lectures of stat -145 (biostatistics) text book biostatistics basic concepts and methodology for the...

Lectures of Stat -145(Biostatistics)

Text bookBiostatistics

Basic Concepts and Methodology for the Health Sciences

ByWayne W. Daniel

Prepared By:Sana A. Abunasrah

Text Book : Basic Concepts and Methodology for the

Health Sciences 2

Chapter 1

Introduction To Biostatistics


Health Sciences 3

Key words :

Statistics , data , Biostatistics, Variable ,Population ,Sample


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4

IntroductionSome Basic concepts

Statistics is a field of study concerned with

1- collection, organization, summarization and analysis of data.

2- drawing of inferences about a body of data when only a part of the data is observed.

Statisticians try to interpret and communicate the results to

others.


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* Biostatistics:The tools of statistics are employed in

many fields:business, education, psychology,

agriculture, economics, … etc.When the data analyzed are derived

from the biological science and medicine,

we use the term biostatistics to distinguish this particular application of statistical tools and concepts.


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6

Data:• The raw material of Statistics is data. • We may define data as figures. Figures

result from the process of counting or from taking a measurement.

•For example: • - When a hospital administrator counts

the number of patients (counting).• - When a nurse weighs a patient

(measurement)


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We search for suitable data to serve as the raw material for our investigation.

Such data are available from one or more of the following sources:

1- Routinely kept records.

For example:- Hospital medical records contain

immense amounts of information on patients.

- Hospital accounting records contain a wealth of data on the facility’s business

- activities.

*Sources of Data:


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8

2- External sources.The data needed to answer a question may already exist in the form ofpublished reports, commercially available data banks, or the research literature, i.e. someone else has already asked the same question.


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3- Surveys:The source may be a survey, if the data

needed is about answering certain questions.

For example: If the administrator of a clinic wishes to

obtain information regarding the mode of transportation used by patients to visit the clinic, then a survey may be conducted among

patients to obtain this information.


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4- Experiments.Frequently the data needed to answer

a question are available only as the result of an experiment.

For example:If a nurse wishes to know which of

several strategies is best for maximizing patient compliance, she might conduct an experiment in which the different strategies of motivating compliance

are tried with different patients.


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*A variable:It is a characteristic that takes on

different values in different persons, places, or things.

For example:- heart rate, - the heights of adult males, - the weights of preschool children,- the ages of patients seen in a dental

clinic.


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Quantitative Variables

It can be measured in the usual sense.

For example: - the heights of

adult males, - the weights of

preschool children,

- the ages of patients seen in a

- dental clinic.

Qualitative VariablesMany characteristics are

not capable of being measured. Some of them can be ordered (called ordinal) and Some of them can’t be ordered (called nominal).

For example:- classification of people

into socio-economic groups

-.hair color

Types of variables

Quantitative Qualitative


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A discrete variableis characterized by

gaps or interruptions in the values that it can assume.

For example:- The number of daily

admissions to a general hospital,

- The number of decayed, missing or filled teeth per child

- in an - elementary - school.

A continuous variablecan assume any value within

a specified relevant interval of values assumed by the variable.

For example:- Height, - weight, - skull circumference.No matter how close

together the observed heights of two people, we can find another person whose height falls somewhere in between.

Types of quantitative variables

Discrete Continuous


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14

As the name implies it consist of “naming” or classifies into various mutually exclusive categories

For example:- Male - female- Sick - well- Married – single -

divorced

.Whenever qualitative observation

Can be ranked or ordered according to some criterion.

For example:- Blood pressure (high-good-low)- Grades (Excellent –

V.good –good –fail)

Types of qualitative variables

Nominal Ordinal


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* A population:

It is the largest collection of It is the largest collection of valuesvalues of a of a ranrandom variabledom variable for which we for which we have an interest at a particular have an interest at a particular time. time.

For example: The weights of all the children

enrolled in a certain elementary school.

Populations may be finite or infinite.


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** A sample: A sample:It is a part of a population. It is a part of a population.

For example:The weights of only a fraction

of these children.


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17

Exercises• Question (6) – • Question (7) – Page 17 “ Situation A , Situation B “

Q6: For each of the following variables indicate whether it is quantitative or qualitative variable:

(a )Class standing of the members of this class relative to each other .

Qualitative ordinal(b) Admitting diagnoses of patients

admitted to a mental health clinic.

Qualitative nominal


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Exercises:

(c )Weights of babies born in a hospital

during a year. Quantitative continues(d )Gender of babies born in a hospital

during a year. Qualitative nominal(e )Range of motion of elbow joint of

students enrolled in a university health sciences curriculum. Quantitative continues

(f )Under-arm temperature of day-old infants born in a hospital. Quantitative continues


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Q7: For each of the following situations ,

answer questions a through d:

(a )What is the population?

(b) What is the sample in the study?

(c )What is the variable of interest?

(d )What is the type of the variable?

Situation A: A study of 300 households in a small southern town revealed that 20 percent had at least one school-age child present.


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All households in a small (a) Population:

southern town .

300 households in a small (b) Sample:

southern town.

(c )Variable: Does households had at least one school age child present.

(d )Variable is qualitative nominal.


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•Situation B: A study of 250 patients admitted to a hospital during the past year revealed that, on the average, the patients lived 15 miles from the

hospital.

(a )Population: All patients admitted to a hospital during the past year.

(b )Sample: 250 patients admitted to a hospital during the past year.


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(c )Variable: Distance the hospital live away from the hospital

Variable is Quantitative continuous. (d)


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Chapter ) 2 (Chapter ) 2 (

Strategies for Strategies for understanding the understanding the meanings of Datameanings of Data

Pages( 19 – 27)Pages( 19 – 27)

Text Book : Basic Concepts and Text Book : Basic Concepts and Methodology for the Health Methodology for the Health

Sciences Sciences 2525

Key wordsKey words

frequency table, bar chart ,rangefrequency table, bar chart ,range

width of interval ,width of interval , mid-intervalmid-interval

Histogram , PolygonHistogram , Polygon

Descriptive StatisticsDescriptive StatisticsFrequency Distribution Frequency Distribution

for Discrete Random Variablesfor Discrete Random VariablesExample:Example:Suppose that we take a Suppose that we take a samplesample of size 16 from of size 16 from children in a primary school children in a primary school and get the following data and get the following data about the number of their about the number of their decayed teeth,decayed teeth,3,5,2,4,0,1,3,5,2,3,2,3,3,2,4,13,5,2,4,0,1,3,5,2,3,2,3,3,2,4,1To construct a To construct a frequencyfrequency table:table:1- 1- OrderOrder the values from the the values from the smallest to the largest.smallest to the largest.0,1,1,2,2,2,2,3,3,3,3,3,4,4,5,50,1,1,2,2,2,2,3,3,3,3,3,4,4,5,52- 2- CountCount how many how many numbers are the same.numbers are the same.

No. of decayed

teeth

FrequencyRelativeFrequency

012345

124522

0.06250.1250.25

0.31250.1250.125

Total161



Representing the Representing the simple frequency table simple frequency table

using the bar chartusing the bar chart

Number of decayed teeth

5.004.003.002.001.00.00

Freq

uenc

y

6

5

4

3

2

1

0

22

5

4

2

1

We can represent the above simple frequency table using the bar chart.



2.3 Frequency Distribution 2.3 Frequency Distribution for Continuous Random Variablesfor Continuous Random Variables

For For large sampleslarge samples, we can’t use the simple frequency table to , we can’t use the simple frequency table to represent the data.represent the data.

We need to We need to dividedivide the data into the data into groupsgroups or or intervals intervals oror classes.classes.

So, we need to determine:So, we need to determine:

1- The number of intervals (k).1- The number of intervals (k).Too fewToo few intervals are not good because information will be intervals are not good because information will be

lost.lost.Too manyToo many intervals are not helpful to summarize the data. intervals are not helpful to summarize the data.A commonly followed rule is that A commonly followed rule is that 6 ≤ k ≤ 15,6 ≤ k ≤ 15,or the following formula may be used,or the following formula may be used,k = 1 + 3.322 (log n)k = 1 + 3.322 (log n)



2- The range (R).2- The range (R).

It is the difference between the It is the difference between the largest and the smallest observation largest and the smallest observation in the data set.in the data set.

3- The Width of the interval (w).3- The Width of the interval (w).ClassClass intervals generally should be of intervals generally should be of

the the same widthsame width. Thus, if we want k . Thus, if we want k intervals, then w is chosen such that intervals, then w is chosen such that

w ≥ R / k.w ≥ R / k.



Example:Example:Assume that the number of observations Assume that the number of observations equal 100, then equal 100, then k = 1+3.322(log 100) k = 1+3.322(log 100) = 1 + 3.3222 (2) = 7.6 = 1 + 3.3222 (2) = 7.6 8. 8.Assume that the smallest value = 5 and the Assume that the smallest value = 5 and the

largest one of the data = 61, then largest one of the data = 61, then R = 61 R = 61 –– 5 = 56 and 5 = 56 andw = 56 / 8 = 7.w = 56 / 8 = 7.

To make the summarization more To make the summarization more comprehensible, the class width may be 5 comprehensible, the class width may be 5 or 10 or the multiples of 10.or 10 or the multiples of 10.



Example 2.3.1Example 2.3.1 We wish to know how many class interval to have We wish to know how many class interval to have

in the frequency distribution of the data in Table in the frequency distribution of the data in Table 1.4.1 Page 9-10 of ages of 189 subjects who 1.4.1 Page 9-10 of ages of 189 subjects who Participated in a study on smoking cessationParticipated in a study on smoking cessation

SolutionSolution : : Since the number of observations Since the number of observations equal 189, then equal 189, then k = 1+3.322(log 169) k = 1+3.322(log 169) = 1 + 3.3222 (2.276) = 1 + 3.3222 (2.276) 9, 9, R = 82 R = 82 –– 30 = 52 and 30 = 52 and w = 52 / 9 = 5.778w = 52 / 9 = 5.778

It is better to let w = 10, then the intervals It is better to let w = 10, then the intervals will be in the form:will be in the form:



Class intervalFrequency

30 – 3911

40 – 4946

50 – 5970

60 – 6945

70 – 7916

80 – 891

Total189

Sum of frequency=sample size=n



The Cumulative FrequencyThe Cumulative Frequency::

It can be computed by adding successive It can be computed by adding successive frequenciesfrequencies..

The Cumulative Relative FrequencyThe Cumulative Relative Frequency::

It can be computed by adding successive relative It can be computed by adding successive relative frequenciesfrequencies..

TheThe Mid-intervalMid-interval::It can be computed by adding the lower bound of It can be computed by adding the lower bound of the interval plus the upper bound of it and then the interval plus the upper bound of it and then divide over 2divide over 2 . .



For the above example, the following table represents the For the above example, the following table represents the cumulative frequency, the relative frequency, the cumulative cumulative frequency, the relative frequency, the cumulative

relative frequency and the mid-intervalrelative frequency and the mid-interval.. Class

intervalMid –

intervalFrequency

Freq )f(Cumulative Frequency

RelativeFrequency

R.f

Cumulative Relative

Frequency

30 – 3934.511110.05820.058240 – 4944.546570.2434-

50 – 5954.5-127-0.6720

60 – 69-45-0.23810.9101

70 – 7974.5161880.08470.9948

80 – 8984.511890.00531

Total1891

R.f= freq/n



ExampleExample : : From the above frequency table, complete the From the above frequency table, complete the

table then answer the following questions:table then answer the following questions: 1-The number of objects with age less than 50 1-The number of objects with age less than 50

years ?years ? 2-The number of objects with age between 40-69 2-The number of objects with age between 40-69

years ?years ? 3-Relative frequency of objects with age between 3-Relative frequency of objects with age between

70-79 years ?70-79 years ? 4-Relative frequency of objects with age more 4-Relative frequency of objects with age more

than 69 years ?than 69 years ? 5-The percentage of objects with age between 40-5-The percentage of objects with age between 40-

49 years ?49 years ?



6-6- The percentage of objects with age less than The percentage of objects with age less than 60 years ?60 years ?

7-The Range (R) ?7-The Range (R) ? 8- Number of intervals (K)?8- Number of intervals (K)? 9- The width of the interval ( W) ?9- The width of the interval ( W) ?



Representing the grouped Representing the grouped frequency table using the frequency table using the

histogramhistogramTo draw the histogram, the To draw the histogram, the true classes limitstrue classes limits should be used. should be used. They can be computed by They can be computed by subtracting subtracting 0.5 from the0.5 from the lower lower limit and limit and adding adding 0.5 to the0.5 to the upper upper limit for each interval.limit for each interval.

True class limitsFrequency

29.5 – <39.511

39.5 – < 49.546

49.5 – < 59.570

59.5 – < 69.545

69.5 – < 79.516

79.5 – < 89.51

Total189



Representing the grouped Representing the grouped frequency table using the frequency table using the

PolygonPolygon



ExercisesExercises

PagesPages : 31 : 31 –– 34 34 QuestionsQuestions: 2.3.2(a) , 2.3.5 (a): 2.3.2(a) , 2.3.5 (a) H.W.H.W. : : 2.3.6 , 2.3.7(a) 2.3.6 , 2.3.7(a)

ExercisesExercises::

Q2.3.2: Q2.3.2: Janardhan et al. (A-2) conducted Janardhan et al. (A-2) conducted a study in which they measured a study in which they measured incidental intracranial aneurysms incidental intracranial aneurysms (IIAs) in 125 patients. The researchers (IIAs) in 125 patients. The researchers examined post procedural examined post procedural complications and concluded that IIAs complications and concluded that IIAs can be safely treated without causing can be safely treated without causing mortality and with a lower mortality and with a lower complications rate than previously complications rate than previously reportedreported..



The following are the sizes (in The following are the sizes (in millimeters) of the 159 IIAs in the millimeters) of the 159 IIAs in the samplesample . .



Class Intervalfrequency

0-429

5-987

10-1426

15-1910

20-244

25-291

30-342

Total159

((aa ) )Use the frequency table to prepareUse the frequency table to prepare::

* *A relative frequency distributionA relative frequency distribution

* *A cumulative frequency distributionA cumulative frequency distribution

* *A cumulative relative frequency A cumulative relative frequency distortiondistortion

* *A histogramA histogram

* *A frequency polygonA frequency polygon



((bb ) )What percentage of the measurements are What percentage of the measurements are between 10 and 14 inclusivebetween 10 and 14 inclusive ? ?

((cc ) )How many observations are less than 20How many observations are less than 20 ? ?

((dd ) )What proportion of the measurements are What proportion of the measurements are greater than or greater than or equal to 25equal to 25??

( ( ee ) )What percentage of the measurements What percentage of the measurements are either less than 10 or greater than 19are either less than 10 or greater than 19??



Q2.3.5: Q2.3.5: The following table shows the The following table shows the number of hours 45 hospital patients number of hours 45 hospital patients slept following the administration of a slept following the administration of a certaincertain

anestheticanesthetic..

((aa ) )From theseFrom these

data constructdata construct::

* *A relativeA relative

frequencyfrequency

distributiondistribution

Class IntervalFrequency

1-521

6-1016

11-156

16-202

Total45




* *A frequency polygonA frequency polygon

((bb ) )How many of the measurements are How many of the measurements are greater than 10? greater than 10? Ans: 8Ans: 8

((cc ) )What percentage of the measurements What percentage of the measurements are between 6-15are between 6-15? ?

Ans: 49%Ans: 49%

((dd ) )What proportion of the What proportion of the measurement is less than or equal measurement is less than or equal 15? 15? Ans: 0.96Ans: 0.96



Q2.3.6: Q2.3.6: The following are the number The following are the number of babies born during a year in 60 of babies born during a year in 60 community hospitalscommunity hospitals..

((aa ) )From theseFrom these

data constructdata construct::

**A relativeA relative

frequencyfrequency

distributiondistribution

**A histogramA histogram

**A frequency polygonA frequency polygonText Book : Basic Concepts and Text Book : Basic Concepts and

Methodology for the Health Methodology for the Health Sciences Sciences 4646

Class IntervalFrequency

20-245

25-296

30-349

35-393

40-445

45-498

50-5411

55-5913

Total60

Q2.3.7: Q2.3.7: In a study ofIn a study of

physical endurancephysical endurance

levels of male collegelevels of male college

freshman, thefreshman, the

following compositefollowing composite

endurance scoresendurance scores

based on severalbased on several

exercise routinesexercise routines

were collectedwere collected..



Class intervalFrequency

115-1346

135-1547

155-17416

175-19431

195-21437

215-23428

235-25418

255-2748

275-2943

295-3141

Total155

((aa ) )From these data constructFrom these data construct::

* *A relative frequency distributionA relative frequency distribution


* *A frequency polygonA frequency polygon..



Section (2.4) :Section (2.4) : Descriptive Statistics Descriptive Statistics

Measures of Central Measures of Central Tendency Tendency

Page 38 - 41Page 38 - 41

Text Book : Basic Concepts and Text Book : Basic Concepts and Methodology for the Health Sciences Methodology for the Health Sciences 5050

key words:

Descriptive Statistic, measure of central tendency ,statistic, parameter, mean )μ( ,median, mode.


The Statistic and The The Statistic and The ParameterParameter • A Statistic:

It is a descriptive measure computed from the data of a sample.

• A Parameter:It is a a descriptive measure computed from

the data of a population.Since it is difficult to measure a parameter from the

population, a sample is drawn of size n, whose values are 1 , 2 , …, n. From this data, we measure the statistic.


Measures of Central Measures of Central TendencyTendency

A measure of central tendency is a measure which indicates where the middle of the data is.

The three most commonly used measures of central tendency are:

The Mean, the Median, and the Mode.

The Mean :It is the average of the data.


The Population Mean:

= which is usually unknown, then we use the

sample mean to estimate or approximate it.

The Sample Mean: =

Example:Here is a random sample of size 10 of ages, where

1 = 42, 2 = 28, 3 = 28, 4 = 61, 5 = 31,

6 = 23, 7 = 50, 8 = 34, 9 = 32, 10 = 37.

= (42 + 28 + … + 37) / 10 = 36.6

x

1

N

ii

N

X

x

1

n

ii

n

x


Properties of the Mean:• Uniqueness. For a given set of data there is

one and only one mean.

• Simplicity. It is easy to understand and to compute.

• Affected by extreme values. Since all values enter into the computation.

Example: Assume the values are 115, 110, 119, 117, 121 and 126. The mean = 118.

But assume that the values are 75, 75, 80, 80 and 280. The mean = 118, a value that is not representative of the set of

data as a whole.


The Median:When ordering the data, it is the observation that divide the

set of observations into two equal parts such that half of the data are before it and the other are after it.

* If n is odd, the median will be the middle of observations. It will be the (n+1)/2 th ordered observation.

When n = 11, then the median is the 6th observation.* If n is even, there are two middle observations. The median

will be the mean of these two middle observations. It will be the mean of the [ (n/2) th , (n/2 +1) th ]ordered observation.

When n = 12, then the median is the 6.5th observation, which is an observation halfway between the 6th and 7th ordered observation.


Example:For the same random sample, the ordered

observations will be as:23, 28, 28, 31, 32, 34, 37, 42, 50, 61.Since n = 10, then the median is the 5.5th

observation, i.e. = (32+34)/2 = 33.

Properties of the Median:• Uniqueness. For a given set of data there is

one and only one median.

• Simplicity. It is easy to calculate.

• It is not affected by extreme values as is the mean.


The Mode:It is the value which occurs most frequently.If all values are different there is no mode.Sometimes, there are more than one mode.

Example:For the same random sample, the value 28 is

repeated two times, so it is the mode.

Properties of the Mode:• Sometimes, it is not unique.• It may be used for describing qualitative

data.

ExamplesExamples

Find the mean and the mode for the following Relative Frequency?

Mode = 7 (has the higher frequency)


n

xfx Age(x)

frequenc

y(f)

X f

567

10

2341

10182810

Total1066

6.610

66x

ExamplesExamplesFind the mean and the mode for the following grouped Frequency table?

Mode :interval( 7 – 9 ) (can't give exact number only the interval with higher Frequency)


n

xfx

4.710

74x

AgeFrequency (f)

Midpoint (X)

X f

1 - 3 4 - 6 7 - 9

10 - 12

2 1 4 3

2 5 8 11

4 5

32 33

Total 10 _ 74


ExamplesExamples

Number of decayed teeth

5.004.003.002.001.00.00

Freq

uenc

y

6

5

4

3

2

1

0

22

5

4

2

1

Find the mean and the mode for the following bar chart?

Solution :

Mode = 3(has the higher frequency)


)225421(

)25()24()53()42()21()10(

xxxxxx

x

687.216

43x

n

xfx

Section (2.5) :Section (2.5) : Descriptive Statistics Descriptive Statistics

Measures of Dispersion Measures of Dispersion Page 43 - 46Page 43 - 46


key words:

Descriptive Statistic, measure of dispersion , range ,variance, coefficient of variation.


2.5. Descriptive Statistics – 2.5. Descriptive Statistics – Measures of Dispersion:Measures of Dispersion:

• A measure of dispersion conveys information regarding the amount of variability present in a set of data.

• Note:1. If all the values are the same → There is no dispersion .2. If all the values are different → There is a dispersion: 3.If the values close to each other →The amount of Dispersion small.b( If the values are widely scattered → The Dispersion is greater.


Ex. Figure 2.5.1 –Page 43Ex. Figure 2.5.1 –Page 43

• ** Measures of Dispersion are :

1.Range )R(.

2. Variance.

3. Standard deviation.

4.Coefficient of variation )C.V(.


1.The Range (R):1.The Range (R):

• Range =Largest value- Smallest value =

• Note: • Range concern only onto two values • Example 2.5.1 Page 40: • Refer to Ex 2.4.2.Page 37 • Data:• 43,66,61,64,65,38,59,57,57,50. • Find Range?• Range=66-38=28

SL xx


2.The Variance:2.The Variance: • It measure dispersion relative to the scatter of the values

a bout there mean.

a) Sample Variance ( ) :• ,where is sample mean

• Example 2.5.2 Page 40: • Refer to Ex 2.4.2.Page 37• Find Sample Variance of ages , = 56 • Solution: • S2= [)43-56( 2 +)66-43( 2+…..+)50-56( 2 ]/ 10• = 900/10 = 90

x

2S

1

)(1

2

2

n

xxS

n

ii

x


• b)Population Variance ( ) :

• where , is Population mean

3.The Standard Deviation:

• is the square root of variance=

a( Sample Standard Deviation = S =

b( Population Standard Deviation = σ =

2

N

xN

ii

1

2

2

)(

Varince2S

2


4.The Coefficient of Variation 4.The Coefficient of Variation (C.V):(C.V):

• Is a measure use to compare the dispersion in two sets of data which is independent of the unit of the measurement .

• where S: Sample standard deviation.

• : Sample mean.

)100(.X

SVC

X


Example 2.5.3 Page 46Example 2.5.3 Page 46::

• Suppose two samples of human males yield the following data:

Sampe1 Sample2

Age 25-year-olds 11year-olds

Mean weight 145 pound 80 pound

Standard deviation 10 pound 10 pound


• We wish to know which is more variable.• Solution:• c.v )Sample1(= )10/145(*100= 6.9

• c.v )Sample2(= )10/80(*100= 12.5

• Then age of 11-years old)sample2( is more variation


ExercisesExercises

• Pages : 52 – 53

• Questions: 2.5.1 , 2.5.2 ,2.5.3

• H.W. :2.5.4 , 2.5.5, 2.5.6, 2.5.14

• * Also you can solve in the review questions page 57:

• Q: 12,13,14,15,16, 19

Exercises :

For each of the data sets in the following exercises compute:

(a )The mean

(b )The median

(c )The mode

(d )The range

(e )The variance

(f )The standard deviation

(g )The coefficient of variation

Text Book : Basic Concepts and Text Book : Basic Concepts and

Methodology for the Health Sciences Methodology for the Health Sciences 7373

Q2.5.1:

Porcellini et al. )A-8( studied 13 HIV-positive patients who were treated with highly active antiretroviral therapy )HAART( for at least 6 months. The CD4 T cell counts ) ( at baseline for the 13 subjects are listed

below .

230 205 313 207 227 245 173

58 103 181 105 301 169


l106


Q2.5.2: Shrair and Jasper )A-9( investigated whether decreasing the venous return in young rats would affect ultrasonic vocalizations )USVs(. Their research showed no significant change in the number of ultrasonic vocalizations when blood was removed from either the superior vena cava or the carotid artery. Another important variable measured was the heart rate )bmp( during the withdrawal of blood. The data below presents the heart rate of

seven rat pups from the experiment involving the carotid artery.

500 570 560 570 450 560 570

(a )The mean )b( The median

Ans: 540 Ans: 560

(c )The mode )d( The range

Ans: 570 Ans: 120

(e )The variance )f( The standard deviation

Ans: 2200.0039 Ans: 46.9042

(g )The coefficient of variation Ans: 8.69%



Q2.5.3:

Butz et al. )A-10( evaluated the duration of benefit derived from the use of noninvasive positive-pressure ventilation by patients with amyotrophic lateral sclerosis on symptoms, quality of life, and survival. One of the variables of interest is partial pressure of arterial carbon dioxide )PaCO2(. The values below ) mm of Hg ( reflect the result of baseline testing on 30 subjects as established by arterial blood gas analyses.


40.0 47.0 34.0 42.0 54.0 48.0 53.6 56.9 58.0 45.0 54.5 54.0 43.0 44.3

53.9 41.8 33.0 43.1 52.4 37.9 34.5

40.1 33.0 59.9 62.6 54.1 45.7 40.6 56.6 59.0


Ans: 47.72 Ans: 46.35


Ans: 33, 54 Ans: 29.6



Ans: 84.135 Ans: 9.17251

(g )The coefficient of variation

Q2.5.4:

According to Starch et al. )A-11(, hamstring tendon grafts have been the “weak link” in anterior cruciate ligament reconstruction. In a controlled laboratory study, they compared two techniques for reconstruction : either an interference screw or a central sleeve and


screw on the tibial side. For eight cadaveric knees, the measurements below represent the required force ) in Newtones( at which initial failure of graft strands occurred for the central sleeve and screw technique.

172.5 216.63 212.62 98.97 66.95 239.76 19.57 195.72


Ans: 152.84 Ans: 184.11


Ans: no mode Ans: 220.19Text Book : Basic Concepts and Text Book : Basic Concepts and



Ans: 6494.732 Ans: 80.5899


Q2.5.5:

Cardosi et al. )A-12( performed a 4 years retrospective review of 102 women undergoing radical hysterectomy for cervical or endometrial cancer. Catheter-associated urinary tract infection was observed in 12 of the subjects. Below are the numbers of


postoperative days until diagnosis of the infection for each subject experiencing an infection.

16 10 49 15 6 15 8 19 11 22 13 17


Ans: 16.75 Ans: 15


Ans: 15 Ans: 43


Ans: 124.0227 Ans: 11.1365




Q2.5.6: The purpose of a study by Nozama et al. )A-13( was to evaluate the outcome of surgical repair of pars interarticularis defect by segmental wire fixation in young adults with lumbar spondylolysis. The authors found that segmental wire fixation historically has been successful in the treatment of nonathletes with spondylolysis, but no information existed on the results of this type of surgery in athletes. In a retrospective study, the authors found 20 subjects who had the surgery between 1993

and 2000. For these subjects, the data below Text Book : Basic Concepts and Text Book : Basic Concepts and


represent the duration in months of follow-up care after the operation.

103 68 62 60 60 54 49 44 42 41 38 36 34 30 19 19 19 19 17 16


Ans: 41.5 Ans: 39.5


Ans: 19 Ans: 87


Ans: 490.264 Ans: 22.1419Text Book : Basic Concepts and Text Book : Basic Concepts and



Q2.5.14:

In a pilot study, Huizinga et al. ) A-14( wanted to gain more insight into the psychosocial consequences for children of a parent with cancer. For the study, 14 families participated in semistructured interviews and completed standardized questionnaires. Below is the age of the sick parent with cancer )in years( for the 14 families.


37 48 53 46 42 49 44 38 32 32 51 51 48 41


Ans: 43.7143 Ans: 45


Ans: 32, 51 Ans: 21


Ans: 48.0659 Ans: 6.93296




Chapter 3Chapter 3

ProbabilityProbability

The Basis of the The Basis of the Statistical inferenceStatistical inference

Text Book : Basic Concepts and Text Book : Basic Concepts and Methodology for the Health Sciences Methodology for the Health Sciences

8888

Key words:Key words:

Probability, objective Probability,Probability, objective Probability,

subjective Probability, equally likelysubjective Probability, equally likely

Mutually exclusive, multiplicative ruleMutually exclusive, multiplicative rule

Conditional Probability, independent events, Conditional Probability, independent events, Bayes theoremBayes theorem


8989

3.13.1 IntroductionIntroduction The concept of probability is frequently encountered in The concept of probability is frequently encountered in

everyday communication. everyday communication. For exampleFor example, a physician may say , a physician may say that a patient has a 50-50 chance of surviving a certain that a patient has a 50-50 chance of surviving a certain operation. operation. Another physician may say that she is 95 Another physician may say that she is 95 percent certain that a patient has a particular disease. percent certain that a patient has a particular disease.

Most people express probabilities in terms of percentages. Most people express probabilities in terms of percentages.

But, it is more convenient to express probabilities as But, it is more convenient to express probabilities as fractions. Thus, we may measure the probability of the fractions. Thus, we may measure the probability of the occurrence of some event by a number between 0 and 1.occurrence of some event by a number between 0 and 1.

The more likely the event, the closer the number is to one. An The more likely the event, the closer the number is to one. An event that can't occur has a probability of zero, and an event event that can't occur has a probability of zero, and an event that is certain to occur has a probability of one.that is certain to occur has a probability of one.


9090

3.23.2 Two views of Probability Two views of Probability objective and subjectiveobjective and subjective::

*** *** Objective ProbabilityObjective Probability ** ** Classical and RelativeClassical and Relative Some definitionsSome definitions::1.Equally likely outcomes: 1.Equally likely outcomes: Are the outcomes that have the same Are the outcomes that have the same

chance of occurring.chance of occurring.2.Mutually exclusive:2.Mutually exclusive:Two events are said to be mutually exclusive Two events are said to be mutually exclusive

if they cannot occur simultaneously such if they cannot occur simultaneously such that A B =Φ .that A B =Φ .


9191

The universal SetThe universal Set (S): The set all (S): The set all possible outcomes.possible outcomes.

The empty setThe empty set Φ Φ : Contain no elements. : Contain no elements. The event ,The event ,EE : is a set of outcomes in S : is a set of outcomes in S

which has a certain characteristic.which has a certain characteristic. Classical ProbabilityClassical Probability : If an event can : If an event can

occur in N mutually exclusive and equally occur in N mutually exclusive and equally likely ways, and if m of these possess a likely ways, and if m of these possess a triat, E, the probability of the occurrence of triat, E, the probability of the occurrence of event E is equal to m/ N .event E is equal to m/ N .

For ExampleFor Example: : in the rolling of the die , in the rolling of the die , each of the six sides is equally likely to be each of the six sides is equally likely to be observed . So, the probability that a 4 will observed . So, the probability that a 4 will be observed is equal to 1/6.be observed is equal to 1/6.


9292

Relative Frequency Probability:Relative Frequency Probability: Def:Def: If some posses is repeated a large If some posses is repeated a large

number of times, n, and if some resulting number of times, n, and if some resulting event E occurs m times , the relative event E occurs m times , the relative frequency of occurrence of E , m/n will be frequency of occurrence of E , m/n will be approximately equal to probability of E . approximately equal to probability of E . P(E) = m/n .P(E) = m/n .

*** *** Subjective ProbabilitySubjective Probability : : Probability measures the confidence that a Probability measures the confidence that a

particular individual has in the truth of a particular individual has in the truth of a particular proposition.particular proposition.

For ExampleFor Example : the probability that a cure : the probability that a cure for cancer will be discovered within the for cancer will be discovered within the next 10 years. next 10 years.


9393

3.33.3 Elementary Properties of Elementary Properties of ProbabilityProbability::

Given some process (or experiment ) Given some process (or experiment ) with n mutually exclusive events Ewith n mutually exclusive events E11, , EE22, E, E33,,……………………, E, Enn, then, then

1-P(E1-P(Eii ) 0, i= 1,2,3, ) 0, i= 1,2,3,…………nn 2- P(E2- P(E1 1 )+ P(E)+ P(E22) +) +…………+P(E+P(Enn )=1 )=1 3- P(E3- P(Eii +E +EJJ )=P(E )=P(Ei i )+ P(E)+ P(EJ J ) )

EEii ,E ,EJJ are mutually exclusive are mutually exclusive


9494

Rules of ProbabilityRules of Probability 1-Addition Rule1-Addition Rule P(A U B)= P(A) + P(B) P(A U B)= P(A) + P(B) –– P (A∩B ) P (A∩B ) 2- If A and B are mutually exclusive 2- If A and B are mutually exclusive

(disjoint) ,then(disjoint) ,then P (A∩B ) = 0P (A∩B ) = 0 Then , addition rule isThen , addition rule is P(A U B)= P(A) + P(B) .P(A U B)= P(A) + P(B) . 3- Complementary Rule3- Complementary Rule P(A' )= 1 P(A' )= 1 –– P(A) P(A) where, A' = = complement eventwhere, A' = = complement event Consider example Consider example 3.4.1 Page 633.4.1 Page 63


9595

Table 3.4.1 in Example 3.4.1Table 3.4.1 in Example 3.4.1

Family history of Mood Disorders

Early = 18( E)

Later >18(L )

Total

Negative)A(283563

Bipolar Disorder)B(

193857

Unipolar )C( 414485

Unipolar and

Bipolar)D(5360113

Total141177318


9696

****Answer the following Answer the following questionsquestions::Suppose we pick a person at random from this sample.Suppose we pick a person at random from this sample.

1-The probability that this person will be 18-years old 1-The probability that this person will be 18-years old or younger?or younger?

2-The probability that this person has family history of 2-The probability that this person has family history of mood orders Unipolar(C)?mood orders Unipolar(C)?

3-The probability that this person has no family history 3-The probability that this person has no family history of mood orders Unipolar( )?of mood orders Unipolar( )?

4-The probability that this person is 18-years old or 4-The probability that this person is 18-years old or younger younger oror has no family history of mood orders has no family history of mood orders Unipolar )C()?)?

5-The probability that this person is more than18-5-The probability that this person is more than18-years old years old andand has family history of mood orders has family history of mood orders Unipolar and Bipolar(D)?Unipolar and Bipolar(D)?

C


9797

Conditional ProbabilityConditional Probability::

P(A\B) is the probability of A assuming P(A\B) is the probability of A assuming that B has happened.that B has happened.

P(A\B)= , P(B)≠ 0P(A\B)= , P(B)≠ 0

P(B\A)= , P(A)≠ 0P(B\A)= , P(A)≠ 0

)(

)(

BP

BAP

)(

)(

AP

BAP


9898

Example 3.4.2 Page 64Example 3.4.2 Page 64From previous example From previous example 3.4.1 Page 633.4.1 Page 63 , ,

answeranswer suppose we pick a person at random and suppose we pick a person at random and

find he is 18 years or younger (E),what is find he is 18 years or younger (E),what is the probability that this person will be one the probability that this person will be one with Negative family history of mood with Negative family history of mood disorders (A)?disorders (A)?

suppose we pick a person at random and suppose we pick a person at random and find he has family history of mood (D) what find he has family history of mood (D) what is the probability that this person will be 18 is the probability that this person will be 18 years or younger (E)? years or younger (E)?


9999

Calculating a joint Calculating a joint ProbabilityProbability: :

Example 3.4.3.Page 64Example 3.4.3.Page 64 Suppose we pick a person at random Suppose we pick a person at random

from the 318 subjects. Find the from the 318 subjects. Find the probability that he will early (E) and probability that he will early (E) and has no family history of mood has no family history of mood disorders (A).disorders (A).


100100

Multiplicative RuleMultiplicative Rule::

P(A∩B)= P(A\B)P(B)P(A∩B)= P(A\B)P(B) P(A∩B)= P(B\A)P(A)P(A∩B)= P(B\A)P(A) Where,Where, P(A): marginal probability of A.P(A): marginal probability of A. P(B): marginal probability of B.P(B): marginal probability of B. P(B\A):The conditional probability.P(B\A):The conditional probability.


101101

Example 3.4.4 Page 65Example 3.4.4 Page 65 From previous example From previous example 3.4.1 Page 633.4.1 Page 63

, we wish to compute the joint , we wish to compute the joint probability of Early age at onset(E) probability of Early age at onset(E) and a negative family history of and a negative family history of mood disorders(A) from a knowledge mood disorders(A) from a knowledge of an appropriate marginal of an appropriate marginal probability and an appropriate probability and an appropriate conditional probability.conditional probability.

Exercise: Example 3.4.5.Page 66Exercise: Example 3.4.5.Page 66 Exercise: Example 3.4.6.Page 67Exercise: Example 3.4.6.Page 67


102102

Independent EventsIndependent Events::

If A has no effect on B, we said that If A has no effect on B, we said that A,B are independent events.A,B are independent events.

Then,Then, 1- P(A∩B)= P(B)P(A)1- P(A∩B)= P(B)P(A) 2- P(A\B)=P(A)2- P(A\B)=P(A) 3- P(B\A)=P(B)3- P(B\A)=P(B)


103103

Example 3.4.7 Page 68Example 3.4.7 Page 68 In a certain high school class consisting of In a certain high school class consisting of

60 girls and 40 boys, it is observed that 24 60 girls and 40 boys, it is observed that 24 girls and 16 boys wear eyeglasses . If a girls and 16 boys wear eyeglasses . If a student is picked at random from this class student is picked at random from this class ,the probability that the student wears ,the probability that the student wears eyeglasses , P(E), is 40/100 or 0.4 .eyeglasses , P(E), is 40/100 or 0.4 .

What is the probability that a student What is the probability that a student picked at random wears eyeglasses given picked at random wears eyeglasses given that the student is a boy?that the student is a boy?

What is the probability of the joint What is the probability of the joint occurrence of the events of wearing eye occurrence of the events of wearing eye glasses and being a boy?glasses and being a boy?


104104

Example 3.4.8 Page 69Example 3.4.8 Page 69

Suppose that of 1200 admission to a Suppose that of 1200 admission to a general hospital during a certain period of general hospital during a certain period of time,750 are private admissions. If we time,750 are private admissions. If we designate these as a set A, then compute designate these as a set A, then compute P(A) , P( ).P(A) , P( ).

Exercise: Example 3.4.9.Page 76Exercise: Example 3.4.9.Page 76

A


105105

Marginal ProbabilityMarginal Probability:: Definition:Definition: Given some variable that can be broken Given some variable that can be broken

down into m categories designated down into m categories designated by and another jointly occurring by and another jointly occurring

variable that is broken down into n variable that is broken down into n categories designated by categories designated by

, the marginal probability of with all the , the marginal probability of with all the categories of B . That is,categories of B . That is,

for all value of jfor all value of j Example 3.4.9.Page 76Example 3.4.9.Page 76 Use data of Table 3.4.1, and rule of Use data of Table 3.4.1, and rule of

marginal Probabilities to calculate P(E). marginal Probabilities to calculate P(E).

),()( jii BAPAP

mi AAAA ,.......,,.......,, 21

nj BBBB ,.......,,.......,, 21

iA


106106

ExerciseExercise::

Page 76-77Page 76-77 Questions :Questions : 3.4.1, 3.4.3,3.4.43.4.1, 3.4.3,3.4.4 H.W.H.W. 3.4.5 , 3.4.73.4.5 , 3.4.7

Q3.4.1: Q3.4.1: In a study of violent victimization of In a study of violent victimization of women and men, Porcelli et al. (A-2) collected women and men, Porcelli et al. (A-2) collected information from 679 women and 345 men information from 679 women and 345 men aged 18 to 64 years at several family practice aged 18 to 64 years at several family practice centers in the metropolitan Detroit area. centers in the metropolitan Detroit area. Patients filled out a health history Patients filled out a health history questionnaire that included a question about questionnaire that included a question about victimization. The following table shows the victimization. The following table shows the sample subjects cross-classified by sex and sample subjects cross-classified by sex and type of violent victimization reported. The type of violent victimization reported. The victimization categories are defined as no victimization categories are defined as no victimization, partner victimization (and not by victimization, partner victimization (and not by others), victimization by persons other thanothers), victimization by persons other than


107107

partners (friends, family members, or partners (friends, family members, or strangers), and those who reported strangers), and those who reported multiple victimizationmultiple victimization..

((aa ) )Suppose we pick a subject at random from Suppose we pick a subject at random from this group. What is the probability that this this group. What is the probability that this

subject will be a womensubject will be a women ? ?Text Book : Basic Concepts and Text Book : Basic Concepts and


No Victimizati

on

Partners

Nonpartners

Multiple Victimizati

on

Total

Women

611341618679

Men308101710345

Total9194433281024

((bb ) )What do we call the probability calculated in part What do we call the probability calculated in part aa ? ?

((cc ) )Show how to calculate the probability asked for Show how to calculate the probability asked for in part a by two additional methodsin part a by two additional methods..

((dd ) )If we pick a subject at random, what is If we pick a subject at random, what is probability that the subject will be a women and probability that the subject will be a women and

have experienced partner abusehave experienced partner abuse??

((ee ) )What do we call the probability calculated in part What do we call the probability calculated in part dd??

((ff ) )Suppose we picked a man at random. Suppose we picked a man at random. Knowing this information, what is the Knowing this information, what is the probability that heprobability that he


109109

experienced abuse from nonpartnersexperienced abuse from nonpartners??

((gg ) )What do we call the probability What do we call the probability calculated in part fcalculated in part f??

((hh ) )Suppose we pick a subject at random. Suppose we pick a subject at random. What is the probability that it is a man or What is the probability that it is a man or someone who experienced abuse from a someone who experienced abuse from a

partnerpartner??

((ii ) )What do we call the method by which you What do we call the method by which you obtained the probability in part hobtained the probability in part h??


110110

Q3.4.3: Q3.4.3: Fernando et al. (A-3) studied drug-Fernando et al. (A-3) studied drug-sharing among injection drug users in the sharing among injection drug users in the South Bronx in New York City. Drug users in South Bronx in New York City. Drug users in New York City use the term “split a bag” or New York City use the term “split a bag” or “get down on a bag” to refer to the practice of “get down on a bag” to refer to the practice of diving a bag of heroin or other injectable diving a bag of heroin or other injectable substances. A common practice includes substances. A common practice includes splitting drugs after they are dissolved in a splitting drugs after they are dissolved in a common cooker, a procedure with considerable common cooker, a procedure with considerable HIV risk. Although this practice is common, HIV risk. Although this practice is common, little is known about the prevalence of such little is known about the prevalence of such practices. The researchers asked injection drug practices. The researchers asked injection drug users in four neighborhoods in the South Bronx users in four neighborhoods in the South Bronx if they everif they ever


111111

““got down on” drugs in bags or shots. got down on” drugs in bags or shots. The results classified by gender and The results classified by gender and splitting practice are given belowsplitting practice are given below::

State theState the

followingfollowing

probabilities inprobabilities in

words and calculatewords and calculate::

((aa ) )Ans: 0.3418Ans: 0.3418

((bb ) )Ans: 0.8746Ans: 0.8746

((cc ) )Ans: 0.6134Ans: 0.6134Text Book : Basic Concepts and Text Book : Basic Concepts and


Gender Split Drugs

Never Split

Drugs

Total

Male349324673

Female 220128348

Total5694521021

)( DrugsSplitMaleP )( DrugsSplitMaleP )( DrugsSplitMaleP

((dd ) )Ans: 0.6592Ans: 0.6592

Q3.4.4: Q3.4.4: Laveist and Nuru-Jeter (A-4) Laveist and Nuru-Jeter (A-4) conducted a study to determine if conducted a study to determine if doctor-patient race concordance was doctor-patient race concordance was associated with greater satisfaction with associated with greater satisfaction with care. Toward that end, they collected a care. Toward that end, they collected a national sample of African-American, national sample of African-American, Caucasian, Hispanic, and Asian-Caucasian, Hispanic, and Asian-American respondents. The following American respondents. The following table classifies the race of the subjects table classifies the race of the subjects as well as the race of their physicianas well as the race of their physician::


113113

)(MaleP


114114

((aa ) )What is the probability that a What is the probability that a randomly selected subject will have an randomly selected subject will have an Asian/Pacific-Islander physician? Asian/Pacific-Islander physician? Ans: 0.1533Ans: 0.1533

Patient Race

Physician’s Race

CaucasianAfrican-American

HispanicAsian-American

Total

White7794364061751796

African-American

14162155196

Hispanic19171282166

Asian/Pacific-Island

687571203417

Other3055564145

Total9107456763892720

((bb ) )What is the probability that an African-American What is the probability that an African-American subject will have an African- American physiciansubject will have an African- American physician??

Ans: 0.2174Ans: 0.2174

((cc ) )What is the probability that a randomly What is the probability that a randomly selected subject in the study will be Asian-selected subject in the study will be Asian-American and have an Asian/Pacific-Islander American and have an Asian/Pacific-Islander physician? physician? Ans: 0.075Ans: 0.075

((dd ) )What is the probability that a subject chosen What is the probability that a subject chosen at random will be Hispanic or have a Hispanic at random will be Hispanic or have a Hispanic physician? physician? Ans: 0.2625Ans: 0.2625

((ee ) )Use the concept of complementary events to Use the concept of complementary events to find the probability that a subject chosen atfind the probability that a subject chosen at


115115

random in the study does not have a random in the study does not have a white physician? white physician? Ans: 0.3397Ans: 0.3397

Q3.4.5Q3.4.5::

If the probability of left-handedness in If the probability of left-handedness in acertain group of people is 0.5, what acertain group of people is 0.5, what is the probability of right-handedness is the probability of right-handedness

(assuming no ambidexterity)(assuming no ambidexterity) ? ?


116116

Q3.4.6Q3.4.6 : :

The probability is 0.6 that a patient The probability is 0.6 that a patient selected at random from the current selected at random from the current residents of a certain hospital will be a residents of a certain hospital will be a male. The probability that the patient male. The probability that the patient will be a male who is in for surgery is will be a male who is in for surgery is 0.2. A patient randomly selected from 0.2. A patient randomly selected from current residents is found to be a current residents is found to be a male; what is the probability that the male; what is the probability that the patient is in the hospital for surgerypatient is in the hospital for surgery??

Ans: 0.3333Ans: 0.3333


117117

Q3.4.7Q3.4.7::

In a certain population of hospital In a certain population of hospital patients the probability is 0.35 that a patients the probability is 0.35 that a randomly selected patient will have randomly selected patient will have heart disease. The probability is 0.86 heart disease. The probability is 0.86 that a patient with heart disease is a that a patient with heart disease is a smoker. What is the probability that a smoker. What is the probability that a patient randomly selected from the patient randomly selected from the population will be a smoker and have population will be a smoker and have heart diseaseheart disease??

Ans: 0.301Ans: 0.301


118118


119119

Baye's Theorem Baye's Theorem Pages 79-83Pages 79-83

In this case if the patient In this case if the patient has to do a blood test in the has to do a blood test in the

laboratory,laboratory,some time the result issome time the result is

PositivePositive(he has the disease) (he has the disease) and if the result is and if the result is negativenegative

(he doesn't has the disease)(he doesn't has the disease)


120120

So, we have the following So, we have the following casescases

The patient has the disease

(D)

The patient doesn't has the disease

( D )

Lab result isNegative

(T)

wrong result

Specificity A symptom

P(T|D)

Lab result is positive

( T )

SensitivityA symptom

P(T|D) wrong result


121121


122122

Definition.1

The sensitivity of the symptomThis is the probability of a positive result given that the subject has the disease. It is denoted by P(T|D)

Definition.2

The specificity of the symptomThis is the probability of negative result given that the subject does not have the disease. It is denoted by P(T|D)

Definition 3Definition 3::

The predictive value positive of the The predictive value positive of the symptomsymptom

This is the probability that the subject has This is the probability that the subject has the disease given that the subject has a the disease given that the subject has a positive screening test resultpositive screening test result..

It is calculated using bayes theorem It is calculated using bayes theorem through the following formulathrough the following formula

Where P(D) is the rate of the diseaseWhere P(D) is the rate of the diseaseText Book : Basic Concepts and Text Book : Basic Concepts and


)()|()()|(

)()|()|(

DPDTPDPDTP

DPDTPTDP

Which is given byWhich is given by

P(D) = 1 – P(D)P(D) = 1 – P(D)

P(T/ D) = 1 - P(T/ D)P(T/ D) = 1 - P(T/ D)

Note that Note that the numerator is equal to the numerator is equal to sensitivity times rate of the disease, sensitivity times rate of the disease, while the denominator is equal to while the denominator is equal to sensitivity times rate of the disease sensitivity times rate of the disease plus 1 minus the specificity times plus 1 minus the specificity times one minus the rate of the diseaseone minus the rate of the disease


124124


125125

Definition.4

The predictive value negative of the symptom

This is the probability that a subject does not have the disease given that the subject has a negative screening test result .It is calculated using Bayes Theorem through the following formula

where,

)()|()()|(

)()|()|(

DPDTPDPDTP

DPDTPTDP

)|(1)|( DTPDTp


126126

Example 3.5.1 page 82

A medical research team wished to evaluate a proposed screening test for Alzheimer’s disease. The test was given to a random sample of 450 patients with Alzheimer’s disease and an independent random sample of 500 patients without symptoms of the disease. The two samples were drawn from populations of subjects who were 65 years or older. The results are as follows.

Test ResultYes (D)No( ) Total

Positive(T)4365441

Negativ( )14495509

Total450500950

T

D


127127

In the context of this examplea)What is a false positive?A false positive is when the test indicates a positive result (T) when the person does not have the disease

b) What is the false negative?A false negative is when a test indicates a negative result ( ) when the person has the disease (D).

c) Compute the sensitivity of the symptom.

d) Compute the specificity of the symptom.

D

T

9689.0450

436)|( DTP

99.0500

495)|( DTP


128128

e) Suppose it is known that the rate of the disease in the general population is 11.3%. What is the predictive value positive of the symptom and the predictive value negative of the symptom The predictive value positive of the symptom is calculated as

The predictive value negative of the symptom is calculated as

996.0.113)(0.0311)(087)(0.99)(0.8

87)(0.99)(0.8

)()|()()|(

)()|()|(

DPDTPDPDTP

DPDTPTDP

925.00.113)-(.01)(1.113)(0.9689)(0

.113)(0.9689)(0

)()|()()|(

)()|()|(

DPDTPDPDTP

DPDTPTDP


129129

ExerciseExercise::

Page 83Page 83 Questions :Questions : 3.5.1, 3.5.23.5.1, 3.5.2 H.W.:H.W.: Page 87 : Q4,Q5,Q7,Q9,Q21Page 87 : Q4,Q5,Q7,Q9,Q21

Q3.5.1; Q3.5.1; A medical research team wishes A medical research team wishes to assess the usefulness of a certain to assess the usefulness of a certain symptom (call it S) in the diagnosis of a symptom (call it S) in the diagnosis of a particular disease. In a random sample particular disease. In a random sample of 775 patients with the disease, 744 of 775 patients with the disease, 744 reported having the symptom. In an reported having the symptom. In an independent random sample of 1380 independent random sample of 1380 subjects without the disease, 21 subjects without the disease, 21 reported that they had the symptomreported that they had the symptom..

((aa ) )In the context of this exercise, what is a In the context of this exercise, what is a false positivefalse positive??

((bb ) )What is a false negativeWhat is a false negative??Text Book : Basic Concepts and Text Book : Basic Concepts and


((cc ) )Compute the sensitivity of the symptomCompute the sensitivity of the symptom..

((dd ) )Compute the specificity of the symptomCompute the specificity of the symptom..

((ee ) )Suppose it is known that the rate of the Suppose it is known that the rate of the diseases in the general population is 0.001. diseases in the general population is 0.001. what is the predictive value positive of the what is the predictive value positive of the

symptomsymptom??

((ff ) )What is the predictive value negative of the What is the predictive value negative of the symptomsymptom??

((gg ) )Find the predictive value positive and Find the predictive value positive and the predictive value negative for the the predictive value negative for the symptom for the following hypothetical symptom for the following hypothetical diseases rates: 0.0001, 0.01 and 0.1diseases rates: 0.0001, 0.01 and 0.1


131131

((hh ) )What do you conclude about the predictive What do you conclude about the predictive value of the symptom on the basis of the value of the symptom on the basis of the results obtained in part gresults obtained in part g??

Q3.5.2Q3.5.2::

Dorsay and Helms (A-6) performed a Dorsay and Helms (A-6) performed a retrospective study of 71 knees scanned retrospective study of 71 knees scanned by MRI. One of the indicators they by MRI. One of the indicators they examined was the absence of the “bow-examined was the absence of the “bow-tie sign” in the MRI as evidence of a tie sign” in the MRI as evidence of a bucket-handle or “bucket-handle type” bucket-handle or “bucket-handle type” tear of the meniscustear of the meniscus . .


132132

In the study, surgery confirmed that 43 In the study, surgery confirmed that 43 of the 71 cases were bucket-handle of the 71 cases were bucket-handle tears. The cases may be cross-tears. The cases may be cross-classified by “bow-tie sign” status classified by “bow-tie sign” status and surgical results as followsand surgical results as follows::


133133

Tear Surgically Confirmed (D)

Tear Surgically Confirmed As

Not Present( )

Total

Positive Test(absent bow-tie

sign( )T)

381048

Negative Test (bow-tie present ()

)

51823

Total432871

D

T

((aa ) )What is the sensitivity of testing to see if the What is the sensitivity of testing to see if the absent bow-tie sign indicates a meniscal tearabsent bow-tie sign indicates a meniscal tear??

Ans: 0.8837Ans: 0.8837

((bb ) )What is the specificity of testing to see if the What is the specificity of testing to see if the absent bow-tie sign indicates a meniscal tearabsent bow-tie sign indicates a meniscal tear??

Ans: 0.6229Ans: 0.6229

((cc ) )What additional information would you need What additional information would you need to determine the predictive value of the testto determine the predictive value of the test??


134134

((dd ) )Suppose it is known that the rate of Suppose it is known that the rate of the disease in the general population the disease in the general population is 0.1, what is the predictive value is 0.1, what is the predictive value positive of the symptom? positive of the symptom? Ans: Ans: 0.206590.20659

((ee ) )What is predictive value negative of What is predictive value negative of the symptom? the symptom? Ans: 0.9797Ans: 0.9797


135135

Chapter 4:Probabilistic features of certain data DistributionsPages 93- 111

Text Book : Basic Concepts and Methodology for the Health Sciences

137

Key words

Probability distribution , random variable ,

Bernolli distribution, Binomail distribution,

Poisson distribution


138

The Random Variable (X):

When the values of a variable )height, weight, or age( can’t be predicted in advance, the variable is called a random variable.

An example is the adult height.

When a child is born, we can’t predict exactly his or her height at maturity.


139

4.2 Probability Distributions for Discrete Random Variables

Definition:

The probability distribution of a discrete random variable is a table, graph, formula, or other device used to specify all possible values of a discrete random variable along with their respective probabilities.


140

The Cumulative Probability Distribution of X, F(x):

It shows the probability that the variable X is less than or equal to a certain value, P)X x(.



Example 4.2.1 page 94Example 4.2.1 page 94::

Number of Number of ProgramsPrograms

frequencfrequencyy

P(X=x)P(X=x)F(x)F(x)==

P(X≤ x)P(X≤ x)

1162620.20880.20880.20880.2088

2247470.15820.15820.36700.3670

3339390.13130.13130.49830.4983

4439390.13130.13130.62960.6296

5558580.19530.19530.82490.8249

6637370.12460.12460.94950.9495

77440.01350.01350.96300.9630

8811110.03700.03701.00001.0000

TotalTotal2972971.00001.0000


142

See figure 4.2.1 page 96

See figure 4.2.2 page 97

Properties of probability distribution of discrete random variable.

1.

2.

3. P(a X b) = P(X b) – P(X a-1)

4. P(X < b) = P(X b-1)

0 ( ) 1P X x ( ) 1P X x


143

Example 4.2.2 page 96: (use table in example 4.2.1)

What is the probability that a randomly selected family will be one who used three assistance programs?


What is the probability that a randomly selected family used either one or two programs?


144


What is the probability that a family picked at random will be one who used two or fewer assistance programs?Example 4.2.5 page 98: (use table in example 4.2.1)

What is the probability that a randomly selected family will be one who used fewer than four programs?Example 4.2.6 page 98: (use table in example 4.2.1)

What is the probability that a randomly selected family used five or more programs?


145


What is the probability that a randomly selected family is one who used between three and five programs, inclusive?


146

4.3 The Binomial Distribution:The binomial distribution is one of the most widely encountered probability distributions in applied statistics. It is derived from a process known as a Bernoulli trial.

Bernoulli trial is :

When a random process or experiment called a trial can result in only one of two mutually exclusive outcomes, such as dead or alive, sick or well, the trial is called a Bernoulli trial.


147

The Bernoulli ProcessA sequence of Bernoulli trials forms a Bernoulli process under the following conditions

1- Each trial results in one of two possible, mutually exclusive, outcomes. One of the possible outcomes is denoted )arbitrarily( as a success, and the other is denoted a failure.

2- The probability of a success, denoted by p, remains constant from trial to trial. The probability of a failure, 1-p, is denoted by q.

3- The trials are independent, that is the outcome of any particular trial is not affected by the outcome of any other trial


148

The probability distribution of the binomial random variable X, the number of successes in n independent trials is:

Where is the number of combinations of n distinct objects taken x of them at a time.

* Note: 0! =1

( ) ( ) , 0,1,2,....,X n Xn

f x P X x p q x nx

n

x

!

!( )!

n n

x n xx

! ( 1)( 2)....(1)x x x x


149

Properties of the binomial distribution

1.

2.

3.The parameters of the binomial distribution are n and p

4.

5.

( ) 0f x ( ) 1f x

( )E X np 2 var( ) (1 )X np p


150

Example 4.3.1 page 100 If we examine all birth records from the North Carolina State Center for Health statistics for year 2001, we find that 85.8 percent of the pregnancies had delivery in week 37 or later )full- term birth(.

If we randomly selected five birth records from this population what is the probability that exactly three of the records will be for full-term births?

Exercise: example 4.3.2 page 104


151

Example 4.3.3 page 104Suppose it is known that in a certain population 10 percent of the population is color blind. If a random sample of 25 people is drawn from this population, find the probability that

a( Five or fewer will be color blind.b( Six or more will be color blindc( Between six and nine inclusive will be color

blind.d( Two, three, or four will be color blind.

Exercise: example 4.3.4 page 106


152

4.4 The Poisson DistributionIf the random variable X is the number of occurrences of some random event in a certain period of time or space )or some volume of matter(.The probability distribution of X is given by:

f )x( =P)X=x( = ,x = 0,1,…..

The symbol e is the constant equal to 2.7183. )Lambda( is called the parameter of the distribution and is the average number of occurrences of the random event in the interval )or volume(

!

x

xe


153

Properties of the Poisson distribution

1.

2.

3.

4.

( ) 0f x

( ) 1f x ( )E X

2 var( )X


154

Example 4.4.1 page 111

In a study of a drug -induced anaphylaxis among patients taking rocuronium bromide as part of their anesthesia, Laake and Rottingen found that the occurrence of anaphylaxis followed a Poisson model with =12 incidents per year in Norway .Find

1- The probability that in the next year, among patients receiving rocuronium, exactly three will experience anaphylaxis?


155

2- The probability that less than two patients receiving rocuronium, in the next year will experience anaphylaxis?3- The probability that more than two patients receiving rocuronium, in the next year will experience anaphylaxis?4- The expected value of patients receiving rocuronium, in the next year who will experience anaphylaxis.5- The variance of patients receiving rocuronium, in the next year who will experience anaphylaxis6- The standard deviation of patients receiving rocuronium, in the next year who will experience anaphylaxis


156

Example 4.4.2 page 111: Refer to example 4.4.1

1-What is the probability that at least three patients in the next year will experience anaphylaxis if rocuronium is administered with anesthesia?2-What is the probability that exactly one patient in the next year will experience anaphylaxis if rocuronium is administered with anesthesia?3-What is the probability that none of the patients in the next year will experience anaphylaxis if rocuronium is administered with anesthesia?


157

4-What is the probability that at most two patients in the next year will experience anaphylaxis if rocuronium is administered with anesthesia?

Exercises: examples 4.4.3, 4.4.4 and 4.4.5 pages111-113

Exercises: Questions 4.3.4 ,4.3.5, 4.3.7 ,4.4.1,4.4.5

Excercices:Q4.3.4: Page 111

The same survey data base cited shows that 32 percent of U.S adults indicated that they have been tested for HIV at some points in their life .Consider a simple random sample of 15 adults selected at that time .Find the probability

that the number of adults who have been

tested for HIV in the sample would be:Text Book : Basic Concepts and Methodology for the

Health Sciences 158

Hint:


159

( ) ( ) , 0,1,2,....,X n Xn

f x P X x p q x nx

(a )Three )Ans. 0.1457(

(b )Less than two )Ans. 0.02477(

(c ) At most one )Ans. 0.02477(

(d )At least three )Ans. 0.9038(

(e )between three and five ,inclusive.


160

Q4.3.5

• refer to Q4.3.4 , find the mean and the variance?

•(Answer: mean = 4.8,

• variance =3.264 )


161

Q 4.4.3 : If the mean number of serious accidents per year in a large factory is five ,find the probability that the current year there will be:

Hint: f)x(=

(a )Exactly seven accidents )Ans. 0.1044(

(b )Ten or more accidents )ans. 0.0318(

(c )No accident )Ans. 0.0067(

(d)fewer than five accidents . )ans. 0.4405(


Health Sciences 162

!x

e x

Q4.4.4

Find mean and variance and standard

deviation for Q 4.4.3


163

4.5 Continuous 4.5 Continuous Probability Probability DistributionDistribution

Pages 114 Pages 114 –– 127 127

4.5 Continuous 4.5 Continuous Probability Probability DistributionDistribution

Pages 114 Pages 114 –– 127 127


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165

• Key words: Continuous random variable,

normal distribution , standard normal distribution , T-distribution


Health Sciences

166

• Now consider distributions of continuous random variables.


Health Sciences

167

1- Area under the curve = 1.2- P(X = a) = 0 , where a is a

constant.3- Area between two points a , b =

P(a<x<b) .

Properties of continuous probability Distributions:


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168

4.6 The normal distribution:

• It is one of the most important probability distributions in statistics.

• The normal density is given by• , - ∞ < x < ∞, - ∞ < µ < ∞, σ

> 0

• π, e : constants• µ: population mean.• σ : Population standard deviation.

2

2

2

)(

2

1)(

x

exf


Health Sciences

169

Characteristics of the normal distribution: Page 111

• The following are some important characteristics of the normal distribution:

1- It is symmetrical about its mean, µ.2- The mean, the median, and the mode

are all equal. 3- The total area under the curve above

the x-axis is one. 4-The normal distribution is completely

determined by the parameters µ and σ.


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170

5- The normal distributiondepends on the twoparameters and . determines the location of the curve.(As seen in figure 4.6.3) ,

But, determines the scale of the curve, i.e. the degree of flatness or peaked ness of the curve.(as seen in figure 4.6.4)

11 22 33

11 < < 22 < < 33

11

22

33

11 < < 22 < < 33


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171

The Standard normal distribution:

• Is a special case of normal distribution with mean equal 0 and a standard deviation of 1.

• The equation for the standard normal distribution is written as

• , - ∞ < z < ∞2

2

2

1)(

z

ezf


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172

Characteristics of the standard normal

distribution

1 -It is symmetrical about 0.

2 -The total area under the curve above the x-axis is one.

3 -We can use table (D) to find the probabilities and areas.


Health Sciences

173

“How to use tables of Z”Note that The cumulative probabilities P(Z z) are given intables for -3.49 < z < 3.49. Thus, P (-3.49 < Z < 3.49) 1.For standard normal distribution, P (Z > 0) = P (Z < 0) = 0.5

Example 4.6.1:If Z is a standard normal distribution, then1) P( Z < 2) = 0.9772is the area to the left to 2 and it equals 0.9772.

2


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174

Example 4.6.2:P(-2.55 < Z < 2.55) is the area between

-2.55 and 2.55, Then it equals

P(-2.55 < Z < 2.55) =0.9946 – 0.0054

= 0.9892.

Example 4.6.2: P(-2.74 < Z < 1.53) is the area between

-2.74 and 1.53.

P(-2.74 < Z < 1.53) =0.9370 – 0.0031

= 0.9339.

-2.74 1.53

-2.55 2.550


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175

Example 4.6.3:P(Z > 2.71) is the area to the right to 2.71.

So,

P(Z > 2.71) =1 – 0.9966 = 0.0034.

Example :

P(Z = 0.84) is the area at z = 0.84.

So,

P(Z = 0.84) = 0

0.84

2.71

Exercise •Given Standard normal distribution by

using the tables:

•4.6.1 :The area to the left of Z=2•4.6.2:

The area under the curve Z =0, Z= 1.43

4.6.3 : P(Z ≥ 0.55)=

4.6.5 : P(Z < - 2.35)=


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176

•4.6.7:

P( -1.95 < Z < 1.95 )=

4.6.10:

P( Z = 1.22)=


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177

Given the following probabilities, find z1

4.6.11

P(Z ≤ z1) = 0.0055 (z1=-2.54)4.6.12

P(-2.67≤ Z ≤ z1) = 0.9718 (z1=1.97)4.6.13

P(Z > z1) = 0.0384 (z1=1.77)4.6.11 :

P(z1 < Z ≤ 2.98) = 0.1117 (z1=1.21)


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178


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179

How to transform normal distribution (X) to standard normal distribution (Z)?

• This is done by the following formula:

• Example:• If X is normal with µ = 3, σ = 2. Find

the value of standard normal Z, If X= 6?

• Answer:

x

z

5.12

36

x

z


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180

4.7 Normal Distribution Applications

The normal distribution can be used to model the distribution of many variables that are of interest. This allow us to answer probability questions about these random variables.

Example 4.7.1:The ‘Uptime ’is a custom-made light weight battery-operated

activity monitor that records the amount of time an individual

spend the upright position. In a study of children ages 8 to 15

years. The researchers found that the amount of time children

spend in the upright position followed a normal distribution with

Mean of 5.4 hours and standard deviation of 1.3.Find


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181

If a child selected at random ,then1-The probability that the child spend less than 3 hours in the upright position 24-hour period

P) X < 3( = P) < ( = P)Z < -1.85( = 0.0322

-------------------------------------------------------------------------2-The probability that the child spend more than 5 hours in the upright position 24-hour period

P) X > 5( = P) > ( = P)Z > -0.31(

= 1- P)Z < - 0.31( = 1- 0.3520= 0.648-----------------------------------------------------------------------

3-The probability that the child spend exactly 6.2 hours in the upright position 24-hour period

P) X = 6.2( = 0

X

3.1

4.53

X

3.1

4.55


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182

4-The probability that the child spend from 4.5 to 7.3 hours in the upright position 24-hour period

P) 4.5 < X < 7.3( = P) < < (

= P) -0.69 < Z < 1.46 ( = P)Z<1.46( – P)Z< -0.69(

= 0.9279 – 0.2451 = 0.6828

• Hw…EX. 4.7.2 – 4.7.3

X

3.1

4.55.4 3.1

4.53.7


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183

• Exercise:

• Questions : 4.7.1, 4.7.2• H.W : 4.7.3, 4.7.4, 4.7.6

Exercises•Q4.7.1 : For another subject (29-

years old male) in the study by Diskin, aceton level were normally distributed with mean of 870 and standard deviation of 211 ppb. Find the probability that in a given day the subjects acetone level is:

•(a )between 600 and 1000 ppb•(b )over 900 ppb

•(c ) under 500 ppb )d( At 700 ppb


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184

• Q4.7.2: In the study of fingerprints an important quantitative characteristic is the total ridge count for the 10 fingers of an individual . Suppose that the total ridge counts of individuals in a certain population are approximately normally distributed with mean of 140 and a standard deviation of 50 .Find the probability that an individual picked at random from this population will have ridge count of:

•( a )200 or more

• (Answer :0.0985)


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185

•(b )less than 200 (Answer :0.8849)

•(c )between 100 and 200•(Answer :0.6982)

•(d )between 200 and 250•(Answer :0.0934)


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186


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187

6.3 The T Distribution:

)167-173(

1- It has mean of zero.2- It is symmetric about the mean.3- It ranges from - to .

0


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188

4- compared to the normal distribution, the t distribution is less peaked in the center and has higher tails.

5- It depends on the degrees of freedom (n-1).

6- The t distribution approaches the standard normal distribution as (n-1) approaches .


Health Sciences

189

Examplest )7, 0.975( = 2.3646

------------------------------t )24, 0.995( = 2.7696

--------------------------

If P )T)18( > t( = 0.975,

then t = -2.1009

-------------------------

If P )T)22( < t( = 0.99,

then t = 2.508

0.005

t (24, 0.995)

0.995

t (7, 0.975)

0.0250.975

t

0.9750.025

0.990.01

t

•Find:

t 0.95,10 = 1.8125

---------------------------------

t 0.975,18 = 2.1009

---------------------------------

t 0.01,20 = - 2.528

---------------------------------

t 0.10,29 = - 1.311

---------------------------------


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190

Chapter 6Using sample data to make estimates about population parameters (P162-172)


Health Sciences 192

Key words:

Point estimate, interval estimate, estimator,

Confident level ,α , Confident interval for mean μ, Confident interval for two means,

Confident interval for population proportion P,

Confident interval for two proportions


Health Sciences 193

6.1 Introduction: Statistical inference is the procedure by which

we reach to a conclusion about a population on the basis of the information contained in a sample drawn from that population.

Suppose that: an administrator of a large hospital is

interested in the mean age of patients admitted to his hospital during a given year.

1. It will be too expensive to go through the records of all patients admitted during that particular year.

2. He consequently elects to examine a sample of the records from which he can compute an estimate of the mean age of patients admitted to his that year.


Health Sciences 194

• To any parameter, we can compute two types of estimate: a point estimate and an interval estimate.

A point estimate is a single numerical value used to estimate the corresponding population parameter.

An interval estimate consists of two numerical values defining a range of values that, with a specified degree of confidence, we feel includes the parameter being estimated.

The Estimate and The Estimator: The estimate is a single computed value, but the

estimator is the rule that tell us how to compute this value, or estimate.

For example, is an estimator of the population mean,. The

single numerical value that results from evaluating this formula is called an estimate of the parameter .

i

ixx


Health Sciences 195

6.2 Confidence Interval for a Population Mean: (C.I) Suppose researchers wish to estimate the mean

of some normally distributed population. They draw a random sample of size n from the

population and compute , which they use as a point estimate of .

Because random sampling involves chance, then can’t be expected to be equal to .

The value of may be greater than or less than .

It would be much more meaningful to estimate by an interval.

x

x


Health Sciences 196

The 1- percent confidence interval (C.I.) for :

We want to find two values L and U between which lies with high probability, i.e.

P( L ≤ ≤ U ) = 1-


Health Sciences 197

For example: When, = 0.01, then 1- = = 0.05, then 1- = = 0.05, then 1- =


Health Sciences 198

We have the following cases

a) When the population is normal

1) When the variance is known and the sample size is large or small, the C.I. has the form:

P( - Z (1- /2) /n < < + Z (1- /2) /n) = 1-

2) When variance is unknown, and the sample size is small, the C.I. has the form:

P( - t (1- /2),n-1 s/n < < + t (1- /2),n-1 s/n) = 1-

x x

xx


Health Sciences 199

b) When the population is not normal and n large (n>30)

1) When the variance is known the C.I. has the form:

P( - Z (1- /2) /n < < + Z (1- /2) /n) = 1-

2) When variance is unknown, the C.I. has the form:

P( - Z (1- /2) s/n < < + Z (1- /2) s/n) = 1-

x x

x x


Health Sciences 200

Case 1: population is normal or approximately normal

σ2 is known σ2 is unknown( n large or small) n large n small

Case2: If population is not normally distributed and n is large

i)If σ2 is known ii) If σ2 is unknown

nZx

2

1

n

SZx

21

n

Stx

n 1,2

1

nZx

2

1

n

SZx

21


Health Sciences 201

Example 6.2.1 Page 167: Suppose a researcher , interested in obtaining

an estimate of the average level of some enzyme in a certain human population, takes a sample of 10 individuals, determines the level of the enzyme in each, and computes a sample mean of approximately

Suppose further it is known that the variable of interest is approximately normally distributed with a variance of 45. We wish to estimate . (=0.05)

22x


Health Sciences 202

Solution: 1- =0.95→ =0.05→ /2=0.025, variance = σ2 = 45 → σ= 45,n=10 95%confidence interval for is given by: P( - Z (1- /2) /n < < + Z (1- /2) /n) = 1- Z (1- /2) = Z 0.975 = 1.96 (refer to table D) Z 0.975(/n) =1.96 ( 45 / 10)=4.1578 22 ± 1.96 ( 45 / 10) → (22-4.1578, 22+4.1578) → (17.84, 26.16) Exercise example 6.2.2 page 169

22x

x x


Health Sciences 203

ExampleThe activity values of a certain enzyme measured

in normal gastric tissue of 35 patients with gastric carcinoma has a mean of 0.718 and a standard deviation of 0.511.We want to construct a 90 % confidence interval for the population mean.

Solution:

Note that the population is not normal, n=35 (n>30) n is large and is

unknown ,s=0.511 1- =0.90→ =0.1 → /2=0.05→ 1-/2=0.95,


Health Sciences 204

Then 90% confident interval for is given by:

P( - Z (1- /2) s/n < < + Z (1- /2) s/n) = 1-

Z (1- /2) = Z0.95 = 1.645 (refer to table D) Z 0.95(s/n) =1.645 (0.511/ 35)=0.1421

0.718 ± 1.645 (0.511) / 35→ (0.718-0.1421, 0.718+0.1421) → (0.576,0.860). Exercise example 6.2.3 page 164:

xx


Health Sciences 205

Example6.3.1 Page 174:

Suppose a researcher , studied the effectiveness of early weight bearing and ankle therapies following acute repair of a ruptured Achilles tendon. One of the variables they measured following treatment the muscle strength. In 19 subjects, the mean of the strength was 250.8 with standard deviation of 130.9

we assume that the sample was taken from is approximately normally distributed population. Calculate 95% confident interval for the mean of the strength ?


Health Sciences 206

Solution: 1- =0.95→ =0.05→ /2=0.025, Standard deviation= S = 130.9 ,n=19 95%confidence interval for is given by: P( - t (1- /2),n-1 s/n < < + t (1- /2),n-1 s/n) = 1- t (1- /2),n-1 = t 0.975,18 = 2.1009 (refer to table E) t 0.975,18(s/n) =2.1009 (130.9 / 19)=63.1 250.8 ± 2.1009 (130.9 / 19) → (250.8- 63.1 , 22+63.1) → (187.7, 313.9) Exercise 6.2.1 ,6.2.2 6.3.2 page 171

8.250x

x x

ExerciseQ6.2.1We wish to estimate the average number of heartbeats per minute for a certain population using a 95% confidence interval . The average number of heartbeats per minute for a sample of 49 subjects was found to be 90 . Assume that these 49 patients is normally distributed with standard deviation of 10.

(answer :( 87.2 , 92.8)Text Book : Basic Concepts and Methodology for the

Health Sciences 207

Q6.2.2:We wish to estimate the mean serum indirect bilirubin level of 4 -day-old infants using a 95% confidence interval . The mean for a sample of 16 infants was found to be 5.98 mg/100 cc .Assume that bilirubin level is approximately normally distributed with variance 12.25 mg/100 cc.

(answer :( 4.5406 , 7.4194)Text Book : Basic Concepts

and Methodology for the Health Sciences 208

Additional Exercise:In a study of the effect of early Alzheimer’s disease on non declarative memory .For a sample of 8 subject was found that mean 8.5 with standard deviation 3. Find 99%

confidence interval for mean?


Health Sciences 209


Health Sciences 210

6.3 Confidence Interval for the difference between two Population Means: (C.I)

If we draw two samples from two independent population

and we want to get the confident interval for thedifference between two population means , then

we havethe following cases :

a) When the population is normal1) When the variance is known and the sample

sizes is large or small, the C.I. has the form: 2

22

1

21

21

21212

22

1

21

21

21 )()(nn

Zxxnn

Zxx


Health Sciences 211

2) When variances are unknown but equal, and the sample size is small, the C.I. has the form:

2

)1()1(

11)(

11)(

21

222

2112

21)2(,

21

212121

)2(,2

121

2121

nn

SnSnS

where

nnStxx

nnStxx

p

pnn

pnn


Health Sciences 212

Example 6.4.1 P174:The researcher team interested in the difference between serum

uricand acid level in a patient with and without Down’s syndrome .In alarge hospital for the treatment of the mentally retarded, a sample

of 12 individual with Down’s Syndrome yielded a mean of mg/100 ml. In a general hospital a sample of 15 normal individual

ofthe same age and sex were found to have a mean value of If it is reasonable to assume that the two population of values arenormally distributed with variances equal to 1 and 1.5,find the

95%C.I for μ1 - μ2

Solution:

1- =0.95→ =0.05→ /2=0.025 → Z (1- /2) = Z0.975 = 1.96

1.1±1.96)0.4282 = (1.1± 0.84 ) = 0.26 , 1.94(

5.41 x

4.32 x

2

22

1

21

21

21 )(nn

Zxx

15

5.1

12

196.1)4.35.4(


Health Sciences 213

Example 6.4.1 P178:The purpose of the study was to determine the effectiveness of anintegrated outpatient dual-diagnosis treatment program formentally ill subject. The authors were addressing the problem of

substance abuseissues among people with sever mental disorder. A retrospective chart

review wascarried out on 50 patient ,the recherché was interested in the number of

inpatienttreatment days for physics disorder during a year following the end of the

program.Among 18 patient with schizophrenia, The mean number of treatment

days was 4.7with standard deviation of 9.3. For 10 subject with bipolar disorder, the

meannumber of treatment days was 8.8 with standard deviation of 11.5. We

wish toconstruct 99% C.I for the difference between the means of the

populationsRepresented by the two samples


Health Sciences 214

Solution: 1-α =0.99 → α = 0.01 → α/2 =0.005 → 1- α/2 = 0.995

n2 – 2 = 18 + 10 -2 = 26+ n1

t (1- /2),(n1+n2-2) = t0.995,26 = 2.7787, then 99% C.I for μ1 – μ2

where

then

(4.7-8.8)± 2.7787 √102.33 √(1/18)+(1/10)- 4.1 ± 11.086 =( - 15.186 , 6.986)Exercises: 6.4.2 , 6.4.6, 6.4.7, 6.4.8 Page

180

21)2(,

21

21

11)(

21 nnStxx p

nn

33.10221018

)5.119()3.917(

2

)1()1( 22

21

222

2112

xx

nn

SnSnS p


Health Sciences 215

6.5 Confidence Interval for a Population proportion (P):

A sample is drawn from the population of interest ,then compute the sample proportion such as

This sample proportion is used as the point estimator of the population proportion . A confident interval is obtained by the following formula

P̂

n

ap

sample in theelement of no. Total

isticcharachtar some with sample in theelement of no.ˆ

n

PPZP

)ˆ1(ˆˆ

21


Health Sciences 216

Example 6.5.1The Pew internet life project reported in 2003 that

18%of internet users have used the internet to search forinformation regarding experimental treatments ormedicine . The sample consist of 1220 adult internetusers, and information was collected from telephoneinterview. We wish to construct 98% C.I for theproportion of internet users who have search forinformation about experimental treatments or

medicine


Health Sciences 217

Solution:

1-α =0.98 → α = 0.02 → α/2 =0.01 → 1- α/2 = 0.99Z 1- α/2 = Z 0.99 =2.33 , n=1220,

The 98% C. I is

0.18 ± 0.0256 = ( 0.1544 , 0.2056 )

Exercises: 6.5.1 , 6.5.3 Page 187

18.0100

18ˆ p

1220

)18.01(18.033.218.0

)ˆ1(ˆˆ

21

n

PPZP

Exercise:Q6.5.1:Luna studied patients who were mechanically ventilated in the intensive care unit of six hospitals in buenos Aires ,Argentina. The researchers found that of 472 mechanically of ventilated patients ,63 had clinical evidence VAP. Construct 95% confidence interval for the proportion of all mechanically ventilated patients at these hospitals who may expected to develop VAP.


Health Sciences 218


Health Sciences 219

6.6 Confidence Interval for the difference between two Population proportions:

Two samples is drawn from two independent population

of interest ,then compute the sample proportion for each

sample for the characteristic of interest. An unbiased

point estimator for the difference between two population

proportionsA 100(1-α)% confident interval for P1 - P2 is given by

21ˆˆ PP

2

22

1

11

21

21

)ˆ1(ˆ)ˆ1(ˆ)ˆˆ(

n

PP

n

PPZPP


Health Sciences 220

Example 6.6.1Connor investigated gender differences in

proactive andreactive aggression in a sample of 323 adults (68

femaleand 255 males ). In the sample ,31 of the female

and 53of the males were using internet in the internet

café. Wewish to construct 99 % confident interval for thedifference between the proportions of adults go tointernet café in the two sampled population .


Health Sciences 221

Solution: 1-α =0.99 → α = 0.01 → α/2 =0.005 → 1- α/2 = 0.995Z 1- α/2 = Z 0.995 =2.58 , nF=68, nM=255,

The 99% C. I is

0.2481 ± 2.58(0.0655) = ( 0.07914 , 0.4171 )

2078.0255

53ˆ,4559.0

68

31ˆ

M

M

MF

F

F n

ap

n

ap

255

)2078.01(2078.0

68

)4559.01(4559.058.2)2078.04559.0(

M

MM

F

FFMF n

PP

n

PPZPP

)ˆ1(ˆ)ˆ1(ˆ)ˆˆ(

21


Health Sciences 222

Exercises: Questions : 6.2.1, 6.2.2,6.2.5 ,6.3.2,6.3.5, 6.4.2 6.5.3 ,6.5.4,6.6.1

Chapter 7Chapter 7Using sample statistics to Using sample statistics to

Test Hypotheses Test Hypotheses about population parametersabout population parameters

PagesPages 215-233 215-233


224224

Key words :Key words :

Null hypothesis HNull hypothesis H0, 0, Alternative hypothesis HAlternative hypothesis HAA , testing , testing hypothesis , test statistic , P-valuehypothesis , test statistic , P-value


225225

Hypothesis TestingHypothesis Testing

One type of statistical inference, estimation, One type of statistical inference, estimation, was discussed in Chapter 6 . was discussed in Chapter 6 .

The other type ,hypothesis testing ,is discussed The other type ,hypothesis testing ,is discussed in this chapter.in this chapter.


226226

Definition of a hypothesisDefinition of a hypothesis

It is a statement about one or more populations . It is a statement about one or more populations .

It is usually concerned with the parameters of It is usually concerned with the parameters of the population. e.g. the hospital administrator the population. e.g. the hospital administrator may want to test the hypothesis that the average may want to test the hypothesis that the average length of stay of patients admitted to the length of stay of patients admitted to the hospital is 5 days hospital is 5 days


227227

Definition of Statistical hypothesesDefinition of Statistical hypotheses

They are hypotheses that are stated in such a way that They are hypotheses that are stated in such a way that they may be evaluated by appropriate statistical they may be evaluated by appropriate statistical techniques. techniques.

There are two hypotheses involved in hypothesis There are two hypotheses involved in hypothesis testing testing

Null hypothesisNull hypothesis H H00: It is the hypothesis to be tested .: It is the hypothesis to be tested . Alternative hypothesisAlternative hypothesis H HAA : It is a statement of what : It is a statement of what

we believe is true if our sample data cause us to reject we believe is true if our sample data cause us to reject the null hypothesisthe null hypothesis


228228

7.27.2 Testing a hypothesis about the Testing a hypothesis about the mean of a populationmean of a population::

We have the following steps:We have the following steps:1.1.DataData:: determine variable, sample size (n), sample determine variable, sample size (n), sample

mean( ) , population standard deviation or sample mean( ) , population standard deviation or sample standard deviation (s) if is unknown standard deviation (s) if is unknown

2. 2. Assumptions :Assumptions : We have two cases: We have two cases: Case1:Case1: Population is normally or approximately Population is normally or approximately

normally distributed with known or unknown normally distributed with known or unknown variance (sample size n may be small or large), variance (sample size n may be small or large),

Case 2:Case 2: Population is not normal with known or Population is not normal with known or unknown variance (n is large i.e. n≥30).unknown variance (n is large i.e. n≥30).

x


229229

3.Hypotheses:3.Hypotheses: we have three caseswe have three cases Case ICase I : : H H00: : μμ==μμ00

HHAA: : μ μμ μ00

e.g. we want to test that the population mean is e.g. we want to test that the population mean is different than 50different than 50

Case IICase II : : H H00: : μ μ = = μμ00

HHAA: : μμ > > μμ00

e.g. we want to test that the population mean is e.g. we want to test that the population mean is greater than 50greater than 50

Case IIICase III : : H H0:0: μ = μ μ = μ00

HHAA: : μμ< < μμ00

e.g. we want to test that the population mean is lesse.g. we want to test that the population mean is less than 50than 50


230230

4.Test Statistic4.Test Statistic:: Case 1:Case 1: population is normalpopulation is normal or or approximately approximately

normalnormal

σσ22 is known σ is known σ22 is unknown is unknown

( n large or small)( n large or small)

n large n smalln large n small

Case2:Case2: If population is If population is not normallynot normally distributed and distributed and n is n is largelarge

i)If σi)If σ22 is known ii) If σ is known ii) If σ22 is unknown is unknown

n

XZ

o-

ns

XZ o-

ns

XT o-

ns

XZ o-

n

XZ

o-


231231

5.Decision Rule:5.Decision Rule:

i) i) If HIf HAA: μ μ: μ μ00

Reject H Reject H 00 if Z >Z if Z >Z1-α/2 1-α/2 or Z< - Zor Z< - Z1-α/21-α/2

(when use Z - test) (when use Z - test)

OrOr Reject H Reject H 00 if T >t if T >t1-α/2,n-1 1-α/2,n-1 or T< - tor T< - t1-α/2,n-11-α/2,n-1

((when use T- testwhen use T- test ) ) ____________________________________________________ ii) If Hii) If HAA: μ> μ: μ> μ00 Reject HReject H00 if Z>Z if Z>Z1-α1-α (when use Z - test) (when use Z - test)

OrOr Reject H Reject H00 if T>t if T>t1-α,n-11-α,n-1 (when use T - test)(when use T - test)


232232

iii) If Hiii) If HAA: μ< μ: μ< μ00

Reject HReject H00 if Z< - Z if Z< - Z1-1-α α (when use Z - test) (when use Z - test) OrOr

Reject HReject H00 if T<- t if T<- t1-1-α,n-1 α,n-1 (when use T - test)(when use T - test)

NoteNote::

ZZ1-α/21-α/2 , Z , Z1-α1-α , Z , Zαα are tabulated values obtained are tabulated values obtained from table Dfrom table D

tt1-α/21-α/2 , t , t1-α1-α , t , tαα are tabulated values obtained from are tabulated values obtained from table E with (n-1) degree of freedom (df)table E with (n-1) degree of freedom (df)


233233

6.Decision :6.Decision : If we reject HIf we reject H00, we can conclude that H, we can conclude that HAA is is

true.true. If ,however ,we do not reject HIf ,however ,we do not reject H00, we may , we may

conclude that Hconclude that H00 is true. is true.


234234

An Alternative Decision Rule using theAn Alternative Decision Rule using the p - value Definition p - value Definition

The The p-valuep-value is defined as the smallest value of is defined as the smallest value of α for which the null hypothesis can be α for which the null hypothesis can be rejected.rejected.

If the p-value is less than or equal to α ,we If the p-value is less than or equal to α ,we reject the null hypothesisreject the null hypothesis (p ≤ (p ≤ αα))

If the p-value is greater than α ,we If the p-value is greater than α ,we do not do not reject the null hypothesis reject the null hypothesis (p > (p > αα))


235235

Example 7.2.1 Page 223Example 7.2.1 Page 223

Researchers are interested in the mean age of a Researchers are interested in the mean age of a certaincertain populationpopulation..

A random sample of 10 individuals drawn from the A random sample of 10 individuals drawn from the population of interest has a mean of 27. population of interest has a mean of 27.

Assuming that the population is approximately Assuming that the population is approximately normally distributed with variance 20,can we normally distributed with variance 20,can we conclude that the mean is different from 30 years ? conclude that the mean is different from 30 years ? (α=0.05) .(α=0.05) .

If the p - value is 0.0340 how can we use it in making If the p - value is 0.0340 how can we use it in making a decision? a decision?


236236

SolutionSolution

1-1-Data:Data: variable is age, n=10, =27 ,σ variable is age, n=10, =27 ,σ22=20,α=0.05=20,α=0.05

2-2-Assumptions:Assumptions: the population is approximately the population is approximately normally distributed with variance 20 normally distributed with variance 20

3-Hypotheses:3-Hypotheses: HH00 : μ=30 : μ=30 HHAA: μ 30: μ 30

x


237237

4-Test Statistic:4-Test Statistic: Z Z = -2.12 = -2.12

5.Decision Rule5.Decision Rule The alternative hypothesis isThe alternative hypothesis is

HHAA: μ ≠ 30: μ ≠ 30

Hence we reject HHence we reject H00 if Z > Z if Z > Z1-0.0251-0.025= Z= Z0.9750.975

or Z< - Zor Z< - Z1-0.0251-0.025 = - Z = - Z0.9750.975

ZZ0.9750.975=1.96(from table D)=1.96(from table D)


238238

6.Decision:6.Decision:

We reject HWe reject H00 ,since -2.12 is in the rejection ,since -2.12 is in the rejection

region .region .

We can conclude that μ is not equal to 30We can conclude that μ is not equal to 30

Using the p value ,we note that p-value Using the p value ,we note that p-value =0.0340< 0.05,therefore we reject H0 =0.0340< 0.05,therefore we reject H0


239239

Example7.2.2 page227Example7.2.2 page227

Referring to example 7.2.1.Suppose that the Referring to example 7.2.1.Suppose that the researchers have asked: Can we conclude researchers have asked: Can we conclude that μ<30.that μ<30.

1.Data.1.Data.see previous examplesee previous example

2. Assumptions .2. Assumptions .see previous examplesee previous example

3.Hypotheses:3.Hypotheses: HH00 μ =30 μ =30

HH ِِAA: μ < 30: μ < 30


240240

4.Test Statistic4.Test Statistic : :

= = = -2.12 = -2.12

5. 5. DecisionDecision RuleRule: : Reject HReject H00 if Z< - Z if Z< - Z 1-1-αα, where , where

- Z - Z 1-1-α α = -1.645. (from table D) = -1.645. (from table D)

6. 6. DecisionDecision: : Reject HReject H00 ,thus we can conclude that the ,thus we can conclude that the population mean is smaller than 30. population mean is smaller than 30.

n

XZ

o-

10

20

3027


241241

Example7.2.4 page232Example7.2.4 page232

Among 157 African-American men ,the mean Among 157 African-American men ,the mean systolic blood pressure was 146 mm Hg with a systolic blood pressure was 146 mm Hg with a standard deviation of 27. We wish to know if standard deviation of 27. We wish to know if on the basis of these data, we may conclude on the basis of these data, we may conclude that the mean systolic blood pressure for a that the mean systolic blood pressure for a population of African-American is greater than population of African-American is greater than 140. Use α=0.01.140. Use α=0.01.


242242

SolutionSolution1. 1. Data:Data: Variable is systolic blood pressure, Variable is systolic blood pressure,

n=157 , =146, s=27, α=0.01.n=157 , =146, s=27, α=0.01.

2. 2. Assumption:Assumption: population is not normal, σ population is not normal, σ22 is is unknownunknown

3. 3. Hypotheses:Hypotheses: HH00 :μ=140 :μ=140

HHAA: μ>140 : μ>140

4.Test Statistic:4.Test Statistic: = = = 2.78= = = 2.78

n

sX

Z o-

157

27140146

1548.2

6


243243

5. Decision Rule:5. Decision Rule:

we reject Hwe reject H00 if Z>Z if Z>Z1-α1-α

= Z= Z0.990.99= 2.33 = 2.33

(from table D)(from table D)

6. 6. Decision:Decision: We reject H We reject H00. .

Hence we may conclude that the mean systolic Hence we may conclude that the mean systolic blood pressure for a population of African-blood pressure for a population of African-American is greater than 140.American is greater than 140.

ExercisesExercisesQ7.2.1Q7.2.1::

Escobar performed a study to validate a translated version Escobar performed a study to validate a translated version of the Western Ontario and McMaster University index of the Western Ontario and McMaster University index (WOMAC) questionnaire used with spanish-speaking (WOMAC) questionnaire used with spanish-speaking patient s with hip or knee osteoarthritis . For the 76 patient s with hip or knee osteoarthritis . For the 76 women classified with sever hip pain. The WOMAC women classified with sever hip pain. The WOMAC mean function score was 70.7 with standard deviation mean function score was 70.7 with standard deviation of 14.6 , we wish to know if we may conclude that the of 14.6 , we wish to know if we may conclude that the mean function score for a population of similar women mean function score for a population of similar women subjects with sever hip pain is less than 75 . Let subjects with sever hip pain is less than 75 . Let αα =0.01 =0.01


244244

SolutionSolution : : 11..DataData: :

22 . .AssumptionAssumption: :

33 . .HypothesisHypothesis: :

44..Test statisticTest statistic: :


245245

55..Decision RuleDecision Rule

66 . .DecisionDecision : :


246246


The purpose of a study by Luglie was to investigate the oral The purpose of a study by Luglie was to investigate the oral status of a group of patients diagnosed with thalassemia status of a group of patients diagnosed with thalassemia major (TM) . One of the outcome measure s was the major (TM) . One of the outcome measure s was the decayed , missing, filled teeth index (DMFT) . In a decayed , missing, filled teeth index (DMFT) . In a sample of 18 patients ,the mean DMFT index value was sample of 18 patients ,the mean DMFT index value was 10.3 with standard deviation of 7.3 . Is this sufficient 10.3 with standard deviation of 7.3 . Is this sufficient evidence to allow us to conclude that the mean DMFT evidence to allow us to conclude that the mean DMFT index is greater than 9 in a population of similar index is greater than 9 in a population of similar subjects? Let subjects? Let αα =0.1 =0.1


247247






248248




249249

For For Q7.2.3Q7.2.3::

Take the p- value = 0.22 , Use the P-value to Take the p- value = 0.22 , Use the P-value to make your decisionmake your decision?? ??


250250


251251

7.37.3 Hypothesis Testing :The Difference Hypothesis Testing :The Difference between two population meanbetween two population mean ::

We have the following steps:We have the following steps:

1.1.DataData:: determine variable, sample size (n), sample means, determine variable, sample size (n), sample means, population standard deviation or samples standard population standard deviation or samples standard deviation (s) if is unknown for two population.deviation (s) if is unknown for two population.

2. 2. Assumptions :Assumptions : We have two cases: We have two cases: Case1:Case1: Population is normally or approximately normally Population is normally or approximately normally

distributed with known or unknown variance (sample size distributed with known or unknown variance (sample size n may be small or large), n may be small or large),

Case 2:Case 2: Population is not normal with known variances (n Population is not normal with known variances (n is large i.e. n≥30).is large i.e. n≥30).


252252

3.Hypotheses:3.Hypotheses: we have three caseswe have three cases Case ICase I : : H H00: : μ μ 11 == μ μ2 → 2 → μ μ 11 - - μμ22 = 0= 0

HHAA: : μ μ 1 1 ≠ ≠ μ μ 2 2 → → μ μ 1 1 -- μ μ 2 2 ≠ 0≠ 0 e.g. we want to test that the mean for first e.g. we want to test that the mean for first

population is different from second population population is different from second population mean.mean.

Case IICase II : : H H00: : μ μ 11 == μ μ2 → 2 → μ μ 11 - - μμ22 = 0= 0

HHAA: : μ μ 1 1 >> μ μ 2 2 →→ μ μ 1 1 -- μ μ 2 2 >> 0 0 e.g. we want to test that the mean for first e.g. we want to test that the mean for first

population is greater than second population mean.population is greater than second population mean. Case IIICase III : : HH00: : μ μ 11 == μ μ2 → 2 → μ μ 11 - - μμ22 = 0= 0

HHAA: : μ μ 1 1 << μ μ 2 2 →→ μ μ 1 1 -- μ μ 2 2 < 0< 0 e.g. we want to test that the mean for first e.g. we want to test that the mean for first

population is greater than second population mean.population is greater than second population mean.


253253

4.Test Statistic4.Test Statistic:: Case 1:Case 1: Two population is normalTwo population is normal or or approximately approximately

normalnormal

σσ22 is known σ is known σ22 is unknown if is unknown if ( n ( n11 ,n ,n22 large or small) large or small)

( n ( n11 ,n ,n22 small) small)

populationpopulation populationpopulation VariancesVariances

Variances equal not equalVariances equal not equal

wherewhere

2

22

1

21

2121 )(- )X-X(

nn

Z

21

2121

11

)(- )X-X(

nnS

T

p

2

22

1

21

2121 )(- )X-X(

nS

nS

T

2

)1(n)1(n

21

222

2112

nn

SSS p


254254

Case2:Case2: If population is If population is not normallynot normally distributed distributed and nand n1, 1, nn2 2 is large(is large(nn1 1 ≥ 0 ,n≥ 0 ,n22≥ 0) ≥ 0) and population variances is known, and population variances is known,

2

22

1

21

2121 )(- )X-X(

nn

Z


255255


i) i) If If HHAA: : μ μ 1 1 ≠ ≠ μ μ 2 2 → → μ μ 1 1 -- μ μ 2 2 ≠ 0≠ 0


(when use Z - test) (when use Z - test)

OrOr Reject H Reject H 00 if T >t if T >t1-α/2 ,(n1-α/2 ,(n11+n+n22 -2) -2) or T< - tor T< - t1-α/2,,(n1-α/2,,(n11+n+n22 -2) -2)

((when use T- testwhen use T- test ) ) ____________________________________________________ ii) ii) HHAA: : μ μ 1 1 >> μ μ 2 2 →→ μ μ 1 1 -- μ μ 2 2 >> 0 0

Reject HReject H00 if Z>Z if Z>Z1-α1-α (when use Z - test) (when use Z - test)

OrOr Reject H Reject H00 if T>t if T>t1-α,(n1-α,(n11+n+n22 -2) -2) (when use T - test)(when use T - test)


256256

iii) If iii) If HHAA: : μ μ 1 1 << μ μ 2 2 →→ μ μ 1 1 -- μ μ 2 2 < 0< 0 Reject Reject HH00 if Z< - Z if Z< - Z1-1-α α (when use Z - test) (when use Z - test)

OrOr

Reject HReject H00 if T<- t if T<- t1-1-α, ,(nα, ,(n11+n+n22 -2) -2) (when use T - test)(when use T - test)

NoteNote::

ZZ1-α/21-α/2 , Z , Z1-α1-α , Z , Zαα are tabulated values obtained are tabulated values obtained from table Dfrom table D

tt1-α/21-α/2 , t , t1-α1-α , t , tαα are tabulated values obtained from are tabulated values obtained from table E with (ntable E with (n11+n+n22 -2) -2) degree of freedom (df)degree of freedom (df)

6.6. Conclusion: Conclusion: reject or fail to reject Hreject or fail to reject H00


257257

Example7.3.1 page238Example7.3.1 page238 Researchers wish to know if the data have collected provide Researchers wish to know if the data have collected provide

sufficient evidence to indicate a difference in mean serum sufficient evidence to indicate a difference in mean serum uric acid levels between normal individuals and individual uric acid levels between normal individuals and individual with Down’s syndrome. The data consist of serum uric with Down’s syndrome. The data consist of serum uric reading on 12 individuals with Down’s syndrome from reading on 12 individuals with Down’s syndrome from normal distribution with variance 1 and 15 normal individuals normal distribution with variance 1 and 15 normal individuals from normal distribution with variance 1.5 . The mean arefrom normal distribution with variance 1.5 . The mean are

andand α=0.05.α=0.05.

Solution:Solution:

1. 1. Data:Data: Variable is Variable is serum uric acid levelsserum uric acid levels, n, n11=12 , n=12 , n22=15, =15,

σσ2211=1, σ=1, σ22

22=1.5 ,α=0.05.=1.5 ,α=0.05.

100/5.41 mgX 100/4.32 mgX


258258

2. 2. Assumption:Assumption: Two population are normal, σ Two population are normal, σ221 1 , σ, σ22

22

are knownare known

3. 3. Hypotheses:Hypotheses: HH00: : μ μ 11 == μ μ2 → 2 → μ μ 11 - - μμ22 = 0= 0

HHAA: : μ μ 1 1 ≠ ≠ μ μ 2 2 → → μ μ 1 1 -- μ μ 2 2 ≠ 0≠ 0

4.Test Statistic:4.Test Statistic: = = 2.57= = 2.57

5. Desicion Rule:5. Desicion Rule:


ZZ1-α/2= 1-α/2= ZZ1-0.05/2= 1-0.05/2= ZZ0.975=0.975=1.96 (from table D)1.96 (from table D)

6-6-Conclusion: Conclusion: Reject Reject HH0 0 sincesince 2.57 > 1.962.57 > 1.96

Or if p-value =0.102→ reject Or if p-value =0.102→ reject HH0 0 if pif p << αα → then reject → then reject HH0 0

2

22

1

21

2121 )(- )X-X(

nn

Z

155.1

121

)0(- 3.4)-(4.5


259259

Example7.3.2 page 240Example7.3.2 page 240The purpose of a study by Tam, was to investigate wheelchairThe purpose of a study by Tam, was to investigate wheelchair

Maneuvering in individuals with over-level spinal cord injury )SCI(Maneuvering in individuals with over-level spinal cord injury )SCI(

And healthy control )C(. Subjects used a modified a wheelchair toAnd healthy control )C(. Subjects used a modified a wheelchair to

incorporate a rigid seat surface to facilitate the specifiedincorporate a rigid seat surface to facilitate the specified

experimental measurements. The data for measurements of theexperimental measurements. The data for measurements of the

left ischial tuerosity left ischial tuerosity ) ) المتحرك الكرسي من وتأثيرها الفخذ المتحرك عظام الكرسي من وتأثيرها الفخذ for for ) )عظامSCI and control C are shown belowSCI and control C are shown below

C13111512413112211788114150169

SCI60150130180163130121119130143


260260

We wish to know if we can conclude, on the We wish to know if we can conclude, on the basis of the above data that the mean of basis of the above data that the mean of left ischial tuberosity for control C lower left ischial tuberosity for control C lower than mean of left ischial tuerosity for SCI, than mean of left ischial tuerosity for SCI, Assume normal populations Assume normal populations equalequal variancesvariances. . αα=0.05, p-value = -1.33=0.05, p-value = -1.33


261261

Solution:Solution:

1. 1. Data:Data:, n, nCC=10 , n=10 , nSCISCI=10, S=10, SCC=21.8, S=21.8, SSCISCI=133.1 ,α=0.05.=133.1 ,α=0.05. ,, (calculated from data)(calculated from data)

2.2.Assumption:Assumption: Two population are normal, σ Two population are normal, σ221 1 , σ, σ22

22 are are

unknown but unknown but equalequal

3. 3. Hypotheses:Hypotheses: HH00: : μ μ CC == μ μ SCISCI → → μ μ CC - - μ μ SCISCI = 0= 0

HHAA: : μ μ C C < < μ μ SCI SCI → → μ μ C C -- μ μ SCI SCI < 0< 0

4.Test Statistic:4.Test Statistic:

Where,Where,

1.126CX 1.133SCIX

569.0

101

101

04.756

0)1.1331.126(

11

)(- )X-X(

21

2121

nnS

T

p

04.75621010

)3.32(9)8.21(9

2

)1(n)1(n 22

21

222

2112

nn

SSS p


262262


Reject H Reject H 00 if T< - T if T< - T1-α,(n1-α,(n11+n+n22 -2) -2)

TT1-α,(n1-α,(n11+n+n22 -2) = -2) = TT0.95,18 =0.95,18 = 1.7341 (from table E) 1.7341 (from table E)

6-6-Conclusion: Conclusion: Fail toFail to reject reject HH0 0 sincesince -0.569 < - -0.569 < - 1.73411.7341OrOr

Fail to reject Fail to reject HH0 0 since p = -1.33 since p = -1.33 >> αα =0.05 =0.05


263263

Example7.3.3 page 241Example7.3.3 page 241Dernellis and Panaretou examined subjects with hypertension Dernellis and Panaretou examined subjects with hypertension and healthy control subjects .One of the variables of interest wasand healthy control subjects .One of the variables of interest wasthe aortic stiffness index. Measures of this variable werethe aortic stiffness index. Measures of this variable werecalculated From the aortic diameter evaluated by M-mode andcalculated From the aortic diameter evaluated by M-mode andblood pressure measured by a sphygmomanometer. Physics wishblood pressure measured by a sphygmomanometer. Physics wishto reduce aortic stiffness. In the 15 patients with hypertensionto reduce aortic stiffness. In the 15 patients with hypertension)Group 1(,the mean aortic stiffness index was 19.16 with a)Group 1(,the mean aortic stiffness index was 19.16 with astandard deviation of 5.29. In the30 control subjects )Group 2(,thestandard deviation of 5.29. In the30 control subjects )Group 2(,themean aortic stiffness index was 9.53 with a standard deviation ofmean aortic stiffness index was 9.53 with a standard deviation of2.69. We wish to determine if the two populations represented by2.69. We wish to determine if the two populations represented bythese samples differ with respect to mean stiffness index .we wishthese samples differ with respect to mean stiffness index .we wishto know if we can conclude that in general a person withto know if we can conclude that in general a person withthrombosis have on the average higher IgG levels than personsthrombosis have on the average higher IgG levels than personswithout thrombosis at without thrombosis at αα=0.01, p-value = 0.0559=0.01, p-value = 0.0559


264264

Solution:Solution:

1. 1. Data:Data:, n, n11=53 , n=53 , n22=54, S=54, S11= = 44.8944.89, S, S22= = 34.8534.85 α=0.01. α=0.01.

2.2.Assumption:Assumption: Two population are not normal, σ Two population are not normal, σ221 1 , σ, σ22

22

are unknown and sample size largeare unknown and sample size large

3. 3. Hypotheses:Hypotheses: HH00: : μ μ 11 == μ μ 2 2 → → μ μ 11 - - μ μ 22 = 0= 0

HHAA: : μ μ 1 1 > > μ μ 2 2 → → μ μ 1 1 -- μ μ 2 2 > 0> 0

4.Test Statistic:4.Test Statistic:

GroupMean LgG levelSample Size

�ٍstandard deviation

Thrombosis59.015344.89

No Thrombosis

46.615434.85

59.1

5485.34

5389.44

0)61.4601.59()(- )X-X(22

2

22

1

21

2121

nS

nS

Z


265265


Reject H Reject H 00 if Z > Z if Z > Z1-α1-α

ZZ1-α = 1-α = ZZ0.99 =0.99 = 2.33 (from table D) 2.33 (from table D)

6-6-Conclusion: Conclusion: Fail toFail to reject reject HH0 0 sincesince 1.59 > 2.33 1.59 > 2.33

OrOr

Fail to reject Fail to reject HH0 0 since p = 0.0559 since p = 0.0559 >> αα =0.01 =0.01


266266

7.57.5 Hypothesis Testing A single Hypothesis Testing A single population proportionpopulation proportion::

Testing hypothesis about population proportion (P) is carried out Testing hypothesis about population proportion (P) is carried out

in much the same way as for mean when condition is necessary forin much the same way as for mean when condition is necessary for

using normal curve are metusing normal curve are met We have the following steps:We have the following steps:

1.1.DataData:: sample size (n), sample proportion( ) , P sample size (n), sample proportion( ) , P00

2. 2. Assumptions :Assumptions :normal distributionnormal distribution , ,

p̂

n

ap

sample in theelement of no. Total

isticcharachtar some with sample in theelement of no.ˆ


267267

3.Hypotheses:3.Hypotheses: we have three caseswe have three cases Case ICase I : : H H00: P = P: P = P00

HHAA: : P ≠ PP ≠ P00

Case IICase II : : H H00: P = P: P = P00

HHAA: : PP > > PP00

Case IIICase III : : HH00: P = P: P = P00

HHAA: : P P < < PP00

4.Test Statistic4.Test Statistic::

Where Where HH00 is true ,is distributed approximately as the is true ,is distributed approximately as the standard normalstandard normal

n

qp

ppZ

00

0ˆ


268268


i) i) If HIf HAA: P ≠ P: P ≠ P00 Reject H Reject H 00 if Z >Z if Z >Z1-α/2 1-α/2 or Z< - Zor Z< - Z1-α/21-α/2

______________________________________________ ii) If Hii) If HAA: P> P: P> P00 Reject HReject H00 if Z>Z if Z>Z1-α1-α __________________________________________________________ iii) If Hiii) If HAA: P< P: P< P00

Reject HReject H00 if Z< - Z if Z< - Z1-1-α α

NoteNote: Z: Z1-α/21-α/2 , Z , Z1-α1-α , Z , Zαα are tabulated values obtained from are tabulated values obtained from table Dtable D

6.6. ConclusionConclusion: : reject or fail to reject Hreject or fail to reject H00


269269

Example7.5.1 page 259Example7.5.1 page 259Wagen collected data on a sample of 301 Hispanic womenWagen collected data on a sample of 301 Hispanic women

Living in Texas .One variable of interest was the percentageLiving in Texas .One variable of interest was the percentage

of subjects with impaired fasting glucose )IFG(. In theof subjects with impaired fasting glucose )IFG(. In the

study,24 women were classified in the )IFG( stage .The articlestudy,24 women were classified in the )IFG( stage .The article

cites population estimates for )IFG( among Hispanic womencites population estimates for )IFG( among Hispanic women

in Texas as 6.3 percent .Is there sufficient evidence toin Texas as 6.3 percent .Is there sufficient evidence to

indicate that the population Hispanic women in Texas has aindicate that the population Hispanic women in Texas has a

prevalence of IFG higher than 6.3 percent ,let prevalence of IFG higher than 6.3 percent ,let αα=0.05=0.05

Solution:Solution:

1.Data:1.Data: n = 301, p n = 301, p00 = 6.3/100=0.063 ,a=24,= 6.3/100=0.063 ,a=24,

qq00 =1- p=1- p00 = 1- 0.063 =0.937, = 1- 0.063 =0.937, αα=0.05=0.05

08.0301

24ˆ

n

ap


270270

2.2. Assumptions : Assumptions : is approximatelyis approximately normaly distributednormaly distributed

3.Hypotheses:3.Hypotheses: we have three caseswe have three cases HH00: P = 0.063: P = 0.063

HHAA: : PP > 0.063 > 0.063

4.Test Statistic 4.Test Statistic ::

5.Decision Rule: 5.Decision Rule: Reject HReject H00 if Z>Z if Z>Z1-α1-α

Where Where ZZ1-α 1-α = Z= Z1-0.051-0.05 =Z =Z0.950.95== 1.6451.645

21.1

301)0.937(063.0

063.008.0ˆ

00

0

nqp

ppZ

p̂


271271

6.6. Conclusion: Conclusion: Fail to reject HFail to reject H00

SinceSince

Z =1.21 > ZZ =1.21 > Z1-α=1-α=1.6451.645

Or , Or ,

If P-value = 0.1131,If P-value = 0.1131,

fail to reject Hfail to reject H0 0 → P > → P > αα


272272

Exercises:Exercises: Questions Questions : Page 234 -237: Page 234 -237 7.2.1,7.8.2 ,7.3.1,7.3.6 ,7.5.2 ,,7.6.17.2.1,7.8.2 ,7.3.1,7.3.6 ,7.5.2 ,,7.6.1

H.WH.W: : 7.2.8,7.2.9, 7.2.11, 7.2.15,7.3.7,7.3.8,7.3.107.2.8,7.2.9, 7.2.11, 7.2.15,7.3.7,7.3.8,7.3.10 7.5.3,7.6.47.5.3,7.6.4


In an article in the journal Health and Place, found In an article in the journal Health and Place, found that among 2428 boys aged from 7 to 12 years, that among 2428 boys aged from 7 to 12 years, 461 were over weight or obese. On the basis of 461 were over weight or obese. On the basis of this study ,can we conclude that more than 15 this study ,can we conclude that more than 15 percent of boys aged from 7 to 12 years in the percent of boys aged from 7 to 12 years in the

sampled population are over weight or obesesampled population are over weight or obese??Let Let αα =0.1 =0.1


273273






274274




275275


276276

7.67.6 Hypothesis Testing :TheHypothesis Testing :The Difference between two Difference between two

population proportionpopulation proportion:: Testing hypothesis about two population proportion (PTesting hypothesis about two population proportion (P1,, 1,, PP2 2 ) is) is

carried out in much the same way as for difference between twocarried out in much the same way as for difference between twomeans when condition is necessary for using normal curve are metmeans when condition is necessary for using normal curve are met We have the following steps:We have the following steps:

1.Data1.Data:: sample size (n sample size (n1 1 ووnn22), sample proportions( ), ), sample proportions( ),

Characteristic in two samples (xCharacteristic in two samples (x11 , x , x22),),

2- Assumption :2- Assumption : Two populations are independent . Two populations are independent .

21ˆ,ˆ PP

21

21

nn

xxp


277277

3.Hypotheses:3.Hypotheses: we have three caseswe have three cases Case ICase I : : H H00: P: P11 = P = P22 → → PP11 - P - P22 = 0 = 0

HHAA: P: P1 1 ≠ ≠ PP2 2 → → PP11 - P - P22 ≠ 0 ≠ 0 Case IICase II : : H H00: P: P1 1 = P = P2 2 → → PP11 - P - P22 = 0 = 0

HHAA: P: P1 1 > P > P2 2 → → PP11 - P - P22 > 0 > 0 Case IIICase III : : HH00: P: P11 = P = P2 2 → → PP11 - P - P22 = 0 = 0

HHAA: P: P11 < P< P2 2 → → PP11 - P - P22 < 0 < 0

4.Test Statistic4.Test Statistic::

Where Where HH00 is true ,is distributed approximately as the is true ,is distributed approximately as the standard normalstandard normal

21

2121

)1()1(

)()ˆˆ(

npp

npp

ppppZ


278278


i) i) If HIf HAA: P: P11 ≠ P ≠ P22 Reject H Reject H 00 if Z >Z if Z >Z1-α/2 1-α/2 or Z< - Zor Z< - Z1-α/21-α/2

______________________________________________ ii) If Hii) If HAA: P: P11 > P > P22 Reject HReject H00 if Z >Z if Z >Z1-α1-α __________________________________________________________ iii) If Hiii) If HAA: P: P11 < P < P22

Reject HReject H00 if Z< - Z if Z< - Z1-1-α α

NoteNote: Z: Z1-α/21-α/2 , Z , Z1-α1-α , Z , Zαα are tabulated values obtained from are tabulated values obtained from table Dtable D

6.6. ConclusionConclusion: : reject or fail to reject Hreject or fail to reject H00


279279

Example7.6.1 page 262Example7.6.1 page 262Noonan is a genetic condition that can affect the heart growth,Noonan is a genetic condition that can affect the heart growth,

blood clotting and mental and physical development. Noonan examinedblood clotting and mental and physical development. Noonan examined

the stature of men and women with Noonan. The study contained 29the stature of men and women with Noonan. The study contained 29

Male and 44 female adults. One of the cut-off values used to assessMale and 44 female adults. One of the cut-off values used to assess

stature was the third percentile of adult height .Eleven of the males fellstature was the third percentile of adult height .Eleven of the males fell

below the third percentile of adult male height ,while 24 of the femalebelow the third percentile of adult male height ,while 24 of the female

fell below the third percentile of female adult height .Does this study fell below the third percentile of female adult height .Does this study

provide sufficient evidence for us to conclude that among subjects with provide sufficient evidence for us to conclude that among subjects with

Noonan ,females are more likely than males to fall below the respectiveNoonan ,females are more likely than males to fall below the respective

of adult height? Let of adult height? Let αα=0.05=0.05

Solution:Solution:

1.Data:1.Data: n n MM = 29, n = 29, n FF = 44 , x = 44 , x MM= 11 , x = 11 , x FF= 24, = 24, αα=0.05=0.05

479.04429

2411

FM

FM

nn

xxp 545.0

44

24ˆ,379.0

29

11ˆ

F

FF

M

mM n

xp

n

xp


280280

2- Assumption :2- Assumption : Two populations are independent . Two populations are independent .3.Hypotheses:3.Hypotheses: Case IICase II : : H H00: P: PF F = P = PM M → → PPFF - P - PMM = 0 = 0

HHAA: P: PF F > P > PM M → → PPFF - P - PMM > 0 > 0 4.Test Statistic4.Test Statistic::


Reject HReject H00 if Z >Z if Z >Z1-α1-α , Where Z , Where Z1-α 1-α = Z= Z1-0.051-0.05 =Z =Z0.950.95== 1.6451.645

6.6. Conclusion: Conclusion: Fail to reject HFail to reject H00

Since Z =1.39 > ZSince Z =1.39 > Z1-α=1-α=1.6451.645

Or , If P-value = 0.0823 → fail to reject HOr , If P-value = 0.0823 → fail to reject H0 0 → P > → P > αα

39.1

29)521.0)(479.0(

44)521.0)(479.0(

0)379.0545.0(

)1()1(

)()ˆˆ(

21

2121

npp

npp

ppppZ


281281

Exercises:Exercises: Questions Questions : Page 234 -237: Page 234 -237 7.2.1,7.8.2 ,7.3.1,7.3.6 ,7.5.2 ,,7.6.17.2.1,7.8.2 ,7.3.1,7.3.6 ,7.5.2 ,,7.6.1

H.WH.W: : 7.2.8,7.2.9, 7.2.11, 7.2.15,7.3.7,7.3.8,7.3.107.2.8,7.2.9, 7.2.11, 7.2.15,7.3.7,7.3.8,7.3.10 7.5.3,7.6.47.5.3,7.6.4

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