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As the head of a social services agency, you believe that your staff of 20 social workers is very much overworked compared to 10 years ago. The case loads for each worker are reported below for each of the two years in question. Has the average caseload increased? What measure of central tendency is most appropriate to answer this question? Why? $$\begin{array}{cccc} {2000} & & {2010} \\ \hline 52 & 55 & 42 & 82 \\ 50 & 49 & 75 & 50 \\ 57 & 50 & 69 & 52 \\ 49 & 52 & 65 & 50 \\ 45 & 59 & 58 & 55 \\ 65 & 60 & 64 & 65 \\ 60 & 65 & 69 & 60 \\ 55 & 68 & 60 & 60 \\ 42 & 60 & 50 & 60 \\ 50 & 42 & 60 & 60 \end{array}$$

Short Answer

Expert verified
The mean caseload increased from 51.7 in 2000 to 57.8 in 2010. The most appropriate measure of central tendency is the mean.

Step by step solution

01

Calculate the Mean Caseload for 2000

To find the average caseload for the year 2000, add up all the caseload numbers for that year and then divide by the number of data points. The caseloads for the year 2000 are: 52, 42, 50, 57, 49, 45, 65, 60, 55, 42.
02

Sum of Caseloads for 2000

Add the caseloads: 52 + 42 + 50 + 57 + 49 + 45 + 65 + 60 + 55 + 42 = 517.
03

Mean Caseload for 2000

Divide the sum by the number of social workers: \( \frac{517}{10} = 51.7 \). The mean caseload for the year 2000 is 51.7.
04

Calculate the Mean Caseload for 2010

To find the average caseload for the year 2010, add up all the caseload numbers for that year and then divide by the number of data points. The caseloads for the year 2010 are: 55, 82, 49, 50, 50, 52, 55, 65, 60, 60.
05

Sum of Caseloads for 2010

Add the caseloads: 55 + 82 + 49 + 50 + 50 + 52 + 55 + 65 + 60 + 60 = 578.
06

Mean Caseload for 2010

Divide the sum by the number of social workers: \( \frac{578}{10} = 57.8 \). The mean caseload for the year 2010 is 57.8.
07

Compare the Means

Compare the mean caseloads for the two years: The mean caseload for 2000 is 51.7 and for 2010 is 57.8. Since 57.8 is greater than 51.7, the average caseload has increased.
08

Determine the Appropriate Measure of Central Tendency

The most appropriate measure of central tendency for this scenario is the mean because it takes into account all the data points and provides an average number of cases per social worker, which is useful for comparing the overall workload between the two years.

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Key Concepts

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

central tendency
Central tendency is a statistical measure that identifies a single value as representative of an entire dataset. This value aims to describe a set of data by identifying the central position within that set. The three most common measures of central tendency are the mean, median, and mode. Each of these measures provides different insights and is useful in various scenarios.
  • Mean: The average of all data points.
  • Median: The middle value when data points are arranged in order.
  • Mode: The most frequently occurring value in the dataset.
By understanding central tendency, you can make informed decisions about data, particularly when you want to summarize and describe the data as a single, central value.
mean calculation
The mean, or average, is a crucial measure of central tendency. To find the mean, you sum up all the values in a dataset and divide by the number of values. This gives an overall idea of the dataset's central value. Let's go through the steps to calculate the mean caseload for 2000 and 2010.
  • First, sum all the caseload values for the year 2000: \( 52 + 42 + 50 + 57 + 49 + 45 + 65 + 60 + 55 + 42 = 517 \).
  • Next, divide the sum by the number of data points, which is 10: \( \frac{517}{10} = 51.7 \).
  • Repeat the process for the year 2010: \( 55 + 82 + 49 + 50 + 50 + 52 + 55 + 65 + 60 + 60 = 578 \).
  • Divide by the number of data points: \( \frac{578}{10} = 57.8 \).
The mean caseloads are 51.7 for 2000 and 57.8 for 2010. This calculation shows how the average caseload can be summarized and easily compared between different years.
workload assessment
Workload assessment involves evaluating the amount of work assigned to individuals or groups over a specific period. In this context, caseload refers to the number of cases handled by each social worker in a given year. Effectively assessing workload helps in understanding if the resources are adequately allocated and if staff are overburdened.
To assess the workload:
  • Calculate the mean caseload for each year.
  • Analyze changes in average workloads over time.
  • Compare the workloads to identify any significant increases or decreases.
It was found that the mean caseload increased from 51.7 in 2000 to 57.8 in 2010. This rise indicates that social workers are handling more cases on average over the decade, signaling an increase in workload.
data analysis
Data analysis involves systematically applying statistical techniques to describe, illustrate, and evaluate data. In this exercise, we analyze caseload numbers to determine trends and patterns in workload over time. Here’s a simple guide to analyzing data like this:
  • Gather your data and organize it clearly (e.g., caseload numbers for each year).
  • Apply statistical measures (mean, median, mode) to summarize the data.
  • Visualize your findings using charts or graphs for better understanding.
  • Interpret the results to derive meaningful insights, such as changes in workload.
Data analysis in this scenario revealed that the average caseload per social worker increased, pointing to a higher demand for social services and possibly an understaffed workforce. Such analysis helps in making informed management decisions.
statistical comparison
Statistical comparison is comparing two or more datasets to identify similarities and differences. It's vital for understanding changes over time and making data-driven decisions.
To compare the 2000 and 2010 caseloads:
  • Calculate the mean for both years.
  • Observe the differences between these means.
  • Determine if the difference is significant or just a minor variation.
In this case, the mean caseload for the year 2000 was 51.7, while for 2010, it was 57.8. The comparison shows a clear increase, indicating that the workload has grown over the decade. Such insights can prompt actions like hiring more staff or redistributing cases to balance the workload more effectively, ensuring that social workers are not overburdened.

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Most popular questions from this chapter

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 $$\begin{array}{lc} {\text { City }} & \text { Annual Person-Hours of Time Lost to Traffic Congestion per Year per Person } \\ \hline \text { Baltimore } & 25 \\ \text { Boston } & 22 \\ \text { Buffalo } & 5 \\ \text { Chicago } & 22 \\ \text { Cleveland } & 7 \\ \text { Dallas } & 32 \\ \text { Detroit } & 29 \\ \text { Houston } & 32 \\ \text { Kansas City } & 8 \\ \text { Los Angeles } & 38 \\ \text { Miami } & 27 \\ \text { Minneapolis } & 22 \\ \text { New Orleans } & 10 \\ \text { New York } & 21 \\ \text { Philadelphia } & 21 \\ \text { Pittsburgh } & 8 \\ \text { Phoenix } & 23 \\ \text { San Antonio } & 21 \\ \text { San Diego } & 29 \\ \text { San Francisco } & 29 \\ \text { Seattle } & 24 \\ \text { Washington, DC } & 31 \end{array}$$ 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.

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: \(0-4=\) actual numbers; \(5=5-10\) partners; \(6=11-20\) partners; \(7=21-100\) 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 \(20-24\). 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?

A sample of 25 freshmen at a major university completed a survey that measured their degree of racial prejudice (the higher the score, the greater the prejudice). a. Compute the median and mean scores for these data. \(\begin{array}{lllll} 10 & 43 & 30 & 30 & 45 \\ 40 & 12 & 40 & 42 & 35 \\ 45 & 25 & 10 & 33 & 50 \\ 42 & 32 & 38 & 11 & 47 \\ 22 & 26 & 37 & 38 & 10 \end{array}\) b. These same 25 students completed the same survey during their senior year. Compute the median and mean for this second set of scores, and compare them to the earlier set. What happened? \(\begin{array}{lllll} 10 & 45 & 35 & 27 & 50 \\ 35 & 10 & 50 & 40 & 30 \\ 40 & 10 & 10 & 37 & 10 \\ 40 & 15 & 30 & 20 & 43 \\ 23 & 25 & 30 & 40 & 10 \end{array}\)

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? $$ \begin{array}{lcc} {\text { Province or Territory }} & 2000 & 2006 \\ \hline \text { Newfoundland and Labrador } & 38,800 & 50,500 \\ \text { Prince Edward Island } & 44,200 & 56,100 \\ \text { Nova Scotia } & 44,500 & 56,400 \\ \text { New Brunswick } & 43,200 & 54,000 \\ \text { Quebec } & 47,700 & 59,000 \\ \text { Ontario } & 55,700 & 66,600 \\ \text { Manitoba } & 47,300 & 58,700 \\ \text { Saskatchewan } & 45,800 & 60,500 \\ \text { Alberta } & 55,200 & 78,400 \\ \text { British Columbia } & 49,100 & 62,600 \\ \text { Yukon } & 56,000 & 76,000 \\ \text { Northwest Territories } & 61,000 & 88,800 \\ \text { Nunavut } & 37,600 & 54,300 \end{array} $$

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. $$ \begin{array}{ccc} \text { Case } & \text { Pretest } & \text { Posttest } \\ \hline \text { A } & 8 & 12 \\ \text { B } & 7 & 13 \\ \text { C } & 10 & 12 \\ \text { D } & 15 & 19 \\ \text { E } & 10 & 8 \\ \text { F } & 10 & 17 \\ \text { G } & 3 & 12 \\ \text { H } & 10 & 11 \\ \text { ? } & 5 & 7 \\ \text { J } & 15 & 12 \\ \text { K } & 13 & 20 \\ \text { L } & 4 & 5 \\ \text { M } & 10 & 15 \\ \text { N } & 8 & 11 \\ \text { O } & 12 & 20 \\ \hline \end{array} $$

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