Displaying quantitative data 1.1
- 1. Displaying Quantitative Data with Graphs Section 1.1
- 2. What you’ll learn To create and interpret the following graphs: Dotplot Stem and leaf Regular Stem and Leaf Split Stem and Leaf Back-to-Back Stem and Leaf Histogram Time Plot Ogive
- 3. To learn how to display and describe quantitative data we will be using some baseball statistics. The following table shows the number of home runs in a single season for three well-known baseball players: Hank Aaron, Barry Bonds, and Babe Ruth. Hank Aaron Barry Bonds Babe Ruth 13 32 16 40 54 46 27 44 25 37 59 41 26 39 24 34 35 34 44 29 19 49 41 22 30 44 33 73 46 39 38 25 25 40 47 34 47 34 34 46 60 45 40 37 54 44 20 33 46 24 42 49
- 4. Dotplot Label the horizontal axis with the name of the variable and title the graph Scale the axis based on the values of the variable Mark a dot (we’ll use x’s) above the number on the axis corresponding to each data value Number of Hom Runs in a Single Season e Dot Plot 20 25 30 35 40 45 50 55 60 Ruth
- 5. Describing a Distribution We describe a distribution (the values the variable takes on and how often it takes these values) using the acronym SOCS Shape– We describe the shape of a distribution in one of two ways: Symmetric/Approx. Symmetric Collection 1 Dot Plot Shape Dot Plot -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 4 Sym etric m Uniform
- 6. Skewed Right Left Shape Dot Plot Shape Dot Plot “tail” “tail” -4 -3 -2 -1 0 1 2 3 4 -3 -2 -1 0 1 2 3 4 RightSkew ed LeftSkew ed Notice that the direction of the “skew” is the same direction as the “tail”
- 7. •Outliers: These are observations that we would consider “unusual”. Pieces of data that don’t “fit” the overall pattern of the data. Babe Ruth had two seasons Number of Home Runs in a Single Season Dot Plot that appear to be somewhat different than the rest of his Unusual observation??? career. These may be “outliers” 20 25 30 35 40 45 50 55 60 65 (We’ll learn a numerical way to Ruth determine if observations are truly “unusual” later) Number of Home Runs in a Single Season Dot Plot Unusual observation??? The season in which Barry Bonds hit 73 home runs does not appear to fit the overall pattern. This piece 10 20 30 40 50 60 70 80 of data may be an outlier. Bonds
- 8. Center: A single value that describes the entire distribution. A “typical” value that gives a concise summary of the whole batch of numbers. Number of Home Runs in a Single Season Dot Plot 20 25 30 35 40 45 50 55 60 65 Ruth A typical season for Babe Ruth appears to be approximately 46 home runs *We’ll learn about three different numerical measures of center in the next section
- 9. Spread: Since we know Number of Home Runs in a Single Season Dot Plot that not everyone is typical, we need to also talk about the variation of a distribution. We need to discuss if the values of the distribution are tightly 20 25 30 35 40 45 50 55 60 65 clustered around the Ruth center making it easy to predict or do the values Babe Ruth’s number of home runs in a vary a great deal from the single season varies from a low of 23 to center making prediction a high of 60. more difficult? *We’ll learn about three different numerical measures of spread in the next section.
- 10. Distribution Description using SOCS The distribution of Babe Ruth’s number of home runs in a single season is approximately symmetric1 with two possible unusual observations at 23 and 25 home runs .2 He typically hits about 463 home runs in a season. Over his career, the number of home runs has varied from a low of 23 to a high of 60. 4 1-Shape 2-Outliers 3-Center 4-Spread
- 11. Stem and Leaf Plot Creating a stem and leaf plot Number of Home Runs in a Single Season Order the data points from least to greatest Separate each observation Hank Aaron into a stem (all but the 1 3 rightmost digit) and a leaf (the final digit)—Ex. 123-> 12 2 04679 (stem): 3 (leaf) In a T-chart, write the stems 3 0244899 vertically in increasing order on the left side of the chart. 4 00444457 On the right side of the chart write each leaf to the right of its stem, spacing the leaves Key equally Include a key and title for the graph 4 6 = 46
- 12. Split Stem and Leaf Plot If the data in a distribution is concentrated in just a few stems, the picture may be more descriptive if we “split” the stems When we “split” stems we want the same number of digits to be possible in each stem. This means that each original stem can be split into 2 or 5 new stems. A good rule of thumb is to have a minimum of 5 stems overall Let’s look at how splitting stems changes the look of the distribution of Hank Aaron’s home run data.
- 13. Split each stem into 2 Number of Home Runs in a new stems. This Single Season means that the first Hank Aaron stem includes the 1 3 leaves 0-4 and the 1 second stem has the 2 04 leaves 5-9 2 679 Splitting the stems 3 0244 helps us to “see” the 3 899 shape of the 4 004444 distribution in this 4 57 Key case. 4 6 = 46
- 14. Back-to-Back Stem and Leaf Number of Home Runs in a Single Back-to-Back stem Season and leaf plots allow Aaron 3 1 Ruth us to quickly 1 compare two 40 2 2 distributions. 976 2 5 4420 3 4 998 3 5 Use SOCS to 444400 4 11 make comparisons 75 4 66679 between 5 449 distributions 5 Key 6 0 4 6 = 46
- 15. Advantages and Disadvantages of dotplots/stem and leaf plots Advantages Disadvantages Preserves each piece If creating by hand, of data large data sets can be cumbersome Shows features of the distribution with Data that is widely regards to shape— varied may be difficult such as clusters, gaps, to graph outliers, etc
- 16. Histograms A histogram is one of the most common graphs used for quantitative variables. Although a histogram looks like a bar chart there are some important differences In a histogram, the “bars” touch each other Histograms do not necessarily preserve individual data pieces Changing the “scale” or “bin width” can drastically alter the picture of the distribution, so caution must be used when describing a distribution when only a histogram has been used
- 17. Creating a histogram Divide the range of Barry Bonds: data into classes of Data Ranges from 16 equal width. Count to 73, so we choose the number of for our classes observations in each class. (Remember 15 ≤ # of HR ≤ 19 . that the width is . . somewhat arbitrary 70 ≤ # of HR ≤ 75 and you might choose We can then a different width than determine the counts someone else) for each “bin”
- 18. So the frequency The horizontal axis distribution looks like: represents the variable values, so Class Frequency using the lower bound 15-19 2 of each class to scale 20-24 1 is appropriate. 25-29 2 The vertical axis can 30-34 4 represent 35-39 2 Frequency 40-44 2 45-49 2 Relative frequency 50-54 0 Cumulative frequency 55-59 0 Relative cumulative 60-64 0 frequency 65-69 0 We’ll use frequency 70-74 1
- 19. Label and scale your axes. Title your graph Draw a bar that represents the frequency for each class. Remember that the bars of the histograms should touch each other. Barry Bonds Histogram 7 6 5 Count 4 3 2 1 10 20 30 40 50 60 70 80 90 HomeRuns
- 20. Interpretation We interpret a histogram in the same way we interpret a dotplot or stem and leaf plot. ALWAYS use SOCS Shape Outliers Center Spread
- 21. Time Plots Sometimes, our data is collected at intervals over time and we are looking for changes or patterns that have occurred. We use a time plot for this type of data A time plot uses both the horizontal and vertical axes. The horizontal axis represents the time intervals The vertical axis represents the variable values
- 22. Creating a Time Plot Barry Bonds Line Scatter Plot Label and scale the 80 70 axes. Title your 60 BondsHR 50 graph. 40 30 Plot a point 20 10 corresponding to the 1986 1990 1994 Year 1998 2002 data taken at each time interval Year 1986 HR 16 Year 1994 HR 37 A line segment drawn 1987 25 1995 33 between each point 1988 1989 24 19 1996 1997 42 40 may be helpful to see 1990 33 1998 37 patterns in the data 1991 1992 25 34 1999 2000 34 49 1993 46 2001 73
- 23. Describing Time Plots When describing time Barry Bonds Line Scatter Plot plots, you should look for 80 trends in the data 70 60 Although the number of BondsHR 50 home runs do not show a 40 constant increase from 30 year to year we note that 20 overall, the number of 10 home runs made by 1986 1990 1994 Year 1998 2002 Barry Bond has increased over time with the most notable increase being between 1999 and 2001.
- 24. Relative frequency, Cumulative frequency, Percentiles, and Ogives Sometimes we are interested in describing the relative position of an observation For example: you have no doubtably been told at one time or another that you scored at the 80th percentile. This means that 80% of the people taking the test score the same or lower than you did. How can we model this?
- 25. Ogive (Relative cumulative frequency graph) We first start # of home Relative runs in a Relative Cumulative Cumulative by creating a season Frequency Frequency Frequency Frequency frequency 15-19 2 0.125 2 0.125 20-24 1 0.0625 3 0.1875 table 25-29 2 0.125 5 0.3125 We’ll look at 30-34 4 0.25 9 0.5625 35-39 2 0.125 11 0.6875 how each 40-44 2 0.125 13 0.8125 column is 45-49 2 0.125 15 0.9375 created in the 50-54 0 0 15 0.9375 next few 55-59 0 0 15 0.9375 60-64 0 0 15 0.9375 slides 65-69 0 0 15 0.9375 70-74 1 0.0625 16 1
- 26. Relative Frequency The # of home runs… and # of home * the frequency are the same runs in a season Frequency Relative Frequency columns as we created for 15-19 2 0.125 the histogram. 20-24 25-29 1 2 0.0625 0.125 To find the values for the 30-34 4 0.25 35-39 2 0.125 “Relative Frequency” 40-44 2 0.125 column find the following: 45-49 2 0.125 50-54 0 0 Frequency Value 55-59 0 0 Total # of = Relative Frequency 60-64 0 0 observations 65-69 0 0 70-74 1 0.0625 * Within rounding, this column should equal 1
- 27. Cumulative Frequency Cumulative frequency # of home simply adds the runs in a Relative Cumulative counts in the season 15-19 Frequency 2 Frequency 0.125 Frequency 2 frequency column that 20-24 1 0.0625 3 fall in or below the 25-29 2 0.125 5 current class level. 30-34 35-39 4 2 0.25 0.125 9 11 For Example: to find 40-44 2 0.125 13 the “13”, add the 45-49 50-54 2 0 0.125 0 15 15 frequencies in the 55-59 0 0 15 oval: 60-64 0 0 15 2+1+2+4+2+2=13 65-69 70-74 0 1 0 0.0625 15 16
- 28. Relative Cumulative Frequency Relative cumulative # of ho m frequency divides the e runs in a Relative Cumulative Relative Cumulative cumulative frequency season Frequency Frequency Frequency Frequency 15-19 2 0.125 2 0.125 by the total number of 20-24 1 0.0625 3 0.1875 observations 25-29 30-34 2 4 0.125 0.25 5 9 0.3125 0.5625 35-39 2 0.125 11 0.6875 40-44 2 0.125 13 0.8125 45-49 2 0.125 15 0.9375 For Example: 50-54 0 0 15 0.9375 55-59 0 0 15 0.9375 .8125 = 13/16 60-64 0 0 15 0.9375 65-69 0 0 15 0.9375 70-74 1 0.0625 16 1 Sum 16 1
- 29. Creating the Ogive Label and scale the axes Horizontal: Variable Vertical: Relative Cumulative Frequency (percentile) Plot a point corresponding to the relative cumulative frequency in each class interval at the left endpoint of the next class interval The last point you should plot should be at a height of 100%
- 30. # of home Relative runs in a Cumulative Barry Bonds Scatter Plot season Frequency * 15-19 0.125 1.2 20-24 0.1875 1.0 Relcumfreq 25-29 0.3125 0.8 30-34 0.5625 35-39 0.6875 0.6 40-44 0.8125 0.4 45-49 0.9375 0.2 50-54 0.9375 55-59 0.9375 0.0 60-64 0.9375 65-69 0.9375 10 20 30 40 50 60 70 80 70-74 1 HR A line segment from point to point can be added for analysis
- 31. Types of Info from Ogives Finding an individual observation within the distribution Find the relative standing of a season in which Barry Bonds hit 40 home runs Barry Bonds Scatter Plot Relcumfreq 1.2 1.0 0.8 0.6 0.4 0.2 0.0 10 20 30 40 50 60 70 80 HR A season with 40 home runs lies at the 60th percentile, meaning that approximately 60% of his seasons had 40 or less home runs
- 32. Locating an observation corresponding to a percentile. How many home runs must be hit in a season to correspond to the 75th percentile? Barry Bonds Scatter Plot Relcumfreq 1.2 1.0 0.8 0.6 0.4 0.2 0.0 10 20 30 40 50 60 70 80 HR To be better than 75% of Mr. Bonds season, approximately 42 home runs must be hit.
- 33. A little History on the word Ogive (sometimes called an Ogee) It was first used by Sir Francis Galton, who borrowed a term from architecture to describe the cumulative normal curve (more about that next chapter). The ogive in architecture was a common decorative element in many of the English Churches around 1400. The picture at right shows the door to the Church of The Holy Cross at the village of Caston in Norfolk. In this image you can see the use of the ogive in the design of the door and repeated in the windows above. Find more about this term at Mathwords.