MM207-Statistics Unit 2 Seminar-Descriptive Statistics Dr Bridgette Stevens AIM:BStevensKaplan (add me to your Buddy list) 1.

MM207-Statistics

Unit 2 Seminar-Descriptive Statistics

Dr Bridgette Stevens

AIM:BStevensKaplan (add me to your Buddy list)

MM207-Statistics

Unit 2 Seminar-Descriptive Statistics

Dr Bridgette Stevens

AIM:BStevensKaplan (add me to your Buddy list)

1

Chapter OutlineChapter Outline

• 2.1 Frequency Distributions and Their Graphs

• 2.2 More Graphs and Displays

• 2.3 Measures of Central Tendency

• 2.4 Measures of Variation

• 2.5 Measures of Position

2Larson/Farber 4th ed.

Section 2.1Section 2.1Section 2.1Section 2.1

Frequency Distributions and Their Graphs

3Larson/Farber 4th ed.

Frequency DistributionFrequency Distribution

Frequency Distribution

• A table that shows classes or intervals of data with a count of the number of entries in each class.

• The frequency, f, of a class is the number of data entries in the class.

Larson/Farber 4th ed. 4

Class Frequency, f

1 – 5 5

6 – 10 8

11 – 15 6

16 – 20 8

21 – 25 5

26 – 30 4

Lower classlimits

Upper classlimits

Class width 6 – 1 = 5

Constructing a Frequency DistributionConstructing a Frequency Distribution


1. Decide on the number of classes. • Usually between 5 and 20; otherwise, it may be difficult to

detect any patterns.

2. Find the class width.• Determine the range of the data.

• Divide the range by the number of classes.

• Round up to the next convenient number.

Constructing a Frequency DistributionConstructing a Frequency Distribution

3. Find the class limits. • You can use the minimum data entry as the lower limit of the

first class.

• Find the remaining lower limits (add the class width to the lower limit of the preceding class).

• Find the upper limit of the first class. Remember that classes cannot overlap.

• Find the remaining upper class limits.


Example: Constructing a Frequency DistributionExample: Constructing a Frequency Distribution

The following sample data set lists the number of minutes 50 Internet subscribers spent on the Internet during their most recent session. Construct a frequency distribution that has seven classes.

50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 8641 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 2018 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44


Solution: Constructing a Frequency DistributionSolution: Constructing a Frequency Distribution

1. Number of classes = 7 (given)

2. Find the class width


max min 86 711.29

#classes 7

Round up to 12

50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 86

41 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 20

18 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44



Lower limit

Upper limit

7Class width = 12

3. Use 7 (minimum value) as first lower limit. Add the class width of 12 to get the lower limit of the next class.

7 + 12 = 19

Find the remaining lower limits.

19

31

43

55

67

79


The upper limit of the first class is 18 (one less than the lower limit of the second class).

Add the class width of 12 to get the upper limit of the next class.

18 + 12 = 30

Find the remaining upper limits.


Lower limit

Upper limit

7

19

31

43

55

67

79

Class width = 1230

42

54

66

78

90

18


4. Make a tally mark for each data entry in the row of the appropriate class.

5. Count the tally marks to find the total frequency f for each class.


Class Tally Frequency, f

7 – 18 IIII I 6

19 – 30 IIII IIII 10

31 – 42 IIII IIII III 13

43 – 54 IIII III 8

55 – 66 IIII 5

67 – 78 IIII I 6

79 – 90 II 2Σf = 50

Determining the MidpointDetermining the Midpoint

Midpoint of a class


(Lower class limit) (Upper class limit)

2

Class Midpoint Frequency, f

7 – 18 6

19 – 30 10

31 – 42 13

7 1812.5

2

19 3024.5

2

31 4236.5

2

Class width = 12

Determining the Relative FrequencyDetermining the Relative Frequency

Relative Frequency of a class

• Portion or percentage of the data that falls in a particular class.


n

f

sizeSample

frequencyclassfrequencyrelative

Class Frequency, f Relative Frequency

7 – 18 6

19 – 30 10

31 – 42 13

60.12

50

100.20

50

130.26

50

•

Graphs of Frequency DistributionsGraphs of Frequency Distributions

Frequency Histogram

• A bar graph that represents the frequency distribution.

• The horizontal scale is quantitative and measures the data values.

• The vertical scale measures the frequencies of the classes.

• Consecutive bars must touch.


data values

freq

uen

cy

Example: Frequency Histogram Example: Frequency Histogram

Construct a frequency histogram for the Internet usage frequency distribution.


ClassClass

boundaries MidpointFrequency, f

7 – 18 6.5 – 18.5 12.5 6

19 – 30 18.5 – 30.5 24.5 10

31 – 42 30.5 – 42.5 36.5 13

43 – 54 42.5 – 54.5 48.5 8

55 – 66 54.5 – 66.5 60.5 5

67 – 78 66.5 – 78.5 72.5 6

79 – 90 78.5 – 90.5 84.5 2

Solution: Frequency Histogram (using Midpoints)Solution: Frequency Histogram (using Midpoints)



Frequency Polygon

• A line graph that emphasizes the continuous change in frequencies.


data values

freq

uen

cy

Solution: Frequency PolygonSolution: Frequency Polygon

02468101214

0.5 12.5 24.5 36.5 48.5 60.5 72.5 84.5 96.5

Freq

uenc

y

Time online (in minutes)

Internet Usage


You can see that the frequency of subscribers increases up to 36.5 minutes and then decreases.

The graph should begin and end on the horizontal axis, so extend the left side to one class width before the first class midpoint and extend the right side to one class width after the last class midpoint.


Relative Frequency Histogram

• Has the same shape and the same horizontal scale as the corresponding frequency histogram.

• The vertical scale measures the relative frequencies, not frequencies.


data values

rela

tive

freq

uen

cy

Example: Relative Frequency Histogram Example: Relative Frequency Histogram

Construct a relative frequency histogram for the Internet usage frequency distribution.


ClassClass

boundariesFrequency,

fRelative

frequency

7 – 18 6.5 – 18.5 6 0.12

19 – 30 18.5 – 30.5 10 0.20

31 – 42 30.5 – 42.5 13 0.26

43 – 54 42.5 – 54.5 8 0.16

55 – 66 54.5 – 66.5 5 0.10

67 – 78 66.5 – 78.5 6 0.12

79 – 90 78.5 – 90.5 2 0.04

Solution: Relative Frequency Histogram Solution: Relative Frequency Histogram

21

6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5

From this graph you can see that 20% of Internet subscribers spent between 18.5 minutes and 30.5 minutes online.


Cumulative Frequency Graph or Ogive

• A line graph that displays the cumulative frequency of each class at its upper class boundary.

• The upper boundaries are marked on the horizontal axis.

• The cumulative frequencies are marked on the vertical axis.


data valuescu

mu

lati

ve

freq

uen

cy

Solution: OgiveSolution: Ogive

0

10

20

30

40

50

60C

um

ula

tive

freq

uen

cy

Time online (in minutes)

Internet Usage

23

6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5

From the ogive, you can see that about 40 subscribers spent 60 minutes or less online during their last session. The greatest increase in usage occurs between 30.5 minutes and 42.5 minutes.


More Graphs and Displays


Section 2.2 ObjectivesSection 2.2 Objectives

• Graph quantitative data using stem-and-leaf plots and dot plots

• Graph qualitative data using pie charts and Pareto charts

• Graph paired data sets using scatter plots and time series charts


Graphing Quantitative Data SetsGraphing Quantitative Data Sets

Stem-and-leaf plot

• Each number is separated into a stem and a leaf.

• Similar to a histogram.

• Still contains original data values.


Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45

26

2 1 5 5 6 7 8

3 0 6 6

4 5

Example: Constructing a Stem-and-Leaf PlotExample: Constructing a Stem-and-Leaf Plot

The following are the numbers of text messages sent last month by the cellular phone users on one floor of a college dormitory. Display the data in a stem-and-leaf plot.


155 159 144 129 105 145 126 116 130 114 122 112 112 142 126156 118 108 122 121 109 140 126 119 113 117 118 109 109 119139 139 122 78 133 126 123 145 121 134 124 119 132 133 124129 112 126 148 147

Solution: Constructing a Stem-and-Leaf PlotSolution: Constructing a Stem-and-Leaf Plot

28

Include a key to identify the values of the data.

From the display, you can conclude that more than 50% of the cellular phone users sent between 110 and 130 text messages.

Graphing Quantitative Data SetsGraphing Quantitative Data Sets

Dot plot

• Each data entry is plotted, using a point, above a horizontal axis

29

Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45

26

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Solution: Constructing a Dot PlotSolution: Constructing a Dot Plot

30

From the dot plot, you can see that most values cluster between 105 and 148 and the value that occurs the most is 126. You can also see that 78 is an unusual data value.

155 159 144 129 105 145 126 116 130 114 122 112 112 142 126156 118 108 122 121 109 140 126 119 113 117 118 109 109 119139 139 122 78 133 126 123 145 121 134 124 119 132 133 124129 112 126 148 147

Graphing Qualitative Data SetsGraphing Qualitative Data Sets

Pie Chart

• A circle is divided into sectors that represent categories.

• The area of each sector is proportional to the frequency of each category.


Example: Constructing a Pie ChartExample: Constructing a Pie Chart

The numbers of motor vehicle occupants killed in crashes in 2005 are shown in the table. Use a pie chart to organize the data. (Source: U.S. Department of Transportation, National Highway Traffic Safety Administration)


Vehicle type

Killed

Cars 18,440

Trucks 13,778

Motorcycles 4,553

Other 823

Solution: Constructing a Pie ChartSolution: Constructing a Pie Chart

• Find the relative frequency (percent) of each category.

33

Vehicle type Frequency, f Relative frequency

Cars 18,440

Trucks 13,778

Motorcycles 4,553

Other 823

37,594

184400.49

37594

137780.37

37594

45530.12

37594

8230.02

37594

Solution: Constructing a Pie ChartSolution: Constructing a Pie Chart

34

Vehicle type

Relative frequency

Central angle

Cars 0.49 176º

Trucks 0.37 133º

Motorcycles 0.12 43º

Other 0.02 7º

From the pie chart, you can see that most fatalities in motor vehicle crashes were those involving the occupants of cars.

Graphing Qualitative Data SetsGraphing Qualitative Data Sets

Pareto Chart

• A vertical bar graph in which the height of each bar represents frequency or relative frequency.

• The bars are positioned in order of decreasing height, with the tallest bar positioned at the left.


Categories

Freq

uen

cy

Constructing a Pareto ChartConstructing a Pareto Chart

36

Cause $ (million)

Admin. error

7.8

Employee theft

15.6

Shoplifting 14.7

Vendor fraud

2.9

From the graph, it is easy to see that the causes of inventory shrinkage that should be addressed first are employee theft and shoplifting.

Graphing Paired Data SetsGraphing Paired Data Sets

Paired Data Sets

• Each entry in one data set corresponds to one entry in a second data set.

• Graph using a scatter plot.• The ordered pairs are graphed as

points in a coordinate plane.

• Used to show the relationship between two quantitative variables.

37

x

y

Example: Interpreting a Scatter PlotExample: Interpreting a Scatter Plot

As the petal length increases, what tends to happen to the petal width?

38

Each point in the scatter plot represents thepetal length and petal width of one flower.

Graphing Paired Data SetsGraphing Paired Data Sets

Time Series

• Data set is composed of quantitative entries taken at regular intervals over a period of time. • e.g., The amount of precipitation measured each day for one

month.

• Use a time series chart to graph.

39

time

Quanti

tati

ve

data

Example: Constructing a Time Series ChartExample: Constructing a Time Series Chart

The table lists the number of cellular telephone subscribers (in millions) for the years 1995 through 2005. Construct a time series chart for the number of cellular subscribers. (Source: Cellular Telecommunication & Internet Association)


Solution: Constructing a Time Series ChartSolution: Constructing a Time Series Chart

41

The graph shows that the number of subscribers has been increasing since 1995, with greater increases recently.


Measures of Central Tendency


Measures of Central TendencyMeasures of Central Tendency

Measure of central tendency

• A value that represents a typical, or central, entry of a data set.

• Most common measures of central tendency:• Mean

• Median

• Mode


Measure of Central Tendency: ModeMeasure of Central Tendency: Mode

Mode

• The data entry that occurs with the greatest frequency.

• If no entry is repeated the data set has no mode.

• If two entries occur with the same greatest frequency, each entry is a mode (bimodal).


Comparing the Mean, Median, and ModeComparing the Mean, Median, and Mode

• All three measures describe a typical entry of a data set.

• Advantage of using the mean:• The mean is a reliable measure because it takes into

account every entry of a data set.

• Disadvantage of using the mean:• Greatly affected by outliers (a data entry that is far

removed from the other entries in the data set).


Example: Comparing the Mean, Median, and ModeExample: Comparing the Mean, Median, and Mode

Find the mean, median, and mode of the sample ages of a class shown. Which measure of central tendency best describes a typical entry of this data set? Are there any outliers?


Ages in a class

20 20 20 20 20 20 21

21 21 21 22 22 22 23

23 23 23 24 24 65

Solution: Comparing the Mean, Median, and ModeSolution: Comparing the Mean, Median, and Mode


Mean: 20 20 ... 24 6523.8 years

20

xx

n

Median: 21 2221.5 years

2

20 years (the entry occurring with thegreatest frequency)

Ages in a class

20 20 20 20 20 20 21

21 21 21 22 22 22 23

23 23 23 24 24 65

Mode:



Mean ≈ 23.8 years Median = 21.5 years Mode = 20 years

• The mean takes every entry into account, but is influenced by the outlier of 65.

• The median also takes every entry into account, and it is not affected by the outlier.

• In this case the mode exists, but it doesn't appear to represent a typical entry.



Sometimes a graphical comparison can help you decide which measure of central tendency best represents a data set.

In this case, it appears that the median best describes the data set.

Weighted MeanWeighted Mean

Weighted Mean

• The mean of a data set whose entries have varying weights.

• where w is the weight of each entry x


( )x wx

w

Example: Finding a Weighted MeanExample: Finding a Weighted Mean

You are taking a class in which your grade is determined from five sources: 50% from your test mean, 15% from your midterm, 20% from your final exam, 10% from your computer lab work, and 5% from your homework. Your scores are 86 (test mean), 96 (midterm), 82 (final exam), 98 (computer lab), and 100 (homework). What is the weighted mean of your scores? If the minimum average for an A is 90, did you get an A?


Solution: Finding a Weighted MeanSolution: Finding a Weighted Mean


Source Score, x Weight, w x∙w

Test Mean 86 0.50 86(0.50)= 43.0

Midterm 96 0.15 96(0.15) = 14.4

Final Exam 82 0.20 82(0.20) = 16.4

Computer Lab 98 0.10 98(0.10) = 9.8

Homework 100 0.05 100(0.05) = 5.0

Σw = 1 Σ(x∙w) = 88.6

( ) 88.688.6

1

x wx

w

Your weighted mean for the course is 88.6. You did not get an A.

Mean of Grouped DataMean of Grouped Data

Mean of a Frequency Distribution

• Approximated by

where x and f are the midpoints and frequencies of a class, respectively


( )x fx n f

n

Finding the Mean of a Frequency DistributionFinding the Mean of a Frequency Distribution

In Words In Symbols


( )x fx

n

(lower limit)+(upper limit)

2x

( )x f

n f

1. Find the midpoint of each class.

2. Find the sum of the products of the midpoints and the frequencies.

3. Find the sum of the frequencies.

4. Find the mean of the frequency distribution.

Example: Find the Mean of a Frequency DistributionExample: Find the Mean of a Frequency Distribution

Use the frequency distribution to approximate the mean number of minutes that a sample of Internet subscribers spent online during their most recent session.


Class Midpoint Frequency, f

7 – 18 12.5 6

19 – 30 24.5 10

31 – 42 36.5 13

43 – 54 48.5 8

55 – 66 60.5 5

67 – 78 72.5 6

79 – 90 84.5 2

Solution: Find the Mean of a Frequency DistributionSolution: Find the Mean of a Frequency Distribution


Class Midpoint, x Frequency, f (x∙f)

7 – 18 12.5 6 12.5∙6 = 75.0

19 – 30 24.5 10 24.5∙10 = 245.0

31 – 42 36.5 13 36.5∙13 = 474.5

43 – 54 48.5 8 48.5∙8 = 388.0

55 – 66 60.5 5 60.5∙5 = 302.5

67 – 78 72.5 6 72.5∙6 = 435.0

79 – 90 84.5 2 84.5∙2 = 169.0

n = 50 Σ(x∙f) = 2089.0

( ) 208941.8 minutes

50

x fx

n

The Shape of DistributionsThe Shape of Distributions


Symmetric Distribution

• A vertical line can be drawn through the middle of a graph of the distribution and the resulting halves are approximately mirror images.



Uniform Distribution (rectangular)

• All entries or classes in the distribution have equal or approximately equal frequencies.

• Symmetric.



Skewed Left Distribution (negatively skewed)

• The “tail” of the graph elongates more to the left.

• The mean is to the left of the median.



Skewed Right Distribution (positively skewed)

• The “tail” of the graph elongates more to the right.

• The mean is to the right of the median.

Section 2.3 SummarySection 2.3 Summary

• Determined the mean, median, and mode of a population and of a sample

• Determined the weighted mean of a data set and the mean of a frequency distribution

• Described the shape of a distribution as symmetric, uniform, or skewed and compared the mean and median for each



Measures of Variation


Deviation, Variance, and Standard DeviationDeviation, Variance, and Standard Deviation

Population Variance

•

Population Standard Deviation

•


22 ( )x

N

Sum of squares, SSx

22 ( )x

N

Example: Using Technology to Find the Standard DeviationExample: Using Technology to Find the Standard Deviation

Sample office rental rates (in dollars per square foot per year) for Miami’s central business district are shown in the table. Use a calculator or a computer to find the mean rental rate and the sample standard deviation. (Adapted from: Cushman & Wakefield Inc.)


Office Rental Rates

35.00 33.50 37.00

23.75 26.50 31.25

36.50 40.00 32.00

39.25 37.50 34.75

37.75 37.25 36.75

27.00 35.75 26.00

37.00 29.00 40.50

24.50 33.00 38.00

Solution: Using Technology to Find the Standard DeviationSolution: Using Technology to Find the Standard Deviation


Sample MeanSample Standard Deviation

Interpreting Standard DeviationInterpreting Standard Deviation

• Standard deviation is a measure of the typical amount an entry deviates from the mean.

• The more the entries are spread out, the greater the standard deviation.

66

Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)

For data with a (symmetric) bell-shaped distribution, the standard deviation has the following characteristics:

67

• About 68% of the data lie within one standard deviation of the mean.

• About 95% of the data lie within two standard deviations of the mean.

• About 99.7% of the data lie within three standard deviations of the mean.

Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)


3x s x s 2x s 3x sx s x2x s

68% within 1 SD

34% 34%

99.7% within 3 SD

2.35% 2.35%

95% within 2 SD

13.5% 13.5%

Example: Using the Empirical RuleExample: Using the Empirical Rule

In a survey conducted by the National Center for Health Statistics, the sample mean height of women in the United States (ages 20-29) was 64 inches, with a sample standard deviation of 2.71 inches. Estimate the percent of the women whose heights are between 64 inches and 69.42 inches.


Solution: Using the Empirical RuleSolution: Using the Empirical Rule


3x s x s 2x s 3x sx s x2x s55.87 58.58 61.29 64 66.71 69.42 72.13

34%

13.5%

• Because the distribution is bell-shaped, you can use the Empirical Rule.

34% + 13.5% = 47.5% of women are between 64 and 69.42 inches tall.

Chebychev’s TheoremChebychev’s Theorem

• The portion of any data set lying within k standard deviations (k > 1) of the mean is at least:


2

11

k

• k = 2: In any data set, at least2

1 31 or 75%

2 4

of the data lie within 2 standard deviations of the mean.

• k = 3: In any data set, at least2

1 81 or 88.9%

3 9

of the data lie within 3 standard deviations of the mean.

Box-and-Whisker PlotBox-and-Whisker Plot

Box-and-whisker plot

• Exploratory data analysis tool.

• Highlights important features of a data set.

• Requires (five-number summary):• Minimum entry

• First quartile Q1

• Median Q2

• Third quartile Q3

• Maximum entry


Drawing a Box-and-Whisker PlotDrawing a Box-and-Whisker Plot

1. Find the five-number summary of the data set.

2. Construct a horizontal scale that spans the range of the data.

3. Plot the five numbers above the horizontal scale.

4. Draw a box above the horizontal scale from Q1 to Q3 and draw a vertical line in the box at Q2.

5. Draw whiskers from the box to the minimum and maximum entries.


Whisker

Whisker

Maximum entry

Minimum entry

Box

Median, Q2

Q3Q1

MM207-Statistics Unit 2 Seminar-Descriptive Statistics Dr Bridgette Stevens AIM:BStevensKaplan (add me to your Buddy list) 1.

Documents

appropriate class

class widthlarsonfarber

preceding class

remaining upper class

frequency distributions

frequency distributionmake

frequency distributionfind

total frequency f