ICT 119-Elementary Statistics - WordPress.com 119-Elementary Statistics ... • Direct personal interview ... Class Frequency, f 1 – 5 5 6 – 10 8 11 – 15 6 16 – 20 8 21 –

ICT 119-Elementary Statistics

For all B.SC.ICT-B & B.SC.ITSStudents

By:

Masoud Amri KomunteDepartment of mathematics and Statistics Studies

(MSS)

Mzumbe UniversityOffice: Block B 209Mobile: 0713227333

Email: [email protected]

Introduction to Statistics

• The systematic and scientific treatment of quantitativemeasurement is precisely known as statistics.

• Statistics may be called as science of counting.• Statistics is concerned with the collection, classification

(or organization), presentation and analysis of data whichare measurable in numerical terms.

Statistical Investigation

What is a statistical investigation ?

statistical investigation

What is a statistical investigation ?The gathering of data and analysing information to helppeople make decisions when faced with uncertainty

Stages of Statistical Investigation

Define the objectives

Collection of Data

Organization of data

Presentation of data

Analysis of data

Interpretation of Results

Planning For An Investigation

• At the planning stage you have to consider all thestages involved right from the first stage to the laststage.

• The most important stage of all is to clearly define theobjectives of the investigation.

• After defining the objectives of the investigation thenmake sure that only data relevant to the objectives arecollected.

Types Of Statistical Investigation

There are two types:-• General purpose statistical investigation.In this type the objective is to address a number ofproblems in just one investigation these problems may besocial, economic, political etc.• Specific purpose statistical investigation.In this type we try to collect information relevant to aspecific purpose.

Division of Statistics

• It is divided into two major parts: Descriptive andInferential Statistics.

• Descriptive statistics, is a set of methods to describedata that we have collected. i.e. summarization of data.

• Inferential statistics, is a set of methods used to make ageneralization, estimate, prediction or decision. Whenwe want to draw conclusions about a distribution.

Collection of Data

• Data can be collected by two ways:>>> Primary Data CollectionIt is the data collected by a particular person or organizationfor his own use.>>> Secondary Data CollectionIt is the data collected by some other person or organization,but the investigator also get it for his use.

Methods of Primary data collection

• Direct personal interview• Data through questionnaire• Indirect investigationEtc.

Methods of Secondary data collection

• Data collected through newspapers & periodicals.• Data collected from research papers.• Data collected from government officials.• Data collected from various NGO, UN, UNESCO,

WHO, ILO, UNICEF etc.• Other published resources

Questionnaire

A questionnaire is a form in hard/soft copy which asksfor certain information and spaces are provided forrecording responses.• A questionnaire may be self-administered or may be

administered by an interviewer.• A self-administered questionnaire may be sent by

mail or may be delivered in personal.

Mail Questionnaire

• Practicable because it is cheaper to collect theinformation as there are no interviewerinvolved.

• Also practicable when a vast area with very many respondents is to be covered.

Limitations Of Mail Questionnaire

• Need a literate population. That is therespondents should be able to read thequestions and fill in the required responses.

• Need an efficient postal/internet system.

Questionnaire Design

The guiding rules for questionnaire design are basically:-• AppearanceThe questionnaire need to be attractive especially if itis a mail questionnaire.• Length of the questionnaireAs far as possible a questionnaire should be short. Avery long questionnaire will tend to discourage arespondent, especially if it is a mail questionnaire.


The guiding rules• Wording-Simple, clear and precise questions.Avoid ambiguous questions. The respondents should be able to interpret aquestion in one and only one way. A question like “how big is your house?”has an element of ambiguity, some respondents may interpret the question interm of dimension and yet others in terms of number of rooms.

-Questions should be presented in a logical order.If there is a known or established order of things then that order should befollowed.


The guiding rules• Wording-Avoid very personal or sensitive questions.-Avoid leading quetions.-Start with the simplest questions and continue according to the

complexity of the issues.

Classification of data

• Classification is a process of arranging data intosequences and groups according to their commoncharacteristics or separating them into different butrelated parts.

• It is a process of arranging data into varioushomogeneous classes and subclasses according tosome common characteristics.

Scrutiny of data

• Collected data should be checked carefully before they aresubjected to any statistical treatment. For however usefulstatistical method may be when properly applied, then can notgive reliable information from faulty and unreliable data.

• In some cases the errors may be obvious for exampleimpossible values in measurements. If you can see a figure ofsay 200 kilograms for a weight of a school child then that isclearly impossible.

Scrutiny of data

• In some cases number might not be impossible butmay seem very unlikely. If for example it is recordedthat a household of 2 people it consumes 5kgs of ricea day, this should rouse suspicion.

Presentation of Data

• Data should be presented in such a manner, so that it maybe easily understood and grasped, and the conclusion maybe drawn promptly from the data presented. e.g.

>>> Histogram>>> Frequency polygon & curve>>> Pie Chart>>> Ogives>>> Bar Chart>>> etc

Variables

• Discrete Variablee.g. Number of books, table, chairs • Continuous Variablee.g. Height, Weight• Quantitative VariableThat can be measured on a scale, that is, it can be expressed as number• Qualitative VariableThat can not be measured on a scale, that is, it can not be expressed as number but they can be put in categories.

ExampleSuppose you survey potential voters among the people onmain street during lunch to determine their political affiliationand age, as well as their opinion on the ballot measure.Classify the variables as quantitative or qualitative.

23

• Political affiliation is a qualitative variable (categories).• Age is a quantitative variable (numbers).• Opinion on the ballot measure is a qualitative variable

(categories).

Solution

Statistics functions & Uses

• It simplifies complex data• It provides techniques for comparison• It studies relationship• It helps in formulating policies• It helps in forecasting• It is helpful for investigation purpose• A quickest way in decision making since Statistical

methods merges with speed of computer through SPSS, STATA, MATLAB, MINITAB etc.

Scope of Statistics

• In Business Decision Making• In Medical Sciences• In Actuarial Science• In Economic Planning• In Agricultural Sciences• In Banking & Insurance• In Politics & Social Science

Distrust & Misuse of Statistics

• Statistics is like a clay of which one can make a God or Devil.

• Statisticians are the liars of first order.• Statistics can prove or disprove anything.

According to Aaron Levenstein:

“Statistics are like bikinis. What they reveal is suggestive, but what they conceal is vital.”

Descriptive Statistics

• Frequency Distributions and Their Graphs

• More Graphs and Displays

• Measures of Central Tendency

• Measures of Variation

• Measures of Position

27

Frequency Distribution

Frequency Distribution• A table that shows

observations, 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.

28

Class Frequency, f1 – 5 56 – 10 8

11 – 15 616 – 20 8

21 – 25 526 – 30 4

Lower classlimits

Upper classlimits

Class width 6 – 1 = 5

Constructing a Frequency Distribution

29

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 (class size). Determine the range of the data. Divide the range by the number of classes. Round up to the next convenient number.


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.

30


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 ffor each class.

31

Example: Constructing a Frequency Distribution

The following sample data set lists the number ofminutes that 50 Internet subscribers spent on theInternet during their most recent session. Construct afrequency 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

32

Solution: Constructing a Frequency Distribution

1. Number of classes = 7 (given) -if not given you may use the following formula; Number of classes=1+3.322log N where N is total number of observation.

2. Find the class width

33

max min 86 7 11.29#classes 7

Round up to 12

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


34

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 = 19Find the remaining lower limits.

193143556779


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 = 30Find the remaining upper limits.

35

Lower limit

Upper limit

7193143556779

Class width = 1230

4254667890

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 ffor each class.

36

Class Tally Frequency, f7 – 18 IIII I 6

19 – 30 IIII IIII 1031 – 42 IIII IIII III 1343 – 54 IIII III 855 – 66 IIII 567 – 78 IIII I 679 – 90 II 2

Σf = 50

Determining the Midpoint (Class Mark)

Midpoint of a class

37

(Lower class limit) (Upper class limit)2

Class Midpoint Frequency, f

7 – 18 6

19 – 30 10

31 – 42 13

7 18 12.52

19 30 24.52

31 42 36.52

Class width = 12

Determining the Relative FrequencyRelative Frequency of a class • Portion or percentage of the data that falls in a

particular class.

38

nf

sizeSample

frequencyclassfrequencyrelative

Class Frequency, f Relative Frequency

7 – 18 6

19 – 30 10

31 – 42 13

6 0.1250

10 0.2050

13 0.2650

•

Determining the Cumulative Frequency

Cumulative frequency of a class• The sum of the frequency for that class and all

previous classes.

39

Class Frequency, f Cumulative frequency

7 – 18 6

19 – 30 10

31 – 42 13

+

+

6

16

29

Expanded Frequency Distribution

40

Class Frequency, f MidpointRelative

frequencyCumulative frequency

7 – 18 6 12.5 0.12 619 – 30 10 24.5 0.20 1631 – 42 13 36.5 0.26 2943 – 54 8 48.5 0.16 3755 – 66 5 60.5 0.10 4267 – 78 6 72.5 0.12 4879 – 90 2 84.5 0.04 50

Σf = 50 1nf

Graphs 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.

41

data valuesfr

eque

ncy

Class BoundariesClass boundaries• The numbers that separate classes without forming

gaps between them.

42

ClassClass

BoundariesFrequency,

f7 – 18 6

19 – 30 1031 – 42 13

• The distance from the upper limit of the first class to the lower limit of the second class is 19 – 18 = 1.

• Half this distance is 0.5.

• First class lower boundary = 7 – 0.5 = 6.5• First class upper boundary = 18 + 0.5 = 18.5

6.5 – 18.5

Class Boundaries

43

ClassClass

boundariesFrequency,

f7 – 18 6.5 – 18.5 6

19 – 30 18.5 – 30.5 1031 – 42 30.5 – 42.5 1343 – 54 42.5 – 54.5 855 – 66 54.5 – 66.5 567 – 78 66.5 – 78.5 679 – 90 78.5 – 90.5 2

Example: Frequency Histogram

Construct a frequency histogram for the Internet usage frequency distribution.

44

ClassClass

boundaries MidpointFrequency,

f7 – 18 6.5 – 18.5 12.5 6

19 – 30 18.5 – 30.5 24.5 1031 – 42 30.5 – 42.5 36.5 1343 – 54 42.5 – 54.5 48.5 855 – 66 54.5 – 66.5 60.5 567 – 78 66.5 – 78.5 72.5 679 – 90 78.5 – 90.5 84.5 2

Solution: Frequency Histogram (using Midpoints)

45

Solution: Frequency Histogram (using class boundaries)

46

6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5

You can see that more than half of the subscribers spent between 19 and 54 minutes on the Internet during their most recent session.


Frequency Polygon• A line graph that emphasizes the continuous change

in frequencies.

47

data values

freq

uenc

y

Example: Frequency Polygon

Construct a frequency polygon for the Internet usage frequency distribution.

48

Class Midpoint Frequency, f7 – 18 12.5 6

19 – 30 24.5 1031 – 42 36.5 1343 – 54 48.5 855 – 66 60.5 567 – 78 72.5 679 – 90 84.5 2

Solution: Frequency Polygon

02468

101214

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

49

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.

50

data valuesre

lativ

e fr

eque

ncy

Example: Relative Frequency Histogram

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

51

ClassClass


fRelative

frequency7 – 18 6.5 – 18.5 6 0.12

19 – 30 18.5 – 30.5 10 0.2031 – 42 30.5 – 42.5 13 0.2643 – 54 42.5 – 54.5 8 0.1655 – 66 54.5 – 66.5 5 0.1067 – 78 66.5 – 78.5 6 0.1279 – 90 78.5 – 90.5 2 0.04

Solution: Relative Frequency Histogram

52

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.

53

data valuescu

mul

ativ

e fr

eque

ncy

Constructing an Ogive

1. Construct a frequency distribution that includes cumulative frequencies as one of the columns.

2. Specify the horizontal and vertical scales. The horizontal scale consists of the upper class

boundaries. The vertical scale measures cumulative

frequencies.3. Plot points that represent the upper class boundaries

and their corresponding cumulative frequencies.

54

Constructing an Ogive

4. Connect the points in order from left to right.5. The graph should start at the lower boundary of the

first class (cumulative frequency is zero) and should end at the upper boundary of the last class (cumulative frequency is equal to the sample size).

55

Example: Ogive

Construct an ogive for the Internet usage frequency distribution.

56

ClassClass


fCumulative frequency

7 – 18 6.5 – 18.5 6 619 – 30 18.5 – 30.5 10 1631 – 42 30.5 – 42.5 13 2943 – 54 42.5 – 54.5 8 3755 – 66 54.5 – 66.5 5 4267 – 78 66.5 – 78.5 6 4879 – 90 78.5 – 90.5 2 50

Solution: Ogive

0

10

20

30

40

50

60

Cum

ulat

ive f

requ

ency

Time online (in minutes)

Internet Usage

57

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

• 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

58

Graphing Quantitative Data SetsStem-and-leaf plot• Each number is separated into a stem and a leaf.• Similar to a histogram.• Still contains original data values.• For decimals its better to approximate

59

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

26

2 1 5 5 6 7 83 0 6 6 4 5

Example: 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.

60

155 159 144 129 105 145 126 116 130 114 122 112 112 142 126118 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 Plot

61

• The data entries go from a low of 78 to a high of 159.• Use the rightmost digit as the leaf. For instance,

78 = 7 | 8 and 159 = 15 | 9• List the stems, 7 to 15, to the left of a vertical line.• For each data entry, list a leaf to the right of its stem.

155 159 144 129 105 145 126 116 130 114 122 112 112 142 126118 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 Plot

62

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 Sets

Dot plot• Each data entry is plotted, using a point, above a

horizontal axis

63

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

Example: Constructing a Dot Plot

Use a dot plot organize the text messaging data.

64

• So that each data entry is included in the dot plot, the horizontal axis should include numbers between 70 and 160.

• To represent a data entry, plot a point above the entry's position on the axis.

• If an entry is repeated, plot another point above the previous point.

155 159 144 129 105 145 126 116 130 114 122 112 112 142 126118 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 Dot Plot

65

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 126118 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 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.

66

Example: 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.

67

Vehicle type KilledCars 18,440Trucks 13,778Motorcycles 4,553Other 823

Solution: Constructing a Pie Chart

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

68

Vehicle type Frequency, f Relative frequency

Cars 18,440

Trucks 13,778

Motorcycles 4,553

Other 823

37,594

18440 0.4937594

13778 0.3737594

4553 0.1237594

823 0.0237594


• Construct the pie chart using the central angle that corresponds to each category. To find the central angle, multiply 360º by the

category's relative frequency. For example, the central angle for cars is

360(0.49) ≈ 176º

69


70

Vehicle type Frequency, fRelative

frequency Central angle

Cars 18,440 0.49

Trucks 13,778 0.37

Motorcycles 4,553 0.12

Other 823 0.02

360º(0.49)≈176º

360º(0.37)≈133º

360º(0.12)≈43º

360º(0.02)≈7º


71

Vehicle typeRelative

frequencyCentral angle

Cars 0.49 176ºTrucks 0.37 133ºMotorcycles 0.12 43ºOther 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 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.

72

Categories

Freq

uenc

y

Example: Constructing a Pareto Chart

In a recent year, the retail industry lost $41.0 million ininventory shrinkage. Inventory shrinkage is the loss ofinventory through breakage, pilferage, shoplifting, andso on. The causes of the inventory shrinkage areadministrative error ($7.8 million), employee theft($15.6 million), shoplifting ($14.7 million), and vendorfraud ($2.9 million). Use a Pareto chart to organize thisdata.

73

Solution: Constructing a Pareto Chart

74

Cause $ (million)

Admin. error 7.8Employee theft 15.6

Shoplifting 14.7Vendor 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 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.

75

x

y

Graphing 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.

76

timeQ

uant

itativ

e da

ta

Example: Constructing a Time Series Chart

The table lists the number of cellulartelephone subscribers (in millions)for the years 1995 through 2005.Construct a time series chart for thenumber of cellular subscribers.

77

Solution: Constructing a Time Series Chart

• Let the horizontal axis represent the years.

• Let the vertical axis represent the number of subscribers (in millions).

• Plot the paired data and connect them with line segments.

78

Solution: Constructing a Time Series Chart

79

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

Measures of Central Tendency

• Determine the mean, median, and mode of apopulation and of a sample

• Determine the weighted mean of a data set and themean of a frequency distribution

• Describe the shape of a distribution as symmetric,uniform, or skewed and compare the mean andmedian for each

80

Measures 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

81

Measure of Central Tendency: Mean

Mean (average)• The sum of all the data entries divided by the number

of entries.• Sigma notation: Σx = add all of the data entries (x)

in the data set.• Population mean:

• Sample mean:

82

xN

xxn

Example: Finding a Sample Mean

The prices (in dollars) for a sample of roundtrip flightsfrom Dar es salaam to Nairobi, Arusha, Mtwara,Mwanza, Mbeya, Kigoma and Iringa are listed. What isthe mean price of the flights?

872 432 397 427 388 782 397

83

Solution: Finding a Sample Mean

872 432 397 427 388 782 397

84

• The sum of the flight prices isΣx = 872 + 432 + 397 + 427 + 388 + 782 + 397 = 3695

• To find the mean price, divide the sum of the prices by the number of prices in the sample

3695 527.97

xxn

The mean price of the flights is about $527.90.

Measure of Central Tendency: Median

Median• The value that lies in the middle of the data when the

data set is ordered.• Measures the center of an ordered data set by dividing

it into two equal parts.• If the data set has an odd number of entries: median is the middle data

entry. even number of entries: median is the mean of

the two middle data entries.85

Example: Finding the Median

The prices (in dollars) for a sample of roundtrip flightsfrom Dar es salaam to Nairobi, Arusha, Mtwara,Mwanza, Mbeya, Kigoma and Iringa are listed. Find themedian of the flight prices.

872 432 397 427 388 782 397

86

Solution: Finding the Median

872 432 397 427 388 782 397

87

• First order the data.388 397 397 427 432 782 872

• There are seven entries (an odd number), the median is the middle, or fourth, data entry.

The median price of the flights is $427.

Example: Finding the Median

The flight priced at $432 is no longer available. What is the median price of the remaining flights?

872 397 427 388 782 397

88

Solution: Finding the Median

872 397 427 388 782 397

89

• First order the data.388 397 397 427 782 872

• There are six entries (an even number), the median is the mean of the two middle entries.

The median price of the flights is $412.

397 427Median 4122

Measure 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).

90

Example: Finding the ModeThe prices (in dollars) for a sample of roundtrip flightsfrom Dar es salaam to Nairobi, Arusha, Mtwara,Mwanza, Mbeya, Kigoma and Iringa are listed.Find the mode of the flight prices.

872 432 397 427 388 782 397

91

Solution: Finding the Mode

872 432 397 427 388 782 397

92

• Ordering the data helps to find the mode.388 397 397 427 432 782 872

• The entry of 397 occurs twice, whereas the other data entries occur only once.

The mode of the flight prices is $397.

Example: Finding the Mode

At a political debate a sample of audience members was asked to name the political party to which they belong. Their responses are shown in the table. What is the mode of the responses?

93

Political Party Frequency, fCcm 34Chadema 56Other 21Did not respond 9

Solution: Finding the Mode

94

Political Party Frequency, fCcm 34Chadema 56Other 21Did not respond 9

The mode is Chadema (the response occurring with thegreatest frequency). In this sample there were moreChadema people than people of any other single affiliation.

Comparing 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).

95

Example: 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?

96

Ages in a class20 20 20 20 20 20 2121 21 21 22 22 22 2323 23 23 24 24 65

Solution: Comparing the Mean, Median, and Mode

97

Mean: 20 20 ... 24 65 23.8 years20

xxn

Median: 21 22 21.5 years2

20 years (the entry occurring with thegreatest frequency)

Ages in a class20 20 20 20 20 20 2121 21 21 22 22 22 2323 23 23 24 24 65

Mode:


98

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.


99

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 Mean

Weighted Mean• The mean of a data set whose entries have varying

weights.

• where w is the weight of each entry x

100

( )x wxw

Example: Finding a Weighted Mean

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

101

Solution: Finding a Weighted Mean

102

Source Score, x Weight, w x∙wTest Mean 86 0.50 86(0.50)= 43.0Midterm 96 0.15 96(0.15) = 14.4Final Exam 82 0.20 82(0.20) = 16.4Computer Lab 98 0.10 98(0.10) = 9.8Homework 100 0.05 100(0.05) = 5.0

Σw = 1 Σ(x∙w) = 88.6

( ) 8 8 .6 8 8 .61

x wxw

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

Mean of Grouped Data

Mean of a Frequency Distribution• Approximated by

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

103

( )x fx n fn

Note:

Make sure you find out how to find other measures of central tendency for the case of grouped data.

Finding the Mean of a Frequency Distribution

In Words In Symbols

104

( )x fxn

(lower limit)+(upper limit)2

x

( )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 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.

105

Class Midpoint Frequency, f7 – 18 12.5 6

19 – 30 24.5 1031 – 42 36.5 1343 – 54 48.5 855 – 66 60.5 567 – 78 72.5 679 – 90 84.5 2

Solution: Find the Mean of a Frequency Distribution

106

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 = 245.031 – 42 36.5 13 36.5∙13 = 474.543 – 54 48.5 8 48.5∙8 = 388.055 – 66 60.5 5 60.5∙5 = 302.567 – 78 72.5 6 72.5∙6 = 435.079 – 90 84.5 2 84.5∙2 = 169.0

n = 50 Σ(x∙f) = 2089.0

( ) 2089 41.8 minutes50

x fxn

The Shape of Distributions

107

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.


108

Uniform Distribution (rectangular)• All entries or classes in the distribution have equal

or approximately equal frequencies.• Symmetric.


109

Skewed Left Distribution (negatively skewed)• The “tail” of the graph elongates more to the left.• The mean is to the left of the median.


110

Skewed Right Distribution (positively skewed)• The “tail” of the graph elongates more to the right.• The mean is to the right of the median.

Measures of Variation

• Determine the range of a data set• Determine the variance and standard deviation of a

population and of a sample• Use the Empirical Rule and Chebychev’s Theorem to

interpret standard deviation• Approximate the sample standard deviation for

grouped data

111

Range

Range• The difference between the maximum and minimum

data entries in the set.• The data must be quantitative.• Range = (Max. data entry) – (Min. data entry)

112

Example: Finding the Range

A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the range of the starting salaries.

Starting salaries (x100 of dollars)41 38 39 45 47 41 44 41 37 42

113

Solution: Finding the Range

• Ordering the data helps to find the least and greatest salaries.

37 38 39 41 41 41 42 44 45 47

• Range = (Max. salary) – (Min. salary)= 47 – 37 = 10

The range of starting salaries is 10 or $10,00.

114

minimum maximum

Deviation, Variance, and Standard Deviation

Deviation• The difference between the data entry, x, and the

mean of the data set.• Population data set:

-Deviation of x = x – μ• Sample data set:

-Deviation of x = x – x

115

Example: Finding the Deviation

A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the total deviation of the starting salaries.


116

Solution:• First determine the mean starting salary.

415 41.510

xN

Solution: Finding the Deviation

117

• Determine the deviation for each data entry.

Salary ($1000s), x Deviation: x – μ41 41 – 41.5 = –0.538 38 – 41.5 = –3.539 39 – 41.5 = –2.545 45 – 41.5 = 3.547 47 – 41.5 = 5.541 41 – 41.5 = –0.544 44 – 41.5 = 2.541 41 – 41.5 = –0.537 37 – 41.5 = –4.542 42 – 41.5 = 0.5

Σx = 415 Σ(x – μ) = 0


Population Variance

•

Population Standard Deviation

•

118

22 ( )x

N

Sum of squares, SSx

22 ( )x

N

Finding the Population Variance & Standard Deviation

In Words In Symbols

119

1. Find the mean of the population data set.

2. Find deviation of each entry.

3. Square each deviation.

4. Add to get the sum of squares.

xN

x – μ

(x – μ)2

SSx = Σ(x – μ)2

Finding the Population Variance & Standard Deviation

120

5. Divide by N to get the population variance.

6. Find the square root to get the population standard deviation.

22 ( )x

N

2( )xN

In Words In Symbols

Example: Finding the Population Standard Deviation

A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the population variance and standard deviation of the starting salaries.


Recall μ = 41.5.

121

Solution: Finding the Population Standard Deviation

122

• Determine SSx

• N = 10Salary, x Deviation: x – μ Squares: (x – μ)2

41 41 – 41.5 = –0.5 (–0.5)2 = 0.2538 38 – 41.5 = –3.5 (–3.5)2 = 12.2539 39 – 41.5 = –2.5 (–2.5)2 = 6.2545 45 – 41.5 = 3.5 (3.5)2 = 12.2547 47 – 41.5 = 5.5 (5.5)2 = 30.2541 41 – 41.5 = –0.5 (–0.5)2 = 0.2544 44 – 41.5 = 2.5 (2.5)2 = 6.2541 41 – 41.5 = –0.5 (–0.5)2 = 0.2537 37 – 41.5 = –4.5 (–4.5)2 = 20.2542 42 – 41.5 = 0.5 (0.5)2 = 0.25

Σ(x – μ) = 0 SSx = 88.5

Solution: Finding the Population Standard Deviation

123

Population Variance

•

Population Standard Deviation

•

22 ( ) 88.5 8.9

10x

N

2 8.85 3.0

The population standard deviation is about 3.0, or $300.


Sample Variance

•

Sample Standard Deviation

•

124

22 ( )

1x xsn

22 ( )

1x xs sn

Finding the Sample Variance & Standard Deviation

In Words In Symbols

125

1. Find the mean of the sample data set.

2. Find deviation of each entry.

3. Square each deviation.

4. Add to get the sum of squares.

xxn

2( )xSS x x

2( )x x

x x

Finding the Sample Variance & Standard Deviation

126

5. Divide by n – 1 to get the sample variance.

6. Find the square root to get the sample standard deviation.

In Words In Symbols2

2 ( )1

x xsn

2( )1

x xsn

Example: Finding the Sample Standard Deviation

The starting salaries are for the CRDB branches of aCRDB Bank PLC. The Bank has several other branches,and you plan to use the starting salaries of the CRDBbranches to estimate the starting salaries for the largerpopulation of Banks. Find the sample standarddeviation of the starting salaries.


127

Solution: Finding the Sample Standard Deviation

128

• Determine SSx

• n = 10Salary, x Deviation: x – μ Squares: (x – μ)2

41 41 – 41.5 = –0.5 (–0.5)2 = 0.2538 38 – 41.5 = –3.5 (–3.5)2 = 12.2539 39 – 41.5 = –2.5 (–2.5)2 = 6.2545 45 – 41.5 = 3.5 (3.5)2 = 12.2547 47 – 41.5 = 5.5 (5.5)2 = 30.2541 41 – 41.5 = –0.5 (–0.5)2 = 0.2544 44 – 41.5 = 2.5 (2.5)2 = 6.2541 41 – 41.5 = –0.5 (–0.5)2 = 0.2537 37 – 41.5 = –4.5 (–4.5)2 = 20.2542 42 – 41.5 = 0.5 (0.5)2 = 0.25

Σ(x – μ) = 0 SSx = 88.5

Solution: Finding the Sample Standard Deviation

129

Sample Variance

•

Sample Standard Deviation

•

22 ( ) 88.5 9.8

1 10 1x xsn

2 88.5 3.19

s s

The sample standard deviation is about 3.1, or $3100.

Example: Using Technology to Find the Standard Deviation

Sample office rental rates (indollars per square foot per year)for Kongo’s business street areshown in the table.(a) find the mean rental rate andthe sample standard deviation.(b)Use a calculator to find themean rental rate and the samplestandard deviation. (try it)

130

Office Rental Rates35.00 33.50 37.0023.75 26.50 31.2536.50 40.00 32.0039.25 37.50 34.7537.75 37.25 36.7527.00 35.75 26.0037.00 29.00 40.5024.50 33.00 38.00

Interpreting 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.

131

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:

132

• 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)

133

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

68% within 1 standard deviation

34% 34%

99.7% within 3 standard deviations

2.35% 2.35%

95% within 2 standard deviations

13.5% 13.5%

Example: Using the Empirical Rule

In a survey conducted by the National Center for HealthStatistics, the sample mean height of women in Morogoro(ages 20-29) was 64 inches, with a sample standarddeviation of 2.71 inches. Estimate the percent of thewomen whose heights are between 64 inches and 69.42inches.

134

Solution: Using the Empirical Rule

135

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 Theorem

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

136

2

11k

• k = 2: In any data set, at least 2

1 31 or 75%2 4

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

• k = 3: In any data set, at least 2

1 81 or 88.9%3 9

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

Example: Using Chebychev’s Theorem

The age distribution for Tabora is shown in the histogram. Apply Chebychev’s Theorem to the data using k = 2. What can you conclude?

137

Solution: Using Chebychev’s Theorem

k = 2: μ – 2σ = 39.2 – 2(24.8) = -10.4 (use 0 since age can’t be negative)

μ + 2σ = 39.2 + 2(24.8) = 88.8

138

At least 75% of the population of Tabora is between 0 and 88.8 years old.

Standard Deviation for Grouped Data

Sample standard deviation for a frequency distribution

•

• When a frequency distribution has classes, estimate the sample mean and standard deviation by using the midpoint of each class.

139

2( )1

x x fsn

where n= Σf (the number of entries in the data set)

Example: Finding the Standard Deviation for Grouped Data

140

You collect a random sample of the number of children per household in a region. Find the sample mean and the sample standard deviation of the data set.

Number of Children in50 Households

1 3 1 1 11 2 2 1 01 1 0 0 01 5 0 3 63 0 3 1 11 1 6 0 13 6 6 1 22 3 0 1 14 1 1 2 20 3 0 2 4

x f xf0 10 0(10) = 01 19 1(19) = 192 7 2(7) = 143 7 3(7) =214 2 4(2) = 85 1 5(1) = 56 4 6(4) = 24

Solution: Finding the Standard Deviation for Grouped Data

• First construct a frequency distribution.

• Find the mean of the frequency distribution.

141

Σf = 50 Σ(xf )= 91

91 1.850

xfxn

The sample mean is about 1.8 children.


• Determine the sum of squares.

142

x f0 10 0 – 1.8 = –1.8 (–1.8)2 = 3.24 3.24(10) = 32.401 19 1 – 1.8 = –0.8 (–0.8)2 = 0.64 0.64(19) = 12.162 7 2 – 1.8 = 0.2 (0.2)2 = 0.04 0.04(7) = 0.283 7 3 – 1.8 = 1.2 (1.2)2 = 1.44 1.44(7) = 10.084 2 4 – 1.8 = 2.2 (2.2)2 = 4.84 4.84(2) = 9.685 1 5 – 1.8 = 3.2 (3.2)2 = 10.24 10.24(1) = 10.246 4 6 – 1.8 = 4.2 (4.2)2 = 17.64 17.64(4) = 70.56

x x 2( )x x 2( )x x f

2( ) 145.40x x f


• Find the sample standard deviation.

143

x x 2( )x x 2( )x x f2( ) 145.40 1.71 50 1

x x fsn

The standard deviation is about 1.7 children.

Measures of Position

• Determine the quartiles of a data set• Determine the interquartile range of a data set• Create a box-and-whisker plot• Interpret other fractiles such as percentiles• Determine and interpret the standard score (z-score)

144

Quartiles• Fractiles are numbers that partition (divide) an

ordered data set into equal parts.• Quartiles approximately divide an ordered data set

into four equal parts. First quartile, Q1: About one quarter of the data

fall on or below Q1. Second quartile, Q2: About one half of the data

fall on or below Q2 (median). Third quartile, Q3: About three quarters of the

data fall on or below Q3.

145

Example: Finding Quartiles

The test scores of 15 employees enrolled in a CPR training course are listed. Find the first, second, and third quartiles of the test scores.13 9 18 15 14 21 7 10 11 20 5 18 37 16 17

146

Solution:• Q2 divides the data set into two halves.

5 7 9 10 11 13 14 15 16 17 18 18 20 21 37

Q2

Lower half Upper half

Solution: Finding Quartiles• The first and third quartiles are the medians of the

lower and upper halves of the data set.

5 7 9 10 11 13 14 15 16 17 18 18 20 21 37

147

Q2

Lower half Upper half

Q1 Q3

About one fourth of the employees scored 10 or less, about one half scored 15 or less; and about three fourths scored 18 or less.

Interquartile Range

Interquartile Range (IQR)• The difference between the third and first quartiles.• IQR = Q3 – Q1

148

Example: Finding the Interquartile Range

Find the interquartile range of the test scores.Recall Q1 = 10, Q2 = 15, and Q3 = 18

149

Solution:• IQR = Q3 – Q1 = 18 – 10 = 8

The test scores in the middle portion of the data set vary by at most 8 points.

Box-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

150

Drawing 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.

151

Whisker Whisker

Maximum entry

Minimum entry

Box

Median, Q2 Q3Q1

Example: Drawing a Box-and-Whisker Plot

Draw a box-and-whisker plot that represents the 15 test scores. Recall Min = 5 Q1 = 10 Q2 = 15 Q3 = 18 Max = 37

152

5 10 15 18 37

Solution:

About half the scores are between 10 and 18. By looking at the length of the right whisker, you can conclude 37 is a possible outlier.

Percentiles and Other Fractiles

Fractiles Summary SymbolsQuartiles Divides data into 4 equal

partsQ1, Q2, Q3

Deciles Divides data into 10 equal parts

D1, D2, D3,…, D9

Percentiles Divides data into 100 equal parts

P1, P2, P3,…, P99

153

Example: Interpreting Percentiles

The ogive represents the cumulative frequency distribution for SAT test scores of college-bound students in a recent year. What test score represents the 72nd

percentile? How should you interpret this?

154

Solution: Interpreting Percentiles

The 72nd percentile corresponds to a test score of 1700. This means that 72% of the students had an SAT score of 1700 or less.

155

Mean

Median

Mode

Summary Measures

Variation

Variance

Standard Deviation

Range

Location/Position

Maximum Minimum

Central Tendency

Percentile Quartile Decile

The Standard Score

Standard Score (z-score)• Represents the number of standard deviations a given

value x falls from the mean μ.

•

157

value - meanstandard deviation

xz

Example: Comparing z-Scores from Different Data Sets

In 2007, Forest Whitaker won the Best Actor Oscar atage 45 for his role in the movie The Last King ofScotland. Helen Mirren won the Best Actress Oscar atage 61 for her role in The Queen. The mean age of allbest actor winners is 43.7, with a standard deviation of8.8. The mean age of all best actress winners is 36, witha standard deviation of 11.5. Find the z-score thatcorresponds to the age for each actor or actress. Thencompare your results.

158

Solution: Comparing z-Scores from Different Data Sets

159

• Forest Whitaker45 43.7 0.15

8.8xz

• Helen Mirren61 36 2.17

11.5xz

0.15 standard deviations above the mean

2.17 standard deviations above the mean

Solution: Comparing z-Scores from Different Data Sets

160

The z-score corresponding to the age of Helen Mirrenis more than two standard deviations from the mean,so it is considered unusual. Compared to other BestActress winners, she is relatively older, whereas theage of Forest Whitaker is only slightly higher than theaverage age of other Best Actor winners.

z = 0.15 z = 2.17

Documents

ICT 119-Elementary Statistics - WordPress.com 119-Elementary Statistics ... • Direct personal interview ... Class Frequency, f 1 – 5 5 6 – 10 8 11 – 15 6 16 – 20 8 21 –