2. Comparing Data Sets
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Chapter 4
2.
Comparing Data Sets
Data is everywhere, and understanding the nuances between different sets is paramount. One effective way to compare data is through histograms, which visually represent frequency distributions, highlighting patterns and outliers. While histograms provide a snapshot, standard deviation gives a numerical measure of how spread out the values in a dataset are. A lower standard deviation indicates values are closer to the mean, while a higher one suggests greater variability. By employing both histograms and understanding standard deviation, one can gain a comprehensive view of multiple data sets, making more informed decisions in various domains.
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| | 11 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Comparing Data Sets
Slide of 11
This lesson investigates how characteristics like the center and spread of data distributions help compare data sets. The learner will be able to make meaning of real-life data sets on several topics like animals, weather patterns, sports, and even a part of the U.S. stock market!
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
- These are measures of the center of a data set.
- These are measures of the spread of a data set.
12.7,4.7,14.4,12.2,12.1,8.1,6.8,16.6,15.7,13.3,5.5,12.6,11.5,11.1,8.8
a Find the mean and give the answer as a decimal rounded to one decimal place.
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b Find the median and give the answer as a decimal rounded to one decimal place.
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c Find the range and give the answer as a decimal rounded to one decimal place.
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d Find the interquartile range and give the answer as a decimal rounded to one decimal place.
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e Find the mean absolute deviation and give the answer as a decimal rounded to two decimal places.
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Challenge
Analyzing Data Sets of Finnish Fish Species
In 1917, Pekka Brofeld published the measurements of various fish species caught in a lake near Tampere, Finland. Representing part of the findings is a data set about the length and height of four of the fish species caught. Move the slider to see the complete data set for these four fish.
Pictured are the four species and their Finnish names. Note that the scale of the four drawings are not equal to one another.
The answer may not be so clear. This lesson will explain how analyzing a data set using various methods can help identify pairs, such as the challenge, correctly. To do so, measures of center — like the median and mean — will be presented and applied to various data sets throughout the lesson.
Norssi
Hauki
Pasuri
Särki
Match the Finnish name with the Latin name.
{"type":"pair","form":{"alts":[[{"id":0,"text":"Abramis bjorkna"},{"id":1,"text":"Esox lucius"},{"id":2,"text":"Leuciscus rutilus"},{"id":3,"text":"Osmerus eperlanus"}],[{"id":0,"text":"Pasuri"},{"id":1,"text":"Hauki"},{"id":2,"text":"S\u00e4rki"},{"id":3,"text":"Norssi"}]],"lockLeft":true,"lockRight":false},"formTextBefore":"","formTextAfter":"","answer":[[0,1,2,3],[0,1,2,3]]}
Example
Pairing Teams to Sports Using Median Heights
The following box plots show the distribution of the heights (in feet and inches) of the players on the Ohio State Buckeyes men's basketball and football teams in the 2020–2021 season.
{"type":"pair","form":{"alts":[[{"id":0,"text":"Team A"},{"id":1,"text":"Team B"}],[{"id":0,"text":"Football"},{"id":1,"text":"Basketball"}]],"lockLeft":true,"lockRight":false},"formTextBefore":"","formTextAfter":"","answer":[[0,1],[0,1]]}
Hint
Of the two sports, which tends to place more importance on a player's height? Identify the maximum and the median heights of each team.
Solution
The box plots show that the range of heights is similar for both teams, but Team B, on average, has taller players.
- The tallest player on Team B is 6 feet 10 inches tall, and the median height is 6 feet 6 inches.
- The tallest player on Team A is 6 feet 8 inches tall, and the median height is 6 feet 2 inches.
Height tends to be more advantageous in basketball than in football. Therefore, it is reasonable to conclude from the box plots that Team B is the basketball team and Team A is the football team.
Extra
Represented using a histogram, the same data set is used to show the distribution of the height of the players on the two teams.
Example
Analyzing a Data Set to Determine Geographical Locations
The table below shows the average monthly high temperatures across three small towns.
![Temperatures in Noma=[63,67,75,80,86,90,95,92,88,80,73,69], Mekoryuk=[-9,3,25,43,57,67,70,65,53,29,5,-5], Nehawka=[34,48,58,66,75,87,87,83,82,69,58,40]](/mediawiki/images/8/85/Mljsx_Slide_A1_U4_C2_Example2_1.svg)
One town is located in the State of Alaska, another in Florida, and the other in Nebraska.
Analyzing the data set and map, try to match each town with the correct corresponding state. Note that, generally, northern states tend to be colder than southern states.
{"type":"pair","form":{"alts":[[{"id":0,"text":"Nehawka"},{"id":1,"text":"Noma"},{"id":2,"text":"Mekoryuk"}],[{"id":0,"text":"Nebraska"},{"id":1,"text":"Florida"},{"id":2,"text":"Alaska"}]],"lockLeft":true,"lockRight":false},"formTextBefore":"","formTextAfter":"","answer":[[0,1,2],[0,1,2]]}
Hint
Think about the relationship between the location and climate of each state of Alaska, Nebraska, and Florida. Then, consider each month's average high temperatures as shown in the data set. Which town is the warmest? Which town is the coldest?
Solution
Investigating the given data set and map, the following observations can be made.
- When comparing each town's average monthly high temperatures, the weather is coldest in Mekoryuk and warmest in Noma.
- Next, observing the map with a keen eye, Florida is the furthest south of these three states. It is likely — on average — to be the warmest. Therefore, it would make sense to match Florida with the town with the highest average monthly temperatures as described in the table.
- Again, observing the map, Alaska is the furthest north of the three states. It is likely — on average — to be the coldest. Therefore, it would make sense to match Alaska with the lowest average monthly temperatures as described in the table.
Considering each observation from the data set and map, it is likely that Noma is in Florida, Mekoryuk is in Alaska, and Nehawka is in Nebraska.
Pop Quiz
Comparing Means in Histograms
The following applet shows the histograms of two data sets. Move the slider to investigate the observations separately.
Discussion
Mean Absolute Deviation and Standard Deviation
When two data sets have similar centers, investigating the spread of each data set can be useful in highlighting their differences. One way of measuring spread is to calculate the average of each data value's distance from the mean. This measure is called the mean absolute deviation.
n∣x1−xˉ∣+∣x2−xˉ∣+⋯+∣xn−xˉ∣
Due to the difficulty of making calculations using the absolute value, a commonly used alternate approach to measuring the spread is to calculate the standard deviation. Concept
Standard Deviation
The standard deviation is a measure of spread of a data set that measures how much the data elements differ from the mean. The standard deviation, often represented by the Greek letter σ (sigma), is calculated by taking the square root of the variance of the data set. Let x1,x2,…,xn be the data values in a set and x their mean.
As shown, finding the standard deviation involves calculating the average of the squared differences between each data point and the mean, and then taking the square root of that average. The sum of squares can also be written in sigma notation.
σ=n(x1−x)2+(x2−x)2+⋯+(xn−x)2
The applet below calculates the standard deviation for the data set on the number line. Move the points around to change the data.
σ=n i=1∑n(xi−x)2
Standard deviation is sensitive to outliers because of the squaring of differences. It is commonly used when analyzing a data set that exhibits a normal distribution.Extra
Comparison of Standard Deviation and Mean Absolute DeviationThe standard deviation is greater or equal to the mean absolute deviation.
The proof of this relationship is based on the following inequality, which is true for any two positive numbers.
The proof of the inequality involving the means of more than two numbers is similar.
2a+b≤2a2+b2
The inequality can be proved as follows.
0≤(a−b)2
ExpandNegPerfectSquare
(a−b)2=a2−2ab+b2
0≤a2−2ab+b2
AddIneq
LHS+a2+2ab+b2≤RHS+a2+2ab+b2
a2+2ab+b2≤2a2+2b2
DivIneq
LHS/4≤RHS/4
4a2+2ab+b2≤42a2+2b2
FactorOut
Factor out 2
4a2+2ab+b2≤42(a2+b2)
FacPosPerfectSquare
a2+2ab+b2=(a+b)2
4(a+b)2≤42(a2+b2)
SimpQuot
Simplify quotient
4(a+b)2≤2a2+b2
SqrtIneq
LHS≤RHS
2a+b≤2a2+b2
Example
Using Measures of Spread in Geography
The table below shows the average monthly low temperatures of two cities — Kansas City and Seattle. The two cities given are not necessarily in order.
![Temperatures: city A = [36,37,39,43,47,52,54,55,52,47,41,38], city B = [18,20,29,40,50,60,66,65,58,47,37,28]](/mediawiki/images/d/d6/Mljsx_Slide_A1_U4_C2_Example3_1.svg)
According to the data set, City A and B have annual average low temperatures around 45∘F and 43∘F, respectively. Referencing the map below, Seattle is located much further north than Kansas City. It is typical that northern states — on average — are colder than southern states. Nevertheless, Seattle experiences less variance in temperature changes during each season due to the ocean's tempering effect on the climate.
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Hint
Think of the climate similarities and differences of coastal areas compared to inland areas. Find the range and standard deviation of the temperatures.
Solution
Using the range and standard deviation — measures of spread — will help compare the two cities' average low temperatures. A graphing calculator can be used to find the standard deviation.
| Range | Standard Deviation | |
|---|---|---|
| City A | 55−36=19 | 6.7 |
| City B | 66−18=48 | 16.4 |
These measures of spread show that the temperature throughout the year changes much less in City A than in City B. Based on that analysis, and considering the information given about the tempering effect of the ocean, it is reasonable to conclude that City A is Seattle and City B is Kansas City. What a cool conclusion to make.
Example
Stock Price Volatility
In the US stock market, a measure of how much a stock price fluctuates during a certain period of time is called historical volatility. The following data set from the year 2020 contains information about the daily closing stock price (in dollars) of two companies.
| Low | Mean | High | Standard Deviation | |
|---|---|---|---|---|
| APDN | $2.52 | $6.89 | $15.21 | $2.24 |
| DSS | $4.04 | $6.90 | $10.89 | $1.69 |
Which stock price was less volatile in 2020?
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Hint
Which part of the table gives information about the spread of the stock price?
Solution
The numbers in the given table can be interpreted into the following sentences.
- The two companies' mean stock price was very close for the year — separated merely by one-cent!
- The range for APDN is the high of $15.21 minus the low of $2.52, which is $12.69. In comparison, the DSS range is the high of $10.89, minus the low of $4.04, which is $6.85.
- The standard deviation for DSS, $1.69, is smaller than the standard deviation for APDN, $2.24.
Both the range and standard deviation are smaller for DSS. These interpretations indicate that DSS's stock price fluctuated less over the year than the stock price of APDN. Therefore, it can be concluded that the stock price of DSS was less volatile in 2020.
Extra
The graph below shows the change of the closing stock prices of the two companies during 2020 along with the ranges and the means.
The histogram below shows the distribution of the stock prices. 
Pop Quiz
Visually Inspecting the Standard Deviation in Histograms
The following applet shows the histograms of two data sets. Move the slider to investigate the data sets separately. Then, answer the given questions based on visual observations. 
Sometimes, instead of calculating the numerical statistics, visual inspection of the data distribution can tell the difference between data sets. This skill can come in handy when reading magazine articles and posts online, for example.
Example
Analyzing Histograms by Their Shapes
Consider the following two histograms where neither the labels nor scales are specified.
Both of these histograms represent different distributions, and both have 26 columns.
- One histogram shows the distribution of the numbers drawn in the Powerball lottery game since the latest version was introduced in 2015. In this version of the game, the randomly drawn Powerball number is between 1 and 26.
- The other histogram shows the distribution of AMC 8 (American Mathematics Competition) results in 2020. In this competition, participants had 25 multiple-choice questions in which a student could answer between 0 and 25 correctly. The histogram shows how many correct answers the participants received.
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Hint
In a lottery, all numbers are drawn with equal probability.
Solution
In a mathematics competition, only a few students answer all questions correctly. Still, a lot of students will be able to answer at least some of the questions correctly. Consequently, it is likely that the histogram is shaped like a mountain, with a peak in the center and low ends. The shape of Histogram A reflects this behavior.
In a lottery, all numbers are drawn with equal probability, so in the long run, it can be expected that there is little difference between the frequencies. The shape of Histogram B reflects this.
There is even more fascinating information to be discovered from the shapes of the histograms.
The height of the lone tall bar furthest to the left in Histogram A shows that in the AMC 8 competition, there were plenty of participants in 2020 who did not answer a single question correctly! Well, it is much more likely, however, that these participants registered but did not attend the competition.
Histogram A's peak shows that in 2020, on average, students in the AMC 8 competition answered less than half of the questions correctly.
The fluctuation of the bar heights in Histogram B shows that although an even distribution of the numbers is expected on the Powerball draw, some numbers historically came out fewer times.
The bar corresponding to 24 is more than twice as high as the bar corresponding to 16. However, this does not mean that 24 is twice as likely to come out in a draw. Nor does this mean that players should now play 16 because it will eventually catch up. The data is historical; it does not have any effect on the next draw.
Closure
Applying Various Methods to Analyze Data
At the beginning of the lesson, a data set of fish species was presented. The task was to match the Latin and Finnish names of four fish species by analyzing and comparing the fish drawings with a data set of fish lengths and heights. 
Since the fish drawings are not equal in scale, inspecting the length and height of the fish directly from the images is not helpful by itself. Using the data set, however, the ratio of length to height for each species can be computed to provide information about the fish shape. Then it can be compared to an analysis of the drawings.
The mean of the length to height ratios for all species should provide a good understanding of the shape of the species. The mean can be found using a graphing calculator, a spreadsheet software, or by hand. To do so by hand, add the values in each column and divide by the number of data points. The results, in increasing order, are summarized in the following table.
Each pair of the longest fish and tallest fish are too close in value to reliably distinguish between the species. Therefore, using the tables, matching the names in corresponding order seems like a natural method to find the conclusion.
This example shows that while data contains valuable information, it needs to be treated carefully. Conclusions based on the evaluation of some data are not always accurate. Therefore, statisticians will typically use additional measures to calculate how confident they can be in their conclusion(s) gathered from the methods used in this lesson.
| Mean Length to Height Ratio | |
|---|---|
| Abramis Bjorkna | 2.55 |
| Leuciscus Rutilus | 3.75 |
| Osmerus Eperlanus | 5.95 |
| Esox Lucius | 6.33 |
Next, the actual drawings can be used to find their length to height ratios. This measurement, however, is in pixels instead of centimeters. Most image software on a standard computer can show these measurements. Here, they are given. Recall that the drawings use the Finnish names.
External credits: Couch, Jonathan (Freshwater and Marine Image Bank)
External credits: Couch, Jonathan (Freshwater and Marine Image Bank)
External credits: Seeley, H.G. (Freshwater and Marine Image Bank)
External credits: Seeley, H. G. (Freshwater and Marine Image Bank)
The results, in increasing order, can be summarized as follows.
| Length to Height Ratio (Images) | |
|---|---|
| Pasuri | 120361≈3.01 |
| Särki | 106393≈3.71 |
| Hauki | 59358≈6.07 |
| Norssi | 55358≈6.51 |
The numbers in the two tables do not match exactly, which would make sense given that they are measured using different measurements, and the images are not matching in scale. Still, in both tables, two species have a ratio above 5 and two species have a ratio below 4. That means the following distinction can be made.
| Latin Name (Data Set) | Finnish Name (Images) | |
|---|---|---|
| Longer Fishes | Osmerus eperlanus and Esox lucius | Hauki and Norssi |
| Taller Fishes | Abramis bjorkna and Leuciscus rutilus | Pasuri and Särki |
Abramis BjorknaLeuciscus RutilusOsmerus EperlanusEsox Lucius⟷⟷⟷⟷PasuriSa¨rkiHaukiNorsi✓✓××
These pairings, however, are not entirely correct. While the taller fishes are paired correctly, the longer fishes are not matched correctly. The correct matches are shown in the table below, which just for reference also includes the English names. | Latin Name | Finnish Name | English Name |
|---|---|---|
| Abramis Bjorkna | Pasuri | Bream |
| Leuciscus Rutilus | Särki | Roach |
| Osmerus Eperlanus | Norssi | Smelt |
| Esox Lucius | Hauki | Pike |
Comparing Data Sets
Exercises
A company is interested in knowing how many miles their employees have to walk to and from work during a given week. They conduct a survey with 100 men and 100 women and present the results with two box plots.
One of the assistant managers, Tearrik, is asked to make a single boxplot showing every observation. The next morning, he produces the following box plot to show upper management.
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Notice that each box plot contains 100 observations, which is an even number. This means the median will be the mean of the 50^(th) and the 51^(st) observations. Similarly, the lower quartile is the mean of the 25^(th) and 26^(st) observations, and the upper quartile is the mean of the 75^(th) and 76^(st) observations. Therefore, each section of our box plots contains 25 observations.
By using this information we can identify how many observations should fall in the different sections of our two boxplots.
Notice that the combined boxplot contains 200 observations. Therefore, each of its four sections must contain 50 observations. Let's replace the two boxplots in the diagram above with the combined boxplot presented by Tearrik. We will keep the three intervals we have identified.
From the women's boxplot, at least 75 observations are less than or equal to 11.5. From the men's boxplot, at least 50 observations are less than or equal to 11.5. Combined, we have that at least 125 are less than or equal to 11.5 — so less than 12. This implies that the median of the combined data cannot be 12 as Tearrik's box plot states. Therefore, Vincenzo is correct.
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Consider three numbers and add four to each number.
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We will call the three numbers x_1, x_2, and x_3. Now let's add 4 to each of these numbers, which gives us a new data set. x_1+4, x_2+4, x_3+4 We will first write two expressions — one for the old mean x_O, and another for the new mean x_N. x_O &= x_1+x_2+x_3/3 [0.5em] x_N &= (x_1+4)+(x_2+4)+(x_3+4)/3 Next, we will simplify the right-hand side of the second expression.
If we examine the right-hand side, we see that we have the sum of a fraction that matches the old mean and 4.
Therefore, adding four to each number increased the mean by 4.
Let's write the standard deviation for the original data set. We have three numbers, so the denominator is 3.
We will also write an expression for the new standard deviation. Remember that the new mean is x_N=x_O+4.
As we can see, σ_N is the same expression as σ_O. This means the standard deviation remains unchanged by adding the same number to each observation from the data set.
Why did the mean increase by 4? Well, when we add 4 to each to each number, every observation in the data set is pushed to the right by 4 units. Therefore, it makes sense that the mean increased by 4.
However, since every number increased by the same amount, their relative distance to the mean remains unchanged. Therefore the standard deviation, which measures spread, is unchanged.
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Comparing Data Sets
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