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The administration is considering a total ban on student automobiles. You have conducted a poll on this issue of 20 fellow students and 20 of the neighbors who live around the campus and have calculated scores for your respondents. On the scale you used, a high score indicates strong opposition to the proposed ban. The scores are presented here for both groups. Calculate an appropriate measure of central tendency and compare the two groups in a sentence or two.  Students  Neighbors 10110710916108001011129874101111097005111052109010100

Short Answer

Expert verified
Students have a higher mean score (7.8) compared to neighbors (5.1). The median score for students is also higher (9) compared to neighbors (4.5).

Step by step solution

01

Organize the Data

Separate the scores into two lists: one for students and one for neighbors.Students: [10, 10, 10, 10, 9, 10, 9, 5, 5, 0]Neighbors: [11, 9, 8, 11, 8, 11, 7, 1, 2, 10]
02

Calculate the Mean Score

Calculate the average score for each group.Students: Mean=10+10+10+10+9+10+9+5+5+010=7810=7.8Neighbors: Mean=11+9+0+1+7+11+0+2+0+1010=5110=5.1
03

Calculate the Median Score

Find the median score for each group by ordering the numbers and picking the middle one.Students:- Sorted scores: [0, 5, 5, 9, 9, 10, 10, 10, 10, 10]- Median: (9 + 9) / 2 = 9Neighbors:- Sorted scores: [0, 0, 0, 1, 2, 7, 9, 10, 11, 11]- Median: (2 + 7) / 2 = 4.5
04

Compare the Measures of Central Tendency

Compare the means and medians of the two groups in one or two sentences.The average score for students is higher than for neighbors (7.8 vs. 5.1). The medians show a similar trend with students having a median score of 9 compared to 4.5 for neighbors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Calculation
A central tendency measure helps summarize a set of data with a single value. The mean or average is a crucial example of such measures. Here’s how you can calculate it:

1. **Add all the numbers in your data set**. For example, the student survey scores are: [10, 10, 10, 10, 9, 10, 9, 5, 5, 0].
2. **Divide the total by the number of values**. For students, the total score is 78, and there are 10 students. So, the mean score would be: Mean=7810=7.8

Similarly, for neighbors, their scores are: [11, 9, 0, 1, 7, 11, 0, 2, 0, 10]. The total score is 51, and there are 10 neighbors:
Mean=5110=5.1

The mean score shows the average attitude towards the student automobile ban. Students have a higher average score, suggesting greater opposition compared to neighbors.
Median Calculation
The median is another measure of central tendency which shows the middle value in an ordered data set.

To find the median:
1. **Arrange the numbers in order**. For students, this becomes: [0, 5, 5, 9, 9, 10, 10, 10, 10, 10].
2. **Find the middle number**. If there’s an even number of values, calculate the average of the two middle numbers. For students, the median is: Median=9+92=9

For neighbors, the sorted set is: [0, 0, 0, 1, 2, 7, 9, 10, 11, 11]. Here the median is:
Median=2+72=4.5

The median gives a better sense of the typical value especially when a data set contains outliers. For this survey, the median shows that students generally have stronger opposition to the ban compared to neighbors.
Data Analysis
Data analysis involves systematically applying statistical methods to describe and interpret data.

1. **Organize your data**: Separate the scores into groups like 'students' and 'neighbors'.
2. **Calculate necessary measures**: Here it means finding the mean and median for each group.
3. **Interpret the results**: The average (mean) and middle value (median) for students are both higher than for neighbors, indicating stronger opposition to the ban among students.

Through such analysis, you can better understand the underlying trends and sentiments represented by raw data. The method revealed that students are more opposed to the proposed ban than the neighboring community.
Student Survey
Conducting a student survey helps gather opinions on important issues within a school community. Steps include:

1. **Create a questionnaire**: Formulate questions that measure attitudes or opinions. In this case, a question about the students' stance on the automobile ban was used.
2. **Distribute the survey**: Get responses from a representative group. Twenty students took part in this exercise.
3. **Analyze the results**: Calculate measures like the mean and median scores. For students, the mean score was 7.8, and the median was 9, showing significant opposition to the ban.

Using surveys and these calculations allow schools to make informed decisions based on the views of their student bodies.
Neighbor Survey
Surveys aren't just for students; involving the neighboring community provides a broader perspective. Consider the following steps:

1. **Design the survey**: Like for students, create targeted questions to understand neighbors' views.
2. **Collect responses**: In our example, 20 neighbors participated. Their scores were noted and helped gauge opposition or support.
3. **Analyze and compare**: Neighbors had a mean score of 5.1 and a median of 4.5, indicating less opposition compared to students.

This comparison between surveys from different groups can highlight differing community perspectives. In this case, it helped show that neighbors are less opposed to the proposed automobile ban than students.

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Most popular questions from this chapter

The following table lists the median family incomes for 13 Canadian provinces and territories in 2000 and 2006. Compute the mean and median for each year and compare the two measures of central tendency. Which measure of central tendency is greater for each year? Are the distributions skewed? In which direction?  Province or Territory 20002006 Newfoundland and Labrador 38,80050,500 Prince Edward Island 44,20056,100 Nova Scotia 44,50056,400 New Brunswick 43,20054,000 Quebec 47,70059,000 Ontario 55,70066,600 Manitoba 47,30058,700 Saskatchewan 45,80060,500 Alberta 55,20078,400 British Columbia 49,10062,600 Yukon 56,00076,000 Northwest Territories 61,00088,800 Nunavut 37,60054,300

Following are four variables for 30 cases from the General Social Survey. Age is reported in years. The variable happiness consists of answers to the question "Taken all together, would you say that you are (1) very happy (2) pretty happy, or (3) not too happy?" Respondents were asked how many sex partners they had over the past five years. Responses were measured on the following scale: 04= actual numbers; 5=510 partners; 6=1120 partners; 7=21100 partners ;8= more than 100 . a. For each variable, find the appropriate measure of central tendency and write a sentence reporting this statistical information as you would in a research report. b. (This problem is optional.) For the variable age, construct a frequency distribution with interval size equal to 5 and the first interval set at 2024. Compute the mean and median for the grouped data and compare with the values computed for the ungrouped data. How accurate are the estimates based on the grouped data?

The following table presents the annual personhours of time lost due to traffic congestion for a group of cities for 2007. This statistic is a measure of traffic congestion  City  Annual Person-Hours of Time Lost to TrafficCongestion per Year per Person  Baltimore 25 Boston 22 Buffalo 5 Chicago 22 Cleveland 7 Dallas 32 Detroit 29 Houston 32 Kansas City 8 Los Angeles 38 Miami 27 Minneapolis 22 New Orleans 10 New York 21 Philadelphia 21 Pittsburgh 8 Phoenix 23 San Antonio 21 San Diego 29 San Francisco 29 Seattle 24 Washington, DC 31 a. Calculate the mean and median of this distribution. b. Compare the mean and median. Which is the higher value? Why? c. If you removed Los Angeles from this distribution and recalculated, what would happen to the mean? To the median? Why? d. Report the mean and median as you would in a formal research report.

For the test scores first presented in problem 2.6 and reproduced here, compute a median and mean for both the pretest and posttest. Interpret these statistics.  Case  Pretest  Posttest  A 812 B 713 C 1012 D 1519 E 108 F 1017 G 312 H 1011 ? 57 J 1512 K 1320 L 45 M 1015 N 811 O 1220

You have compiled the following information on each of the graduates voted "most likely to succeed" by a local high school for a 10 -year period. For each variable, find the appropriate measure of central tendency.  Case Present Income  Marital Status  Owns a Home?  Years of Education Post-High School  A 104,000 Divorced  Yes 8 B 68,000 Divorced  No 4 C 54,000 Married  Yes 4 D 45,000 Married  No 4 E 40,000 Single  No 4 F 85,000 Separated  Yes 8 G 30,000 Married  No 3 H 27,000 Married  No 1 I 93,000 Married  Yes 6 J 48,000 Single  Yes 4

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