Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.
Chapter 11: The Chi-Square Distribution
61 problems · 2 figures11-1 §11.1 source ↗
Answer
11-3 §11.1 source ↗
Answer
11-5 §11.1 source ↗
Answer
11-6 §11.2 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
11-8 §11.2 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
11-10 §11.2 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
11-12 §11.2 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
11-16 §11.3 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
11-18 §11.3 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 11.22.
The actual distribution for a class of 20 is in Table 11.23.
______
| 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
11-20 §11.3 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 11.22.
The actual distribution for a class of 20 is in Table 11.23.
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
11-21 §11.3 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 11.22.
The actual distribution for a class of 20 is in Table 11.23.
At the 5% 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
11-23 §11.3 source ↗
Use the following information to answer the next eight 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 11.24.
The percentage of each ethnic group in Santa Clara County is as in Table 11.25.
H0: _______
| Ethnicity | Number of cases |
|---|---|
| White | 2,229 |
| Hispanic/Latino | 1,157 |
| Black/African-American | 457 |
| Asian, Pacific Islander | 232 |
| Total = 4,075 |
| 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
11-25 §11.3 source ↗
Use the following information to answer the next eight 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 11.24.
The percentage of each ethnic group in Santa Clara County is as in Table 11.25.
Is this a right-tailed, left-tailed, or two-tailed test?
| Ethnicity | Number of cases |
|---|---|
| White | 2,229 |
| Hispanic/Latino | 1,157 |
| Black/African-American | 457 |
| Asian, Pacific Islander | 232 |
| Total = 4,075 |
| 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
11-27 §11.3 source ↗
Use the following information to answer the next eight 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 11.24.
The percentage of each ethnic group in Santa Clara County is as in Table 11.25.
χ2 test statistic = _______
| Ethnicity | Number of cases |
|---|---|
| White | 2,229 |
| Hispanic/Latino | 1,157 |
| Black/African-American | 457 |
| Asian, Pacific Islander | 232 |
| Total = 4,075 |
| 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
11-28 §11.3 source ↗
Use the following information to answer the next eight 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 11.24.
The percentage of each ethnic group in Santa Clara County is as in Table 11.25.
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the confidence level.
Let α = 0.05
Decision: ________________
Reason for the Decision: ________________
Conclusion (write out in complete sentences): ________________
| Ethnicity | Number of cases |
|---|---|
| White | 2,229 |
| Hispanic/Latino | 1,157 |
| Black/African-American | 457 |
| Asian, Pacific Islander | 232 |
| Total = 4,075 |
| 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
Graph: Answers may vary.
Decision: Cannot accept the null hypothesis.
Reason for the Decision: Calculated value of test statistics is either in or out of the tail of the distribution.
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.
11-30 §11.4 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
11-32 §11.4 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
11-34 §11.4 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 11.26 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 = _______
| 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
11-36 §11.4 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 11.26 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?
| 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
11-39 §11.4 source ↗
| Smoking level per day | Black | Native Hawaiian | Hispanic/Latinos | 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 Hispanic/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 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 |
| Smoking level per day | Black | Native Hawaiian | Hispanic/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 |
11-41 §11.4 source ↗
| Smoking level per day | Black | Native Hawaiian | Hispanic/Latinos | 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 Hispanic/Latino people, 2,305 Japanese American people, and 3,970 White people.
Enter expected values in Table 11.27. Round to two decimal places.
Calculate the following values:
Answer
| Smoking level per day | Black | Native Hawaiian | Hispanic/Latino | Japanese Americans | 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 |
| Smoking level per day | Black | Native Hawaiian | Hispanic/Latino | Japanese Americans | 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 |
11-43 §11.4 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 Hispanic/Latino people, 2,305 Japanese American people, and 3,970 White people.
test statistic = ______
Answer
11-44 §11.4 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 Hispanic/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
11-46 §11.4 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. Cannot accept the null hypothesis.
b. Calculated value of test statistics is either in or out of the tail of the distribution.
c. There is sufficient evidence to conclude that smoking level is dependent on ethnic group.
11-48 §11.5 source ↗
Answer
11-50 §11.5 source ↗
Answer
11-52 §11.5 source ↗
Answer
11-54 §11.5 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 11.28.
df = _______
| 20–30 | 30–40 | 40–50 | 50–60 | |
|---|---|---|---|---|
| Private practice | 16 | 40 | 38 | 6 |
| Hospital | 8 | 44 | 59 | 39 |
Answer
11-57 §11.6 source ↗
Answer
11-59 §11.6 source ↗
Answer
11-61 §11.6 source ↗
Answer
11-63 §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
11-65 §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
11-67 §11.2 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
11-69 §11.2 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
11-71 §11.2 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
11-72 §11.2 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 area associated with the level of confidence.
Answer
11-74 §11.2 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
11-76 §11.2 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
11-78 §11.2 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.
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. Answers may vary.
g.
i. Alpha = 0.05
ii. Decision: Cannot reject null hypothesis.
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.
11-80 §11.2 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.
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. Answers may vary.
g.
i. Alpha: 0.05
ii. Decision: Cannot accept the null hypothesis.
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.
11-82 §11.2 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.
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 distribution with df = 14
e. chi-square test statistic = 5.2094
f. Answers may vary.
g.
i. Alpha = 0.05
ii. Decision: Cannot accept the null hypothesis
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.
11-84 §11.2 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.
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: ;
Alpha: 0.05
Decision: Cannot reject the null hypothesis.
Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.
11-87 §11.3 source ↗
The marital status distribution of the U.S. male population, ages 15 and older, is as shown in Table 11.31.
Suppose that a random sample of 400 U.S. young adult men, 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 11.31, rounding to two decimal places.
| Marital status | Percent | 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 | Percent | 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 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. Answers may vary.
h.
i. Alpha = 0.05
ii. Decision: Cannot accept null hypothesis at the 5% level of significance
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: Data does not fit the distribution.
| 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 |
11-89 §11.3 source ↗
Use the following information to answer the next two exercises: The columns in Table 11.33 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 who use mass transit, based on ethnicity.
| 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. Answers may vary.
g.
i. Alpha = 0.05
ii. Decision: Cannot accept null when a = 0.05
iii. Reason for Decision: Calculated value of test statistics is either in or out of the tail of the distribution.
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 students who use mass transit distribution.
11-91 §11.3 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 a certain year. 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 11.91 and Exercise 11.92. 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.
| 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. Answers may vary.
g.
i. Alpha = 0.05
ii. Decision: Cannot reject null when a = 0.05 and a = 0.01
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating women fits the distribution of their expected majors.
11-94 §11.3 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
11-96 §11.3 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
11-98 §11.3 source ↗
Read the statement and decide whether it is true or false.
Table 11.38 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.
| 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.
11-100 §11.4 source ↗
| 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. Answers may vary.
g.
i. Alpha: 0.05
ii. Decision: Cannot reject the null hypothesis.
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent.
11-102 §11.4 source ↗
| 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. Answers may vary.
g.
i. Alpha: 0.05
ii. Decision: Cannot reject the null hypothesis.
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent.
11-104 §11.4 source ↗
| 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. Answers may vary.
g.
i. Alpha: 0.05
ii. Decision: Cannot accept the null hypothesis.
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: At the 5% significance level, There is sufficient evidence that types of fries and location are dependent.
11-106 §11.4 source ↗
| 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. Answers may vary.
g.
Alpha: 0.05
Decision: Cannot accept the null hypothesis.
Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent.
11-108 §11.4 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
11-110 §11.4 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
11-112 §11.4 source ↗
Read the statement and decide whether it is true or false.
Table 11.48 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.
| 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. Answers may vary.
g.
i. Alpha: 0.05
ii. Decision: Cannot reject the null hypothesis.
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
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.
11-114 §11.5 source ↗
| 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. Answers may vary.
g.
i. Alpha: 0.05
ii. Decision: Cannot reject the null hypothesis.
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors.
11-116 §11.5 source ↗
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. Answers may vary.
g.
i. Alpha: 0.05
ii. Decision: Cannot accept the null hypothesis.
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
iv. Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake
11-118 §11.5 source ↗
| 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. Answers may vary.
g.
i. Alpha: 0.05
ii. Decision: Cannot reject the null hypothesis.
iii. Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
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.
11-120 §11.6 source ↗
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: Cannot accept 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.
11-122 §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 Business Statistics 2e by Alexander Holmes, 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-business-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.