Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.
Chapter 8: Confidence Intervals
67 problems · 11 figures8-1 §8.1 source ↗
Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 lb. We wish to construct a 95 percent confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 lb. The sample standard deviation is 11 lb.
Identify the following:
a. = _____
b. σ = _____
c. n = _____
Answer
a. 244
b. 15
c. 50
8-3 §8.1 source ↗
Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 lb. We wish to construct a 95 percent confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 lb. The sample standard deviation is 11 lb.
Which distribution should you use for this problem?
Answer
8-5 §8.1 source ↗
Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 lb. We wish to construct a 95 percent confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 lb. The sample standard deviation is 11 lb.
What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why?
Answer
8-7 §8.1 source ↗
Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal.
In words, define the random variables X and .
Answer
8-9 §8.1 source ↗
Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal.
Construct a 90 percent confidence interval for the population mean time to complete the forms. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
CI: (7.9441, 8.4559)
EBM = 0.26
8-11 §8.1 source ↗
Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal.
If the Census did another survey, kept the error bound the same, and surveyed only 50 people instead of 200, what would happen to the level of confidence? Why?
Answer
8-13 §8.1 source ↗
Use the following information to answer the next 10 exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 lb, with a standard deviation of 0.1 lb. The population standard deviation is known to be 0.2 lb.
Identify the following:
a. = ______
b. σ = ______
c. n = ______
Answer
a. = 2.2
b. σ = 0.2
c. n = 20
8-15 §8.1 source ↗
Use the following information to answer the next 10 exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 lb, with a standard deviation of 0.1 lb. The population standard deviation is known to be 0.2 lb.
In words, define the random variable .
Answer
8-17 §8.1 source ↗
Use the following information to answer the next 10 exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 lb, with a standard deviation of 0.1 lb. The population standard deviation is known to be 0.2 lb.
Construct a 90 percent confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
CI: (2.1264, 2.2736)
8-19 §8.1 source ↗
Use the following information to answer the next 10 exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 lb, with a standard deviation of 0.1 lb. The population standard deviation is known to be 0.2 lb.
In complete sentences, explain why the confidence interval in Exercise 8.17 is larger than in Exercise 8.18.
Answer
8-21 §8.1 source ↗
Use the following information to answer the next 10 exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 lb, with a standard deviation of 0.1 lb. The population standard deviation is known to be 0.2 lb.
What would happen if 40 heads of lettuce were sampled instead of 20 and the error bound remained the same?
Answer
8-23 §8.1 source ↗
Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that 25 winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for winter Foothill College students. Let X = the age of a winter Foothill College student.
= _____
Answer
8-25 §8.1 source ↗
Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that 25 winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for winter Foothill College students. Let X = the age of a winter Foothill College student.
________ = 15
Answer
8-27 §8.1 source ↗
Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that 25 winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for winter Foothill College students. Let X = the age of a winter Foothill College student.
What is estimating?
Answer
8-29 §8.1 source ↗
Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that 25 winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for winter Foothill College students. Let X = the age of a winter Foothill College student.
As a result of your answer to Exercise 8.26, state the exact distribution to use when calculating the confidence interval.
Answer
8-31 §8.1 source ↗
Construct a 95 percent confidence interval for the true mean age of winter Foothill College students by working out and then answering the next eight exercises.
How much area is in each tail? =________
Answer
8-33 §8.1 source ↗
Construct a 95 percent confidence interval for the true mean age of winter Foothill College students by working out and then answering the next eight exercises.
The 95 percent confidence interval is __________________.
Answer
8-35 §8.1 source ↗
Construct a 95 percent confidence interval for the true mean age of winter Foothill College students by working out and then answering the next eight exercises.
In one complete sentence, explain what the interval means.
Answer
8-37 §8.1 source ↗
Construct a 95 percent confidence interval for the true mean age of winter Foothill College students by working out and then answering the next eight exercises.
Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90 percent? Why?
Answer
8-39 §8.2 source ↗
Use the following information to answer the next five exercises: A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hr, with a sample standard deviation of 0.5 hr.
Define the random variables X and in words.
Answer
8-41 §8.2 source ↗
Use the following information to answer the next five exercises: A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hr, with a sample standard deviation of 0.5 hr.
Construct a 95 percent confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
CI: (1.3808, 1.6192)
EBM = 0.12
8-43 §8.2 source ↗
Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watch an average of 151 hours each month, with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Identify the following:
a. =_______
b. =_______
c. n =_______
d. n – 1 =_______
Answer
a. = 151
b. = 32
c. n = 108
d. n – 1 = 107
8-45 §8.2 source ↗
Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watch an average of 151 hours each month, with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Define the random variable in words.
Answer
8-47 §8.2 source ↗
Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watch an average of 151 hours each month, with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Construct a 99 percent confidence interval for the population mean hours spent watching television per month. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
CI: (142.92, 159.08)
EBM = 8.08
8-49 §8.2 source ↗
Use the following information to answer the next 13 exercises: The data in Table 8.10 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag.
Calculate the following:
a. =______
b. =______
c. n =______
| X | Freq. |
|---|---|
| 1 | 1 |
| 2 | 7 |
| 3 | 18 |
| 4 | 7 |
| 5 | 6 |
Answer
a. 3.26
b. 1.02
c. 39
8-51 §8.2 source ↗
Use the following information to answer the next 13 exercises: The data in Table 8.10 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag.
What is estimating?
| X | Freq. |
|---|---|
| 1 | 1 |
| 2 | 7 |
| 3 | 18 |
| 4 | 7 |
| 5 | 6 |
Answer
8-53 §8.2 source ↗
Use the following information to answer the next 13 exercises: The data in Table 8.10 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag.
As a result of your answer to Exercise 8.52, state the exact distribution to use when calculating the confidence interval.
| X | Freq. |
|---|---|
| 1 | 1 |
| 2 | 7 |
| 3 | 18 |
| 4 | 7 |
| 5 | 6 |
Answer
8-55 §8.2 source ↗
Construct a 95 percent confidence interval for the true mean number of colors on national flags.
How much area is in each tail?
Answer
8-57 §8.2 source ↗
Construct a 95 percent confidence interval for the true mean number of colors on national flags.
The 95 percent confidence interval is_____.
Answer
8-59 §8.2 source ↗
Construct a 95 percent confidence interval for the true mean number of colors on national flags.
In one complete sentence, explain what the interval means.
Answer
8-60 §8.2 source ↗
Construct a 95 percent confidence interval for the true mean number of colors on national flags.
Using the same , , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know?
Answer
8-63 §8.3 source ↗
Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percentage of women who make the majority of household purchasing decisions.
If it were later determined that it was important to be more than 90 percent confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why?
Answer
8-65 §8.3 source ↗
Use the following information to answer the next five exercises: Suppose a marketing company conducted a survey. It randomly surveyed 200 households and found that in 120 of them, the women made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions.
Define the random variables X and P′ in words.
Answer
8-67 §8.3 source ↗
Use the following information to answer the next five exercises: Suppose a marketing company conducted a survey. It randomly surveyed 200 households and found that in 120 of them, the women made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions.
Construct a 95 percent confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
CI: (0.5321, 0.6679)
EBM: 0.0679
8-69 §8.3 source ↗
Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82 percent of manual laborers preferred trucks, 62 percent of non-manual wage earners preferred trucks, 54 percent of mid-level managers preferred trucks, and 26 percent of executives preferred trucks.
We are interested in finding the 95 percent confidence interval for the percentage of executives who prefer trucks. Define random variables X and P′ in words.
Answer
8-71 §8.3 source ↗
Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82 percent of manual laborers preferred trucks, 62 percent of non-manual wage earners preferred trucks, 54 percent of mid-level managers preferred trucks, and 26 percent of executives preferred trucks.
Construct a 95 percent confidence interval. State the confidence interval, sketch the graph, and calculate the error bound.
Answer
CI: (0.19432, 0.33068)
EBM: 0.0707
8-73 §8.3 source ↗
Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82 percent of manual laborers preferred trucks, 62 percent of non-manual wage earners preferred trucks, 54 percent of mid-level managers preferred trucks, and 26 percent of executives preferred trucks.
The sampling error given in the survey is ±2 percent. Explain what the ±2 percent means.
Answer
8-75 §8.3 source ↗
Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered "the economy." We are interested in the population proportion of voters who believe the economy is the most important.
Define the random variable P′ in words.
Answer
8-77 §8.3 source ↗
Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered "the economy." We are interested in the population proportion of voters who believe the economy is the most important.
Construct a 90 percent confidence interval, and state the confidence interval and the error bound.
Answer
CI: (0.62735, 0.67265);
EBM: 0.02265
8-79 §8.3 source ↗
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.
What is being counted?
Answer
8-81 §8.3 source ↗
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.
Calculate the following:
a. x = _______
b. n = _______
c. p′ = _______
Answer
a. x = 64
b. n = 80
c. p′ = 0.8
8-83 §8.3 source ↗
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.
Define a new random variable P′. What is p′ estimating?
Answer
8-85 §8.3 source ↗
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.
State the estimated distribution of P′. Construct a 92 percent confidence interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet.
Answer
CI = (0.72171, 0.87829).
8-87 §8.3 source ↗
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.
How much area is in each tail?
Answer
8-89 §8.3 source ↗
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.
The 92 percent confidence interval is _______.
Answer
8-91 §8.3 source ↗
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.
In one complete sentence, explain what the interval means.
Answer
8-93 §8.3 source ↗
Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 p.m., Monday night, ages 8 to 12, beginning ice-skating class is picked. In that class are 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.
Using the same p′ and n = 80, how would the error bound change if the confidence level were increased to 98 percent? Why?
Answer
8-95 §8.1 source ↗
Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95 percent confidence interval for the mean height of male Swedes. 48 male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 in.
a.
i. =________
ii. σ =________
iii. n =________
b. In words, define the random variables X and .
c. Which distribution should you use for this problem? Explain your choice.
d. Construct a 95 percent confidence interval for the population mean height of male Swedes.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
e. What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why?
Answer
a.
i. 71
ii. 3
iii. 48
b. X is the height of a Swedish male, and is the mean height from a sample of 48 Swedish males.
c. Normal. We know the standard deviation for the population, and the sample size is greater than 30.
d.
i. CI: (70.151, 71.49)
ii.
iii. EBM = 0.849
e. The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.
8-97 §8.1 source ↗
Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal.
a.
i. =________
ii. σ =________
iii. n =________
b. In words, define the random variables X and .
c. Which distribution should you use for this problem? Explain your choice.
d. Construct a 90 percent confidence interval for the population mean time to complete the tax forms.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
e. If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, which changes should it make?
f. If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why?
g. Suppose that the firm decided that it needed to be at least 96 percent confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why?
Answer
a.
i. = 23.6
ii. = 7
iii. n = 100
b. X is the time needed to complete an individual tax form. is the mean time to complete tax forms from a sample of 100 customers.
c. because we know sigma.
d.
i. (22.228, 24.972)
ii.
iii. EBM = 1.372
e. It will need to change the sample size. The firm needs to determine what the confidence level should be and then apply the error bound formula to determine the necessary sample size.
f. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval.
g. According to the error bound formula, the firm needs to survey 206 people. Because we increase the confidence level, we need to increase either our error bound or the sample size.
8-99 §8.1 source ↗
A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9, with a sample standard deviation of 2.8.
a.
i. =________
ii. σ =________
iii. n =________
b. Define the random variables X and in words.
c. Which distribution should you use for this problem? Explain your choice.
d. Construct a 90 percent confidence interval for the population mean number of letters campers send home.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
e. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why?
Answer
a.
i. 7.9
ii. 2.5
iii. 20
b. X is the number of letters a single camper will send home. is the mean number of letters sent home from a sample of 20 campers.
c.
N
d.
i. CI: (6.98, 8.82)
ii.
iii. EBM: 0.92
e. The error bound and confidence interval will decrease.
8-101 §8.1 source ↗
The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees during each election cycle. During the 2012 campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Table 8.11 shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is σ = $909,200.
a. Find the point estimate for the population mean.
b. Using 95 percent confidence, calculate the error bound.
c. Create a 95 percent confidence interval for the mean total individual contributions.
d. Interpret the confidence interval in the context of the problem.
| $3,600 | $1,243,900 | $10,900 | $385,200 | $581,500 |
| $7,400 | $2,900 | $400 | $3,714,500 | $632,500 |
| $391,000 | $467,400 | $56,800 | $5,800 | $405,200 |
| $733,200 | $8,000 | $468,700 | $75,200 | $41,000 |
| $13,300 | $9,500 | $953,800 | $1,113,500 | $1,109,300 |
| $353,900 | $986,100 | $88,600 | $378,200 | $13,200 |
| $3,800 | $745,100 | $5,800 | $3,072,100 | $1,626,700 |
| $512,900 | $2,309,200 | $6,600 | $202,400 | $15,800 |
Answer
a. = $568,873
b. CL = 0.95, α = 1 – 0.95 = 0.05, = 1.96
EBM = = 1.96 = $281,764
c. − EBM = 568,873 − 281,764 = 287,109
+ EBM = 568,873 + 281,764 = 850,637
Alternate solution:
Using the TI-83, 83+, 84, 84+ Calculator1. Press STAT and arrow over to TESTS.
2. Arrow down to 7:ZInterval.
3. Press ENTER.
4. Arrow to Stats and press ENTER.
5. Arrow down and enter the following values:
• σ : 909,200
• : 568,873
• n: 40
• CL: 0.95
6. Arrow down to Calculate and press ENTER.
7. The confidence interval is ($287,114, $850,632).
8. Notice the small difference between the two solutions—these differences are simply due to rounding error in the hand calculations.
d. We estimate with 95 percent confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637.
8-103 §8.1 source ↗
Answer
Use the formula for EBM, solved for n:
From the statement of the problem, you know that σ = 2.5, and you need EBM = 1.
z = z0.035 = 1.812.
(This is the value of z for which the area under the density curve to the right of z is 0.035.)
You need to measure at least 21 male students to achieve your goal.
8-105 §8.2 source ↗
A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414, 1,550, 2,109, 9,350, 21,828, 4,300, 5,944, 5,722, 2,825, 2,044, 5,481, 5,200, 5,853, 2,750, 10,012, 6,357, 27,000, 9,414, 7,681, 3,200, 17,500, 9,200, 7,380, 18,314, 6,557, 13,713, 17,768, 7,493, 2,771, 2,861, 1,263, 7,285, 28,165, 5,080, 11,622. Assume the underlying population is normal.
a.
i. = __________
ii. = __________
iii. n = __________
iv. n – 1 = __________
b. Define the random variables and in words.
c. Which distribution should you use for this problem? Explain your choice.
d. Construct a 95 percent confidence interval for the population mean enrollment at community colleges in the United States.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
e. What will happen to the error bound and confidence interval if 500 community colleges are surveyed? Why?
Answer
a.
i. 8,629
ii. 6,944
iii. 35
iv. 34
b.
c.
i. CI: (6244, 11,014)
ii.
iii. EB = 2385
d. It will become smaller.
8-107 §8.2 source ↗
A pharmaceutical company makes a drug used during surgery. It is assumed that the distribution for the length of time the drug lasts is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the antibiotic drug for each patient (in hours) was as follows: 2.7, 2.8, 3.0, 2.3, 2.3, 2.2, 2.8, 2.1, and 2.4.
a.
i. = __________
ii. = __________
iii. n = __________
iv. n – 1 = __________
b. Define the random variable in words.
c. Define the random variable in words.
d. Which distribution should you use for this problem? Explain your choice.
e. Construct a 95 percent confidence interval for the population mean length of time.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
f. What does it mean to be 95 percent confident in this problem?
Answer
a.
i. = 2.51
ii. = 0.318
iii. n = 9
iv. n - 1 = 8
b. The effective length of time for a tranquilizer
c. The mean effective length of time of tranquilizers from a sample of nine patients
d. We need to use a Student’s t-distribution, because we do not know the population standard deviation.
e.
i. CI: (2.27, 2.76)
ii. Check student's solution.
iii. EBM: 0.25
f. If we were to sample many groups of nine patients, 95 percent of the samples would contain the true population mean length of time.
8-109 §8.2 source ↗
The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees during each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns.
The FEC has reported financial information for 556 Leadership PACs that operated during the 2011–2012 election cycle. The following table shows the total receipts during this cycle for a random selection of 20 Leadership PACs.
Use the sample data to construct a 96 percent confidence interval for the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle. Use the Student's t-distribution.
| $46,500.00 | $0 | $40,966.50 | $105,887.20 | $5,175.00 |
| $29,050.00 | $19,500.00 | $181,557.20 | $31,500.00 | $149,970.80 |
| $2,555,363.20 | $12,025.00 | $409,000.00 | $60,521.70 | $18,000.00 |
| $61,810.20 | $76,530.80 | $119,459.20 | $0 | $63,520.00 |
| $6,500.00 | $502,578.00 | $705,061.10 | $708,258.90 | $135,810.00 |
| $2,000.00 | $2,000.00 | $0 | $1,287,933.80 | $219,148.30 |
Answer
Note that we are not given the population standard deviation, only the standard deviation of the sample.
There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29.
CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04.
= 2.150.
- EBM = $251,854.23 - $204,561.66 = $47,292.57.
+ EBM = $251,854.23 + $204,561.66 = $456,415.89.
We estimate with 96 percent confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89.
Alternate Solution
Using the TI-83, 83+, 84, 84+ CalculatorEnter the data as a list.
Press
STATand arrow over to
TESTS.
Arrow down to
8:TInterval.
Press
ENTER.
Arrow to Data and press
ENTER.
Arrow down and enter the name of the list where the data are stored.
Enter
Freq: 1
Enter
C-Level: 0.96
Arrow down to
Calculateand press
Enter.
The 96 percent confidence interval is ($47,262, $456,447).
The difference between solutions arises from rounding differences.
8-111 §8.2 source ↗
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected, and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats, and the sample standard deviation is 4.1 seats.
a.
i. = __________
ii. = __________
iii. n = __________
iv. n – 1 = __________
b. Define the random variables and in words.
c. Which distribution should you use for this problem? Explain your choice.
d. Construct a 92 percent confidence interval for the population mean number of unoccupied seats per flight.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
Answer
a.
i. =
ii. =
iii. n =
iv. n - 1 =
b. X is the number of unoccupied seats on a single flight. is the mean number of unoccupied seats from a sample of 225 flights.
c. We will use a Student’s t-distribution, because we do not know the population standard deviation.
d.
i. CI: (11.12 , 12.08)
ii. Check student's solution.
iii. EBM: 0.48
8-113 §8.2 source ↗
Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8, 8, 10, 7, 9, 9. Assume the underlying distribution is approximately normal.
a. Construct a 90 percent confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
b. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done?
c. Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies.
d. Calculate the mean.
e. Is the mean within the interval you calculated in Part a? Did you expect it to be? Why or why not?
Answer
a.
i. CI: (7.64, 9.36)
ii.
iii. EBM: 0.86
b. The sample should have been increased.
c. Answers will vary.
d. Answers will vary.
e. Answers will vary.
8-115 §8.2 source ↗
Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16-oz serving size. The sample mean is 13.30, with a sample standard deviation of 1.55. Assume the underlying population is normally distributed.
Find the 95 percent confidence interval for the true population mean for the amount of soda served.
a. (12.42, 14.18)
b. (12.32, 14.29)
c. (12.50, 14.10)
d. Impossible to determine
Answer
8-117 §8.3 source ↗
Insurance companies are interested in knowing the population percentage of drivers who always buckle up before riding in a car.
a. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95 percent confident that the population proportion is estimated to within 0.03?
b. If it were later determined that it was important to be more than 95 percent confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why?
Answer
a. 1,068
b. The sample size would need to be increased, because the critical value increases as the confidence level increases.
8-119 §8.3 source ↗
According to a recent survey of 1,200 people, 61 percent believe that the president is doing an acceptable job. We are interested in the population proportion of people who believe the president is doing an acceptable job.
a. Define the random variables X and P′ in words.
b. Which distribution should you use for this problem? Explain your choice.
c. Construct a 90 percent confidence interval for the population proportion of people who believe the president is doing an acceptable job.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
Answer
a.
X = the number of people who believe that the president is doing an acceptable job;
P′ = the proportion of people in a sample who believe that the president is doing an acceptable job.
b.
c.
i. CI: (0.59, 0.63)
ii. Check student’s solution.
iii. EBM: 0.02
8-121 §8.3 source ↗
Refer to the information in Exercise 8.120.
a. Construct three 95 percent confidence intervals:
i. percentage of all ethnicity C who would welcome a person of ethnicity D into their families
ii. percentage of all ethnicity C who would welcome a person of ethnicity A into their families
iii. percentage of all ethnicity C who would welcome a person of ethnicity B into their families
b. Even though the three point estimates are different, do any of the confidence intervals overlap? Which?
c. For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions?
d. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions?
Answer
a.
i. (0.72, 0.82)
ii. (0.65, 0.76)
iii. (0.60, 0.72)
b. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.
c. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.
d. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families.
8-123 §8.3 source ↗
A telephone poll of 1,000 adult Americans was reported in an issue of a national magazine. One of the questions asked, “What is the main problem facing the country?” Twenty percent responded "crime". We are interested in the population proportion of adult Americans who believe that crime is the main problem.
a. Define the random variables X and P′ in words.
b. Which distribution should you use for this problem? Explain your choice.
c. Construct a 95 percent confidence interval for the population proportion of adult Americans who believe that crime is the main problem.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
d. Suppose we want to lower the sampling error. What is one way to accomplish that?
e. The sampling error given by the group of researchers who conducted the poll is ±3 percent. In one to three complete sentences, explain what the ±3 percent represents.
Answer
a. X = the number of adult Americans who believe that crime is the main problem; P′ = the proportion of adult Americans who believe that crime is the main problem.
b. Because we are estimating a proportion, that P′ = 0.2 and n = 1,000, the distribution we should use is .
c.
i. CI: (0.18, 0.22)
ii. Check student’s solution.
iii. EBM: 0.02
d. One way to lower the sampling error is to increase the sample size.
e. The stated ± 3 percent represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3 percent. Thus, they estimate the percentage of adult Americans who the percentage of adult Americans who that crime is the main problem to be between 18 percent and 22 percent.
8-125 §8.3 source ↗
Use the following information to answer the next three exercises: According to a Field Poll, 79 percent of California adults (actual results are 400 out of 506 surveyed) believe that education and our schools is one of the top issues facing California. We wish to construct a 90 percent confidence interval for the true proportion of California adults who believe that education and the schools is one of the top issues facing California.
A point estimate for the true population proportion is _______.
a. 0.90
b. 1.27
c. 0.79
d. 400
Answer
8-127 §8.3 source ↗
Use the following information to answer the next three exercises: According to a Field Poll, 79 percent of California adults (actual results are 400 out of 506 surveyed) believe that education and our schools is one of the top issues facing California. We wish to construct a 90 percent confidence interval for the true proportion of California adults who believe that education and the schools is one of the top issues facing California.
The error bound is approximately _____.
a. 1.581
b. 0.791
c. 0.059
d. 0.030
Answer
8-129 §8.3 source ↗
Use the following information to answer the next two exercises: Five hundred eleven (511) homes in a certain southern California community are randomly surveyed to determine whether they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed meet the minimum recommendations for earthquake preparedness, and 338 do not.
The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______.
a. 0.6614
b. 0.3386
c. 173
d. 338
Answer
8-131 §8.3 source ↗
Use the following information to answer the next two exercises: Five hundred eleven (511) homes in a certain southern California community are randomly surveyed to determine whether they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed meet the minimum recommendations for earthquake preparedness, and 338 do not.
A national survey of 1,000 adults was conducted on May 13, 2013, by a group of researchers. It concluded with 95 percent confidence that 49 percent to 55 percent of Americans believe that big-time college sports programs corrupt the process of higher education.
a. Find the point estimate and the error bound for this confidence interval.
b. Can we (with 95 percent confidence) conclude that more than half of all American adults believe this?
c. Use the point estimate from Part a and n = 1,000 to calculate a 75 percent confidence interval for the proportion of American adults who believe that major college sports programs corrupt higher education.
d. Can we (with 75 percent confidence) conclude that at least half of all American adults believe this?
Answer
a. p′ = = 0.52; EBP = 0.55 – 0.52 = 0.03
b. No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.
c. CL = 0.75, so α = 1 – 0.75 = 0.25 and . (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.)
(p′ - EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538)
Alternate Solution
Using the TI-83, 83+, 84, 84+ CalculatorSTAT TESTS A: 1-PropZinterval with x = (0.52)(1,000), n = 1,000, CL = 0.75.
Answer is (0.502, 0.538).
d. Yes, this interval does not fall below 0.50, so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75 percent confidence.
8-133 §8.3 source ↗
Use the following information to answer the next two exercises: Five hundred eleven (511) homes in a certain southern California community are randomly surveyed to determine whether they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed meet the minimum recommendations for earthquake preparedness, and 338 do not.
You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95 percent confidence and a margin of error no greater than 5 percent. How many students must you interview?
Answer
CL = 0.95; α = 1 – 0.95 = 0.05; = 0.025; = 1.96. Use p′ = q′ = 0.5.
You need to interview at least 385 students to estimate the proportion to within 5 percent at 95 percent confidence.
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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).