Question

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

Short answer

2000: Mean > Median, Right-skewed. 2006: Mean > Median, Right-skewed.

Step-by-step solution

List the Median Incomes for Each Year

First, write down the median family incomes provided in the table for the years 2000 and 2006. This helps in visualizing the data clearly.

Calculate the Mean for the Year 2000

Add all the median incomes for the year 2000 and divide by the number of provinces and territories (13). Mean (2000) = $$\frac{38,800+44,200+44,500+43,200+47,700+55,700+47,300+45,800+55,200+49,100+56,000+61,000+37,600}{13}$$Mean (2000) = $$\frac{626,100}{13}$$Mean (2000) = 48,161.54

Calculate the Mean for the Year 2006

Add all the median incomes for the year 2006 and divide by the number of provinces and territories (13). Mean (2006) = $$\frac{50,500+56,100+56,400+54,000+59,000+66,600+58,700+60,500+78,400+62,600+76,000+88,800+54,300}{13}$$Mean (2006) = $$\frac{821,900}{13}$$Mean (2006) = 63,223.08

Calculate the Median for the Year 2000

To find the median, sort the incomes from the year 2000 in ascending order and find the middle value.Sorted 2000: 37,600, 38,800, 43,200, 44,200, 44,500, 45,800, 47,300, 47,700, 49,100, 55,200, 55,700, 56,000, 61,000Median (2000) = 47,300

Calculate the Median for the Year 2006

Sort the incomes from the year 2006 in ascending order and find the middle value.Sorted 2006: 50,500, 54,000, 54,300, 56,100, 56,400, 58,700, 59,000, 60,500, 62,600, 66,600, 76,000, 78,400, 88,800Median (2006) = 60,500

Compare the Means and Medians

For the year 2000:- Mean = 48,161.54- Median = 47,300For the year 2006:- Mean = 63,223.08- Median = 60,500In both years, the mean is higher than the median, indicating the distribution is right-skewed.

Key concepts

Mean

The mean, or average, is a measure of central tendency frequently used in data analysis. It is calculated by adding up all the values in a dataset and then dividing by the number of values. In this exercise, we calculated the mean family income for the years 2000 and 2006. For 2000, the mean was calculated as follows:

$$\text{Mean (2000)}=\frac{38,800+44,200+44,500+43,200+47,700+55,700+47,300+45,800+55,200+49,100+56,000+61,000+37,600}{13}\text{Mean (2000)}=\frac{626,100}{13}=48,161.54$$
Similarly, for 2006, the mean was:

$$\text{Mean (2006)}=\frac{50,500+56,100+56,400+54,000+59,000+66,600+58,700+60,500+78,400+62,600+76,000+88,800+54,300}{13}\text{Mean (2006)}=\frac{821,900}{13}=63,223.08$$
The mean is useful for understanding the overall level of family income, but it can be affected by very high or very low values, known as outliers.

Median

The median is another measure of central tendency, and it represents the middle value in a dataset when the values are arranged in ascending order. Unlike the mean, the median is not affected by outliers.

Here's how we calculated the median family income for the years 2000 and 2006:

For 2000, the values in ascending order were:
$$\text{37,600, 38,800, 43,200, 44,200, 44,500, 45,800, 47,300, 47,700, 49,100, 55,200, 55,700, 56,000, 61,000}$$
The middle value is 47,300, so the median for 2000 is 47,300.

For 2006, the values in ascending order were:
$$\text{50,500, 54,000, 54,300, 56,100, 56,400, 58,700, 59,000, 60,500, 62,600, 66,600, 76,000, 78,400, 88,800}$$
The middle value is 60,500, so the median for 2006 is 60,500.

The median is a good measure when you need a value that is representative of the central tendency, even if your data has outliers.

Income Distribution

Income distribution refers to how income is spread among different groups or individuals in an economy. In the provided exercise, we examined the data from Canadian provinces and territories, allowing us to see how median family incomes differ across regions and over time.

The distribution of family income can provide insights into the economic well-being of households:

• Higher variance in income distribution signifies more inequality.
• Graphs such as histograms can show the spread and shape of income distribution.
• Both the mean and median can give indications of the central tendency but may imply different things when the distribution is skewed.

These insights are critical for policymakers, economists, and researchers to identify and address income inequality and economic disparities among different regions.

Statistical Analysis

Statistical analysis involves collecting, organizing, interpreting, and presenting data to uncover patterns and trends. The exercise we analyzed is an example of introductory statistical analysis:

The steps we took included

• Listing the incomes
• Calculating measures of central tendency (mean and median)
• Comparing these measures to assess the nature of the distribution

Statistical analysis is essential in a variety of fields like economics, psychology, marketing, and more. It's used to make informed decisions based on data. Central tendency measures, like the mean and median, play foundational roles in this analysis, helping to summarize and understand the central values in datasets.

Right-Skewed Distribution

A right-skewed distribution, also known as positively skewed, occurs when the mean is greater than the median. This happens because the distribution has a long tail on the right side. In the exercise, for both years 2000 and 2006, the mean family income was higher than the median, indicating right-skewed distributions.

Characteristics of right-skewed distributions include:

• The bulk of the data values lie to the left of the mean.
• The tail on the right side is longer due to higher income outliers.
• Usage in real-world scenarios, such as income distributions, where a few high-income values can shift the mean higher.

Recognizing the skewness of data distributions is important in statistical analysis as it affects how we interpret measures of central tendency and make subsequent decisions or inferences from the data.