It is important to determine if generalizations are valid. We can consider the following two things:
1. Was the sample random? Did all members of the population have an equal chance of being selected?
2. Is the sample size large enough based on the population size?
(3 points)
Valid or Invalid?
1. The principal at a middle school wants to know the average number of clubs students participate in. She uses a random number generator from 0-500 representing all students. She chooses 100 numbers at random and surveys these students. She concludes that students participate on average in 2 clubs each.
2. Jorge wants to know how many students in his school are going to the basketball game. Jorge asks 5 of his friends, and all of them say they are going. He concludes that all students will attend the basketball game.
3. Nick wants to determine the most popular sport in his school of 500 students.
He puts the names of all the students in the box. draws 100 of them, and surveys them all. He determines that soccer is the most popular sport in the school.
1. Valid - The sample was random and a sufficient size for the population size of the middle school.
2. Invalid - The sample size is too small and not randomly selected.
3. Invalid - While the sample was randomly selected, the sample size is too small to accurately determine the most popular sport in the school.
These two box plots show the distances of a standing jump, in inches, for a random sample of
10-year-olds and a random sample of 15-year-olds. Use these box plots to answer the questions below.
[—-|———|—-|———]
10 year olds: 51, 53, 54, 56, 57, 58, 59
[—-|——————|——|———-]
15 year olds: 64, 65, 66, 67, 68, 69, 70,……..80,
The difference between the two medians is____
The box plot for the_____-year-olds has more spread.
The box plots above show us how far a group of students jumped. What can you infer about the differences between the two age groups? Show your work (2 points)
The difference between the two medians is calculated by finding the median of each group and then subtracting the median of the 15-year-olds from the median of the 10-year-olds.
For the 10-year-olds:
Median = (56 + 57) / 2 = 56.5
For the 15-year-olds:
Median = (67 + 68) / 2 = 67.5
Difference between the two medians: 67.5 - 56.5 = 11
The box plot for the 15-year-olds has more spread because the interquartile range (IQR) is larger. The IQR for the 10-year-olds is 6 (59 - 53), and the IQR for the 15-year-olds is 6 (69 - 63). Since the IQR for the 15-year-olds is larger, the spread of their distances jumped is more varied compared to the 10-year-olds.
What about the last question “The box plots above show us how far a group of students jumped. What can you infer about the differences between the two age groups? Show your work (2 points)”
To infer about the differences between the two age groups based on the box plots provided, we can consider the following observations:
1. Median: The median distance jumped by the 15-year-olds (67.5) is higher than the median distance jumped by the 10-year-olds (56.5). This suggests that, on average, 15-year-olds are able to jump further than 10-year-olds.
2. Interquartile Range (IQR): The IQR for the 15-year-olds (6) is the same as the IQR for the 10-year-olds (6). This indicates that the middle 50% of the distances jumped by both age groups have a similar range of values.
3. Spread: However, if we look at the range of values in each group, we can see that the 15-year-olds have a wider range of distances jumped (from 64 to 80) compared to the 10-year-olds (from 51 to 59). This suggests that there is more variability in the distances jumped by 15-year-olds than by 10-year-olds.
Overall, we can infer that while the median distance jumped by the two age groups is different, the variability in distances jumped is higher among the 15-year-olds compared to the 10-year-olds.
The mean weight for pugs is 7 kilograms, and the MAD is 0.5 kilogram.
The mean weight for beagles is 10 kilograms, and the MAD is 0.5 kilogram.
Which statement below accurately compares the mean and Mean Absolute Deviation (level of variability of each data set?
• Beagles on average are lighter than pugs and have a higher variability than pugs.
• Pugs on average are lighter than beagles, but have equal variability
• Beagles on average are heavier and have a lower variability than pugs.
• Pugs on average are the same weight as beagles and have equal variability
Based on the given information:
1. The mean weight for pugs is 7 kilograms with a MAD of 0.5 kilograms.
2. The mean weight for beagles is 10 kilograms with a MAD of 0.5 kilograms.
From this information, we can make the following comparison:
- Beagles on average are heavier than pugs and have an equal level of variability.
Therefore, the statement that accurately compares the mean and Mean Absolute Deviation (level of variability) of each data set is:
• Beagles on average are heavier and have an equal variability as pugs.
That is not one of the answers
I apologize for the oversight. Let's reevaluate the comparison based on the given information:
The mean weight for pugs is 7 kilograms with a MAD of 0.5 kilograms.
The mean weight for beagles is 10 kilograms with a MAD of 0.5 kilograms.
Comparing the two breeds:
- Pugs have a lower average weight compared to beagles (7 kg < 10 kg).
- Both breeds have an equal level of variability, as indicated by the same MAD for both (0.5 kg).
Therefore, the statement that accurately compares the mean and Mean Absolute Deviation (level of variability) of each data set is:
• Pugs on average are lighter than beagles, but have equal variability.