chapter 3 displaying and summarizing quantitative...

Chapter 3 Displaying and SummarizingQuantitative Data

Dot Plots

DefinitionA dot plot is the representation of a set of data over a number line.The number of dots over a number represents the relative quantity ofthe value.

ExampleThe following ate the test scores for a particular high school studentin their math class over the course of an academic year.

64 73 85 74 8371 56 83 76 8583 87 92 84 9592 95 92 91

Solution

Grades of a High School Student

Grades

Freq

uenc

y

1

2

3

50 60 70 80 90 100

• • • •• ••

•

•

•

•

• •• •

•

••

•

Pros and Cons

What’s GoodGives a good idea of distribution

Preserves all of the data points

What’s Not GoodTedious to plot

Can be hard to read

Not practical for large data sets

Distributions

DefinitionA distribution is a representation of data vs. frequency. It shows allpossible values and how often they occur.

Now we want to concern ourselves with the analysis of the graphs.We can analyze these in a much more constructive way that we couldwith the graphs of categorical variables. Here we are analyzing thedistribution represented by the graph.

Distribution Analysis

1 Center: Which class contains the central element(s)

2 Shape: Number of peaks, skewness3 Spread: Range=max-min

In our example, we can see a couple of things:

Range: Highest value - lowest valueHere, the range would be 95− 56 = 39.Center: The central value(s) is the center. It could be a value or aclass, depending on the type of graph.Here, the center is the 10th value, since there are 19 data points inthe set. The value we seek is 84.Shape: How many peaks are there? Is it roughly in the middle orto one side?Here we have one peak, so we would say the distribution isunimodal. That peak is to the right, so the tail stretches out to theleft. We would say this graph is left skewed.


1 Center: Which class contains the central element(s)2 Shape: Number of peaks, skewness

3 Spread: Range=max-min




1 Center: Which class contains the central element(s)2 Shape: Number of peaks, skewness3 Spread: Range=max-min






Range: Highest value - lowest valueHere, the range would be 95− 56 = 39.

Center: The central value(s) is the center. It could be a value or aclass, depending on the type of graph.Here, the center is the 10th value, since there are 19 data points inthe set. The value we seek is 84.Shape: How many peaks are there? Is it roughly in the middle orto one side?Here we have one peak, so we would say the distribution isunimodal. That peak is to the right, so the tail stretches out to theleft. We would say this graph is left skewed.




Range: Highest value - lowest valueHere, the range would be 95− 56 = 39.Center: The central value(s) is the center. It could be a value or aclass, depending on the type of graph.Here, the center is the 10th value, since there are 19 data points inthe set. The value we seek is 84.

Shape: How many peaks are there? Is it roughly in the middle orto one side?Here we have one peak, so we would say the distribution isunimodal. That peak is to the right, so the tail stretches out to theleft. We would say this graph is left skewed.

Stem-and-Leaf Plots

Similarities to Dot Plots

Gives idea of distribution

Preserves data

Not practical for large data sets

Differences from Dot Plots

Used for quantitative variables

Easier to read actual data elements

Can be used for comparisons of two data sets

Stem-and-Leaf Plot Example

ExampleUsing the same data set as we did for the dot plot, construct astem-and-leaf plot.

First thing we need to do is order the data elements.

56 64 71 73 7476 83 83 83 8485 85 87 91 9292 92 95 95


Grades for a High School Student98765

These would be the stems for our plot.

Note: Repetition is extremely important.


Grades for a High School Student9 1 2 2 2 5 58 3 3 3 4 5 5 77 1 3 4 66 45 6

Here we get the exact same answer for the range and the center,although we only give the class in which the center lies, so we wouldsay that the center is in the 80’s. We get that the shape is againunimodal and skewed left. It may look different, but since itrepresents the same distribution, we expect similar answers.

Notice that the values on the right are essentially in columns - this iswhat allows us to quickly see which classes have more elements.

More Stem-and-Leaf Plots

What if we had a 3 digit number? Suppose the student got a 100 onthe next exam?

Grades for a High School Student10 09 1 2 2 2 5 58 3 3 3 4 5 5 77 1 3 4 66 45 6

Stem-and-Leaf Plots for Comparisons

ExampleSuppose we wanted to compare the careers of Babe Ruth and MarkMcGwire in terms of their yearly home run totals to determine whichplayer was the more consistent long ball hitter. Make a back-to-backstem-and-leaf plot to make the is determination.

Ruth: 54, 59, 35, 41, 46, 25, 47, 60, 54, 46, 49, 46, 41, 34, 22McGwire: 49, 32, 33, 39, 22, 42, 9, 9, 39, 52, 58, 70, 65, 32, 29


Ruth v. McGwire76543210

We set up the graph with one set of data increasing out to the rightand the other increasing out to the left. This way we have aside-by-side comparison of the data sets.


Ruth v. McGwire7 0

0 6 59 4 4 5 2 8

9 7 6 6 6 1 1 4 2 95 4 3 2 2 3 9 95 2 2 2 9

10 9 9

Who is more consistent and why?

Histograms

Used for quantitative variables

Tracks frequency and shows distribution

Does not preserve individual values

Good for a large number of values

Bars must be vertical and must touch

Histograms

ExampleFor our test scores example, construct a histogram and analyze thedistribution.

It is easier if the values are in order as we will be grouping them intoclasses.

56 64 71 73 7476 83 83 83 8485 85 87 91 9292 92 95 95

Histograms

We first want to create a frequency table. This is a collection ofnon-overlapping classes and the frequency of observation in each ofthose classes. We need to determine the following in this order:

Number of classesThe rule of thumb with the number of classes is to use the square rootof the number of observations in the data set.

√19 ≈ 4.36

So, we can use 4 or 5 classes. I tend to go up to the next integer to besure I have enough classes. So we will use 5 for our graph.

Histograms

Size of each classWe want them to be the same width so that the taller classes will beknown to have the most elements. If not then we have to find the areaof each rectangle to determine relative size.To find the size, we divide the ‘range’ by the number of classes.

size =95− 56 + 1

5=

385

= 7.6

We could use 7.6 for the class width or we can go to the next largestinteger. Where we may have extra if we round up, it is better than nothaving enough of a range in the classes to cover all of the data. Forthe sake of simplicity, we will use 8.

Histograms

Endpoints of each classWe start the smallest class with a left endpoint of 56, since that wasour minimum. Then, to find the next left endpoint, add 8 to 56.Continue in this manner until we have 5 classes.

Grade Range Frequency56-64-72-80-88-

Histograms

Then, we subtract 1 from each left endpoint to find the right endpointof the previous class.

Grade Range Frequency56-6364-7172-7980-8788-95

Finally, we count how many elements go in each class.

Grade Range Frequency56-63 164-71 272-79 380-87 788-95 6

Histograms

Grades of a High School Student

Grades

Freq

uenc

y

2

4

6

8

56 64 72 80 88 96

We see the same range and shape. Here, we’d have no choice but togive the class only for the center as we would lose the ability to seeindividual values.

Using The Calculator

We can make some graphs on the TI-series graphing calculator. Oneof the options we have is to make a histogram.

The advantages to using technology are that we don’t have to makefrequency tables or figure out how many classes we need, etc.

We do have to keep in mind, however, that the number of classes maybe different than when we make the graph by hand. We are usingmore approximations when we work by hand than when we usetechnology. But, this is an acceptable difference as long as the methodwe use is valid.

How To Make Histograms On The TI

1 In the STAT menu, select EDIT

2 Input all of the data in the same column

3 Press 2nd and then MODE to quit to a blank screen

4 Press 2nd and Y= to get into the STATPLOT menu5 Make sure all of the plots are off (if need be, use option 4)6 Press ENTER on one of the plots7 Turn the plot ON with the ENTER key8 Select the histogram, which is the third graph in the top row9 Make sure the XList is correct and then press ZOOM and then

9 , which is the option for ZOOMSTAT

If you want to get data from the graph, like endpoints for classes orfor the frequency for a class, you can press TRACE and then use thearrows to scroll around.


1 In the STAT menu, select EDIT2 Input all of the data in the same column








4 Press 2nd and Y= to get into the STATPLOT menu

5 Make sure all of the plots are off (if need be, use option 4)6 Press ENTER on one of the plots7 Turn the plot ON with the ENTER key8 Select the histogram, which is the third graph in the top row9 Make sure the XList is correct and then press ZOOM and then






4 Press 2nd and Y= to get into the STATPLOT menu5 Make sure all of the plots are off (if need be, use option 4)

6 Press ENTER on one of the plots7 Turn the plot ON with the ENTER key8 Select the histogram, which is the third graph in the top row9 Make sure the XList is correct and then press ZOOM and then






4 Press 2nd and Y= to get into the STATPLOT menu5 Make sure all of the plots are off (if need be, use option 4)6 Press ENTER on one of the plots

7 Turn the plot ON with the ENTER key8 Select the histogram, which is the third graph in the top row9 Make sure the XList is correct and then press ZOOM and then






4 Press 2nd and Y= to get into the STATPLOT menu5 Make sure all of the plots are off (if need be, use option 4)6 Press ENTER on one of the plots7 Turn the plot ON with the ENTER key

8 Select the histogram, which is the third graph in the top row9 Make sure the XList is correct and then press ZOOM and then






4 Press 2nd and Y= to get into the STATPLOT menu5 Make sure all of the plots are off (if need be, use option 4)6 Press ENTER on one of the plots7 Turn the plot ON with the ENTER key8 Select the histogram, which is the third graph in the top row

9 Make sure the XList is correct and then press ZOOM and then9 , which is the option for ZOOMSTAT


Histograms

ExampleThe EPA lists most sports cars in its “two-seater” category. The tablebelow gives the city mileage in miles per gallon. Make and analyze ahistogram for the the city mileage.

Model Mileage Model MileageAcura NSX 17 Insight 57Audi Quattro 20 S2000 20Audi Roadster 22 Lamborghini 9BMW M Coupe 17 Mazda 22BMW Z3 Coupe 19 SL500 16BMW Z3 Roadster 20 SL600 13BMW Z8 13 SLK230 23Corvette 18 SLK 320 20Prowler 18 911 15Ferrari 360 11 Boxster 19Thunderbird 17 MR2 25

Histograms

There are 22 cars, so we would use 4 <√

22 < 5 classes, so here Iwill choose 5. The size of each class would be

57− 9 + 15

=495

= 9.8

So we will use 10.

Mileage Frequency9 -18 1119-28 1029-38 039-48 049-58 1

Histograms

MPG for Sports Cars

MPG

Freq

uenc

y

3

6

9

12

9 19 29 39 49 59

Center: Boundary between the first two classesRange: 58− 9 = 49Shape: Unimodal, skewed right

Histograms

MPG for Sports Cars

MPG

Freq

uenc

y

3

6

9

12

9 19 29 39 49 59

Center:

Boundary between the first two classesRange: 58− 9 = 49Shape: Unimodal, skewed right

Histograms

MPG for Sports Cars

MPG

Freq

uenc

y

3

6

9

12

9 19 29 39 49 59

Center: Boundary between the first two classesRange:

58− 9 = 49Shape: Unimodal, skewed right

Histograms

MPG for Sports Cars

MPG

Freq

uenc

y

3

6

9

12

9 19 29 39 49 59

Center: Boundary between the first two classesRange: 58− 9 = 49Shape:

Unimodal, skewed right

Histograms

MPG for Sports Cars

MPG

Freq

uenc

y

3

6

9

12

9 19 29 39 49 59

Center: Boundary between the first two classesRange: 58− 9 = 49Shape: Unimodal, skewed right

Central Tendency

We will use three methods of measuring central tendency:1 mean2 median3 mode

Example

ExampleFind the mean, median and mode for the following data set.

11 010987 46 05 0 0 0 543 0 0 0 0 1 2 52 510 4

Solution

Mean xThis the arithmetic center.

x =1n

n∑k=1

xk

This is just a fancy way of saying to add the 16 values togetherand divide by 16. When we do we get

x = 43.5

Solution

Median MThis is the geometric center. To find, we line up all of the valuesin order and find the middle one. If there is an odd number ofobservations, then the median is the one in the middle. If there isan even number of observations, the median is the mean of thetwo ‘middle’ values. Here, we have

M =32 + 35

2= 33.5

ModeThis is the value(s) that occur most often, unless all values occurthe same number of times, in which case there is no mode. Here,

mode = 30

The Relationship Between Mean and Median

This picture indicates a serious drawback to using means: outliers.The median is what we call resistant; an extreme value does not affectthe median. The mean, however, is not resistant.

When We Use Mean v. Median

1 If distribution is symmetric, then mean = median, and we use themean

2 If there are outliers or strong skewness, we use the median

Using the Mean

ExampleSuppose you got an 84, 72 and 78 on your first 3 exams and wanted toknow what grade you needed to get on the fourth exam to have at leastan 80 average?

We want an average of 80 for the 4 grades. So, we need to solve for xin

84 + 72 + 78 + x4

=234 + x

4= 80

So, we get

234 + x4

= 80⇒ 234 + x = 320⇒ x = 86

Another Mean Example

Example

Suppose you had a 75 average through 4 tests and got an 85 on the 5th

test. What is your average now?

If we have a 75 average through 4 exams, then we have accumulated75× 4 = 300 points. So, if we wanted to know the average with this5th grade, we’d have

x =300 + 85

5=

3855

= 77

Yet Another Mean Example

ExampleSuppose you had a group of 11 people and the average age was 27. Ifone of those people left, the average age of the remaining 10 was 29.What is the age of the person who left?

Total age of the 11 people: 11× 27 = 297.

Total age of the 10 people : 10× 29 = 290

Difference is 297− 290 = 7

Means From Frequency Tables

ExampleFind the mean of the following values.

Age Frequency21 522 823 424 125 2

We first count the total number of observations, which is 20. Then ...

x =21 ∗ 5 + 22 ∗ 8 + 23 ∗ 4 + 24 ∗ 1 + 25 ∗ 2

20=

44720

= 22.35

Box Plots and the 5-Number Summary

When dealing with the median, we measure variation with the5-number summary. These 5 numbers indicate the maximum andminimum, the median and the quartiles.

In order to find the five number summary, we first line the dataelements in order. Then we find the minimum and maximum, andthen the median.

minimum smallest value of the setmaximum largest value of the set

median central(s) value of the setfirst quartile Q1 median of all values smaller than the medianthird quartile Q3 median of all values larger than the median




minimum smallest value of the set

maximum largest value of the setmedian central(s) value of the set

first quartile Q1 median of all values smaller than the medianthird quartile Q3 median of all values larger than the median





median central(s) value of the set

first quartile Q1 median of all values smaller than the medianthird quartile Q3 median of all values larger than the median





median central(s) value of the setfirst quartile Q1 median of all values smaller than the median

third quartile Q3 median of all values larger than the median

5-Number Summary Example

ExampleFind the 5-number summary for the data from the first example.

11 010987 46 05 0 0 0 543 0 0 0 0 1 2 52 510 4

5-Number Summary Example

Since the values are already in order, we only need to calculate thevalues.

minimum 4Q1 30

Median 33.5Q3 52.5

maximum 110

Teddy Ballgame

ExampleTed Williams yearly RBI totals:145, 113, 120, 137, 123, 114, 127, 159, 97, 126, 3, 34, 89, 83, 82, 87,85, 43, 72Find the 5-number summary for this set of data,

What do we do first?

We put the values in order:3, 34, 43, 72, 82, 83, 85, 87, 89, 97, 113, 114, 120, 123, 126, 127,137, 145, 159.Then ...

Teddy Ballgame



We put the values in order:3, 34, 43, 72, 82, 83, 85, 87, 89, 97, 113, 114, 120, 123, 126, 127,137, 145, 159.

Then ...

Teddy Ballgame



We put the values in order:3, 34, 43, 72, 82, 83, 85, 87, 89, 97, 113, 114, 120, 123, 126, 127,137, 145, 159.Then ...

Solution

Minimum 3Q1 82

Median 97Q3 126

Maximum 159

Box-and-Whisker Plots

How can we visually represent this summary of the data? We use boxplots, or box-and-whisker plots.

Ted Williams’ RBI Totals

RBIs

Teddy Ballgame

20 40 60 80 100

120

140

160

Box-and-Whisker Plots

How can we visually represent this summary of the data? We use boxplots, or box-and-whisker plots.

Ted Williams’ RBI Totals

RBIs

Teddy Ballgame20 40 60 80 10

0

120

140

160

Using Technology

The box plot is another that we can construct using the TI-seriesgraphing calculator. We do everything the same as when constructinga histogram until we reach the point where we choose the type ofgraph.

There are two options for box plots.

1 Second row, first graph shows outliers (we will get to those soon)2 Second row, second graph does not show outliers

We again use ZOOM and 9 to produce the graph.

Using Technology



1 Second row, first graph shows outliers (we will get to those soon)

2 Second row, second graph does not show outliers


Using Technology



1 Second row, first graph shows outliers (we will get to those soon)2 Second row, second graph does not show outliers


Getting Statistics

We can also find the statistics we need using the calculator relativelyeasily.

1 Input the data in the usual way

2 Press 2nd and MODE to quit to a blank screen

3 Press STAT , scroll to CALC, and select 1-Var Stats4 You will see 1-Var Stats on the screen; now select which list the

data is in by pressing 2nd and then the appropriate number 1-6,followed by the ENTER key

On this screen are some statistics we need x and Sx and if we scrolldown, we will see the 5-number summary.

Getting Statistics




3 Press STAT , scroll to CALC, and select 1-Var Stats

4 You will see 1-Var Stats on the screen; now select which list thedata is in by pressing 2nd and then the appropriate number 1-6,followed by the ENTER key


Getting Statistics




3 Press STAT , scroll to CALC, and select 1-Var Stats4 You will see 1-Var Stats on the screen; now select which list the

data is in by pressing 2nd and then the appropriate number 1-6,followed by the ENTER key


The Geometric View

From the minimum to Q1 is the bottom 25% of the observations

From Q1 to Q3 is the middle 50% of the observations

From Q3 to the maximum of the top 25% of the observations

We can look at this in other ways too:

The top half lies above the median

The top 75% lies above Q1

The bottom 75% lies below Q3

Box-and-Whisker Plot Example

ExampleConstruct a box-and-whisker plot for the data from the first example.

minimum 4Q1 30

Median 33.5Q3 52.5

maximum 110

Solution

Some Data Set

Values

Set 125 50 75 10

0

125

Analysis of Box-and-Whisker Plots

We can also look at the distribution like we did with histograms, butin a limited way as we cannot really tell how many peaks. But we canlook at the spread and center (directly from the table) and we can lookat the skewness.

What is the range here? 106

What do we know about the distribution? Skewed right distribution.

Further, this right endpoint seems to be pretty far away, so we maythink it is an outlier. But how do we determine if it is analytically?



What is the range here?

106






What do we know about the distribution?

Skewed right distribution.


IQR Criterion

DefinitionThe IQR Criterion is an analytic way for us to determine if data pointsare outliers based on a 5-number summary. To determine outliers, weuse

Q1 − 1.5IQR

andQ3 + 1.5IQR

to give us endpoints of the acceptable data range, where IQR is theInterquartile Range and

IQR = Q3 − Q1

These new endpoints are sometimes referred to as fences.

Using the IQR Criterion

So, basically what we are doing is saying that any values no furtheraway from the middle 50% than 1.5 times the range of the middle50% are acceptable. Anything outside that range is an outlier.

ExampleAre there any outliers in the previous data set?

First we find the IQR, which is

Q3 − Q1 = 52.5− 30 = 22.5

and then we consider the new endpoints (fences).

Q1 − 1.5IQR = 30− 1.5(22.5) = 30− 33.75 = −3.75

Q3 + 1.5IQR = 52.5 + 1.5(22.5) = 52.5 + 33.75 = 86.25

Since 110 is larger than this upper threshhold, we would say it is anoutlier.

Standard Deviation

The standard deviation measures the variation in data by measuringthe distance that the observations are from the mean. The standarddeviation tells us how far we can expect the average observation to befrom the mean.

Absolute Deviation

1n

n∑i=1

|xi − x|

Standard Deviation

s =

√∑(x− xi)2

n− 1

Standard Deviation and Variance

Standard Deviation

s =

√∑(xi − x)2

n− 1

Whereas it won’t have a lot of use for our purposes

Variance

s2 =

∑(xi − x)2

n− 1

Finding the Standard Deviation

ExampleFind the standard deviation of the daily caloric intake for a personover the course of a week.

{1792, 1666, 1362, 1614, 1460, 1867, 1439}

First we find the mean.

x =11200

7= 1600


Then, we need to find the difference between each of these values andthe mean, then square that differences and then sum them.

xi (xi − x)2 square contribution1792 (1792− 1600)2 1922 36864

1666 (1666− 1600)2 662 43561362 (1362− 1600)2 (−238)2 566441614 (1614− 1600)2 142 1961460 (1460− 1600)2 (−140)2 196001867 (1867− 1600)2 2672 712891439 (1439− 1600)2 (−161)2 25921



xi (xi − x)2 square contribution1792 (1792− 1600)2 1922 368641666 (1666− 1600)2 662 43561362 (1362− 1600)2 (−238)2 566441614 (1614− 1600)2 142 1961460 (1460− 1600)2 (−140)2 196001867 (1867− 1600)2 2672 712891439 (1439− 1600)2 (−161)2 25921



xi (xi − x)2 square contribution1792 (1792− 1600)2 1922 368641666 (1666− 1600)2 662 43561362 (1362− 1600)2 (−238)2 566441614 (1614− 1600)2 142 1961460 (1460− 1600)2 (−140)2 196001867 (1867− 1600)2 2672 712891439 (1439− 1600)2 (−161)2 25921

sum 214870


Next, we divide by 6.

s2 =214870

6≈ 35811.67

s2 is the ...

variance.

Now we take the square root.

s =√

35811.67 ≈ 189.24

So, the average value of the caloric intake is approximately 189calories from the mean. Notice that we only care about magnitude andnot whether we are above or below the mean.


Next, we divide by 6.

s2 =214870

6≈ 35811.67

s2 is the ...variance.

Now we take the square root.

s =√

35811.67 ≈ 189.24

So, the average value of the caloric intake is approximately 189calories from the mean. Notice that we only care about magnitude andnot whether we are above or below the mean.

Mean and Standard Deviation

So what can we do with mean and standard deviation?

We can use them to relate individuals within our data set to thedistribution of the sample.

This is related to probability.

The total area underneath a distribution curve is always 1, so thearea under the curve is the same as the percent of observationsfalling in the region.

We will see this better when we get to Normal distributions.

Uniform Distributions

For these, all values have the same probability of occurring. So, theshape is that of a rectangle.

Random Number Between 0 and 2

x

f (x) 12

1 2


ExampleIf we have a uniform distribution for a random number to be chosenbetween 0 and 2, what is the probability that the number selected isbetween .5 and 1.1?


x

f (x) 12

1 2



x

f (x) 12

1 2

What is the area of a rectangle? length× width.What are our dimensions? .5× .6 = .3.So, there is a 30% chance that the number randomly selected falls inthis region.



x

f (x) 12

1 2

What is the area of a rectangle?

length× width.What are our dimensions? .5× .6 = .3.So, there is a 30% chance that the number randomly selected falls inthis region.



x

f (x) 12

1 2

What is the area of a rectangle? length× width.

What are our dimensions? .5× .6 = .3.So, there is a 30% chance that the number randomly selected falls inthis region.



x

f (x) 12

1 2

What is the area of a rectangle? length× width.What are our dimensions?

.5× .6 = .3.So, there is a 30% chance that the number randomly selected falls inthis region.



x

f (x) 12

1 2

What is the area of a rectangle? length× width.What are our dimensions? .5× .6 = .3.

So, there is a 30% chance that the number randomly selected falls inthis region.



x

f (x) 12

1 2

What is the area of a rectangle? length× width.What are our dimensions? .5× .6 = .3.So, there is a 30% chance that the number randomly selected falls inthis region.

chapter 3 displaying and summarizing quantitative...

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