lectures of stat -145 (biostatistics) text book biostatistics basic concepts and methodology for the...
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Lectures of Stat -145(Biostatistics)
Text bookBiostatistics
Basic Concepts and Methodology for the Health Sciences
ByWayne W. Daniel
Prepared By:Sana A. Abunasrah
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Health Sciences 2
Chapter 1
Introduction To Biostatistics
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Health Sciences 3
Key words :
Statistics , data , Biostatistics, Variable ,Population ,Sample
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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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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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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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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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Exercises• Question (6) – Page 17• 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)
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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
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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.
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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)
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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.
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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.
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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:
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Class intervalFrequency
30 – 3911
40 – 4946
50 – 5970
60 – 6945
70 – 7916
80 – 891
Total189
Sum of frequency=sample size=n
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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 . .
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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
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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 ?
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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) ?
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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
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Representing the grouped Representing the grouped frequency table using the frequency table using the
PolygonPolygon
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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..
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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 . .
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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
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((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??
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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
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* *A histogramA histogram
* *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
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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..
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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 histogramA histogram
* *A frequency polygonA frequency polygon..
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Section (2.4) :Section (2.4) : Descriptive Statistics Descriptive Statistics
Measures of Central Measures of Central Tendency Tendency
Page 38 - 41Page 38 - 41
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key words:
Descriptive Statistic, measure of central tendency ,statistic, parameter, mean )μ( ,median, mode.
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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.
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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.
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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
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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 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.
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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.
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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 5.5th
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.
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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)
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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)
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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
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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)
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)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
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key words:
Descriptive Statistic, measure of dispersion , range ,variance, coefficient of variation.
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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.
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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(.
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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
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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
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• 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
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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
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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
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• 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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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
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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
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l106
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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%
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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 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.
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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
(a )The mean )b( The median
Ans: 47.72 Ans: 46.35
(c )The mode )d( The range
Ans: 33, 54 Ans: 29.6
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(e )The variance )f( The standard deviation
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
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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
(a )The mean )b( The median
Ans: 152.84 Ans: 184.11
(c )The mode )d( The range
Ans: no mode Ans: 220.19Text Book : Basic Concepts and Text Book : Basic Concepts and
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(e )The variance )f( The standard deviation
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 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
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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
(a )The mean )b( The median
Ans: 16.75 Ans: 15
(c )The mode )d( The range
Ans: 15 Ans: 43
(e )The variance )f( The standard deviation
Ans: 124.0227 Ans: 11.1365
(g )The coefficient of variation Ans: 66.49%
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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 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
Methodology for the Health Sciences Methodology for the Health Sciences 8383
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
(a )The mean )b( The median
Ans: 41.5 Ans: 39.5
(c )The mode )d( The range
Ans: 19 Ans: 87
(e )The variance )f( The standard deviation
Ans: 490.264 Ans: 22.1419Text Book : Basic Concepts and Text Book : Basic Concepts and
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(g )The coefficient of variation Ans: 53.35%
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.
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37 48 53 46 42 49 44 38 32 32 51 51 48 41
(a )The mean )b( The median
Ans: 43.7143 Ans: 45
(c )The mode )d( The range
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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Chapter 3Chapter 3
ProbabilityProbability
The Basis of the The Basis of the Statistical inferenceStatistical inference
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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
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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.
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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 .
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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.
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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.
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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
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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
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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
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****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
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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
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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)?
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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).
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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.
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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
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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)
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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?
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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
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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
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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
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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
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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
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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??
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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
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““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
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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::
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)(MaleP
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((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
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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) ? ?
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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
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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
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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)
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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
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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
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)()|()()|(
)()|()|(
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
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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 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
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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
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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 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
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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
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((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
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((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 . .
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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::
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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??
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((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
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Chapter 4:Probabilistic features of certain data DistributionsPages 93- 111
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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( 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.
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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.
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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, P)X x(.
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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
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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
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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?
Example 4.2.3 page 96: (use table in example 4.2.1)
What is the probability that a randomly selected family used either one or two programs?
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Example 4.2.4 page 98: (use table in example 4.2.1)
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?
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Example 4.2.7 page 98: (use table in example 4.2.1)
What is the probability that a randomly selected family is one who used between three and five programs, inclusive?
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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.
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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
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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
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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
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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 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
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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
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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
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Properties of the Poisson distribution
1.
2.
3.
4.
( ) 0f x
( ) 1f x ( )E X
2 var( )X
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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?
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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
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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?
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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
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Hint:
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( ) ( ) , 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.
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Q4.3.5
• refer to Q4.3.4 , find the mean and the variance?
•(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 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(
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!x
e x
Q4.4.4
Find mean and variance and standard
deviation for Q 4.4.3
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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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• Key words: Continuous random variable,
normal distribution , standard normal distribution , T-distribution
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• Now consider distributions of continuous random variables.
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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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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
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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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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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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 the x-axis is one.
3 -We can use table (D) to find the probabilities and areas.
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“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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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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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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•4.6.7:
P( -1.95 < Z < 1.95 )=
4.6.10:
P( Z = 1.22)=
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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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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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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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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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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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• 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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• 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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•(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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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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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 .
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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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Chapter 6Using sample data to make estimates about population parameters (P162-172)
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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
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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.
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• 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
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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
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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-
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For example: When, = 0.01, then 1- = = 0.05, then 1- = = 0.05, then 1- =
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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
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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
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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
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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
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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
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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,
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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
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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 ?
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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
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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?
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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
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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
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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(
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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
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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
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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
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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
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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.
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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
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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 .
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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
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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
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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
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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.
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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
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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
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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
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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
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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-
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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)
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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)
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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.
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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))
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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?
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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
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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)
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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
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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
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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
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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.
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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
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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
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SolutionSolution : : 11..DataData: :
22 . .AssumptionAssumption: :
33 . .HypothesisHypothesis: :
44..Test statisticTest statistic: :
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55..Decision RuleDecision Rule
66 . .DecisionDecision : :
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ExercisesExercisesQ7.2.3Q7.2.3::
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
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SolutionSolution : : 11..DataData: :
22 . .AssumptionAssumption: :
33 . .HypothesisHypothesis: :
44..Test statisticTest statistic: :
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55..Decision RuleDecision Rule
66 . .DecisionDecision : :
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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?? ??
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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).
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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.
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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
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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
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5.Decision Rule:5.Decision Rule:
i) i) If If HHAA: : μ μ 1 1 ≠ ≠ μ μ 2 2 → → μ μ 1 1 -- μ μ 2 2 ≠ 0≠ 0
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 ,(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)
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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
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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
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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:
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
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
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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
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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
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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
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5. Decision Rule:5. Decision Rule:
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
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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
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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
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5. Decision Rule:5. Decision Rule:
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
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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
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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
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5.Decision Rule:5.Decision Rule:
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
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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
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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
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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
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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
ExercisesExercisesQ7.5.2Q7.5.2::
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
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SolutionSolution : : 11..DataData: :
22 . .AssumptionAssumption: :
33 . .HypothesisHypothesis: :
44..Test statisticTest statistic: :
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55..Decision RuleDecision Rule
66 . .DecisionDecision : :
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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
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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
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5.Decision Rule:5.Decision Rule:
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
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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
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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::
5.Decision Rule:5.Decision Rule:
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
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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