A store manager wants to know the percentage of shoppers who use coupons. He surveys everyone who enters the meat aisle. He finds that 60 percent of shoppers use coupons. Does the sampling method the store manager used support a valid inference?
No, because the sample is not chosen randomly.
No, because not all people who enter the meat aisle use coupons.
Yes, because the sample contains coupon users and non-users.
Yes, because the manager asked everyone who entered the meat aisle.
No, because the sample is not chosen randomly.
A quality control manager wants to determine the percentage of defective light bulbs in the production line. Which of the following sampling methods would most likely lead to a valid inference?
Random sampling, where each light bulb has an equal chance of being selected, would most likely lead to a valid inference.
A quality control manager wants to determine the percentage of defective light bulbs in the production line. Which of the following sampling methods would most likely lead to a valid inference?(1 point)
Responses
testing the first 2,000 light bulbs in the production line
testing the first 2,000 light bulbs in the production line
testing every fiftieth light bulb in the production line
testing every fiftieth light bulb in the production line
testing the last 4,000 bulbs in the production line
testing the last 4,000 bulbs in the production line
retesting 100 light bulbs that have already been tested
Testing every fiftieth light bulb in the production line, where a random starting point is chosen and every fiftieth bulb on the line is selected for testing, would most likely lead to a valid inference.
Avery randomly selects 50 colored pencils from a basket and finds that 2 are not sharpened. If the basket contains 300 colored pencils, how many would you expect to be unsharpened?(1 point)
Responses
20
20
2
2
12
12
120
You would expect approximately 12 colored pencils to be unsharpened.
To find out, you can set up a proportion:
unsharpened pencils in sample/total sample = unsharpened pencils in population/total population
Substituting the given values, 2/50 = x/300.
Solving for x:
2/50 = x/300
x = (2/50) * 300
x = 12
There are 628 students in Tan’s high school. Tan surveys 120 randomly chosen students about their favorite movie genre. The results are as follows.
I'm sorry, but you have not provided any results for me to work with. Please provide the results of the survey, or let me know how I can assist you.
Random Sampling Quick Check
4 of 54 of 5 Items
Question
There are 628 students in Tan’s high school. Tan surveys 120 randomly chosen students about their favorite movie genre. The results are as follows.
Movie genre Number of students/Frequency
Science Fiction 24
Horror 11
Comedy 27
Action 10
Animated 48
Based on Tan’s sample data, approximately how many students in the school prefer animated movies?
(1 point)
Responses
125
125
251
251
141
141
52
Based on Tan’s sample data, approximately 141 students in the school prefer animated movies.
To find out, you can set up the proportion:
Number of students who prefer animated movies in sample/total sample = Number of students who prefer animated movies in population/total population
Substituting the given values, we have:
48/120 = x/628
Solving for x:
48/120 = x/628
x = (48/120) * 628
x ≈ 250.9333
Rounding to the nearest whole number, we get approximately 251 students who prefer animated movies in the population.
Random Sampling Quick Check
5 of 55 of 5 Items
Question
There are 528 students in Pablo’s high school. Pablo surveys 130 randomly chosen students about their favorite ice cream flavor. The results are as follows.
Ice cream flavor Number of students/Frequency
Strawberry 29
Butter Pecan 31
Chocolate 43
Orange Sherbet 10
Vanilla 17
Based on Pablo’s sample data, approximately how many students in the school prefer orange sherbet?
(1 point)
Responses
110
110
45
45
41
41
195
Based on Pablo's sample data, approximately 41 students in the school prefer orange sherbet.
To find out, you can set up the proportion:
Number of students who prefer orange sherbet in sample/total sample = Number of students who prefer orange sherbet in population/total population
Substituting the given values, we have:
10/130 = x/528
Solving for x:
10/130 = x/528
x = (10/130) * 528
x ≈ 40.92
Rounding to the nearest whole number, we get approximately 41 students who prefer orange sherbet in the population.
The temperatures (in °F) in a city on randomly selected days during the summer are 89, 85, 90, 87, 86, 91, 88, and 89. What is the mean temperature? Round your answer to one decimal place.
The low temperatures during one week in January were 2°, −3°, −9°, −11°, 1°, −8°, and −9°. What was the median low temperature that week?
First, we need to sort the temperatures in order:
-11°, -9°, -9°, -8°, -3°, 1°, 2°
There are 7 temperatures, so the middle is between the 3rd and 4th values.
-11°, -9°, -9°, [-8°, -3°], 1°, 2°
The median is the average of the two middle values, so:
median = (-8° - 3°)/2 = -5.5°
Therefore, the median low temperature that week was -5.5°.
The stem-and-leaf plot shows the ages of a group of friends. Find the median age.
Stem Leaf
2 8
3 1 2 4
4 2 7 9
5 4
Key: 2|8=28 years
To find the median age, we need to arrange the ages in order from lowest to highest. Using the stem-and-leaf plot, we can write down the following list of ages:
28, 31, 32, 34, 42, 47, 49, 54
There are 8 ages in the list, so the median is the middle value.
The middle is between the 4th and 5th values:
28, 31, 32, [34, 42], 47, 49, 54
The median age is the average of the two middle values:
median = (34 + 42)/2
median = 38
Therefore, the median age is 38 years.
Here are the math test scores of 10 randomly selected students from Class 1 and Class 2.
Class 1: 75 80 84 79 92 86 88 86 86 90
Class 2: 83 85 91 93 72 85 76 80 82 89
Find the mean score of each sample. Based on the mean, which class generally performed better on the math test?
(2 points)
The mean score for Class 1 is
. The mean score for Class 2 is
. Based on the mean, Class
generally performed better on the math test.
The mean score for Class 1 is:
(75 + 80 + 84 + 79 + 92 + 86 + 88 + 86 + 86 + 90)/10 = 85
The mean score for Class 2 is:
(83 + 85 + 91 + 93 + 72 + 85 + 76 + 80 + 82 + 89)/10 = 83.6
Based on the mean score, Class 1 generally performed better on the math test than Class 2.
Kat and Joshua want to know which pizza shop has the fastest delivery time. Kat took a random sample of 8 delivery times from pizza shop 1 over a period of six months. Joshua did the same for pizza shop 2. The delivery times (in minutes) for each pizza shop are shown in the following table.
Pizza Shop 1 Pizza Shop 2
18 20
23 17
15 16
22 38
21 21
24 20
28 18
20 25
Find the median delivery time for each pizza shop. Based on the median, which pizza shop has the fastest delivery time?
(2 points)
The median delivery time of Pizza Shop 1 is
minutes. The median delivery time of Pizza Shop 2 is
minutes. Based on the median, Pizza Shop
has a faster delivery time.
To find the median delivery time for each pizza shop, we need to first order the data from smallest to largest:
Pizza Shop 1: 15, 18, 20, 21, 22, 23, 24, 28
Pizza Shop 2: 16, 17, 18, 20, 21, 25, 38
For Pizza Shop 1, the median delivery time is the middle value of the ordered list. Since there are an even number of values, we take the average of the two middle values:
(21 + 22)/2 = 21.5
The median delivery time of Pizza Shop 1 is 21.5 minutes.
For Pizza Shop 2, the median delivery time is the middle value of the ordered list. Since there are an odd number of values, we simply find the middle value:
20
The median delivery time of Pizza Shop 2 is 20 minutes.
From the medians, we can see that Pizza Shop 2 has a faster delivery time.
The ages, in years, of randomly selected employees are 34, 41, 28, 35, 27, 44, 48, 29, 33, and 39. Determine the mean age.(1 point)
Responses
36.8
36.8
39.8
39.8
37.8
37.8
35.8
To find the mean age, we can add up all the ages and divide by the total number of employees:
mean age = (34 + 41 + 28 + 35 + 27 + 44 + 48 + 29 + 33 + 39)/10
mean age = 348/10
mean age = 34.8
Therefore, the mean age is 34.8 years.
The closest option is 35.8, but the correct answer is 34.8.
The graph shows the number of hours randomly selected students spent on math homework. What is the mean number of hours students spent on math homework?
(1 point)
Responses
2.14 hours
2.14 hours
1.4 hours
1.4 hours
1 hour
1 hour
9 hours
We cannot see the graph you are referencing. Please provide the graph or the data in order for me to calculate the mean number of hours spent on math homework.
Measures of Center Quick Check
3 of 53 of 5 Items
Question
For 10 consecutive days, Anthony recorded the daily high temperatures in Dayton and Laredo. The table shows the data.
Daily High Temperatures (in °F)
Dayton 77 82 85 87 90 85 79 86 91 88
Laredo 81 79 80 86 88 92 82 88 84 87
Which city has a higher median daily high temperature? How much higher is this city’s median than the other city’s median daily high temperature?
(1 point)
Responses
Laredo has a higher median daily high temperature than Dayton by 0.5°F.
Laredo has a higher median daily high temperature than Dayton by , 0.5 degree , F.
Dayton has a higher median daily high temperature than Laredo by 0.3°F.
Dayton has a higher median daily high temperature than Laredo by , 0.3 degree , F.
Dayton has a higher median daily high temperature than Laredo by 0.5°F.
Dayton has a higher median daily high temperature than Laredo by , 0.5 degree , F.
Laredo has a higher median daily high temperature than Dayton by 1°F
To find the median daily high temperature for each city, we need to order the data from lowest to highest:
Dayton: 77, 79, 82, 85, 85, 86, 87, 88, 90, 91
Laredo: 79, 80, 81, 82, 84, 86, 87, 88, 88, 92
For Dayton, the median is the middle value of the ordered list. There are 10 temperatures, so the middle is between the 5th and 6th values:
77, 79, 82, 85, [85, 86], 87, 88, 90, 91
The median is the average of these two values:
median = (85 + 86)/2
median = 85.5°F for Dayton.
For Laredo, the median is the middle value of the ordered list. There are 10 temperatures, so the middle is between the 5th and 6th values:
79, 80, 81, 82, [84, 86], 87, 88, 88, 92
The median is the average of these two values:
median = (84 + 86)/2
median = 85°F for Laredo.
Therefore, Laredo has a higher median daily high temperature than Dayton, but only by 0.5°F.
The closest option is "Laredo has a higher median daily high temperature than Dayton by 0.5°F", which is the correct answer.
The following data set represents the highway fuel efficiency in miles per gallon (mpg) of randomly selected cars from two car companies.
Car Company A: 35, 28, 35, 30, 31, 38, 35, 30
Car Company B: 29, 33, 36, 27, 34, 34, 34, 25
Which of the following statements is true?
(1 point)
Responses
The cars from both companies have the same mean, median, and modal highway fuel efficiency.
The cars from both companies have the same mean, median, and modal highway fuel efficiency.
The cars from Company A have higher median highway fuel efficiency than Company B.
The cars from Company A have higher median highway fuel efficiency than Company B.
The cars from Company B have higher modal highway fuel efficiency than Company A.
The cars from Company B have higher modal highway fuel efficiency than Company A.
The cars from Company A have a higher mean highway fuel efficiency that Company B.
To find the mean fuel efficiency for each car company, we need to add up all the values and divide by the total number of cars in each company:
mean for Company A = (35 + 28 + 35 + 30 + 31 + 38 + 35 + 30)/8 = 32.375 mpg
mean for Company B = (29 + 33 + 36 + 27 + 34 + 34 + 34 + 25)/8 = 31.75 mpg
Therefore, the statement "The cars from Company A have a higher mean highway fuel efficiency than Company B" is true.
To find the median fuel efficiency for each car company, we need to order the data from lowest to highest:
Car Company A: 28, 30, 30, 31, 35, 35, 35, 38
Car Company B: 25, 27, 29, 33, 34, 34, 34, 36
For Company A, the median is the middle value of the ordered list. There are 8 values, so the middle is between the 4th and 5th values:
28, 30, 30, [31, 35], 35, 35, 38
The median is the average of these two values:
median for Company A = (31 + 35)/2 = 33 mpg
For Company B, the median is the middle value of the ordered list. There are 8 values, so the middle is between the 4th and 5th values:
25, 27, 29, [33, 34], 34, 34, 36
The median is the average of these two values:
median for Company B = (33 + 34)/2 = 33.5 mpg
Therefore, the statement "The cars from Company A have higher median highway fuel efficiency than Company B" is false.
To find the mode of each car company, we need to find the most frequently occurring value.
For Company A, the mode is 35 mpg, since it appears three times and no other value appears more than that.
For Company B, the mode is 34 mpg, since it appears three times and no other value appears more than that.
Therefore, the statement "The cars from both companies have the same mean, median, and modal highway fuel efficiency" is false.
The only statement that is true is "The cars from Company A have a higher mean highway fuel efficiency than Company B."
The points scored by a football team are 17, 23, 15, 21, 7, and 13. What is the range?
To find the range, we need to subtract the lowest score from the highest score:
Highest score: 23
Lowest score: 7
Range = Highest score - Lowest score
Range = 23 - 7
Range = 16
Therefore, the range is 16 points.
A poll is conducted to determine who is most likely to win the election for town mayor. Two random samples of 500 residents each are polled. The data is as follows.
Candidate A Candidate B Total
Sample 1 346 154 500
Sample 2 248 252 500
What percentage of the total polled residents would vote for Candidate A as the town mayor?
(1 point)
To determine the average height of trees in a national forest, 10 trees are randomly selected. Their heights in feet are 85.4, 87.6, 87.1, 85.6, 85.9, 85.3, 84.9, 83.7, 84.0, and 85.2. To ensure the accuracy of the data, another set of 10 trees is randomly selected. Their heights are 87.1, 88.0, 84.2, 85.3, 86.5, 84.2, 83.2, 84.1, 85.2, and 87.3. What is the difference between the mean heights of the trees of these two random samples?
The numbers of miles Ron biked are 31, 35, 49, 32, 45, 29, 33, 28, and 40.
Determine the range of this data set.(1 point)
Responses
35.8 miles
35.8 miles
9 miles
9 miles
18 miles
18 miles
21 miles