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 figures11-1 §11.1 source ↗
Answer
11-3 §11.1 source ↗
Answer
11-5 §11.1 source ↗
Answer
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. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened.
Answer
11-9 §11.2 source ↗
Use the following information to answer the next five exercises. A teacher predicts the distribution of grades on the final exam. The predictions are shown in Table 11.27.
The actual distribution for a class of 20 is in Table 11.28.
______
| Grade | Proportion |
|---|---|
| A | 0.25 |
| B | 0.30 |
| C | 0.35 |
| D | 0.10 |
| Grade | Frequency |
|---|---|
| A | 7 |
| B | 7 |
| C | 5 |
| D | 1 |
Answer
11-11 §11.2 source ↗
Use the following information to answer the next five exercises. A teacher predicts the distribution of grades on the final exam. The predictions are shown in Table 11.27.
The actual distribution for a class of 20 is in Table 11.28.
χ2 test statistic = ______
| Grade | Proportion |
|---|---|
| A | 0.25 |
| B | 0.30 |
| C | 0.35 |
| D | 0.10 |
| Grade | Frequency |
|---|---|
| A | 7 |
| B | 7 |
| C | 5 |
| D | 1 |
Answer
11-13 §11.2 source ↗
Use the following information to answer the next five exercises. A teacher predicts the distribution of grades on the final exam. The predictions are shown in Table 11.27.
The actual distribution for a class of 20 is in Table 11.28.
At the 5 percent significance level, what can you conclude?
| Grade | Proportion |
|---|---|
| A | 0.25 |
| B | 0.30 |
| C | 0.35 |
| D | 0.10 |
| Grade | Frequency |
|---|---|
| A | 7 |
| B | 7 |
| C | 5 |
| D | 1 |
Answer
11-15 §11.2 source ↗
Use the following information to answer the next nine exercises. The cumulative number of cases of a chronic disease reported for Santa Clara County 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: _______
| Ethnicity | Number of Cases |
|---|---|
| White | 2,229 |
| Hispanic | 1,157 |
| Black/African American | 457 |
| Asian, Pacific Islander | 232 |
| Total = 4,075 |
| Ethnicity | % of Total County Population | Number Expected (round to two decimal places) |
|---|---|---|
| White | 42.9% | 1,748.18 |
| Hispanic | 26.7% | |
| Black/African American | 2.6% | |
| Asian, Pacific Islander | 27.8% | |
| Total = 100% |
Answer
11-17 §11.2 source ↗
Use the following information to answer the next nine exercises. The cumulative number of cases of a chronic disease reported for Santa Clara County 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?
| Ethnicity | Number of Cases |
|---|---|
| White | 2,229 |
| Hispanic | 1,157 |
| Black/African American | 457 |
| Asian, Pacific Islander | 232 |
| Total = 4,075 |
| Ethnicity | % of Total County Population | Number Expected (round to two decimal places) |
|---|---|---|
| White | 42.9% | 1,748.18 |
| Hispanic | 26.7% | |
| Black/African American | 2.6% | |
| Asian, Pacific Islander | 27.8% | |
| Total = 100% |
Answer
11-19 §11.2 source ↗
Use the following information to answer the next nine exercises. The cumulative number of cases of a chronic disease reported for Santa Clara County 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 = _______
| Ethnicity | Number of Cases |
|---|---|
| White | 2,229 |
| Hispanic | 1,157 |
| Black/African American | 457 |
| Asian, Pacific Islander | 232 |
| Total = 4,075 |
| Ethnicity | % of Total County Population | Number Expected (round to two decimal places) |
|---|---|---|
| White | 42.9% | 1,748.18 |
| Hispanic | 26.7% | |
| Black/African American | 2.6% | |
| Asian, Pacific Islander | 27.8% | |
| Total = 100% |
Answer
11-21 §11.2 source ↗
Use the following information to answer the next nine exercises. The cumulative number of cases of a chronic disease reported for Santa Clara County 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): ________________
| Ethnicity | Number of Cases |
|---|---|
| White | 2,229 |
| Hispanic | 1,157 |
| Black/African American | 457 |
| Asian, Pacific Islander | 232 |
| Total = 4,075 |
| Ethnicity | % of Total County Population | Number Expected (round to two decimal places) |
|---|---|---|
| White | 42.9% | 1,748.18 |
| Hispanic | 26.7% | |
| Black/African American | 2.6% | |
| Asian, Pacific Islander | 27.8% | |
| Total = 100% |
Answer
Graph: Check student’s solution.
Decision: Reject the null hypothesis.
Reason for decision: p-value < alpha
Conclusion: The make-up of 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
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. She takes a random sample of 50 runners and records their run times and the brand of shoes they were wearing.
Answer
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 the passenger must travel.
df = _______
| 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
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 the passenger must travel.
How many passengers are expected to travel between 401 and 500 miles and purchase first-class tickets?
| 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
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 the passenger must travel.
What is the p-value?
| 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
11-33 §11.3 source ↗
| Product use Per Day | African American | Native Hawaiian | Latino | Japanese American | White | TOTALS |
|---|---|---|---|---|---|---|
| 1–10 | ||||||
| 11–20 | ||||||
| 21–30 | ||||||
| 31+ | ||||||
| TOTALS |
Use the following information to answer the next ten exercises. An article in the New England Journal of Medicine discussed a study on people who used a certain product in California and Hawaii. In one part of the report, the self-reported ethnicity and product-use levels per day were given. Of the people using the product at most 10 times per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 whites. Of the people using the product 11 to 20 times per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people using the product 21 to 30 times per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 whites. Of the people using the product at least 31 times per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites.
Complete the table.
Answer
| Product-use Per Day | African American | Native Hawaiian | Latino | Japanese Americans | 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 |
| Product-use Per Day | African American | Native Hawaiian | Latino | Japanese Americans | 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 ↗
| Product use Per Day | African American | Native Hawaiian | Latino | Japanese American | White | TOTALS |
|---|---|---|---|---|---|---|
| 1–10 | ||||||
| 11–20 | ||||||
| 21–30 | ||||||
| 31+ | ||||||
| TOTALS |
Use the following information to answer the next ten exercises. An article in the New England Journal of Medicine discussed a study on people who used a certain product in California and Hawaii. In one part of the report, the self-reported ethnicity and product-use levels per day were given. Of the people using the product at most 10 times per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 whites. Of the people using the product 11 to 20 times per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people using the product 21 to 30 times per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 whites. Of the people using the product at least 31 times per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites.
Enter expected values in Table 11.32. Round to two decimal places.
Calculate the following values.
Answer
| Product Use Per Day | African American | Native Hawaiian | Latino | Japanese Americans | White |
|---|---|---|---|---|---|
| 1-10 | 7,777.57 | 3,310.11 | 8,248.02 | 10,771.29 | 11,383.01 |
| 11-20 | 6,573.16 | 2797.52 | 6970.76 | 9,103.29 | 9,620.27 |
| 21-30 | 2,863.02 | 1,218.49 | 3,036.20 | 3,965.05 | 4,190.23 |
| 31+ | 1,616.25 | 687.87 | 1,714.01 | 2,238.37 | 2,365.49 |
| Product Use Per Day | African American | Native Hawaiian | Latino | Japanese Americans | White |
|---|---|---|---|---|---|
| 1-10 | 7,777.57 | 3,310.11 | 8,248.02 | 10,771.29 | 11,383.01 |
| 11-20 | 6,573.16 | 2797.52 | 6970.76 | 9,103.29 | 9,620.27 |
| 21-30 | 2,863.02 | 1,218.49 | 3,036.20 | 3,965.05 | 4,190.23 |
| 31+ | 1,616.25 | 687.87 | 1,714.01 | 2,238.37 | 2,365.49 |
11-37 §11.3 source ↗
Use the following information to answer the next ten exercises. An article in the New England Journal of Medicine discussed a study on people who used a certain product in California and Hawaii. In one part of the report, the self-reported ethnicity and product-use levels per day were given. Of the people using the product at most 10 times per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 whites. Of the people using the product 11 to 20 times per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people using the product 21 to 30 times per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 whites. Of the people using the product at least 31 times per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites.
test statistic = ______
Answer
11-39 §11.3 source ↗
Use the following information to answer the next ten exercises. An article in the New England Journal of Medicine discussed a study on people who used a certain product in California and Hawaii. In one part of the report, the self-reported ethnicity and product-use levels per day were given. Of the people using the product at most 10 times per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 whites. Of the people using the product 11 to 20 times per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people using the product 21 to 30 times per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 whites. Of the people using the product at least 31 times per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites.
Is this a right-tailed, left-tailed, or two-tailed test? Explain why.
Answer
11-41 §11.3 source ↗
State the decision and conclusion (in a complete sentence) for the following 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 product use is dependent on ethnic group.
11-43 §11.4 source ↗
Answer
11-45 §11.4 source ↗
Answer
11-47 §11.4 source ↗
Answer
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 = _______
| 20–30 | 30–40 | 40–50 | 50–60 | |
|---|---|---|---|---|
| Private Practice | 16 | 40 | 38 | 6 |
| Hospital | 8 | 44 | 59 | 39 |
Answer
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?
| 20–30 | 30–40 | 40–50 | 50–60 | |
|---|---|---|---|---|
| Private Practice | 16 | 40 | 38 | 6 |
| Hospital | 8 | 44 | 59 | 39 |
Answer
11-53 §11.5 source ↗
Answer
11-55 §11.5 source ↗
Answer
11-57 §11.5 source ↗
Answer
11-59 §11.6 source ↗
Use the following information to answer the next three exercises. An archer’s standard deviation for his hits is six, where the data are measured in distance from the center of the target. An observer claims the standard deviation is less than six.
What type of test should be used?
Answer
11-61 §11.6 source ↗
Use the following information to answer the next three exercises. An archer’s standard deviation for his hits is six, where the data are measured in distance from the center of the target. An observer claims the standard deviation is less than six.
Is this a right-tailed, left-tailed, or two-tailed test?
Answer
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
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
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
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
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 U.S. male population, ages 15 and older, is as shown in Table 11.35.
Suppose that a random sample of 400 U.S. males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males 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.
| Marital Status | % | Expected Frequency |
|---|---|---|
| Never Married | 31.3% | |
| Married | 56.1% | |
| Widowed | 2.5% | |
| Divorced/Separated | 10.1% |
| Marital Status | Frequency |
|---|---|
| Never Married | 140 |
| Married | 238 |
| Widowed | 2 |
| Divorced/Separated | 20 |
Answer
| Marital Status | % | Expected Frequency |
|---|---|---|
| Never Married | 31.3% | 125.2 |
| Married | 56.1% | 224.4 |
| Widowed | 2.5% | 10 |
| Divorced/Separated | 10.1% | 40.4 |
a. The data fit the distribution.
b. The data do 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 hypothesis.
iii. Reason for decision: p-value < alpha
iv. Conclusion: Data do not fit the distribution.
| Marital Status | % | 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 the Advanced Placement Examinee Population for that class, and the Overall Student Population. Suppose the right column contains the results of a survey of 1,000 local students from that year who took an AP exam.
Perform a goodness-of-fit test to determine whether the local results follow the distribution of U.S. AP examinee population, based on ethnicity.
| Race/Ethnicity | AP Examinee Population | 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 U.S. AP examinee population.
b. Ha: The local results do not follow the distribution of the U.S. AP examinee population.
c. df = 5
d. chi-square distribution with df = 5
e. chi-square test statistic = 13.4
f. p-value = 0.0199
g. Check student’s solution.
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 AP examinee 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 U.S. AP examinee distribution.
11-77 §11.2 source ↗
Use the following information to answer the next two exercises. UCLA conducted a survey of more than 263,000 college freshmen 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 females and 5,000 graduating males 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 percent because of rounding.
Conduct a goodness-of-fit test to determine if the actual college majors of graduating females fit the distribution of their expected majors.
| Major | Females—Expected Major | Females—Actual Major |
|---|---|---|
| Arts & Humanities | 14% | 670 |
| Biological Sciences | 8.4% | 410 |
| Business | 13.1% | 685 |
| Education | 13% | 650 |
| Engineering | 2.6% | 145 |
| Physical Sciences | 2.6% | 125 |
| Professional | 18.9% | 975 |
| Social Sciences | 13% | 605 |
| Technical | 0.4% | 15 |
| Other | 5.8% | 300 |
| Undecided | 8% | 420 |
Answer
a. H0: The actual college majors of graduating females fit the distribution of their expected majors.
b. Ha: The actual college majors of graduating females 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. Check student’s solution.
h.
i. Alpha = 0.05
ii. Decision: Do not reject null hypothesis 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 females 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
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.
Answer
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
11-85 §11.2 source ↗
Read the statement and decide whether it is true or false.
Table 11.42 contains information from a survey of 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 United States. Perform a hypothesis test at the 5 percent significance level to determine whether the survey participants are a representative sample of the USA obese population.
| Age Class (years) | Obese (Percentage) | Expected USA Average (Percentage) |
|---|---|---|
| 20–30 | 75 | 32.6 |
| 31–40 | 26.5 | 32.6 |
| 41–50 | 13.6 | 36.6 |
| 51–60 | 21.9 | 36.6 |
| 61–70 | 21 | 39.7 |
Answer
The hypotheses for the goodness-of-fit test are:
• H0: Surveyed obese fit the distribution of expected obese.
• 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 χ2 = 9.85
• The p-value = 0.0431
• At the 5% significance level, α = 0.05. For this data, p < α.
• 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.
| 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. Check student’s solution.
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 percent 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 hotspots 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.
| Location | 20–29 | 30–39 | 40–49 | 50+ |
|---|---|---|---|---|
| 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. Check student’s solution.
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 percent 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.
| 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. Check student’s solution.
h.
i. Alpha: 0.05
ii. Decision: Reject the null hypothesis.
iii. Reason for decision: p-value < alpha
iv. Conclusion: At the 5 percent 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 there is a relationship between the level of education an individual has and salary. Conduct a test of independence.
| 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. Check student’s solution.
h.
Alpha: 0.05
Decision: Reject the null hypothesis.
Reason for decision: p-value < alpha
Conclusion: At the 5 percent 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
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
11-99 §11.3 source ↗
Read the statement and decide whether it is true or false.
Table 11.52 provides results of 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 percent significance level.
| Age Group/Net Worth Value (in millions of U.S. 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. Check student’s solution.
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 percent 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 percent level of significance.
| 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. Check student’s solution.
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 fisherman 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 percent 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. Check student’s solution.
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 United States and the joint European Union countries (EU) for the six-year period 2005 to 2010. Do the energy use values in these two areas come from the same distribution? Perform the analysis at the 5 percent significance level.
| Year | European Union | United States | Row Total |
|---|---|---|---|
| 2010 | 3,413 | 7,164 | 10,557 |
| 2009 | 3,302 | 7,057 | 10,359 |
| 2008 | 3,505 | 7,488 | 10,993 |
| 2007 | 3,537 | 7,758 | 11,295 |
| 2006 | 3,595 | 7,697 | 11,292 |
| 2005 | 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 United States is the same as in Europe between 2005 and 2010.
b. Ha: The distribution of average energy use in the United States is not the same as in Europe between 2005 and 2010.
c. df = 4
d. chi-square with df = 4
e. test statistic = 2.7434
f. p-value = 0.7395
g. Check student’s solution.
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 percent significance level, there is insufficient evidence to conclude that the average energy use values in the United States and EU are not derived from different distributions for the period from 2005 to 2010.
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. Check student’s solution.
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 12 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
11-112 §11.6 source ↗
Use the following information to answer the next 12 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
11-114 §11.6 source ↗
Use the following information to answer the next 12 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
11-116 §11.6 source ↗
Use the following information to answer the next 12 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
11-118 §11.6 source ↗
Use the following information to answer the next 12 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
11-120 §11.6 source ↗
Use the following information to answer the next 12 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
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. Check student’s solution.
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. Check student’s solution.
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. She counts 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; and 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. Check student’s solution.
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 two.
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; and $1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5 percent 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: ;
p-value:
Alpha: 0.05
Decision: Do not reject the null hypothesis.
Reason for decision: p-value > alpha
Conclusion: At the 5 percent 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 Statistics by Barbara Illowsky and Susan Dean, attributed to the Texas Education Agency (TEA) and published by OpenStax, and are used under the CC BY 4.0 licence. Access for free at https://openstax.org/books/statistics/pages/1-introduction. The original material is available at https://www.texasgateway.org/book/tea-statistics. Changes were made to the original material, including updates to art, structure, and other content updates. Art is credited in its own caption wherever the book credits it. This adaptation is neither affiliated with nor endorsed by OpenStax, Rice University or the Texas Education Agency (TEA).