You have probably already come across methods of analysing and interpreting data in given data distributions. In many real-world applications, we are required to compare information between multiple data sets. Let's look at how to compare data between data distributions.
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Sign up for freeWhich of the following is appropriate to use along with standard deviation for comparison?
Which of the following should you use for comparing a data set with extreme values?
Which of the following is appropriate to use along with interquartile range for comparison?
Which of the following is appropriate to use along with median for comparison?
Which of the following is appropriate to use along with mean for comparison?
Which of the following is appropriate to use along with standard deviation for comparison?
Which of the following should you use for comparing a data set with extreme values?
Which of the following is appropriate to use along with interquartile range for comparison?
Which of the following is appropriate to use along with median for comparison?
Which of the following is appropriate to use along with mean for comparison?
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Team Comparing Data Teachers
When comparing multiple data distributions, you can comment on
A measure of location – a measure of location is used to summarise an entire data set with a single value. For example, mean and median are measures of location.
A measure of spread – a measure of spread provides us information regarding the variability of data in a given data set, i.e. how close or far away the different points in a data set are from each other. Standard deviation and interquartile range are examples of measures of spread.
You can compare different data distributions using the mean and standard deviation, or using the median and interquartile ranges. In cases where data sets contain extreme values and/or outliers, median and interquartile ranges are usually more appropriate to use.
Do not use the median and standard deviation together or the mean and interquartile ranges together.
Let's explore the concept further with the help of examples.
The daily mean temperatures during August is recorded at Heathrow and Leeming. For Heathrow, ∑x=562, ∑x²=10301.2. For Leeming, the mean temperature was 15.6°C with a standard deviation of 2.01° C
a) Calculate the mean and standard deviation for Heathrow. b) Compare the data for Heathrow with that of Leeming.
Solutions
For Heathrow,
\(\begin{align} mean &= \frac {\sum{x}}{n} \\ &= \frac{562}{31} = 18.1ºC \end{align}\)
\(Standard \quad deviation = \sqrt{\frac{\sum{x^2}}{n} - (\frac{\sum{x}}{n})^2} = \sqrt {\frac{10301.2}{31} - (\frac{562}{31})^2} = 1.91ºC\)
b) From the above information, we see that the mean temperature at Heathrow during August was higher than Leeming, and the spread/variability of temperatures was less than Leeming.
A company collects the delivery times in minutes for suppliers A and B for a period of 20 days. The following is the result of the data collected. Compare the performance of the two suppliers.
| suppliers | ∑x | ∑x² |
| A | 360 | 18000 |
| B | 300 | 29000 |
solutions
For supplier A,
\(\begin{align} mean_A &= \frac {\sum x}{n} \\ &= \frac{360}{20} = 18 \end{align}\)
For supplier B,
\(\begin{align} mean_B &= \frac {\sum x}{n} \\ &= \frac{300}{20} = 15 \end{align}\)
From the above information, we see that supplier A has a longer delivery time, while supplier B has a greater spread in delivery time.
Consider the above example in a real-world context. If the company wants to keep one of its suppliers and let go of the other, it could compare the above data just like we have. If the priority of the company is to reduce delivery times on average, it would favour supplier B. If the priority on the other hand is greater reliability, it would favour the supplier with less variability, and that would be supplier A.
The students of two different sections sit for an exam. The quartile and median marks of each section is provided. Compare the performance of the 2 sections.
| Section | median | ||
| Section 1 | 58 | 71 | 87 |
| Section 2 | 62 | 74 | 83 |
Solutions
The interquartile range for Section 1 = Q3 - Q1 = 87-58 = 29
The interquartile range for Section 2 = Q3 - Q1 = 83-62 = 21
From the given data, we see that the median marks is higher for section 2, while the variability of marks is higher in section 1.
A company collects the delivery times for suppliers, A and B, for a period of 20 days. The median delivery time was 4 hours for supplier A, and 3 hours for supplier B. The interquartile range for supplier A was 0.8 hours and for supplier B was 1.5 hours.
Compare the performance of the suppliers in terms of speed and reliability.
Solutions
Supplier B appears to be the more efficient performing better in terms of speed with a lower median delivery time. Supplier A appears to be more reliable with a lower spread/variability in delivery time.
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Why are bar graphs useful for comparing data?
Bar graphs allow you to easily visualise the measures of location and spread.
What is the importance of comparing data?
In many real-world applications, we are required to compare information between multiple data sets to make better-informed decisions.
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