wayne daniel ppt

8/10/2019 Wayne Daniel Ppt

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Lectures of Stat -145

(Biostatistics)

Text bookBiostatisticsBasic Concepts and Methodology for the Health Sciences

By

Wayne W. Daniel

Prepared By:

Sana A. [email protected]

Chapter 1

Introduction To

Biostatistics

Text Book : Basic Concepts andMethodology for the Health Sciences 2


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Key words :

Statistics , data , Biostatistics,

Variable ,Population ,Sample


IntroductionSome Basic concepts

Statisticsis a field of studyconcerned

with

1-collection, organization, summarizationand analysisof data.

2-drawing of inferencesabout a body of

data when only a part of the data is

observed.

Statisticianstry to interpretand

communicatethe results to others.



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Data:

The raw material of Statisticsis data.We may define data as figures. Figures

result from the process of countingorfrom taking a measurement.

For example

- When a hospital administrator counts thenumber of patients (counting).

- When a nurse weighs a patient

(measurement)


* A variable:

It is a characteristic that takes on differentvalues 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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* Biostatistics:

The tools of statisticsare employed in manyfields:

business, education, psychology, agriculture,

economics, etc.

When the data analyzed are derived from

the biological scienceand medicine,

we use the term biostatisticsto distinguish

this particular application of statistical

tools and concepts.


Quantitative VariablesIt can be measured in

the usual sense.

For example:

- the heights of adultmales,

- the weights ofpreschool children,

- the ages of patientsseen in a

- dental clinic.

Qualitative Variables

Many characteristics are not

capable of being measured.

Some of them can be

ordered (called ordinal)

and Some of them cant be

ordered (called nominal).

For example:

- classification of people into

socio-economic groups

-.hair color


Types of variablesQuantitative

Qualitative


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A discrete variableis characterized by gaps

or interruptions in thevalues that it canassume.

For example:

- The number of dailyadmissions to ageneral hospital,

- The number of

decayed, missing orfilled teeth per child

in an elementary

- school.

A continuous variablecan assume any value within a

specified relevant interval ofvalues 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 personwhose height falls somewhere inbetween.


Types of quantitative variables

Discrete

Continuous

As the name implies itconsist of namingor classifies intovarious mutuallyexclusive categories

For example:

- Male - female- Sick - well

- Marriedsingle -divorced

.Whenever qualitativeobservation

Can be ranked or orderedaccording to somecriterion.

For example:- Blood pressure level

(high-good-low)

- Grades

(ExcellentV.goodgoodfail)


Types of qualitative variables

Nominal

Ordinal


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*A population:

It is the largest collection of valuesof arandom variablefor which we have aninterest at a particular time.

For example:

The weights of all the children enrolled in acertain elementary school.

Populations may be finiteor infinite.


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

For example:

The weights of only a fraction of

these children.



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Exercises

Question (6)Page 17 Question (7)Page 17

Situation A , Situation B


Exercises:Q6: For each of the following variables

indicate whether it is quantitative orqualitative variable:

(a)The blood type of some patient in the

hospital. Qualitative Nominal(b) Blood pressure level of a patient.

(Qualitative ordinal)



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(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)The distance between the hospital to thehouseQuantitative continues

(f) Under-arm temperature of day-old infants

born in a hospital. Quantitativecontinues


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 asmall southern town revealed that if she hasschool-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 school agechild present.

(d)Variable is qualitative nominal.


Situation B:A study of 250 patients admitted

to a hospital during the past year revealedthat, the patients lived from the hospital.

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

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



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(c) Variable: Distancethe patient lived away

from the hospital

Variable is Quantitative continuous.(d)


Choose the right answer:1-The variable is a

a. subset of the population.

b. parameter of the population.

c. relative frequency.

d. characteristic of the population to be measured.

e. class interval.

2-Which of the following is an example of discrete variablea. the number of students taking statistics in this term at ksu.

b. the time to exercise daily.

c. whether or not someone has a disease

d. height of certain buildings

e. Level of education


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3.Which of the following is not an example of discrete

variable

a. the number of students at the class of statistics.b. the number of times a child cry in a certain street.

c. the time to run a certain distance.

d. the number of buildings in a certain street.

e. number of educated persons in a family.

4.Which of the following is an example of nominal

qualitative variable

a. blood pressure level.

b. the number of times a child brush his/her teeth.

c. whether or not someone fail in an exam.

d. Weight of babies at birth.

e. the time to run a certain distance.

5.The continuous variable is a

a. variable with a specific number of values.

b. variable which cant be measured.

c. variable takes on values within intervals.

d. variable with no mode.

e. qualitative variable.

6. which of the following is an example of continuous variable

a. The number of visitors of the clinic yesterday.

b. The time to finish the exam.

c. The number of patients suffering from certain disease.

d. Whether or not the answer is true.


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7. The discrete variable is

a-qualitative variable.

b-variable takes on values within interval.C-variable with a specific number of values.d-variable with no mode.

8-Which of the following is an example of nominalvariable :

a-age of visitors of a clinic.b-The time to finish the exam.

c-Whether or not a person is infected by influenza.

d-Weight for a sample of girls .

9-The nominal variable is a

a-A variable with a specific number of values

b-Qualitative variable that cant be ordered.

c-variable takes on values within interval.

d-Quantitative variable .

10-Which of the following is an example of

nominal variable :a-The number of persons who are injured in accident.

b-The time to finish the exam.

c-Whether or not the medicine is effective.

d-Socio-economic level.


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11-The ordinal variable is :

a-variable with a specific number of values.

b-variable takes on values within interval.

c-Qualitative variable that can be ordered.

d-Variable that has more than mode.

Chapter ( 2 )Strategies for understanding the

meanings of Data

Pages 19 27)


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Key words

frequency table, bar chart ,range

width of interval ,mid-intervalHistogram , Polygon


Descriptive StatisticsFrequency Distribution

for Discrete Random VariablesExample:

Suppose that we take sampleof size 16 from children in aprimary school and get thefollowing data about thenumber of their decayed teeth,

3,5,2,4,0,1,3,5,2,3,2,3,3,2,4,1

To construct a frequency table

We need three columns:

1.Variable name

2.Frequency (f):how manmber3.Relative frequency(R.f)=

Frequency / n

Relative

Frequenc

y

Frequenc

y

No. of

decayed

teeth

0.0625

0.1250.25

0.3125

0.125

0.125

1

24

5

2

2

0

12

3

4

5

116=nTotal


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Representing the simple frequency table

using the bar chart

Number of decayed teeth

5.004.003.002.001.00.00

6

5

4

3

2

1

0

22

5

4

2

1


We can represent theabove simplefrequency table usingthe bar chart.

We can get :

1. The sample size?

2. Number of childrenwith decayedteeth=2?

3. Relative frequencyof children withdecayed teeth =4?

2.3 Frequency Distributionfor Continuous Random Variables

For large samples, we cant use the simple frequency tableto represent the data.

We need to dividethe data into groupsor intervals orclasses.

So, we need to determine:

1- The number of intervals (k).

Too fewintervals are not good because information will belost.

Too manyintervals are not helpful to summarize the data.



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2- The range (R).

It is the difference between the largest and

the smallest observation in the dataset.[R=MaxMin]

3- The Width of the interval (w).

Classintervals generally should be of the same

width.


Frequency

(f)

Class interval

113039

464049

705059

456069

167079

18089

189Total


Sum of frequency=sample size=n


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Cumulative Frequency:The

It can be computed by adding successivefrequencies.

The Cumulative Relative Frequency:

It can be computed by adding successive relativefrequencies.

interval:-MidTheIt can be computed by adding the lower bound of

the interval plus the upper bound of it and thendivide over 2.[(Lower bound + Upper bound)/2]

For the above example, the

following table represents the

cumulative frequency, the

relative frequency, the

cumulative relative frequencyand the mid-interval.



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Cumulative

RelativeFrequency

Relative

FrequencyR.f

Cumulative

Frequency

Frequency

Freq (f)

Mid

interval

Class

interval

0.05820.0582111134.53039

-0.2434574644.54049

0.6720-127-54.55059

0.91010.2381-45-6069

0.99480.08471881674.57079

10.0053189184.58089

1189Total


R.f= freq/n

Example :

From the above frequency table, complete the tablethen answer the following questions:

1-The number of objects with age less than 50 years ?

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

3-Relative frequency of objects with age between 70-79

years ? 4-Relative frequency of objects with age more than 69

years ?

5-The percentage of objects with age between 40-49years ?



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

7-The Range (R) ?

8- Number of intervals (K)?

9- The width of the interval ( W) ?


Representing the grouped frequency table using the histogram

To draw the histogram, the true classes limitsshould be used. They can becomputed by subtracting 0.5 from thelower limit and adding 0.5 to theupper limit for each interval.

FrequencyTrue class

limits

1129.5


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Representing the grouped frequency table

using the Polygon

0

10

20

30

40

50

60

7080

345 445 545 645 745 845


Q1] For a sample of patients, we obtain the following graph forapproximated hours spent without pain after a certain surgery


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1) The type of the graph is:

(e)Bar chart (f) polygon(d) histogram(c) line(a) curve (b) not known

2) The number of patients stayed the longest time without pain is:(f) 10(e) 15(d) 6(c) 5(b) 80(a) 1

3)The percent of patients spent 3.5 hours or more without pain is:

(f) 37.5%(e) 68.75%(d) 18.75%(c) 50%(a) 25% (b) 30%

4)The lowest number of hours spent without pain is:

(f))10(e) 1(d) 0.5(c) 5(b) 25 (a) 6.5

6)The mode equals

(f) 80(d) 3(d) 15(c) 2,4(b) 6(a) we can't find it

the following histogram show the frequency distribution of pathologic][H.Wtumor size ( in cm) for a sample of 110 cancer patients:


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1. The percent of cancer patients with approximate level of pathologic tumor size =2

cm is:

(a)

18% (b) 50% (c) 16.36% (d) 32.72% (e) 36% (f) 0%2. The number of cancer patients with lowest pathologic tumor size is:

(a) 0.5 (b) 3 (c) 11 (d) 13 (e) 15 (f) 24

3. The approximate size of pathologic tumorwith highest percentage of patients is:

(a) 3 (b) 1 (c) 36 (d) 110 (e) 32.72% (f) 16.36%

4. What the approximate value of the sample mean

(a) 18.33 (b) 36 (c) 1 (d) 1.75 (e) 1.586 (f) we can't find it

5. The mode equals

(a)

1 (b) 3 (c) 36 (d) 55 (e) 110 (f) we can't find it

6. The approximate value of the sample variance

(a) 3 (b) 0.6 (c) 0.774 (d) 1.586 (e) 110 (f) we can't find it

Exercises

Pages : 3134

Questions: 2.3.2(a) , 2.3.5 (a)

H.W. : 2.3.6 , 2.3.7(a)



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Exercises:

Q2.3.2:Janardhan et al. (A-2) conducted a study

in which they measured incidental intracranialaneurysms (IIAs) in 125 patients. Theresearchers examined post proceduralcomplications and concluded that IIAs can besafely treated without causing mortality andwith a lower complications rate than previouslyreported.


The following are the sizes (in millimeters) of the159 IIAs in the sample.


frequencyClass Interval

290-4

875-9

2610-14

1015-19

420-24

125-29

230-34

159Total


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(a)Use the frequency table to prepare:

* A relative frequency distribution* A cumulative frequency distribution

* A cumulative relative frequency distortion


(b) What percentage of the measurements arebetween 10 and 14 inclusive?

(c) How many observations are less than 20?

(d) What proportion of the measurements are

greater than orequal to 25?(e) What percentage of the measurements areeither less than 10 or greater than 19?

Text Book : Basic Concepts and

Methodology for the Health Sciences 48


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Q2.3.5:The following table shows the number of

hours 45 hospital patients slept following the

administration of a certainanesthetic.(a) From thesedata construct:* A relativefrequencydistribution

FrequencyClass Interval

211-5

166-10

611-15

216-20

45Total


(b) How many of the measurements are greaterthan 10? Ans: 8

(c) What percentage of the measurements arebetween 6-15 ?

Ans: 49%

(d) What proportion of the measurement is lessthan or equal 15? Ans: 0.96



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Q2.3.6:The following are the number of babies

born during a year in 60 community hospitals.

(a) From thesedata construct:

*A relative

frequency

distribution

*


FrequencyClass Interval

520-24

625-29

930-34

335-39

540-44

845-49

1150-54

1355-5960Total

Q2.3.7: In a study ofphysical endurance

levels of male college

freshman, the

following composite

endurance scores

based on severalexercise routines

were collected.


FrequencyClass interval

6115-134

7135-154

16155-174

31175-194

37195-214

28215-234

18235-254

8255-274

3275-2941295-314

155Total


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(a) From these data construct:

* A relative frequency distribution


) :4.2Section (

Descriptive Statistics

Measures of Central

Tendency41-38Page


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key words:

Descriptive Statistic, measure ofcentral tendency ,statistic, parameter,mean () ,median, mode.


The Statistic and The Parameter

A Statistic:It is a descriptive measure computed from the

data of a sample.

A Parameter:It is aa descriptive measure computed from

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

population, a sampleis drawn of size n, whosevalues are

1 , 2 , ,n. From this data, wemeasure the statistic.



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Measures of Central Tendency

A measure of central tendency is a measure whichindicates where the middleof 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 = 36.6x

x


1

N

ii

N

X

1

n

ii

n

x


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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 valuesenter 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. Themean = 118, a value that is not representative of the set of

data as a whole.


The Median:

When orderingthe data, it is the observation that divide theset of observations into two equal partssuch that half of thedata are before it and the other are after it.

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

When n = 11, then the median is the 6thobservation.

* If n is even, there are two middle observations. The medianwill 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 6.5thobservation, which

is an observation halfway between the 6thand 7thorderedobservation.



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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 5.5thobservation,i.e. = (32+34)/2 = 33.

Properties of the Median:

Uniqueness.For a given set of data there isone 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 isrepeated two times, so it is the mode.

Properties of the Mode:

Sometimes, it is notunique.

It may be used fordescribing qualitative data.

It is not affected by extreme values



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Examples

Find the mean and the mode for thefollowing Relative Frequency?

Mode = 7(has the higher frequency)


n

xfx

Age(x)

frequenc

y(f)

X f

56710

2341

10182810

Total 10 66

6.610

66x

ExamplesFind the mean and the mode for the

following grouped

Frequency table?

Mode :interval( 79 )(can't give exact number only the interval

with higher Frequency)Text Book : Basic Concepts and


n

xfx

4.71074 x

Age

Frequency

(f)

Midpoint

(X)

X f

1 - 34 - 67 - 9

10 - 12

2 1

4 3

2 5 8 11

4 5 32 33

Total

10

_

74


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Examples

Number of decayed teeth

5.004.003.002.001.00.00

6

5

4

3

2

1

0

22

5

4

2

1


Find the mean andthe mode for the

following bar

chart?

Solution :

Mode = 3

(has the higherfrequency)

16

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

225421

xxxxxxx

n


687.216

43x

n

xfx


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key words:

Descriptive Statistic, measure ofdispersion , range ,variance, coefficient ofvariation.


2.5. Descriptive StatisticsMeasures of Dispersion:

A measure of dispersion conveys information regarding theamount 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:a).If the valuescloseto each other

The amount of Dispersion small.b)If the values are widely scattered

The Dispersion is greater.



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43Page1.5.2Ex. Figure

**Measures of Dispersion are :1.Range (R).

2.Variance.

3.Standard deviation.

4.Coefficient of variation (C.V).


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


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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+(66-56) 2+..+(50-56) 2]/ (10-1)

= 810/9 = 90

x

x


2S

1

)(1

2

2

n

xx

S

n

i

i

b)PopulationVariance( ) :

where , is Population mean

3.The Standard Deviation:

is the square root of variance=

a) Sample Standard Deviation = S =

b)Population Standard Deviation = =


2

N

xN

i

i

1

2

2

)(

Varince

2S

2


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Note:

You can use the calculator to find : 1.Mean

2.Variance

3. Standard deviation

How to use calculator is in the Appendix))


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

Is a measure use to compare thedispersion in two sets of data which isindependent of the unit of themeasurement .

where S: Sample standard deviation.

: Sample mean.


)100(.

X

SVC

X


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:46Page3.5.2Example

Suppose two samples of human males yield thefollowing 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



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Exercises

Pages : 5253 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 Use calculator if possible))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



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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 amyotrophiclateral sclerosis on symptoms, quality of life, andsurvival. One of the variables of interest is partialpressure of arterial carbon dioxide (PaCO2). Thevalues below ( mm of Hg ) reflect the result of

baseline testing on 30 subjects as established by

arterial blood gas analyses.


seven rat pups from the experiment involving the carotidartery.

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 Ans: 46.90

(g) The coefficient of variation Ans: 8.69%



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H.W :Q2.5.1:

Porcellini et al. (A-8) studied 13 HIV-positive

patients who were treated with highly activeantiretroviral therapy (HAART) for at least 6months. The CD4 T cell counts ( ) atbaseline for the 13 subjects are listed below.

230 205 313 207 227 245 173

58 103 181 105 301 169


l106


Ans: 193.6 Ans: 207


Ans: no mode Ans: 255

(e) The variance (f) The standard deviation

Ans: 5568.09 Ans: 74.62




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Q2.5.2:H.W Shrair and Jasper (A-9) investigatedwhether decreasing the venous return in young

rats would affect ultrasonic vocalizations(USVs). Their research showed no significantchange in the number of ultrasonicvocalizations when blood was removed fromeither the superior vena cava or the carotidartery. Another important variable measuredwas the heart rate (bmp) during the withdrawal

of blood. The data below presents the heartrate of


40.0 47.0 34.0 42.0 54.0 48.0 53.6 56.958.0 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 56.659.0


Ans: 47.41 Ans: 46.35(c) The mode (d) The rangeAns: 33, 54 Ans: 29.6



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Ans: 46.53 Ans: 8.75

(g) The coefficient of variation

(H.W)Q2.5.4:

According to Starch et al. (A-11), hamstring tendongrafts have been the weak link in anteriorcruciate ligament reconstruction. In a controlledlaboratory study, they compared two techniquesfor reconstruction : either an interference screwor a central sleeve and


screw on the tibial side. For eight cadavericknees, the measurements below represent therequired force ( in Newtones) at which initialfailure of graft strands occurred for thecentral sleeve and screw technique.

172.5 216.63 212.62 98.97 66.95239.76 19.57 195.72

(a) The mean (b) The medianAns: 152.84 Ans: 184.11


Ans: no mode Ans: 220.19




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Ans: 6494.732 Ans: 80.5899

(g) The coefficient of variation Ans: 52.73%Q2.5.5:

Cardosi et al. (A-12) performed a 4 yearsretrospective review of 102 women undergoingradical hysterectomy for cervical or endometrialcancer. Catheter-associated urinary tract infection

was observed in 12 of the subjects. Below are thenumbers of


postoperative days until diagnosis of the infection foreach subject experiencing an infection.

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


Ans: 16.75 Ans: 15(c) The mode (d) The range

Ans: 15 Ans: 43

(e) The variance (f) The standard deviationAns: 124.0227 Ans: 11.1365




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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 wirefixation in young adults with lumbar spondylolysis.The authors found that segmental wire fixationhistorically has been successful in the treatment ofnonathletes with spondylolysis, but no informationexisted on the results of this type of surgery inathletes. In a retrospective study, the authorsfound 20 subjects who had the surgery between

1993 and 2000. For these subjects, the data below


represent the duration in months of follow-up careafter the operation.

103 68 62 60 60 54 49 44 42 41 3836 34 30 19 19 19 19 17 16


Ans: 41.5 Ans: 39.5

(c) The mode (d) The rangeAns: 19 Ans: 87


Ans: 490.264 Ans: 22.1419



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Q2.5.14:

In a pilot study, Huizinga et al. ( A-14) wanted togain more insight into the psychosocialconsequences for children of a parent with cancer.For the study, 14 families participated insemistructured interviews and completedstandardized questionnaires. Below is the age ofthe sick parent with cancer (in years) for the 14families.


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

(a) The mean (b) The medianAns: 43.7143 Ans: 45


Ans: 32, 51 Ans: 21(e) The variance (f) The standard deviation

Ans: 48.0659 Ans: 6.93296(g) The coefficient of variation Ans: 15.8597%



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3Chapter

Probability

The Basis of the Statistical

inference

Key words:

Probability, objective Probability,

subjective Probability, equally likely

Mutually exclusive, multiplicative ruleConditional Probability, independent events, Bayes

theorem



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Introduction1.3

The concept of probability is frequently encountered ineveryday communication. For example, a physician maysay that a patient has a 50-50 chance of surviving a certainoperation.

Another physician may say that she is 95 percent certainthat a patient has a particular disease.

Most people express probabilities in terms of percentages.

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

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


Some definitions of:2.3probability

1.Equally likely outcomes:

Are the outcomes that have the same chance ofoccurring.

2.Mutually exclusive:

Two events are said to be mutually exclusive ifthey cannot occur simultaneously such that

A B =.



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The universal Set(S): The set all possibleoutcomes.

The empty set : Contain no elements. The event ,E: is a set of outcomes in S which has a

certain characteristic.

Classical Probability: If an event can occur in Nmutually exclusive and equally likely ways, and if m ofthese possess a triat, E, the probability of theoccurrence of event E is equal to m/ N .

P(A)=n(A)/n(S)

For Example: in the rolling of the die , each of the

six sides is equally likely to be observed . So, theprobability that a 4 will be observed is equal to 1/6.


Relative Frequency Probability:

Def:If some posses is repeated a large number oftimes, n, and if some resulting event E occurs m times, the relative frequency of occurrence of E , m/n willbe approximately equal to probability of E .

P(E) = m/n .



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Elementary Properties of3.3:Probability

Given some process (or experiment )with n mutually exclusive events E1, E2,E3,, En, then

1-P(Ei) 0, i= 1,2,3,n 2- P(E1 )+ P(E2) ++P(En)=1

3- P(Ei+EJ)=P(Ei )+ P(EJ )

Ei,EJare mutually exclusive


Rules of Probability1-Addition Rule P(A U B)= P(A) + P(B)P (AB )

2- If A and B are mutually exclusive (disjoint) ,then P (AB ) = 0 Then , addition rule is

P(A U B)= P(A) + P(B) . 3- Complementary Rule P(A' )= 1P(A) where, A' = = complement event Consider example 3.4.1 Page 63



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Table 3.4.1 in Example 3.4.1TotalLater >18

(L)Early = 18

(E)

Family history of

Mood Disorders

633528Negative(A)

573819Bipolar

Disorder(B)

854441Unipolar (C)

1136053Unipolar and

Bipolar(D)318177141Total


Find the following probabilities:

1. P(C) =

2.P(L) =

(=3.P(AcE) =3.P(A

4.P(L U D)=

B) =5.P(AE) =6.P(L

(=Ac7.P(E

(=U Ac8.P(L


A


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Answer the following questions:**

Suppose we pick a person at random from this sample.1-The probability that this person will be 18-years old?

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

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

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

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


C

Conditional Probability:

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

happened.

P(A\B)= , P(B) 0

P(B\A)= , P(A) 0

)(

)(

)(

)(

BP

BAPor

Bn

BAn

)(

)(

)(

)(

AP

BAPor

An

BAn



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64Page2.4.3ExampleFrom previous example 3.4.1 Page 63 , answer

suppose we pick a person at random and find he is 18

years (E),what is the probability that this person will

be one with Negative family history of mood

disorders (A)?

suppose we pick a person at random and find he has

family history of mood (D) what is the probability that

this person will be 18 years (E)?


:Calculating a joint Probability

Example 3.4.3.Page 64

Suppose we pick a person at randomfrom the 318 subjects. Find theprobability that he will early (E) and has

Negative family history of mooddisorders (A).

Exercise: Example 3.4.5.Page 66




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Independent Events:

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

Then,

1- P(AB)= P(B)P(A) 2- P(A\B)=P(A)

3- P(B\A)=P(B)


68Page7.4.3Example In a certain high school class consisting of 100

students60 , 40 of them are boys, it is observed that24 girls and 16 boys wear eyeglasses . If a student ispicked at random from this class

1.The probability that the student wears eyeglasses ,

2.What is the probability that a student picked at

random wears eyeglasses given that the student is aboy?

3.What is the probability of the joint occurrence ofthe events of wearing eye glasses and being a girl?



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SolutionTake the following symbols:

not a boy = girl:B : is boy. BC

not were glasses:G:were glasses GC

n(s)=100 ,n(B)=40,n(GB)=16,

n(GBC)=24


TotalBCB

n(G)n(BCG)

n(BG)G

n(GC

)n(BC

GC

)n(BGC

)GC

n(S)(n(BCn(B)Total

1.P(G)=

=2.P(G/B)

(=3.P(GBC

TotalBCB

2416G

GC

10040Total



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69Page8.4.3Example

Suppose that of 1200 admission to a general

hospital during a certain period of time,750 are

private admissions. If we designate these as a set A,

then compute P(A) , P( ).


A


Marginal Probability: Definition: Given some variable that can be broken down into m

categories designatedby and another jointly occurring variable

that is broken down into n categories designated by, the marginal probability of with all the categories of

B . That is,

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

Probabilities to calculate P(E).,P(L),P(A),..


),()( jii BAPAP

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

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

iA


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Exercise:

Page 76-77 Questions :

3.4.1, 3.4.3,3.4.4

H.W.

3.4.5 , 3.4.7


Q3.4.1: In a study of violent victimization of womenand men, Porcelli et al. (A-2) collected informationfrom 679 women and 345 men aged 18 to 64years at several family practice centers in themetropolitan Detroit area. Patients filled out ahealth history questionnaire that included aquestion about victimization. The following tableshows the sample subjects cross-classified by sex

and type of violent victimization reported. Thevictimization categories are defined as novictimization, partner victimization (and not byothers), victimization by persons other than




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partners (friends, family members, or strangers), andthose who reported multiple victimization.

(a) Suppose we pick a subject at random from this

group. What is the probability that this subject willbe a women?


TotalMultipleVictimization

(T)

Nonpartners

(N)

Partners

(P)

NoVictimization

(V)

679181634611Women(W)

345101710308Men(M)

1024283344919Total

(b) What do we call the probability calculated inpart a?

(c) Show how to calculate the probability asked forin part a by two additional methods.

(d) If we pick a subject at random, what isprobability that the subject will be a women andhave experienced partner abuse?

(e) What do we call the probability calculated inpart d?

(f) Suppose we picked a man at random. Knowingthis information, what is the probability that he



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experienced abuse from nonpartners?

(g) What do we call the probability calculated in

part f?(h) Suppose we pick a subject at random. Whatis the probability that it is a man or someonewho experienced abuse from a partner?

(i) What do we call the method by which youobtained the probability in part h?


H.WQ3.4.3: Fernando et al. (A-3) studied drug-sharingamong injection drug users in the South Bronx inNew York City. Drug users in New York City usethe term split a bag or get down on a bag torefer to the practice of diving a bag of heroin orother injectable substances. A common practiceincludes splitting drugs after they are dissolved ina common cooker, a procedure with considerable

HIV risk. Although this practice is common, little isknown about the prevalence of such practices. Theresearchers asked injection drug users in fourneighborhoods in the South Bronx if they ever



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got down on drugs in bags or shots. The resultsclassified by gender and splitting practice are

given below:State the

following

probabilities in

words and calculate:

(a) Ans: 0.3418

(b) Ans: 0.8746(c) Ans: 0.6134



TotalNever SplitDrugs

Split DrugsGender

673324349Male

348128220Female

1021452569Total

)( DrugsSplitMaleP

)( DrugsSplitMaleP

)( DrugsSplitMaleP

(d) Ans: 0.6592

Q3.4.4: Laveist and Nuru-Jeter (A-4) conducteda study to determine if doctor-patient raceconcordance was associated with greatersatisfaction with care. Toward that end, theycollected a national sample of African-

American, Caucasian, Hispanic, and Asian-American respondents. The following tableclassifies the race of the subjects as well as therace of their physician:



)(MaleP


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(a) What is the probability that a randomly

selected subject will have an Asian/Pacific-Islander physician? Ans: 0.1533


Patient Race

TotalAsian-American

HispanicAfrican-American

CaucasianPhysiciansRace

1796175406436779White19651516214African-

American

16621281719Hispanic

417203717568Asian/Pacific-Island

1454565530Other

2720389676745910Total

(b) What is the probability that an African-Americansubject will have an African- American physician?

Ans: 0.2174

(c) What is the probability that a randomly selectedsubject in the study will be Asian-American andhave an Asian/Pacific-Islander physician?Ans: 0.075

(d) What is the probability that a subject chosen atrandom will be Hispanic or have a Hispanicphysician? Ans: 0.2625

(e) Use the concept of complementary events tofind the probability that a subject chosen at



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

Q3.4.5:

If the probability of left-handedness in a certaingroup of people is 0.5, what is the probabilityof right-handedness (assuming noambidexterity)?


Q3.4.6:The probability is 0.6 that a patient selected atrandom from the current residents of acertain hospital will be a male. The probabilitythat the patient will be a male who is in forsurgery is 0.2. A patient randomly selectedfrom current residents is found to be a male;

what is the probability that the patient is inthe hospital for surgery?

Ans: 0.3333



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Q3.4.7:

In a certain population of hospital patients the

probability is 0.35 that a randomly selectedpatient will have heart disease. The probabilityis 0.86 that a patient with heart disease is asmoker. What is the probability that a patientrandomly selected from the population will bea smoker and have heart disease?

Ans: 0.301



Baye's TheoremPages 79-83


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In this case if the patient has to do ablood test in the laboratory,some time the result is

(he has the disease) and ifPositivenegativethe result is

(he doesn't has the disease)


So, we have the following cases

The patient has the

disease(D)

The patient doesn't has

the disease( D )

Lab result is

Negative(T)

wrong result

Specificity

A symptom

P(T|D)

Lab result is

positive

( T )

Sensitivity

A symptom

P(T|D)

wrong result



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Definition.1

The sensitivity of the symptom

This is the probability of a positive result given that the

subject has the disease. It is denoted by P(T|D)

2Definition.

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)

:3Definition

The predictive value positive of the symptom

This is the probability that the subject has thedisease given that the subject has a positivescreening test result.

It is calculated using bayes theorem through thefollowing formula

Where P(D) is the rate of the disease



)()|()()|()()|()|(

DPDTPDPDTPDPDTPTDP


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Which is given by

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

the numerator is equal to sensitivityNote thattimes rate of the disease, while thedenominator is equal to sensitivity times rateof the disease plus 1 minus the specificity

times one minus the rate of the disease



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


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Example 3 5 1 page 82

A medical research team wished to evaluate a proposed screening test for

Alzheimers disease. The test was given to a random sample of 450 patients

with Alzheimers 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 yearsor older. The results are as follows.

Test Result Yes (D) No ( ) Total

Positive(T) 436 5 441

Negativ( ) 14 495 509

Total 450 500 950

T

D


In the context of this example

a)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


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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 symptomThe 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

Exercise:

Page 83

Questions :

3.5.1, 3.5.2

H.W.:

Page 87 : Q4,Q5,Q7,Q9,Q21



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Q3.5.1;A medical research team wishes toassess the usefulness of a certain symptom

(call it S) in the diagnosis of a particulardisease. In a random sample of 775 patientswith the disease, 744 reported having thesymptom. In an independent random sampleof 1380 subjects without the disease, 21reported that they had the symptom.

(a) In the context of this exercise, what is a false

positive?(b) What is a false negative?



(c) Compute the sensitivity of the symptom.

(d) Compute the specificity of the symptom.

(e) Suppose it is known that the rate of the diseasesin the general population is 0.001. what is thepredictive value positive of the symptom?

(f) What is the predictive value negative of thesymptom?



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(h) What do you conclude about the predictivevalue of the symptom on the basis of the results

obtained in part g?Q3.5.2:

Dorsay and Helms (A-6) performed aretrospective study of 71 knees scanned by MRI.One of the indicators they examined was theabsence of the bow-tie sign in the MRI asevidence of a bucket-handle or bucket-handle

type tear of the meniscus.


In the study, surgery confirmed that 43 of the71 cases were bucket-handle tears. The casesmay be cross-classified by bow-tie signstatus and surgical results as follows:


TotalTear SurgicallyConfirmed As Not

Present ( )

Tear SurgicallyConfirmed (D)

481038Positive Test(absent bow-tie sign)

(T)

23185Negative Test(bow-tie present)( )

712843Total

D

T


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

Ans: 0.8837(b) What is the specificity of testing to see if theabsent bow-tie sign indicates a meniscal tear?

Ans: 0.6229

(c) What additional information would you need todetermine the predictive value of the test?


(d) Suppose it is known that the rate of thedisease in the general population is 0.1, what isthe predictive value positive of the symptom?Ans: 0.20659

(e) What is predictive value negative of the

symptom? Ans: 0.9797



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Chapter 4:Probabilistic features of

certain data Distributions

Pages 93- 111

Key words

Probability distribution , random variable ,

Bernolli distribution, Binomail distribution,

Poisson distribution



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The Random Variable (X):

When the values of a variable (height, weight,or age) cant be predicted in advance, thevariable is called a random variable.

An example is the adult height.

When a child is born, we cant predict exactly

his or her height at maturity.


4.2 Probability Distributions forDiscrete Random Variables

Definition:

The probability distribution of a discreterandom variable is a table, graph, formula,or other device used to specify all

possible values of a discrete randomvariable along with their respectiveprobabilities.



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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,

F(x)=P(X x).


Example 4.2.1 page 94:

F(x)=

P(X x)

P(X=x)

(f/n)=

Frequeny

(f)

Number ofPrograms

0.20880.2088621

0.36700.1582472

0.49830.1313393

0.62960.13133940.82490.1953585

0.94950.1246376

0.96300.013547

1.00000.0370118

1.0000297TotalText Book : Basic Concepts and



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Properties of probability distribution

of discrete random variable. 1. 0 P(X = x) 1

2. P(X = x) =1

3. P(a X b)=P(X b)- P(X a-1)

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


Example 4.2.2 page 96: (use table inexample 4.2.1)

What is the probability that a randomlyselected family will be one who usedthree assistance programs?

Example 4.2.3 page 96: (use table in

example 4.2.1)

What is the probability that a randomlyselected family used either one or twoprograms?



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What is the probability that a family picked at

random will be one who used two or fewerassistance programs? Example 4.2.5 page 98: (use table in

example 4.2.1)What is the probability that a randomlyselected family will be one who used fewerthan four programs?


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


Example 4.2.7 page 98: (use table in

example 4.2.1)

What is the probability that a randomlyselected family is one who used betweenthree and five programs, inclusive?



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Find the following probability

For the following probability distribution table:

P(X= 15) =

P(X = 10)=

P( X = 11)=

P(X 10)=

P( 5 X 15)=

P(X 10.5) =

Mean =

Text Book : Basic Concepts andMethodology for the Health Sciences

153

P(X=x)X

0.235

0.3010

-------15

0.4420

For the following probability distribution table:

Find:

P(X 6) = P(X < 5)=

P( X = 7)=

P(X =5)=

P(X > 4)=

P( 4 X 6)= Mean=


P(X x)X

0.103

0.254

0.555

0.606

0.78718


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The Binomial Distribution:3.4 The binomial distribution is one of the most

widely encountered probability distributions inapplied 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.


The Bernoulli Process A sequence of Bernoulli trials forms a Bernoulli

process under the following conditions

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

2- The probability of a success, denoted by p, remainsconstant from trial to trial. The probability of afailure, 1-p, is denoted by q.

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



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The probability distribution of the binomial

random variableX, the number of successes in

nindependenttrials is:

Where is the number of combinations of ndistinct objects taken x ofthemata time.

* Note: 0! =1


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

nf x P X x p q x n

x

n

x

!

!( )!

n n

x n xx

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

Properties of the binomial distribution

1. f(x) 0 2. f(x) =1 3.The parameters of the binomial

distribution are n and p

4. = E(X)= n p 5.2= Var(X)= npq



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Example 4.3.1 page 100 If we examine all birth records from the North

Carolina State Center for Health statistics for year2001, we find that 85.8 percent of the pregnancieshad delivery in week 37 or later (full- term birth).

If we randomly selected five birth records from thispopulation what is the probability that exactly threeof the records will be for full-term births?

Exercise: example 4.3.2 page 104


Example 4.3.3 page 104 Suppose it is known that in a certain

population 10 percent of the population iscolor blind. If a random sample of 25 people isdrawn from this population, find the probabilitythat

a) Two or fewer will be color blind.

b) 24 or more will be color blindc) Between two and five inclusive will be color

blind.

d) At most one will be color blind.

Exercise: example 4.3.4 page 106



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The Poisson Distribution4.4 If the random variable X is the number of

occurrences of some random event in a certainperiod 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 distributionand is the average number of occurrences of the

random event in the interval (or volume)

!

x

x

e


Properties of the Poisson distribution

1. f(x) 0 2. f(x) =1 3.The parameters of the poisson

distribution is

4. = E(X)= 5.2= Var(X)=



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Example 4.4.1 page 111

In a study of a drug -induced anaphylaxis amongpatients taking rocuronium bromide as part oftheir anesthesia, Laake and Rottingen foundthat the occurrence of anaphylaxis followed aPoisson model with =12 incidents per yearin Norway .Find

1- The probability that in the next year, among

patients receiving rocuronium, exactly threewill experience anaphylaxis?


2- The probability that less than two patientsreceiving rocuronium, in the next year willexperience anaphylaxis?

3- The probability that more than two patientsreceiving rocuronium, in the next 6-month

will experience anaphylaxis?

4- The expected value of patients receivingrocuronium, in the next year who will experienceanaphylaxis.

5- The variance of patients receiving rocuronium,in the next year who will experience anaphylaxis

6- The standard deviation of patients receivingrocuronium, in the next year who will experienceanaphylaxis



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Example 4.4.2 page 111: Refer to

example 4.4.1

1-What is the probability that at least threepatients in the next year will experienceanaphylaxis if rocuronium is administered withanesthesia?

2-What is the probability that exactly one patientin the next month will experience anaphylaxis ifrocuronium is administered with anesthesia?

3-What is the probability that none of thepatients in the next month will experienceanaphylaxis if rocuronium is administered withanesthesia?


4-What is the probability that at mosttwo patients in the next year willexperience anaphylaxis if rocuronium isadministered 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



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Excercices:

111Page:4.3.4QThe same survey data base cited shows that 32 percent ofU.S adults indicated that they have been tested for HIV atsome points in their life .Consider a simple random sampleof 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:


Hint:

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

nx P X x p q x n

x



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(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.



Q4.3.5

refer to Q4.3.4 , find the mean and thevariance?

(Answer: mean = 4.8 ,

variance =3.264 )



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Q 4.4.3 :If the mean number of serious accidents per year

in a large factory is five ,find the probability thatthe current year there will be:

Hint: f(x)=

per yeara) Exactly seven accidents(

per yearb) Ten or more accidents(months6perc) No accident(

.per yeard)fewer than one accidents(


!x

e x

For Q 4.4.3:Q4.4.4

Find:

months-6per.mean1

2. variance and standard

months3deviation per



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4.5 Continuous

Probability Distribution

Pages 114127

Key words:

Continuous random variable, normaldistribution , standard normal distribution

, T-distribution



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Now consider distributions of

continuous random variables.


Properties of continuousprobability Distributions:

1- Area under the curve = 1.

2- P(X = a) = 0 , where ais a constant.

3- Area between two points a , b

= P(a


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4.6 The normal distribution:

It is one of the most important probabilitydistributions in statistics.

The normal density is given by

, - < x < , - 0

, e : constants

: population mean.

: Population standard deviation.


2

2

2

)(

2

1)(

x

exf

Characteristics of the normal distribution:Page 111

The following are some important characteristicsof the normal distribution:

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

3- The total area under the curve above the x-axis isone.

4-The normal distribution is completely determinedby the parameters and .



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5- The normal distribution

depends on the two

parameters

and

.

determines thelocation 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)


1 2 3

1< 2< 3

1

2

3

1< 2< 3

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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Characteristics of the standard

normal distribution

1- It is symmetrical about 0.2- The total area under the curve above thex-axis is one.

3- We can use table (D) in the Appendix tofind the probabilities and areas.


How to use tables of ZNote that

The cumulative probabilities P(Z z)are given in

tables 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, then

1) P( Z < 2) =0.9772

is the area to the left to 2

and it equals 0.9772.


2


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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)=0.99460.0054

= 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)=0.93700.0031

= 0.9339.


-2.74 1.53

-2.55 2.550

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

P(Z > 2.71)=10.9966 = 0.0034.

Example :

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

So,

P(Z = 0.84)= 0


0.84

2.71


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Exercise

Given Standard normal distribution by usingthe tables :

2:The area to the left of Z=1.6.4

:2.6.4

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

)=55.0: P(Z 3.6.4

)=35.2-: P(Z


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1Given the following probabilities, find z11.6.4

P(Z z1) = 0.0055 (z1=-2.54)12.6.4

P(-2.67Z z1) = 0.9718 (z1=1.97)13.6.4

P(Z > z1) = 0.0384 (z1=1.77):11.6.4

P(z1 < Z 2.98) = 0.1117 (z1=1.21)


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

(Z)?

This is done by the following formula:

Example:

If X is normal with = 3, = 2. Find the value of

standard normal Z, If X= 6?

Answer:


x

z

5.12

36

xz


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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 withMean of 5.4 hours and standard deviation of 1.3.Find


If a child selected at random ,then1-The probability that the child spend less than 3hours 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 5hours 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 6.2hours 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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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


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Exercises

oldyears-29For another subject (:1.7.4Qmale) 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


In the study of fingerprints an important:2.7.4Qquantitative 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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(b) less than 200 (Answer :0.8849)

(c) between 100 and 200

(Answer :0.6982)

(d) between 200 and 250

(Answer :0.0934)


The T Distribution:

)7367

1- It has mean of zero.

2- It is symmetric about the

mean.3- It ranges from -to .


0


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4- compared to the normal distribution, the t

distribution is less peaked in the center andhas higher tails.

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

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

7- Can find values of t in the table (E) in

the Appendix.


Examples

t (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

0.99

0.01

t


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0

Find :

t0.95,10

= 1.8125

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

t 0.975,18 = 2.1009

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

t 0.01,20 = - 2.528

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

t 0.10,29 = - 1.311

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


Sampling4.6Distributions:

Definition:1.4.6

The probability distribution of a statistic is called a

sampling distribution.

Sampling Distributions of1.1.4.6

Means:

2

2( ) , ( ) (3)X X

E X V Xn


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1

Definition:2.4.6

Central Limit Theorem:1.2.4.6

If is the mean of a random sample of size ntaken from a population with mean and finitevariance , then the limiting form of the

distribution of:

is approximately the standard normal

distribution;

(4)/

XZ as n

n

X

2

( ) ~ (0,1) ( ) , ( ) ,

~ ( , ) (5)

XZ N E X V X

n n

X Nn

:Example

An electrical firm manufactures light bulbs that have a

length of life that is approximately normally

distributed with mean equal to 800hours and a

standard deviation of 40hours. Find :

probability that a random sample of 16

bulbs will have an average life of lessthan 775hours.


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2

:Solution

Let Xbe the length of life and is theaverage life;

X

16, 800, 40

775 800( 775) ( ( 2.5) 0.0062

40/ 16

n

P X P Z P Z

Sampling Distribution of the sampleProportion:

Let X= no. of elements of type Ain the

sample

P= population proportion = no. of elements

of type A in the population / N = sample proportion= no. of elements of

type A in the sample / n = x/n

p


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3

3. For large n, we have:

~ ( , )

~ (0,1)

pqp N p

n

p pZ N

pq

n

~ ( , ) ( ) , ( )

1. ( ) ( )

2. ( ) ( ) , 1

x binomial n p E x np V x npq

xE p E pn

x pqV p V q p

n n

Example:

If X is Binomial distribution with n=10,P=0.1, find P

Solution:


)2.0( P

85314.0)05.1(

)009.0

1.0()

10

)9.0)(1.0(

1.02.0()2.0(

009.0

10

)9.0)(1.0()(

1.0)(

ZP

ZP

n

pq

PPPPP

n

pqPV

PPE


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4

6Chapter

Using sample data to make

estimates about population

)172-162parameters (P

Key words:

Point estimate, interval estimate, estimator,

Confident level ,, Confident interval for mean ,

Confident interval for two means,

Confident interval for population proportion P,

Confident interval for two proportions



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5

6.1 Introduction: Statistical inference is the procedure by which we reach toa conclusion about a population on the basis of the

information contained in a sample drawn from thatpopulation.

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

Apoint estimateis a single numerical value used toestimate the corresponding population parameter.

Note:

Point estimate for ( ) is

Point estimate for ( ) is S


X


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6

Definition:

An interval estimateconsists of two

numerical values defining a range of valuesthat, with a specified degree ofconfidence, we feel includes theparameter being estimated.


aConfidence Interval for2.6Population Mean: (C.I)

Suppose researchers wish to estimate the mean ofsome normally distributed population.

They draw a random sample of size n from thepopulation and compute , which they use as a pointestimate of .

Because random sampling involves chance, thencant be expected to be equal to .

The value of may be greater than or less than.

It would be much more meaningful to estimate byan interval.

x


x


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7

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

We want to find two values L :Lower bound andU:Upper bound between which lies with highprobability, i.e.

P( L U ) = 1-


For example:

When,

= 0.01,

then 1- =

= 0.05,

then 1- =

= 0.05,then 1- =



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8

For example:

When, (1- )100%:Level of confident

(1- )100% = 90%,

then =

99%

then =

80%

then =


(1- )100% confident interval for the mean : When the value of sample size (n):

population is normal or not normal population is normal

( n 30 ) (n< 30)

is known is not k

wayne daniel ppt

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