622
Totals 18,830 8,014 19,969 26,078 27,559 10,0450

11-35 §11.3 source ↗

Table 11.32 Smoking Levels by Ethnicity (Observed) in the book ↗
Smoking Level Per Day Black Native Hawaiian Hispanic/Latino Japanese American White TOTALS
1-10
11-20
21-30
31+
TOTALS


Use the following information to answer the next eight exercises: An article in the New England Journal of Medicine, discussed a study on smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 Black people, 2,745 Native Hawaiian people, 12,831 Hispanic/Latino people, 8,378 Japanese American people and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 Black people, 3,062 Native Hawaiian people, 4,932 Hispanic/Latino people, 10,680 Japanese American people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 Black people, 1,419 Native Hawaiian people, 1,406 Hispanic/Latino people, 4,715 Japanese American people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 Black people, 788 Native Hawaiian people, 800 Latino people, 2,305 Japanese American people, and 3,970 White people.

Enter expected values in Table 11.32. Round to two decimal places.

Calculate the following values:

Answer

Smoking Level Per Day Black Native Hawaiian Hispanic/Latino Japanese White
1-10 7777.57 3310.11 8248.02 10771.29 11383.01
11-20 6573.16 2797.52 6970.76 9103.29 9620.27
21-30 2863.02 1218.49 3036.20 3965.05 4190.23
31+ 1616.25 687.87 1714.01 2238.37 2365.49
Table 11.61
Table 11.61 in the book ↗
Smoking Level Per Day Black Native Hawaiian Hispanic/Latino Japanese White
1-10 7777.57 3310.11 8248.02 10771.29 11383.01
11-20 6573.16 2797.52 6970.76 9103.29 9620.27
21-30 2863.02 1218.49 3036.20 3965.05 4190.23
31+ 1616.25 687.87 1714.01 2238.37 2365.49

11-37 §11.3 source ↗


Use the following information to answer the next eight exercises: An article in the New England Journal of Medicine, discussed a study on smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 Black people, 2,745 Native Hawaiian people, 12,831 Hispanic/Latino people, 8,378 Japanese American people and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 Black people, 3,062 Native Hawaiian people, 4,932 Hispanic/Latino people, 10,680 Japanese American people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 Black people, 1,419 Native Hawaiian people, 1,406 Hispanic/Latino people, 4,715 Japanese American people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 Black people, 788 Native Hawaiian people, 800 Latino people, 2,305 Japanese American people, and 3,970 White people.

χ 2 test statistic = ______

Answer

10,301.8

11-39 §11.3 source ↗


Use the following information to answer the next eight exercises: An article in the New England Journal of Medicine, discussed a study on smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 Black people, 2,745 Native Hawaiian people, 12,831 Hispanic/Latino people, 8,378 Japanese American people and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 Black people, 3,062 Native Hawaiian people, 4,932 Hispanic/Latino people, 10,680 Japanese American people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 Black people, 1,419 Native Hawaiian people, 1,406 Hispanic/Latino people, 4,715 Japanese American people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 Black people, 788 Native Hawaiian people, 800 Latino people, 2,305 Japanese American people, and 3,970 White people.

Is this a right-tailed, left-tailed, or two-tailed test? Explain why.

Answer

right

11-41 §11.3 source ↗

State the decision and conclusion (in a complete sentence) for the following preconceived levels of α.

α = 0.05

a. Decision: ___________________

b. Reason for the decision: ___________________

c. Conclusion (write out in a complete sentence): ___________________

Answer

a. Reject the null hypothesis.

b. p-value < alpha

c. There is sufficient evidence to conclude that smoking level is dependent on ethnic group.

11-43 §11.4 source ↗

A math teacher wants to see if two of their classes have the same distribution of test scores. What test should they use?

Answer

test for homogeneity

11-45 §11.4 source ↗

A market researcher wants to see if two different stores have the same distribution of sales throughout the year. What type of test should they use?

Answer

test for homogeneity

11-47 §11.4 source ↗

What condition must be met to use the test for homogeneity?

Answer

All values in the table must be greater than or equal to five.

11-49 §11.4 source ↗

Use the following information to answer the next five exercises: Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in Table 11.33.

df = _______

Table 11.33 in the book ↗
20–30 30–40 40–50 50–60
Private Practice 16 40 38 6
Hospital 8 44 59 39

Answer

3

11-51 §11.4 source ↗

Use the following information to answer the next five exercises: Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in Table 11.33.

What is the p-value?

Table 11.33 in the book ↗
20–30 30–40 40–50 50–60
Private Practice 16 40 38 6
Hospital 8 44 59 39

Answer

0.00005

11-53 §11.5 source ↗

Which test do you use to decide whether an observed distribution is the same as an expected distribution?

Answer

a goodness-of-fit test

11-55 §11.5 source ↗

Which test would you use to decide whether two factors have a relationship?

Answer

a test for independence

11-57 §11.5 source ↗

How are tests of independence similar to tests for homogeneity?

Answer

Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way ∑ (ij) (O-E) 2 E . In addition, all values must be greater than or equal to five.

11-59 §11.6 source ↗

Use the following information to answer the next three exercises: An archer’s standard deviation for 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?

Answer

a test of a single variance

11-61 §11.6 source ↗

Use the following information to answer the next three exercises: An archer’s standard deviation for hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less.

Is this a right-tailed, left-tailed, or two-tailed test?

Answer

a left-tailed test

11-63 §11.6 source ↗


Use the following information to answer the next three exercises: 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.

Answer

H0: σ2 = 0.812;

Ha: σ2 > 0.812

11-65 §11.6 source ↗


Use the following information to answer the next four exercises: 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?

Answer

a test of a single variance

11-67 §11.6 source ↗


Use the following information to answer the next four exercises: 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 p-value?

Answer

0.0542

11-69 §11.1 source ↗

Decide whether the following statements are true or false.

As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical.

Answer

true

11-71 §11.1 source ↗

Decide whether the following statements are true or false.

The mean and the median of the chi-square distribution are the same if df = 24.

Answer

false

11-73 §11.2 source ↗

For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E Solution Sheets for the chi-square solution sheet. Round expected frequency to two decimal places.

The marital status distribution of the population of U.S. men, ages 15 and older, is as shown in Table 11.35.

Suppose that a random sample of 400 U.S. young adults, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in Table 11.35, rounding to two decimal places.

Table 11.35 in the book ↗
Marital Status Percent Expected Frequency
never married 31.3
married 56.1
widowed 2.5
divorced/separated 10.1
Table 11.36 in the book ↗
Marital Status Frequency
never married 140
married 238
widowed 2
divorced/separated 20

Answer

Marital Status Percent Expected Frequency
never married 31.3 125.2
married 56.1 224.4
widowed 2.5 10
divorced/separated 10.1 40.4
Table 11.62

a. The data fits the distribution.

b. The data does not fit the distribution.

c. 3

d. chi-square distribution with df = 3

e. 19.27

f. 0.0002

g. Check student’s solution.

h.

i. Alpha = 0.05

ii. Decision: Reject null

iii. Reason for decision: p-value < alpha

iv. Conclusion: Data does not fit the distribution.

Table 11.62 in the book ↗
Marital Status Percent Expected Frequency
never married 31.3 125.2
married 56.1 224.4
widowed 2.5 10
divorced/separated 10.1 40.4

11-75 §11.2 source ↗

Use the following information to answer the next two exercises: The columns in Table 11.37 contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for students using mass transit to get to school, and the Overall Student Population. Suppose the right column contains the result of a survey of 1,000 local students from that year who used mass transit to get to school.

Perform a goodness-of-fit test to determine whether the local results follow the distribution of those that use mass transit, based on ethnicity.

Table 11.37 in the book ↗
Race/Ethnicity Percentage Who Use Mass Transit to Get to School Overall Student Population Survey Frequency
Asian, Asian American, or Pacific Islander 10.2% 5.4% 113
Black or African-American 8.2% 14.5% 94
Hispanic or Latino 15.5% 15.9% 136
American Indian or Alaska Native 0.6% 1.2% 10
White 59.4% 61.6% 604
Not reported/other 6.1% 1.4% 43

Answer

a. H0: The local results follow the distribution of the percentage of students who use mass transit to get to school

b. Ha: The local results do not follow the distribution of the percentage of students who use mass transit to get to school

c. df = 5

d. chi-square distribution with df = 5

e. chi-square test statistic = 13.4

f. p-value = 0.0199

g. Answers may vary.

h.

i. Alpha = 0.05

ii. Decision: Reject null when a = 0.05

iii. Reason for Decision: p-value < alpha

iv. Conclusion: Local data do not fit the mass transit Distribution.

v. Decision: Do not reject null when a = 0.01

vi. Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the of students who use mass transit.

11-77 §11.2 source ↗

Use the following information to answer the next two exercises: UCLA conducted a survey of more than 263,000 first-year students from 385 colleges in fall 2005. The results of students' expected majors by gender were reported in The Chronicle of Higher Education (2/2/2006). Suppose a survey of 5,000 graduating women and 5,000 graduating men was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables for Exercise 11.77 and Exercise 11.78. The second column in each table does not add to 100% because of rounding.

Conduct a goodness-of-fit test to determine if the actual college majors of graduating women fit the distribution of their expected majors.

Table 11.39 in the book ↗
Major Women - Expected Major Women - Actual Major
Arts & Humanities 14.0% 670
Biological Sciences 8.4% 410
Business 13.1% 685
Education 13.0% 650
Engineering 2.6% 145
Physical Sciences 2.6% 125
Professional 18.9% 975
Social Sciences 13.0% 605
Technical 0.4% 15
Other 5.8% 300
Undecided 8.0% 420

Answer

a. H0: The actual college majors of graduating women fit the distribution of their expected majors

b. Ha: The actual college majors of graduating women do not fit the distribution of their expected majors

c. df = 10

d. chi-square distribution with df = 10

e. test statistic = 11.48

f. p-value = 0.3211

g. Answers may vary.

h.

i. Alpha = 0.05

ii. Decision: Do not reject null when a = 0.05 and a = 0.01

iii. Reason for decision: p-value > alpha

iv. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating women do not fit the distribution of their expected majors.

11-79 §11.2 source ↗

Read the statement and decide whether it is true or false.

In a goodness-of-fit test, the expected values are the values we would expect if the null hypothesis were true.

Answer

true

11-81 §11.2 source ↗

Read the statement and decide whether it is true or false.

Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week or not.

Answer

true

11-83 §11.2 source ↗

Read the statement and decide whether it is true or false.

In a goodness-of fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis.

Answer

false

11-85 §11.2 source ↗

Read the statement and decide whether it is true or false.

Table 11.42 contains information from a survey among 499 participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the USA. Perform a hypothesis test at the 5% significance level to determine whether the survey participants are a representative sample of the USA obese population.

Table 11.42 in the book ↗
Age class (years) Obese (percentage) Expected USA average (percentage)
20–30 15.0 32.6
31–40 26.5 32.6
41–50 13.6 36.6
51–60 21.9 36.6
61–70 21.0 39.7

Answer

The hypotheses for the goodness-of-fit test are:

a. H0: Surveyed obese fit the distribution of expected obese

b. Ha: Surveyed obese do not fit the distribution of expected obese

Use a chi-square distribution with df = 4 to evaluate the data.

The test statistic is X2 = 9.85

The P-value = 0.0431

At 5% significance level, α = 0.05. For this data, P < α. Reject the null hypothesis.

At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.

11-87 §11.3 source ↗

For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places.

Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver’s family (that is, whether car size and family size are independent). To test this, suppose that 800 car owners were randomly surveyed with the results in Table 11.44. Conduct a test of independence.

Table 11.44 in the book ↗
Family Size Sub & Compact Mid-size Full-size Van & Truck
1 20 35 40 35
2 20 50 70 80
3–4 20 50 100 90
5+ 20 30 70 70

Answer

a. H0: Car size is independent of family size.

b. Ha: Car size is dependent on family size.

c. df = 9

d. chi-square distribution with df = 9

e. test statistic = 15.8284

f. p-value = 0.0706

g. Answers may vary.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: p-value > alpha

iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent.

11-89 §11.3 source ↗

For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places.

Some travel agents claim that honeymoon hot spots vary according to age of the bride. Suppose that 280 recent brides were interviewed as to where they spent their honeymoons. The information is given in Table 11.46. Conduct a test of independence.

Table 11.46 in the book ↗
Location 20–29 30–39 40–49 50 and over
Niagara Falls 15 25 25 20
Poconos 15 25 25 10
Europe 10 25 15 5
Virgin Islands 20 25 15 5

Answer

a. H0: Honeymoon locations are independent of bride’s age.

b. Ha: Honeymoon locations are dependent on bride’s age.

c. df = 9

d. chi-square distribution with df = 9

e. test statistic = 15.7027

f. p-value = 0.0734

g. Answers may vary.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: p-value > alpha

iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent.

11-91 §11.3 source ↗

For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places.

A major food manufacturer is concerned that the sales for its skinny french fries have been decreasing. As a part of a feasibility study, the company conducts research into the types of fries sold across the country to determine if the type of fries sold is independent of the area of the country. The results of the study are shown in Table 11.48. Conduct a test of independence.

Table 11.48 in the book ↗
Type of Fries Northeast South Central West
skinny fries 70 50 20 25
curly fries 100 60 15 30
steak fries 20 40 10 10

Answer

a. H0: The types of fries sold are independent of the location.

b. Ha: The types of fries sold are dependent on the location.

c. df = 6

d. chi-square distribution with df = 6

e. test statistic =18.8369

f. p-value = 0.0044

g. Answers may vary.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: p-value < alpha

iv. Conclusion: At the 5% significance level, There is sufficient evidence that types of fries and location are dependent.

11-93 §11.3 source ↗

For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places.

Suppose that 600 thirty-year-olds were surveyed to determine whether or not there is a relationship between the level of education an individual has and salary. Conduct a test of independence.

Table 11.50 in the book ↗
Annual Salary Not a high school graduate High school graduate College graduate Masters or doctorate
< $30,000 15 25 10 5
$30,000–$40,000 20 40 70 30
$40,000–$50,000 10 20 40 55
$50,000–$60,000 5 10 20 60
$60,000+ 0 5 10 150

Answer

a. H0: Salary is independent of level of education.

b. Ha: Salary is dependent on level of education.

c. df = 12

d. chi-square distribution with df = 12

e. test statistic = 255.7704

f. p-value = 0

g. Answers may vary.

h.

Alpha: 0.05

Decision: Reject the null hypothesis.

Reason for decision: p-value < alpha

Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent.

11-95 §11.3 source ↗

Read the statement and decide whether it is true or false.

The test for independence uses tables of observed and expected data values.

Answer

true

11-97 §11.3 source ↗

Read the statement and decide whether it is true or false.

In a test of independence, the expected number is equal to the row total multiplied by the column total divided by the total surveyed.

Answer

true

11-99 §11.3 source ↗

Read the statement and decide whether it is true or false.

Table 11.52 provides a recent survey of the youngest online entrepreneurs whose net worth is estimated at one million dollars or more. Their ages range from 17 to 30. Each cell in the table illustrates the number of entrepreneurs who correspond to the specific age group and their net worth. Are the ages and net worth independent? Perform a test of independence at the 5% significance level.

Table 11.52 in the book ↗
Age Group\ Net Worth Value (in millions of US dollars) 1–5 6–24 ≥25 Row Total
17–25 8 7 5 20
26–30 6 5 9 20
Column Total 14 12 14 40

Answer

a. H0: Age is independent of the youngest online entrepreneurs’ net worth.

b. Ha: Age is dependent on the net worth of the youngest online entrepreneurs.

c. df = 2

d. chi-square distribution with df = 2

e. test statistic = 1.76

f. p-value 0.4144

g. Answers may vary.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: p-value > alpha

iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that age and net worth for the youngest online entrepreneurs are dependent.

11-101 §11.4 source ↗

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E Solution Sheets for the chi-square solution sheet. Round expected frequency to two decimal places.

A psychologist is interested in testing whether there is a difference in the distribution of personality types for business majors and social science majors. The results of the study are shown in Table 11.54. Conduct a test of homogeneity. Test at a 5% level of significance.

Table 11.54 in the book ↗
Open Conscientious Extrovert Agreeable Neurotic
Business 41 52 46 61 58
Social Science 72 75 63 80 65

Answer

a. H0: The distribution for personality types is the same for both majors

b. Ha: The distribution for personality types is not the same for both majors

c. df = 4

d. chi-square with df = 4

e. test statistic = 3.01

f. p-value = 0.5568

g. Answers may vary.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: p-value > alpha

iv. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors.

11-103 §11.4 source ↗

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E Solution Sheets for the chi-square solution sheet. Round expected frequency to two decimal places.

A fish and wildlife technician is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform a test for homogeneity at a 5% level of significance.

Answer

a. H0: The distribution for fish caught is the same in Green Valley Lake and in Echo Lake.

b. Ha: The distribution for fish caught is not the same in Green Valley Lake and in Echo Lake.

c. 3

d. chi-square with df = 3

e. 11.75

f. p-value = 0.0083

g. Answers may vary.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: p-value < alpha

iv. Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake

11-105 §11.4 source ↗

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E Solution Sheets for the chi-square solution sheet. Round expected frequency to two decimal places.

When looking at energy consumption, we are often interested in detecting trends over time and how they correlate among different countries. The information in Table 11.57 shows the average energy use (in units of kg of oil equivalent per capita) in the USA and the joint European Union countries (EU) for a specific six-year period. Do the energy use values in these two areas come from the same distribution? Perform the analysis at the 5% significance level.

Table 11.57 in the book ↗
Year European Union United States Row Total
6 3,413 7,164 10,557
5 3,302 7,057 10,359
4 3,505 7,488 10,993
3 3,537 7,758 11,295
2 3,595 7,697 11,292
1 3,613 7,847 11,460
Column Total 20,965 45,011 65,976

Answer

a. H0: The distribution of average energy use in the USA is the same as in Europe between Year 1 and Year 6.

b. Ha: The distribution of average energy use in the USA is not the same as in Europe between Year 1 and Year 6.

c. df = 4

d. chi-square with df = 4

e. test statistic = 2.7434

f. p-value = 0.7395

g. Answers may vary.

h.

i. Alpha: 0.05

ii. Decision: Do not reject the null hypothesis.

iii. Reason for decision: p-value > alpha

iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from Year 1 to Year 6.

11-107 §11.5 source ↗

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E Solution Sheets for the chi-square solution sheet. Round expected frequency to two decimal places.

Is there a difference between the distribution of community college statistics students and the distribution of university statistics students in what technology they use on their homework? Of some randomly selected community college students, 43 used a computer, 102 used a calculator with built in statistics functions, and 65 used a table from the textbook. Of some randomly selected university students, 28 used a computer, 33 used a calculator with built in statistics functions, and 40 used a table from the textbook. Conduct an appropriate hypothesis test using a 0.05 level of significance.

Answer

a. H0: The distribution for technology use is the same for community college students and university students.

b. Ha: The distribution for technology use is not the same for community college students and university students.

c. 2

d. chi-square with df = 2

e. 7.05

f. p-value = 0.0294

g. Answers may vary.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: p-value < alpha

iv. Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities.

11-110 §11.6 source ↗

Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.

A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes.

Answer

225

11-112 §11.6 source ↗

Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.

H0: __________

Answer

H0: σ2 ≤ 150

11-114 §11.6 source ↗

Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.

chi-square test statistic = ________

Answer

36

11-116 §11.6 source ↗

Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.

Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the p-value.

Answer

Answers may vary.

11-118 §11.6 source ↗

Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.

How did you know to test the variance instead of the mean?

Answer

The claim is that the variance is no more than 150 minutes.

11-120 §11.6 source ↗

Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.

If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use?

Answer

a Student's t- or normal distribution

11-122 §11.6 source ↗

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E Solution Sheets for the chi-square solution sheet. Round expected frequency to two decimal places.

Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of $84 and a sample standard deviation of $12, test the claim that the standard deviation is greater than $15.

Answer

a. H0: σ = 15

b. Ha: σ > 15

c. df = 42

d. chi-square with df = 42

e. test statistic = 26.88

f. p-value = 0.9663

g. Answers may vary.

h.

i. Alpha = 0.05

ii. Decision: Do not reject null hypothesis.

iii. Reason for decision: p-value > alpha

iv. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.

11-124 §11.6 source ↗

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E Solution Sheets for the chi-square solution sheet. Round expected frequency to two decimal places.

Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief.

Answer

a. H0: σ ≤ 3

b. Ha: σ > 3

c. df = 17

d. chi-square distribution with df = 17

e. test statistic = 28.73

f. p-value = 0.0371

g. Answers may vary.

h.

i. Alpha: 0.05

ii. Decision: Reject the null hypothesis.

iii. Reason for decision: p-value < alpha

iv. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.

11-126 §11.6 source ↗

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E Solution Sheets for the chi-square solution sheet. Round expected frequency to two decimal places.

According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. They count the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11

Answer

a. H0: σ = 2

b. Ha: σ ≠ 2

c. df = 14

d. chi-square distiribution with df = 14

e. chi-square test statistic = 5.2094

f. p-value = 0.0346

g. Answers may vary.

h.

i. Alpha = 0.05

ii. Decision: Reject the null hypothesis

iii. Reason for decision: p-value < alpha

iv. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.

11-128 §11.6 source ↗

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E Solution Sheets for the chi-square solution sheet. Round expected frequency to two decimal places.

You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $1,249 with a very narrow standard deviation of $25. You find a website that has a price comparison for the same computer at a series of stores as follows: $1,299; $1,229.99; $1,193.08; $1,279; $1,224.95; $1,229.99; $1,269.95; $1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5% significance level. As a potential buyer, what would be the practical conclusion from your analysis?

Answer

The sample standard deviation is $34.29.

H0 : σ2 = 252
Ha : σ2 > 252


df = n – 1 = 7.


test statistic: x 2 =   x 7 2 =   (n–1) s 2 25 2 =   (8–1) (34.29) 2 25 2 = 13.169;


p-value: P( x 7 2 >13.169)= 1–P( x 7 2  ≤13.169)= 0.0681


Alpha: 0.05


Decision: Do not reject the null hypothesis.


Reason for decision: p-value > alpha


Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.

11-130 §11.BIT source ↗

a. Explain why a goodness-of-fit test and a test of independence are generally right-tailed tests.

b. If you did a left-tailed test, what would you be testing?

Answer

a. The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected.

b. Testing to see if the data fits the distribution “too well” or is too perfect.

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Problems and answers are adapted from Introductory Statistics 2e by Barbara Illowsky and Susan Dean, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/introductory-statistics-2e/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.