Section 10.1 Exercises
CLARIFYING THE CONCEPTS
Question 10.1
1. When are two samples considered independent? (p. 576)
10.1.1
When the subjects selected for the first sample do not determine the subjects in the second sample.
Question 10.2
2. When are two samples considered dependent? (p. 576)
Question 10.3
3. What do we call the data obtained from dependent sampling? (p. 576)
10.1.3
Matched pairs or paired samples
Question 10.4
4. How do we interpret the meaning of ? (p. 578)
PRACTICING THE TECHNIQUES
CHECK IT OUT!
| To do | Check out | Topic |
|---|---|---|
| Exercises 5–8 | Example 1 | Dependent or independent sampling? |
| Exercises 9–14 | Example 2 | Calculating and |
| Exercises 15–17 | Example 3 | Paired test for : critical-value method |
| Exercises 18–20 | Example 4 | Paired test for : -value method |
| Exercises 21–26 | Example 5 | confidence interval for |
| Exercises 27–30 | Example 6 | Using a interval for to perform tests about |
Determine whether the experiments in Exercises 5–8 represent an independent sampling method or a dependent sampling method. Explain your answer.
Question 10.5
5. The Jacksonville Jaguars are interested in comparing the performance of their first-year players. For each player, a sample is taken of their games from their last year in college and compared with a sample of games taken from their first year in the pros.
10.1.5
Since both samples of games were based on the same players, this is an example of dependent sampling.
Question 10.6
6. For her senior project, an exercise science major takes a sample of females majoring in exercise science and a sample of females from her college who are not majoring in exercise science. She records the body mass index for each subject.
Question 10.7
7. Before the first lecture, an algebra instructor gives a pretest to his students to determine the students' algebra readiness. At the end of the course, the instructor gives a post-test to the same students and compares the results with the pretest.
10.1.7
Since the same students are taking both tests, this is an example of dependent sampling.
Question 10.8
8. The sheriff's department takes a sample of vehicle speeds on a certain stretch of road and compares the results to a sample of vehicle speeds on a certain stretch of a different road. Both roads have the same posted speed limit.
In Exercises 9–14, assume that samples of differences are obtained through dependent sampling and follow a normal distribution. Calculate and .
Question 10.9
9.
| Subject | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Sample 1 | 3.0 | 2.5 | 3.5 | 3.0 | 4.0 |
| Sample 2 | 2.5 | 2.5 | 2.0 | 2.0 | 1.5 |
10.1.9
Question 10.10
10.
| Subject | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Sample 1 | 10 | 12 | 9 | 14 | 15 | 8 |
| Sample 2 | 8 | 11 | 10 | 12 | 14 | 9 |
587
Question 10.11
11.
| Subject | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Sample 1 | 20 | 25 | 15 | 10 | 20 | 30 | 15 |
| Sample 2 | 30 | 30 | 20 | 20 | 25 | 35 | 25 |
10.1.11
Question 10.12
12.
| Subject | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Sample 1 | 1.5 | 1.8 | 2.0 | 2.5 | 3.0 | 3.2 | 4.0 |
| Sample 2 | 1.0 | 1.7 | 2.1 | 2.0 | 2.7 | 2.9 | 3.3 |
Question 10.13
13.
| Subject | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Sample 1 | 0 | 0.5 | 0.75 | 1.25 | 1.9 | 2.5 | 3.2 | 3.3 |
| Sample 2 | 0.25 | 0.25 | 0.75 | 1.5 | 1.8 | 2.2 | 3.3 | 3.4 |
10.1.13
Question 10.14
14.
| Subject | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Sample 1 | 105 | 88 | 103 | 97 | 115 | 125 | 122 | 92 |
| Sample 2 | 110 | 95 | 108 | 97 | 116 | 127 | 125 | 95 |
Question 10.15
15. For the data in Exercise 9, test whether , using the critical-value method and level of significance .
10.1.15
. Reject if . . Since is , we reject . There is evidence at the level of significance that the population mean difference is greater than 0.
Question 10.16
16. For the data in Exercise 10, test whether , using the critical-value method and level of significance .
Question 10.17
17. For the data in Exercise 11, test whether , using the critical-value method and level of significance .
10.1.17
vs. . Reject if . . Since is , we reject . There is evidence at the level of significance that the population mean difference is less than 0.
Question 10.18
18. For the data in Exercise 12, test whether , using the -value method and level of significance .
Question 10.19
19. For the data in Exercise 13, test whether , using the -value method and level of significance .
10.1.19
. Reject if the . . . Since the is not , we do not reject . There is insufficient evidence at the level of significance that the population mean difference is not equal to 0.
Question 10.20
20. For the data in Exercise 14, test whether , using the -value method and level of significance .
Question 10.21
21. Using the data from Exercise 9, construct a 95% confidence interval for .
10.1.21
(–0.0940, 2.294). We are 95% confident that the population mean difference lies between −0.0940 and 2.294.
Question 10.22
22. Using the data from Exercise 10, construct a 99% confidence interval for .
Question 10.23
23. Using the data from Exercise 11, construct a 90% confidence interval for .
10.1.23
(–9.106, −5.180). We are 90% confident that the population mean difference lies between −9.106 and −5.180.
Question 10.24
24. Using the data from Exercise 12, construct a 99% confidence interval for .
Question 10.25
25. Using the data from Exercise 13, construct a 95% confidence interval for .
10.1.25
(–0.181, 0.169). We are 95% confident that the population mean difference lies between −0.181 and 0.169.
Question 10.26
26. Using the data from Exercise 14, construct a 90% confidence interval for .
For Exercises 27–30 a confidence interval for is given. Use the confidence interval to test, using level of significance , whether differs from each of the indicated hypothesized values.
Question 10.27
27. A 95% confidence interval for is (–5, 5).
Hypothesized values are
- 0
- −6
- 4
10.1.27
(a) . lies inside of the interval (–5, 5), so we do not reject at the level of significance. (b) . lies outside of the interval (–5, 5), so we reject at the level of significance. (c) . lies inside of the interval (–5, 5), so we do not reject at the level of significance.
Question 10.28
28. A 99% confidence interval for is (−10, −4).
Hypothesized values are
- −12
- 0
- 4
Question 10.29
29. A 90% confidence interval for is (10, 20).
Hypothesized values are
- −10
- 25
- 0
10.1.29
(a) . lies outside of the interval (10, 20), so we reject at the level of significance. (b) . lies outside of the interval (10, 20), so we reject at the level of snificance. (c) . lies outside of the interval (10, 20), so we reject at the level of significance.
Question 10.30
30. A 95% confidence interval for is (0, 1).
Hypothesized values are
- 0.41
- 0.29
- 1.23
APPLYING THE CONCEPTS
Question 10.31
carprice
31. New Car Prices. Kelley's Blue Book (www.kbb.com) publishes data on new and used cars. The following table contains the fair market value for five new 2013 and 2014 vehicles (data recorded July 2014). We are interested in the difference in price between the 2013 models and the 2014 models. Assume that the population of price differences is normally distributed.
- Find the mean of the differences, , and the standard deviation of the differences, .
- Test whether 2014 models are on average more expensive, using level of significance .
| Toyotay Camry |
Honda Civic |
Ford F-150 |
Chevy Corvette |
Tesla Model S |
|
|---|---|---|---|---|---|
|
2014 (sample 1) |
$20,672 | $17,069 | $24,362 | $45,684 | $68,738 |
|
2013 (sample 2) |
$20,284 | $16,499 | $22,674 | $44,021 | $68,674 |
10.1.31
Differences: $388, $570, $1688, $1663, $64 (a) and (b) versus . Reject if the . = 0.0301463228; The is , so we reject . There is evidence at level of significance that the population mean of the differences in price of cars is greater than 0. That is, there is evidence at level of significance that the 2014 models of cars are on average more expensive than the 2013 models of cars.
Question 10.32
fridaythe13th
32. Does Friday the 13th Change Human Behavior? In Example 4 of Chapter 1, we discussed whether Friday the 13th changed human behavior. The researchers obtained data kept by the British Department of Transport on the traffic flow through certain junctions of the M25 motorway in England.4
- Find the mean of the differences, , and the standard deviation of the differences, .
- Perform the appropriate hypothesis test for determining whether the mean traffic flow is smaller for Friday the 13th compared with Friday the 6th, using level of significance . Assume normality.
| Friday the 6th | Friday the 13th | Difference |
|---|---|---|
| 139,246 | 138,548 | 698 |
| 134,012 | 132,908 | 1104 |
| 137,055 | 136,018 | 1037 |
| 133,732 | 131,843 | 1889 |
| 123,552 | 121,641 | 1911 |
| 121,139 | 118,723 | 2416 |
| 128,293 | 125,532 | 2761 |
| 124,631 | 120,249 | 4382 |
| 124,609 | 122,770 | 1839 |
| 117,584 | 117,263 | 321 |
588
Question 10.33
waterlootemp
33. High and low Temperatures. The University of Waterloo Weather Station tracks the daily low and high temperatures in Waterloo, Ontario, Canada. The table contains a random sample of the daily high and low temperatures in degrees Celsius for 10 days in calendar year 2010. Assume that the temperature differences are normally distributed.
| Day | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| High | 9.4 | 6.1 | 5.9 | 29.1 | 11.9 | 30.6 | 23.1 | 33.1 | 14.8 | 0.1 |
| Low | 0.8 | −8.9 | −1.3 | 19.3 | 6.7 | 21.5 | 10.5 | 18.7 | 7.4 | −9.9 |
- Find the mean of the differences, , and the standard deviation of the differences, .
- Construct and interpret a 95% confidence interval for , the population mean difference in temperature.
10.1.33
Differences: 8.6, 15, 7.2, 9.8, 5.2, 9.1, 12.6, 14.4, 7.4, 10 (a) and (b) (7.6488, 12.211); We are 95% confident that the population mean of the differences between high and low temperatures in May in Waterloo, Ontario, Canada lies between 7.6488 degrees centigrade and 12.211 degrees centigrade.
Question 10.34
nasdaq72814
34. NASDAQ Stock Prices. The table provides the start-of-trading and end-of trading prices for the eight most active stocks on July 28, 2014. Assume that the differences are normally distributed.
| Stock | End-of-trading price |
Start-of-trading price |
|---|---|---|
| Sirius XM | $3.38 | $3.44 |
| Apple | $99.02 | $97.67 |
| $74.92 | $75.19 | |
| Micron Technology | $31.98 | $33.42 |
| Dollar Tree | $54.87 | $54.22 |
| Intel | $34.23 | $34.25 |
| Microsoft | $43.97 | $44.50 |
| Cisco Systems | $25.92 | $25.97 |
- Find the mean of the differences, , and the standard deviation of the differences, .
- Test whether the population mean difference in share prices differs from zero, using level of significance a .
Question 10.35
35. New Car Prices. Use the information in Exercise 31 to construct and interpret a 95% confidence interval for , the population mean difference in price.
10.1.35
(–60.74,1809.9); We are 95% confident that the population mean of the differences between prices of 2014 model cars and prices of 2013 model cars lies between and $1809.9.
Question 10.36
36. Does Friday the 13th Change Human Behavior? Use the data from Exercise 32 to construct and interpret a 95% confidence interval for , the population mean difference in traffic for Friday the 13th and Friday the 6th.
Question 10.37
37. High and low Temperatures. Use the information in Exercise 33 for the following: Construct and interpret a 99% confidence interval for , the population mean difference in temperature.
10.1.37
(6.6528,13.207); We are 99% confident that the population mean of the differences between high and low temperatures in May in Waterloo, Ontario, Canada lies between 6.6528 degrees centigrade and 13.207 degrees centigrade.
Question 10.38
38. NASDAQ Stock Prices. Use the information in Exercise 34 for the following:
- Construct and interpret a 90% confidence interval for , the population mean difference in price.
- Explain how your confidence interval supports your conclusion to the hypothesis test in Exercise 34.
BRINGING IT ALL TOGETHER
Mathematics Scores Worldwide. The National Center for Educational Statistics publishes the results from the Trends in International Math and Science Study (TIMSS). The table contains the 2007 and 2011 mean mathematics scores for eighth-graders from various countries. Assume that the population of score differences is normally distributed. Use this information for Exercises 39–44.
| Country | 2007 | 2011 |
|---|---|---|
| Singapore | 593 | 611 |
| Japan | 570 | 570 |
| Hong Kong | 572 | 586 |
| England | 513 | 507 |
| United States | 508 | 509 |
| Hungary | 517 | 505 |
| Italy | 480 | 498 |
| Russia | 512 | 539 |
| Ukraine | 462 | 479 |
| Australia | 496 | 505 |
| South Korea | 597 | 613 |
| Slovenia | 501 | 505 |
| Thailand | 441 | 427 |
| Norway | 469 | 475 |
| Indonesia | 397 | 386 |
Question 10.39
mathscores
39. Explain why these are dependent samples and not independent samples.
10.1.39
The mean mathematics score in 2007 and 2011 are given for the same country for each of the 15 different countries.
Question 10.40
mathscores
40. Calculate the following statistics:
Question 10.41
mathscores
41. Using level of significance , test whether the 2011 scores are higher than the 2007 scores, on average.
10.1.41
versus ; Reject if the . ; ; The is , so we reject . There is evidence at level of significance that the population mean of the differences in math scores is greater than 0. That is, there is evidence at level of significance that the mean math score in 2011 is higher than the mean math score in 2007.
Question 10.42
mathscores
42. Construct a 95% confidence interval for , the population mean difference in score.
Question 10.43
mathscores
43. Use your confidence interval from the previous exercise to test, using level of significance , whether the population mean difference equals the following values:
- 15 points
- 5 points
- 0 points
10.1.43
(a) versus ; lies outside the interval, so we reject . (b) versus lies inside the interval, so we do not reject . (c) versus lies inside the interval, so we do not reject .
Question 10.44
mathscores
44. What if we added a certain constant number to each score in the table? How would this change affect the following?
- The conclusion
589
WORKING WITH LARGE DATA SETS
Phosphorus and Potassium. Use technology to solve Exercises 45–47.
Question 10.45
nutrition
45. Open the nutrition data set. Explore the variable phosphor, which lists the amount of phosphorus (in milligrams) for each food item. Generate numerical summary statistics and graphs for the amount of phosphorus in the food. What is the sample mean amount of phosphorus? The sample standard deviation?
10.1.45
,
Question 10.46
nutrition
46. Explore the variable potass, which lists the amount of potassium (in milligrams) for each food item. Generate numerical summary statistics and graphs for the amount of potassium in the food. What is the sample mean amount of potassium? The sample standard deviation?
Question 10.47
nutrition
47. Create a new variable in Excel or Minitab, phos_pot, which equals the amount of phosphorus minus the amount of potassium in each food item. Use a paired sample hypothesis test to test, at level of significance , whether the population mean difference differs from 0.
10.1.47
vs. . Reject if the . . . Since the is ≤ 0.05, we reject . There is evidence at the level of significance that the population mean difference is not equal to 0.
WORKING WITH LARGE DATA SETS
Restaurants. Open the data set, Restaurants. Here, we will examine the difference in the number of fast food restaurants and the number of full service restaurants. Use technology to do the following:
Question 10.48
restaurants
48. Obtain a random sample of size 100 from the data set.
Question 10.49
restaurants
49. For each county, compute the difference (number of fast food restaurants minus number of full service restaurants).
10.1.49
Answers will vary.
Question 10.50
restaurants
50. Test whether the population mean difference equals zero, using level of significance .
Question 10.51
restaurants
51. Find the actual value of the population mean difference. Did your hypothesis test in Exercise 50 make the right decision? Explain.
10.1.51
–3.47; Rest of answer will vary.