ARM Lecture 7

8/14/2019 ARM Lecture 7

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Data AnalysisPart-A

Data AnalysisPart-A

Lecture # 7

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joke

Three statisticians go hunting. Whenthey see a rabbit, the first oneshoots, missing it on the left. Thesecond one shoots and misses it onthe right. The third one shouts:"We've hit it!"

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joke

Two statisticians were travelling in an airplane from LA to

New York. About an hour into the flight, the pilot announcedthat they had lost an engine, but don't worry, there arethree left. However, instead of 5 hours it would take 7 hoursto get to New York. A little later, he announced that asecond engine failed, and they still had two left, but itwould take 10 hours to get to New York. Somewhat later,

the pilot again came on the intercom and announced that athird engine had died...

Never fear, he announced, because the plane could fly on asingle engine. However, it would now take 18 hours to getto New York. At this point, one statistician turned to theother and said, "Gee, I hope we don't lose that last engine,

or we'll be up here forever!"

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Contents

What is data analysis?1.

Descriptive Statistics2.

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Steps for data analysis

(1) preparing the data foranalysis,

(2) analyzing the data, and

(3) interpreting the data (i.e.,testing the research hypothesesand drawing valid inferences)

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LOGO

DATA PREPARATIONDATA PREPARATION

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DATA PREPARATION

Data represent the fruit of researcherslabor because they provide theinformation that will ultimately allowthem to:

describe phenomena,

predict events,

identify and quantify differences between conditions, and

establish the effectiveness of interventions

Because of their critical nature, datashould be treated with the utmostrespect and care.

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Researcher should carefully planhow the data will be: logged,

entered, transformed (as necessary), and

organized into a database that will facilitate

accurate and efficient statistical analysis.

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Logging and Tracking Data

Computer applications to facilitate the

processMicrosoft Access, Microsoft Excel, Claris

FileMaker, SPSS, SAS

The recruitment log is a comprehensive

record of all individuals approachedaboutparticipation in a study.

The log can also serve to record the datesand times that potential participants were

approached, whether they met eligibilitycriteria, and whether they agreed andprovided informed consent to participate inthe study

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Data Screening

Prior to data entry, the researchershould carefully screen all data foraccuracy.

You may need to recontact studyparticipants to address any omissions,errors, or inaccuracies.

Purposes of screening:

(1) responses are legible and understandable,

(2) responses are within an acceptable range,

(3) responses are complete, and

(4) all of the necessary information has been included.

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Constructing a Database

Data should be entered into a well-structured database.

Researcher should carefully consider

the structure of the database andhow it will be used

While designing the generalstructure of the database, the

researcher must carefully considerall of the variables that will need tobe entered.

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The Data Codebook

A data codebook is a written orcomputerized list that provides aclear and comprehensive descriptionof the variables that will be includedin the database.

A detailed codebook is essentialwhen the researcher begins to

analyze the data

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At a bare minimum, a data codebookshould contain the followingelements for each variable: Variable name

Variable description

Variable format (number, data, text)

Instrument or method of collection

Date collected Respondent or group

Variable location (in database)

Notes

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Types of StatisticsTypes of Statistics

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Types of Statistics

Descriptive statistics allow theresearcher to describe the data andexamine relationships betweenvariables,

Inferential statistics allow theresearcher to examine causalrelationships

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Descriptive Statistics

Frequency distribution is simply acomplete list of all possible valuesor scores for a particular variable,along with the number of times(frequency) that each value or scoreappears in the data set.

This information can be delineated

in what is known as a frequencytable,

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Example of Frequency Distribution


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Example of Frequency DistributionTable

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Histogram:

Still another way that thisdistribution may be depicted is inwhat is known as a histogram.

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Central TendencyThe central tendency of a

distribution is a number that

represents the typicalor mostrepresentative value in thedistribution. The most widely usedmeasures of central tendency are

the mean, median, and mode.

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LOGO

Mean, Median, ModeMean, Median, Mode& Range& RangeMean, Median, ModeMean, Median, Mode& Range& Range

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Vocabulary Review

Sum the answer to anaddition problem.

Addend the numbersyou added together to getthe sum.

6 + 9 = 15

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Definition

MeanMean

MeansMeans

AverageAverage

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Definition

MeanMean the average ofa group of numbers.

2, 5, 2, 1, 5

MeanMean ==

(2+5+2+1+5)/5(2+5+2+1+5)/5

= 3= 3 23 M i f d b


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Mean is found byevening out the

numbers2, 5, 2, 1, 5

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numbers2, 5, 2, 1, 5

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M i f d b


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numbers2, 5, 2, 1, 5

mean =

3

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How to Find the Mean of

a Group of NumbersStep 1 Add all the numbers.

8, 10, 12, 18, 22, 26

8+10+12+18+22+26 = 96

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How to Find the Mean ofa Group of Numbers

Step 2 Divide the sum by thenumber of addends.

8, 10, 12, 18, 22, 26

8+10+12+18+22+26 = 96How many addends are there?

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Step 2 Divide the sum by thenumber of addends.

16

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The mean or average ofthese numbers is 16.

8, 10, 12, 18, 22, 26

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What is the mean ofthese numbers?

7, 10, 16

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2, 9, 14, 27

13

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1, 2, 7, 11, 19

8

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26, 33, 41, 52

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D fi iti


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Definition

MeMeddianian

is in theis in the

MiMiddddlele

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D fi iti


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Definition

MedianMedian the middlenumber in a set of orderednumbers.

1, 3, 7, 10, 13

Median = 7Median = 7

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ow o n e e an


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ow o n e e anin a Group of Numbers

Step 1 Arrange the numbers inorder from least to greatest.

21, 18, 24, 19, 27

18, 19, 21, 24, 27

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How to Find the Medianin a Group of Numbers

Step 2 Find themiddle number.

21, 18, 24, 19, 27

18, 19, 21, 24, 27

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How to Find the Median


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Step 2 Find themiddle number.

18, 19, 21, 24, 27

This is your median number.

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H t Fi d th M di


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Step 3 If there are two middle numbers,find the mean of these two numbers.

18, 19, 21, 25, 27,28

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How to Find the Median


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Step 3 If there are two middle numbers,

find the mean of these two numbers.

21+ 25

=

46

46/2 = 23 median

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What is the median of


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What is the median ofthese numbers?

16, 10, 7

10

7, 10, 16

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Wh i h di f


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29, 8, 4, 11, 19

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4, 8, 11, 19, 29

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Wh t i th di f


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31, 7, 2, 12, 14, 19

132, 7, 12, 14, 19, 31

12 + 14= 26 2) 26

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What is the median of


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53, 5, 81, 67, 25, 78

6053 + 67= 120

5, 25, 53, 67, 78, 81

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Definition

ModeMode the mostpopular or that whichis in fashion.

Baseball caps are a mode today.

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Definition


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Definition

ModeMode the number thatappears most frequently in a

set of numbers.

1, 1, 3, 7, 10, 13

Mode = 1Mode = 1

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How to Find the Mode in


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How to Find the Mode ina Group of Numbers


21, 18, 24, 19, 18

18, 18, 19, 21,24

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How to Find the Mode in


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How to Find the Mode ina Group of Numbers

Step 2 Find the number that isrepeated the most.

21, 18, 24, 19, 18

18, 18, 19, 21, 24

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Which number is the


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Which number is themode?

29, 8, 4, 8, 19

8

4, 8, 8, 19, 29

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Which number is the


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1, 2, 2, 9, 9, 4, 9, 10

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1, 2, 2, 4, 9, 9, 9, 10

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Which number is the


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22, 21, 27, 31, 21, 32

21

21, 21, 22, 27, 31, 32

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Calculation of Mode


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Calculation of Mode

Data set (30.0, 32.0, 31.5, 33.5,32.0, 33.0, 29.0, 29.5, 31.0, 32.5,34.5, 33.5, 31.5, 30.5, 30.0, 34.0,32.0, 32.0, 35.0, 32.5.) mg/ L

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Joke


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Joke

Three professors (a physicist, a chemist, and astatistician) are called in to see their dean. Justas they arrive the dean is called out of his office,leaving the three professors there. Theprofessors see with alarm that there is a fire in

the wastebasket.

The physicist says, "I know what to do! We mustcool down the materials until their temperature islower than the ignition temperature and then the

fire will go out."

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The chemist says, "No! No! I know what todo! We must cut off the supply of oxygenso that the fire will go out due to lack ofone of the reactants."

While the physicist and chemist debatewhat course to take, they both arealarmed to see the statistician runningaround the room starting other fires. They

both scream, "What are you doing?"

To which the statistician replies, "Trying toget an adequate sample size."

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LOGO

Normal DistributionNormal Distribution

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Normal Distribution


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Normal Distribution

All values are symmetrically distributed

around the meanA normal distribution is a distribution of the

values of a variable that, when plotted,produces a symmetrical, bell-shaped curve

that rises smoothly from a small number ofcases at each extreme to a large number ofcases in the middle.

Characteristic bell-shaped curve

Assumed for all quality control statistics

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Normal Distribution


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Normal Distribution

B l o o d U r e a m g

0

1

2

3

4

5

2 9 2 9 .5 3 0 3 0. 5 3 1 3 1 .5 3 2 3 2. 5 3 3 3 3 .5 34 3 4 .5 3 5

V a l u

Frequency

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Accuracy and Precision


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Accuracy and Precision

Precision is the closeness ofrepeated measurements to eachother.

Accuracy is the closeness of

measurements to the true value.Quality Control monitors both

precision and the accuracy of theassay in order to provide reliable

results.

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Precise and inaccurate


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Precise and inaccurate

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Imprecise and inaccurate


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Imprecise and inaccurate

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Precise and accurate


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Precise and accurate

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LOGO


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LOGO

Measures ofDispersion

Measures ofDispersion

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Measures of Dispersion


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Measures of Dispersionor Variability

There are several terms thatdescribe the dispersion or variabilityof the data around the mean:

Range

Variance

Standard Deviation

Coefficient of Variation

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Range


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Range

Range is the difference or spreadbetween the highest and lowestobservations.

It is the simplest measure ofdispersion.

It makes no assumption about thecentral tendency of the data.

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Definition


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Definition

RangeRange

is the distanceis the distance

BetweenBetween

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Definition


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Definition

RangeRange the differencebetween the greatest andthe least value in a set of

numbers.

1, 1, 3, 7, 10, 13

Range = 12Range = 12

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How to Find the Range in


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How to Find the Range ina Group of Numbers


21, 18, 24, 19, 27

18, 19, 21, 24, 27

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Step 2 Find the lowestand highest numbers.

21, 18, 24, 19, 27

18, 19, 21, 24, 27

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Step 3 Find the differencebetween these 2 numbers.

18, 19, 21, 24, 27

27 18 = 9The range is 9

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What is the range?


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What is the range?

29, 8, 4, 8, 19

29 4= 25

4, 8, 8, 19, 29

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What is the range?


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What is the range?

22, 21, 27, 31, 21, 32

32 21 = 11

21, 21, 22, 27, 31, 32

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What is the range?


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What is the range?

31, 8, 3, 11, 19

31 3 = 28

3, 8, 11, 19, 31

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What is the range?


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at s t e a ge

23, 7, 9, 41, 19

41 7 = 34

7, 9, 23, 19, 41

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Calculation of Variance


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Variance is the measure ofvariability about the mean.

It is calculated as the averagesquared deviation from the mean.

the sum of the deviations from the mean, squared,divided by the number of observations (correctedfor degrees of freedom)

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Calculation of Variance (S2)


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Ca cu a o o a a ce (S )

2

12

1n

)X(X)(SVariance

=

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Calculation of Variance


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2

12

2.75

52.25/19

1n

)X(X)(SVariance

=

=

=

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Calculation of Standard


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Calculation of StandardDeviation

The standard deviation (SD) is the square rootof the variance

-SD is the square root of the average squareddeviation from the mean

-SD is commonly used due to the same units asthe mean and the original observations

-SD is the principle calculation used to measuredispersion of results around a mean

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Calculation of Standard


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variance

1

2

=

= n

)X(Xs

i

Calculation of StandardDeviations

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Calculation of 1, 2 & 3 Standard


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Calculation of 1, 2 & 3 StandardDeviations

3s = 1.66 x 3 = 4.98

3.322x1.662s

1.662.751s

==

=

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Standard Deviation and


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Probability

68.2%

95.5%

99.7%99.7%

Frequen

cy

-3s-3s -2s-2s -1s-1s MeanMean +1s+1s +2s+2s +3s+3s

X

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Standard Deviation andProbability

For a data set of normal distribution,a value will fall within a range of:

+/- 1 SD 68.2% of the time +/- 2 SD 95.5% of the time

+/- 3 SD 99.7% of the time

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Calculation of Range


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68.2% confidence limit: (1SD)

Mean + s = 32.0+1.66

Mean - s = 32.0-1.66Range 33.66- 30.34

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95. 5% confidence limit:(2SD)

Mean + 2s = 32.0+3.32

Mean - 2s = 32.0-3.32

Range 28.68 35.32

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99. 7 % confidence limit: (3SD)

Mean + 3s = 32.0+4.98

Mean - 3s = 32.0-4.98

Range 27.02 36.98

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Probability

In general, people use the +/- 2 SDcriteria for the limits of theacceptable range for a test

When the measurement falls within

that range, there is 95.5%confidence that the measurement iscorrect

Only 4.5% of the time will a valuefall outside of that range due tochance; more likely it will be due toerror

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Example


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Consider the following threedatasets:(1) 5, 25, 25, 25, 25, 25, 45(2) 5, 15, 20, 25, 30, 35, 45

(3) 5, 5, 5, 25, 45, 45, 45

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Solution


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Case Standard Deviation1 11.55

2 13.23

3 20.00

The standard deviations for the datasets are

11.55, 13.23, and 20. The larger standard

deviations indicate greater variability in the data,

and in general we can say that smaller standarddeviations indicate less variability in the data.

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Example 2


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Canal 1: Average width = 3 ft (max 4 and min 2 ft)

Canal 2: Average width = 3 ft (max 6 and min 1 ft)

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Example 3


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Class 1 score: 25, 36, 45, 53, 69, 89Class 2 score: 45, 46, 47, 48, 50, 52

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Example 4


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For example, each of the three populations{0, 0, 14, 14}, {0, 6, 8, 14} and {6, 6, 8,8} has a mean of 7.

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Solution


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Their standard deviations are 7, 5, and1, respectively.The third population has a much

smaller standard deviation than theother two because its values are all

close to 7.In a loose sense, the standard

deviation tells us how far from themean the data points tend to be.

It will have the same units as the data

points themselves. If, for instance, thedata set {0, 6, 8, 14} represents theages of a population of 4 cows, thestandard deviation is 5 years.

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Example 5


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Consider average temperatures for cities. Whiletwo cities may each have an average temperatureof 15 C, it's helpful to understand that the rangefor cities near the coast is smaller than for citiesinland, which clarifies that, while the average issimilar, the chance for variation is greater inlandthan near the coast.

So, an average of 15 occurs for one city with highsof 25 C and lows of 5 C, and also occurs foranother city with highs of 18 and lows of 12. Thestandard deviation allows us to recognize that theaverage for the city with the wider variation, andthus a higher standard deviation, will not offer asreliable a prediction of temperature as the citywith the smaller variation and lower standarddeviation.

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Sigma


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z

percentage within percentage outside ratio outside

1 68.2689492% 31.7310508% 1 / 3.1514871

1.645 90% 10% 1 / 10

1.960 95% 5% 1 / 20

2 95.4499736% 4.5500264% 1 / 21.977894

2.576 99% 1% 1 / 100

3 99.7300204% 0.2699796% 1 / 370.398

3.2906 99.9% 0.1% 1 / 1000

4 99.993666% 0.006334% 1 / 15,788

5 99.9999426697% 0.0000573303% 1 / 1,744,278

6 99.9999998027% 0.0000001973% 1 / 506,800,000

7 99.9999999997440% 0.0000000002560% 1 / 390,600,000,000

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Coefficient of Variation


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The Coefficient of Variation (CV) isthe standard Deviation (SD)expressed as a percentage of themean

Also known as Relative Standarddeviation (RSD)

CV % = (SD mean) x 100

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Estimation Process


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Mean, , isunknown

Population Random SampleI am 95% confident

that is between40 & 60.Mean

X = 50

Estimation Process

Sample

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Conclusion


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SD is a measure of dispersionaround the mean. In a normaldistribution, 68% of cases fallwithin one standard deviation of

the mean and 95% of cases fallwithin two standard deviations.

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Standard Error of Mean.


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A measure of how much the value of themean may vary from sample to sampletaken from the same distribution.

It can be used to roughly compare the

observed mean to a hypothesized value(that is, you can conclude the two valuesare different if the ratio of the differenceto the standard error is less than -2 or

greater than +2).

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Type I and II Erros


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Analyzing variables that are notnormally distributed can lead to: serious overestimation (Type I error) or

underestimation (Type II error).

So you must examin each variables skewness, which measures the overall lack of

symmetry of the distribution, and whether it looks

the same to the left and right of the center point;

and

kurtosis, which measures whether the data are

peaked or flat relative to a normal distribution

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Kurtosis


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Kurtosis value tells whetherdistribution is peaked, flat, ornormal.

If Kurtosis value is zero, distribution

is normal, if it is positive, thendistribution is more peaked thannormal and if it is negative, thendistribution is flatter than normal.

Kurtosis values ranging from -1 to+1 are considered excellent. (George& Mallery, 2006, p. 98)

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For a normal distribution, the value of the kurtosis

statistic is zeroBell-shaped curves = describe in terms of its kurtosis

(curvature)

1. Leptokurtic = thin distribution

(concentrated at midpoint) (-)

2. Mesokurtic = normal distribution

3. Platykurtic = flat distribution (+)

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The large positive kurtosis tells youthat the distribution of data is morepeaked and has heavier tails thanthe normal distribution.

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Skewness


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Skewness value tells whether

distribution is symmetrical orasymmetrical.

If Skewness value is zero, distribution issymmetrical, if it is positive, then

smaller values are in greater number indistribution and if it is negative, thenlarger values are greater in number indistribution.

Skewness values ranging from -2 to +2are acceptable.

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Non-symmetrical


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1. Positive Skew = highnumber of low scores

2. Negative Skew = highnumber of high scores

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Intro to Statistics ToolboxStatistics Toolbox/Descriptive Statistics


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Examples of Skewness & Kurtosis:

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Skewness value = 0


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Large positive skewness shows thatsale has a long right tail.

That is, the distribution isasymmetric, with some distant

values in a positive direction fromthe center of the distribution.

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month-wise average temp (mm)


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Month Karachi Peshawar

January 30 -1February 31 4

March 32 25

April 33 35

May 34 40

June 35 48

July 35 50

August 34 45

September 33 38

October 32 35

November 31 25

December 30 4

Calculate CoV and see whether meaningful conclusion can be drawn109


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Example - Grapefruit Juice Study


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Descriptive Statistics

8 38 120 621 77.63 8.63 24.401 595.411

8

CRCL

Valid N (listwise)

Statist ic Stat ist ic Statis tic Statist ic Statist ic Std. Error Statist ic Statist ic

N Minimum Maximum Sum Mean Std.

DeviationVariance

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Example - Smoking Status


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SMKSTTS

1990 37.9 37.9 37.91063 20.3 20.3 58.2

609 11.6 11.6 69.8

1332 25.4 25.4 95.2

253 4.8 4.8 100.0

5247 100.0 100.0

Never SmokedQuit > 10 Years Ago

Quit < 10 Years Ago

Current Cigarette Smoker

Other Tobacco User

Total

Valid

Frequency Percent Valid Percent

Cumulative

Percent

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How to improve normality?


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Researchers often rely on one ofseveral transformations topotentially improve the normality ofcertain variables.

The most frequently usedtransformations are the square roottransformation, the logtransformation, and the inversetransformation.

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Square root transformation: Described simply, this type of transformation

involves taking the square root of each value within

a certain variable.

The one caveat is that you cannot take a squareroot of a negative number.

Fortunately, this can be easily remedied by adding

a constant, such as 1, to each item before

computing the square root.

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Log transformation: There is a wide variety of log transformations.

In general, however, a logarithm is the power (also

known as the exponent) to which a base number

has to be raised to get the original number. As with square root transformation, if a variable

contains values less than 1, a constant must be

added to move the minimum value of the

distribution.

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Inverse transformation:

This type of transformation involves taking the

inverse of each value by dividing it into 1.

For example, the inverse of 3 would be computed

as 1/3.

Essentially, this procedure makes very small values

very large, and very large values very small, and it

has the effect of reversing the order of a variables

scores.

Therefore, researchers using this transformationprocedure should be careful not to misinterpret the

scores following their analysis.

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LOGO


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Box Plot

(Box andWhiskers Plot)

Box Plot

(Box andWhiskers Plot)

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Box and Whisker Diagrams.


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4 5 6 7 8 9 10 11 12

Median

LowerQuartile

UpperQuartile

LowestValue

HighestValue

Box WhiskerWhisker

130 140 150 160 170 180 190

Boys

Girlscm

g

Box plots are useful for comparing two or more sets of data like

that shown below for heights of boys and girls in a class.

Anatomy of a Box and Whisker Diagram.

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If one whisker is longer , thedistribution of data is skewed in thedirection of longer whisker

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Drawing a Box Plot.


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LowerQuartile= 5

Q1

UpperQuartile

= 9

Q3

Median= 8

Q2

4 5 6 7 8 9 10 11 12

4, 4, 5, 6, 8, 8, 8, 9, 9, 9, 10, 12

Example 1: Draw a Box plot for the data below

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Drawing a Box Plot.


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UpperQuartile= 10

Q3

LowerQuartile

= 4

Q1

Median= 8

Q2

3, 4, 4, 6, 8, 8, 8, 9, 10, 10, 15,

Example 2: Draw a Box plot for the data below

3 4 5 6 7 8 9 10 11 12 13 14 15

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Drawing a Box Plot.


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UpperQuartile= 180

Qu

LowerQuartile= 158

QL

Median= 171

Q2

Question: Stuart recorded the heights in cm of boys in his

class as shown below. Draw a box plot for this data.

137, 148, 155, 158, 165, 166, 166, 171, 171, 173, 175, 180, 184, 186, 186

130 140 150 160 170 180 190cm

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Drawing a Box Plot.


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2.The boys are taller on average.

Question: Gemma recorded the heights in cm of girls in the same class andconstructed a box plot from the data. The box plots for both boys and girlsare shown below. Use the box plots to choose some correct statementscomparing heights of boys and girls in the class. Justify your answers.

130 140 150 160 170 180 190

Boys

Girls

cm

1.The girls are taller on average.

3.The girls show less variability in height.

4.The boys show less variability in height.

5.The smallest person is a girl

6.The tallest person is a boy123

LOGO


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Measures ofAssociation

Measures of

Association

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Correlations


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Correlations are perhaps the mostbasic and most useful measure ofassociation between two or morevariables.

Expressed in a single number calleda correlation coefficient (r),correlations provide informationabout the direction of therelationship (either positive ornegative) and the intensity of therelationship (1.0 to +1.0).

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In addition to the direction and strength of

a correlation, the coefficient can be usedto determine the proportion of varianceaccounted for by the association. This isknown as the coefficient of determination

(r

2

).R2 is calculated quite easily by squaring the

correlation coefficient. in the followingmanner:

.70 .70 = .49 It explains approximately 49% of the

variance

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The Pearson r

The Pearson r is used to examineassociations between two variablesthat are measured on either ratio or

interval scales.For example, the Pearson r could be

used to examine the correlation

between days of exercise andpounds of weight loss.

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Point-biserial (rpbi): This is used toexamine the relationship between avariable measured on a naturallyoccurring dichotomous nominal

scale and a variable measured on aninterval (or ratio) scale

e.g., a correlation between gender[dichotomous] and SAT scores[interval].

128


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Spearman rank-order (rs):

This is used to examine therelationship between two variablesmeasured on ordinal scales

e.g., a correlation of class rank[ordinal] and socioeconomic status[ordinal]

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Gamma ():This is used to examine the

relationship between one nominalvariable and one variable measured

on an ordinal scalee.g., a correlation of ethnicity

[nominal] and socioeconomic status

[ordinal]

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LOGO


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Testing ofHypothesis

Testing of

Hypothesis

132

Hypothesis Testing


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Goal: Make statement(s) regardingunknown population parameter valuesbased on sample data

Elements of a hypothesis test:

Null hypothesis - Statement regarding the value(s) ofunknown parameter(s). Typically will imply no association

between explanatory and response variables in our

applications (will always contain an equality)

Alternative hypothesis - Statement contradictory to thenull hypothesis (will always contain an inequality)

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Common Statistical Tests

Large sample tests (z test)

Small sample tests (student t test)

Paired t test

Chi-square test

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Determine The Hypothesis:Whether There is an Association

or Not


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or Not

Ho: The two variables are

independent

Ha: The two variables are

associated


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Exposure Outcome

Yes NoYes 2020 55

No 55 2525

Out of 25 men who had cancer, 20 claimed to have used

estrogens. Out of 30 men without cancer 5 claimed tohave used estrogens.

Total

Total

25 30

30

25

55136

4. Calculating Test Statistics


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=

e

eo

F

FF

22 )(

Observed

frequencies

Exp

ecte

d

freq

uenc

y

Exp

ecte

d

frequ

ency

5. Determine Degrees of Freedom


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df= (R-1)(C-1)

138

Numbero

f

levelsin

column

varia

ble

Numberof

levelsinrow

variable

Compare computed teststatistic against a


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statistic against a

tabled/critical valueThe computed value of the Pearson

chi- square statistic is comparedwith the critical value to determine if

the computed value is improbable

The critical tabled values are basedon sampling distributions of the

Pearson chi-square statisticIf calculated 2 is greater than 2

table value, reject Ho

Example


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140

Suppose a researcher is interestedin voting preferences on NRO issue.

A questionnaire was developed andsent to a random sample of 90

voters.The researcher also collects

information about the political partymembership of the sample of 90respondents.


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var a e requency a eor Contingency Table


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142

Favor Neutral Oppose f row

PML 10 10 30 50

PPP 15 15 10 40

fcolumn 25 25 40 n = 90

Obs

erve

d

frequ

encies

var a e requency a eor Contingency Table

Row

frequ


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143


PML 10 10 30 50

PPP 15 15 10 40

fcolumn 25 25 40 n = 90

uency


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Calculating Test Statistics


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145


Democrat f o =10

fe =13.9

fo =10

fe =13.9

fo =30

fe=22.2

50

Republican fo =15

fe

=11.1

fo =15

fe

=11.1

fo =10

fe

=17.8

40

fcolumn 25 25 40 n = 90

= 50*25/90



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PML f o =10

fe =13.9

fo =10

fe =13.9

fo =30

fe=22.2

50

PPP f o =15

fe

=11.1

fo =15

fe

=11.1

fo =10

fe

=17.8

40

fcolumn 25 25 40 n = 90

= 40* 25/90



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147


PML f o =10

fe =13.9

fo =10

fe =13.9

fo =30

fe=22.2

50

PPP f o =15

fe

=11.1

fo =15

fe

=11.1

fo =10

fe

=17.8

40

fcolumn 25 25 40 n = 90



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=

e

eo

F

FF

2

2 )(

Observed

frequencies

Exp

ecte

d

freq

uenc

y

Exp

ecte

d

frequency


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Determine Degrees of Freedom


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df = (R-1)(C-1) =(2-1)(3-1) = 2

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Compare computed test statistic againsta tabled/critical value


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= 0.05

df = 2

Critical tabled value = 5.991

Test statistic, 11.03, exceeds critical

valueNull hypothesis is rejected

PML & PPP differ significantly intheir opinions on gun control issues

LOGO


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ARM Lecture 7

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