Unformatted text preview: STAT 6200  Introduction to Biostatistics
Lecture Notes
Introduction*
Statistics and Biostatistics: The eld of statistics: The study and use of theory and methods for the
analysis of data arising from random processes or phenomena. The study
of how we make sense of data.
The eld of statistics provides some of the most fundamental tools
and techniques of the scienti c method:
{ forming hypotheses,
{ designing experiments and observational studies,
{ gathering data,
{ summarizing data,
{ drawing inferences from data (e.g., testing hypotheses)
A statistic (rather than the eld of \statistics") also refers to a
numerical quantity computed from sample data (e.g., the mean, the
median, the maximum).
Roughly speaking, the eld of statistics can be divided into
Mathematical Statistics: the study and development of statistical
theory and methods in the abstract and
Applied Statistics: the application of statistical methods to solve real
problems involving randomly generated data, and the development
of new statistical methodology motivated by real problems. * Read Ch.1 of our text.
1 Biostatistics is the branch of applied statistics directed toward applications in the health sciences and biology.
Biostatistics is sometimes distinguished from the eld of biometry
based upon whether applications are in the health sciences (biostatistics) or in broader biology (biometry e.g., agriculture, ecology,
wildlife biology).
Other branches of (applied) statistics: psychometrics, econometrics,
chemometrics, astrostatistics, environmetrics, etc.
Why biostatistics? What's the di erence?
Because some statistical methods are more heavily used in health
applications than elsewhere (e.g., survival analysis, longitudinal data
analysis).
Because examples are drawn from health sciences.
{ Makes subject more appealing to those interested in health.
{ Illustrates how to apply methodology to similar problems encountered in real life.
We will emphasize the methods of data analysis, but some basic theory
will also be necessary to enhance understanding of the methods and to
allow further coursework.
Mathematical notation and techniques are necessary! (No apologies.)
We will study what to do and how to do it, but also very important is why
the methods are appropriate and what are the concepts justifying those
methods.
The latter (the why) will get you further than the former (the what). 2 Data* Data Types:
Data are observations of random variables made on the elements of a
population or sample.
Data are the quantities (numbers) or qualities (attributes) measured
or observed that are to be collected and/or analyzed.
The word \data" is plural, \datum" is singular!
A collection of data is often called a data set (singular). Example  Low Birth Weight Infant Data Appendix B of our text contains a data set called lowbwt containing measurements and observed attributes on 100 low birth weight
infants born in two teaching hospitals in Boston, MA.
The variables measured here are
sbp = systolic blood pressure
sex = gender (1=male, 0=female)
tox = maternal diagnosis of toxemia (1=yes, 0=no)
grmhem = whether infant had a germinal matrix hemorrhage (1=yes, 0=no)
gestage = gestational age (weeks)
apgar5 = Apgar score (measures oxygen deprivation) at 5 minutes after birth
Data are reproduced on the top of the following page. * Read Ch.2 of our text.
3 There are 6 variables here (sbp, sex, etc.) measured on 100
units/elements/subjects (the infants) of a random sample of size 100.
An observation can refer to the value of a single variable for a particular subject, but more commonly it refers to the observed values
of all variables measured on a particular subject.
{ There are 100 observations here. 4 Types of Variables:
Variable types can be distinguished based on their scale. Typically, di erent statistical methods are appropriate for variables of di erent scales.
Scale
Nominal Characteristic Question
Is A di erent than B? Ordinal Is A bigger than B? Interval By how many units do A and B di er? Ratio How many times bigger than B is A? Examples
Marital status
Eye color
Gender
Religious a liation
Race
Stage of disease
Severity of pain
Level of satisfaction
Temperature
SAT score
Distance
Length
Time until death
Weight Operations that make sense for variables of di erent scales:
Scale
Nominal
Ordinal
Interval
Ratio Counting
p
p
p
p Operations that make sense
Addition/
Multiplication/
Ranking Subtraction Division
p
p
p p
p p Often, the distinction between interval and ratio scales can be ignored in statistical analyses. Distinction between these two types
and ordinal and nominal are more important.
5 Another way to distinguish between types of variables is as quantitative
or qualitative.
Qualitative variables have values that are intrinsically nonnumeric
(categorical).
{ E.g., Cause of death, nationality, race, gender, severity of pain
(mild, moderate, severe).
{ Qualitative variables generally have either nominal or ordinal
scales.
{ Qualitative variables can be reassigned numeric values (e.g.,
male=0, female=1), but they are still intrinsically qualitative.
Quantitative variables have values that are intrinsically numeric.
{ E.g., survival time, systolic blood pressure, number of children
in a family, height, age, body mass index.
Quantitative variables can be further subdivided into discrete and continuous variables.
Discrete variables have a set of possible values that is either nite or
countably in nite.
{ E.g., number of pregnancies, shoe size, number of missing teeth.
{ For a discrete variable there are gaps between its possible values. Discrete values often take integer (whole numbers) values
(e.g., counts), but some discrete variables can take noninteger
values.
A continuous variable has a set of possible values including all values
in an interval of the real line.
{ E.g., duration of a seizure, body mass index, height.
{ No gaps between possible values. 6 The distinction between discrete and continuous quantitative variables is
typically clear theoretically, but can be fuzzy in practice.
In practice the continuity of a variable is limited by the precision of
the measurement. E.g., height is measured to the nearest centimeter,
or perhaps millimeter, so in practice heights measured in millimeters
only take integer values.
{ Another example: survival time is measured to the nearest day,
but could, theoretically, be measured to any level of precision.
On the other hand, the total annual attendance at UGA football
games is a discrete (inherently integervalued) variable, but, in practice, can be treated as continuous.
In practice, all variables are discrete, but we treat some variables
as continuous based upon whether their distribution can be \well
approximated" by a continuous distribution. Data Sources: Data arise from experimental or observational studies, and it is important
to distinguish the two.
In an experiment, the researcher deliberately imposes a treatment
on one or more subjects or experimental units (not necessarily human). The experimenter then measures or observes the subjects'
response to the treatment.
{ Crucial element is that there is an intervention.
Example: To assess whether or not saccharine is carcinogenic, a researcher feeds 25 mice daily doses of saccharine. After 2 months, 10 of
the 25 mice have developed tumors.
By de nition, this is an experiment, but not a very good one.
In the saccharine example, we don't know whether 10/25 with tumors is
high because there is no control group to which comparison can be made.
Solution: Select 25 more mice and treat them exactly the same but give
them daily doses of an inert substance (a placebo). 7 Suppose that in the control group only 1 mouse develops a tumor. Is this
evidence of a carcinogenic e ect?
Maybe, but there's still a problem:
What if the mice in the 2 groups di er systematically? E.g., group
1 from genetic strain 1, group 2 from genetic strain 2.
Here, we don't know whether saccharine is carcinogenic, or if genetic strain
1 is simply more susceptible to tumors.
We say that the e ects of genetic strain and saccharine are confounded (mixed up).
Solution: Starting with 50 relatively homogeneous (similar) mice, randomly assign 25 to the saccharine treatment, and 25 to the control treatment.
Randomization an extremely important aspect of experimental design.
{ In the saccharine example, we should start out with 50 homogeneous mice, but of course they will di er some. Randomization
ensures that the two experimental groups will be probabilistically alike with respect to all nuisance variables (potential
confounders). E.g., the distribution of body weights should be
about the same in the two groups. 8 Another important concept, especially in human experimentation, is blind ing. An experiment is blind if the subjects don't know which treatment
they receive.
E.g., suppose we randomize 25 of 50 migraine su erers to an active
drug and the remaining 25 to a placebo control treatment.
{ Experiment is blind if pills in the two treatment groups look
and taste identical and subjects are not told which treatment
they receive.
{ This guards against the placebo e ect.
An experiment is doubleblind if the researcher who administers
the treatments and measures the response does not know which treatment is assigned.
{ Guards against experimenter e ects. (Experimenter may
behave di erently toward the subjects in the two groups, or
measure the response di erently in the two groups.)
Experiments are to be contrasted with observational studies.
No intervention.
Data collected on an existing system.
{ Less expensive.
{ Easier logistically.
{ More often ethically practical.
{ Interventions often not possible.
Experiments have many advantages and are strongly preferred when
possible. However, experiments are rarely feasible in public health/epidemiology.
{ In health sciences/medicine, experiments involving humans are
called clinical trials. 9 Types of Observational Studies:
1. Case studies or case series.
{ A descriptive account of interesting characteristics (e.g., symptoms) observed in a single case (subject with disease) or in a
sample of cases.
{ Typically are unplanned and don't involve any research hypotheses. No comparison group.
{ Poor design, but can generate research hypotheses for subsequent investigation.
2. Casecontrol study.
{ Conducted retrospectively (by looking into past).
{ Two types of subjects included:
cases = subjects with the disease/outcome of interest
controls = subjects without the disease/outcome
{ History of two groups is examined to determine which subjects
were exposed to, or otherwise possessed, a prior characteristic.
Association between exposure and disease then quanti ed.
{ Controls are often matched to cases based on similar characteristics.
Advantages:
{ Useful for studying rare disease.
{ Useful for studying diseases with long latency periods.
{ Can explore several potential risk factors (exposures) for disease simultaneously.
{ Can use existing data sources  cheap, quick, easy to conduct.
Disadvantages:
{ Prone to methodological errors and biases.
{ Dependent on high quality records.
{ Di cult to select an appropriate control group.
{ More di cult statistical methods required for proper analysis.
10 3. Crosssectional Studies.
Collect data from a group of subjects at one point in time.
Sometimes called prevalence studies, due to their focus on a single
point in time.
Advantages:
{ Often based on a sample of the general population, not just
people seeking medical care.
{ Can be carried out over a relatively short period of time.
Disadvantages:
{ Di cult to separate cause and e ect because measurement of
exposure and disease are made at one point in time, so it may
not be possible to determine which came rst.
{ Are biased toward detecting cases with disease of long duration
and can involve misclassi cations of cases in remission or under
e ective medical treatment.
{ Snapshot in time can be misleading in a variety of other ways. 11 4. Cohort Studies.
{ Usually conducted prospectively (forward in time).
{ A cohort is a group of people who have something in common
at a particular point in time and who remain part of the group
through time.
{ A cohort of diseasefree subjects are selected and their exposure
status evaluated at the start of the study.
{ They are then followed through time in order to observe who
develops disease. Association between exposures (risk factors)
and disease are then quanti ed.
Advantages:
{ Useful when exposure of interest is rare.
{ Can examine multiple e ects (e.g., diseases) of a single exposure.
{ Can elucidate temporal relationship between exposure and disease, thereby getting closer to causation.
{ Allows direct measurement of incidence of disease.
{ Minimizes bias in ascertainment of exposure.
Disadvantages:
{ Ine cient for studying rare diseases.
{ Generally requires a large number of subjects.
{ Expensive and timeconsuming.
{ Subjects can be lost to followup (drop out of study) leading
to bias.
Cohort studies can also be conducted retrospectively by identifying
a cohort in present, determining exposure status in past, and then
determining subsequent disease occurrence between time of exposure
and present through historical records. 12 Data Presentation: Even quite small data sets are di cult to comprehend without some summarization. Statistical quantities such as the mean and variance can be
extremely helpful in summarizing data, but rst we discuss tabular and
graphical summaries.
Tables:
One of the most important means of summarizing the data from a single
variable is to tabulate the frequency distribution of the variable.
A frequency distribution simply tells how often a variable takes on
each of its possible values. For quantitative variables with many
possible values, the possible values are typically binned or grouped
into intervals. Example  Gender in this Class (Nominal Variable):
Gender
Female
Male
Total Frequency Relative
Frequency
(proportion) Relative
Frequency
(percent) Here, the relative frequency as a proportion is just
Relative frequency (proportion) = Frequency=n
where n =sample size.
The relative frequency as a percent is
Relative Frequency (percent) = Relative frequency (proportion) 100%
It is worth distinguishing between the empirical relative frequency
distribution, which gives the proportion or percentage of observed
values, and the probability distribution, which gives the probability
that a random variable takes each of its possible values.
{ The latter can be thought of as the relative frequency distribution for an in nite sample size.
13 Example  Keypunching Errors (Discrete Quantitative Variable):
A typist entered 156 lines of data into a computer. The following table
gives the number of errors made for each line. Number of
Errors
0
1
2
3 or more
Total Frequency
124
27
5
0 Relative
Frequency (%) Here, it was not necessary to bin the data. 14 Example  Age at Death (in Days) for SIDS Cases:
The following table contains the age at death in days for 78 cases of sudden
infant death syndrome (SIDS, or Crib Death) occuring in King County,
WA, during 1976{1977. Age Interval
(Days)
1{30
31{60
61{90
91{120
121{150
151{180
181{210
211{240
241{270
271{300
301{330 Frequency
6
13
23
18
6
5
3
2
0
1
1 Relative
Frequency (%)
7.69
16.67
29.49
23.08
7.69
6.41
3.85
2.56
0
1.28
1.28 Cumulative
Frequency
6
19
42
60
66
71
74
76
76
77
78 Cumulative
Relative
Frequency (%)
7.69
24.36
53.85
76.92
84.62
91.03
94.87
97.44
97.44
98.72
100.00 Here it is necessary to bin the data. The bins should be
{ Mutually exclusive (nonoverlapping).
{ Exhaustive (every observed value falls in a bin)
{ The handling of cutpoints between bins should be consistent
and clearly de ned.
{ (preferrably) The bins should be of equal width, although it
can be better to violate this rule sometimes, especially for the
smallest and largest bins.
15 In this example we have also tabulated the cumulative frequency and
the cumulative relative frequency. The cumulative frequency simply
counts the number of observations the current value (or current
bin if the data are binned).
{ The cumulative relative frequency expresses the same information as a percent by multiplying by 100% .
Graphs:
Frequency distributions can often be displayed e ectively using graphical
means such as the bar chart, pie chart, or histogram.
Pie charts are useful for displaying the relative frequency distribution
of a nomianl variable. Here is an example created in Minitab of the
relative frequency distribution of the school a liation of students in
this class.
n A legend, or key is important in many di erent graph types, but is
especially crucial in a pie chart. 16 Bar charts display absolute or relative frequency distributions for
categorical variables (ordinal or nominal). Here is a Minitab bar
chart of the school a liations of students in this class. Note that the horizontal axis in a bar chart has no scale. The categories can be reordered arbitrarily without a ecting the information
contained in the plot.
A histogram depicts the frequency distribution of a quantitative random
variable. Below is a histogram of the Age at Death data for the 78 SIDS
cases in King Co., WA. 17 Histograms are sometimes constructed so that the height of each bar
gives the frequency (or relative frequence) in each interval. This is
ok if the intervals all have the same width, but can be misleading
otherwise.
{ Here's an example of what can go wrong with unequal bins
when frequency or relative frequency is plotted. The above example can be xed by making the relative frequency
in each interval equal to the area in each bar, not the height. That
is, the height of each bar should be equal to Rel Freq=Bin Width.
Here's a xed version of the histogram given above. 18 Note that the choice of number of bins and bin width can a ect histograms
dramatically. Here's a di erent choice of bins for the SIDS data (a bad
choice). It is not easy to give a general rule on how many bins should be used
in a histogram, but somewhere between 5 and 20 bins is typically
avisable.
Frequency polygons are formed by plotting lines between the midpoints
of the tops of the bars of a histogram. The histogram should have equal
bin widths and the lines should extend down to 0 at the right and left
extremes of the data.
Here is a frequency polygon for the SIDS data. Its principle advantages are that (i) it is continuous, and (ii) multiple frequency
polygons can be displayed on the same plot. 19 A oneway scatter plot is just a plot of the real line with tick marks, or
sometimes dots, at each observed value of the variable. Here is a oneway
scatter plot, or dotplot, for the SIDS data Another plot useful for summarizing the distribution of a single variable
is the boxplot.
A boxplot summarizes the distribution of a variable by locating the 25th,
50th and 75th percentiles of the data, plus two adjacent values.
A pth percentile of a data set is a number such that at least p% of
the data are this value and at least 100 ; p% of the values are
this value.
{ The median is the 50th percentile.
{ The 25th , 50th and 75th percentiles are sometimes called the
rst, second, and third quartiles of the data.
The box in a boxplot extends from the 25th to the 75th percentiles
of the data. The line in the box locates the median. 20 Here is a boxplot for the SIDS data. This boxplot also includes a oneway
scatterplot of the data. The lines extending on either side of the box are called whiskers.
They indicate roughly the extent of the data.
{ The whiskers sometimes extend to the 10th and 90th percentiles.
{ In Minitab's implementation of a boxplot, however, the whiskers
extend to the adjactent values, which are de ned to be the most
extreme values in the data set that are not more than 1.5 times
the width of the box beyond either quartile.
{ The width of the box is the distance between the rst and third
quartile. This distance is called the interquartile range.
The term outlier is used to refer to data points that are not typical
of the rest of the values. Exactly what constitutes \not typical" is
somewhat controversial, but one way to de ne an outlier is as a point
beyond the adjacent values.
{ Based on this de nition there are four large outliers in the SIDS
data (marked by *'s) and no small outliers. 21 There are a variety of tabular and graphical methods to summarize the
joint frequency distribution of two variables.
For two qualitative variables, a contingency table or crosstabulation
is useful.
This is just a table where the rows represent the values of one variable, the columns the values of the other variable, and the cells give
the frequency with which each combination of values is observed.
Here is a contingency table giving the joint frequency distribution of
grmhem (germinal matrix hemorrhage) and tox (diagnosis of toxemia for
mother) for the low birth weight data:
Germinal
Hemorrhage
No
Yes
Toxemia
No
65
14
79
Yes
20
1
21
85
15
100
Notice that the margins of the table give the univariate frequency
distributions of the two variables.
Crosstabulations can be constructed when one or more of the variables
are quantitative. In this case, it may be necessary to bin the quantitative
variable(s). E.g., here is a crosstab of gestage (gestastional age) and
grmhem:
Germinal
Hemorrhage
No
Yes
23{25
7
4
11
26{28
26
4
30
Gestational Age
29{31
38
6
44
32{34
13
1
14
35
1
0
1
85
15
100
22 For displaying the relationship between a quantitative variable and a qualitative variable, sidebyside boxplots or superimposed frequency polygons
can be useful. The most useful graphical tool for displaying the relationship between two
quantitative variables is a twoway scatterplot. Here is one that displays
systolic blood pressure vs. gestational age for the low birth weight data. Line graphs are useful when a variable is measured at each of many con secutive points in time (or some other dimension like depth of the ocean).
In such a situation it is useful to construct a scatterplot with the measured
variable on the vertical axis and time on the horizontal. Connecting the
points gives a sense of the time trend and any other temporal pattern (e.g.,
seasonality).
Above is a line graph displaying US Life Expectancies over time.
Male and female life expectancies are plotted on the same graph
here. 23 Numerical Summary Measures* In talking about numerical summary measures, it is useful to distinguish
between whether a particular quantity (such as the mean) is computed on
a sample or on an entire population.
Sometimes, we have data on all units (e.g., subjects) in which we have
interest. That is, we have observations on the entire population.
E.g., the diameters of the nine planets of the solar system are
Planet
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto Diameter (miles)
3030
7520
7926
4217
88838
74896
31762
30774
1428 The mean diameter of the 9 planets is
(3030 + 7520 + + 1428)=9 = 27 821:22 miles. Summary measures such as the mean and variance are certainly useful for
such data, but it is important to realize that there is no need to estimate
anything here or to perform statistical inference.
The mean diameter of the 9 planets in our solar system is 27,821.22
miles. This is a population quantity or parameter that can be
computed from direct measurements on all population elements. * Read Ch.3 of our text.
24 In contrast, the more common situation is one in which we can't observe
the entire population. Instead we select a subset of the population called
a sample, chosen to be representative of the population of interest.
Summary measures computed on the sample are used to make statistical
inference on the corresponding population quantities.
That is, we don't know the parameter, so we estimate its value from
a sample, quantify the uncertainty in that estimate, test hypotheses
about the parameter value, etc.
E.g., we don't know the proportion of US registered voters who approve of President Bush's job performance, so we take a representative sample of the population of size 1,000, say, and ask each sample
member whether they approve. The proportion of these 1,000 sample members who approve (a sample statistic) is used to estimate the
corresponding proportion of the total US population (the parameter).
{ Note that this estimate will almost certainly be wrong. One of
the major tasks of statistical inference is in determining how
wrong it is likely to be. 25 Notation:
Random variables will be denoted by Roman letters (e.g., x y).
Sample quantities: Roman letters (e.g., mean=x, variance=s2)
Population quantities: Greek letters (e.g., mean= , variance= 2 ).
Suppose we have a sample on which we measure a random variable that
we'll call x (e.g., age at death for 78 SIDS cases):
225 174 274 164 : : : 32 44
A convenient way to refer to these numbers is as x1 x2 : : : x where n is
the sample size. Here,
n x1 = 225 x2 = 174 x3 = 274 : : : x77 = 32 x78 = 44:
Summation notation: many statistical formulas involve summing a series
of number like this, so it is convenient to have a shorthand notation for
P
x1 + x2 + + x . Such a sum is denoted by =1 x . That is,
n n i X
n x = x1 + x2 +
i =1 i +x :
n i Similarly,
3
X 4(x i =1 ; yi )2 = 4(x1 ; y1 )2 + 4(x2 ; y2)2 + 4(x3 ; y3 )2 : i 26 Measures of Location:
Mean: The sample mean measures the location or central tendency of the observations in the sample. For a sample x1 : : : x , the mean is
denoted by x and is computed via the formula
n 1 Xx :
+x )= n
=1 1
x = n (x1 + x2 + n n i i The mean gives the point of balance for a histogram of the sample
values and is a ected by every value in the sample.
Sample mean for age at death, SIDS cases:
78
1 X x = 1 (225 + 174 +
x = 78
78
=1
i + 44) = 99:29 i The population mean is the same quantity computed on all the elements in the population.
{ In the SIDS example, the population is not clearly de ned. We
may think of the SIDS cases in 1976{77 in King Co. Washington as representative of the entire US or of similar metropolitan
areas in the US at that point in time, or as representative of
King County at points in time other than 1976{77.
The mean is not an appropriate measure for ordinal or nomianl variables. 27 Median: One feature of the mean that is sometimes undesirable is that it is a ected by every value in the data set. In particular, this means that
it is sensitive to extreme values, which at times may not be typical of the
data set as a whole.
The median does not have this feature, and is therefore sometimes
more appropriate for conveying the \typical" value in a data set.
The median is de ned as the 50th percentile or middle value of a data set.
That is, the median is a value such that at least half of the data are greater
than or equal to it and at least half are less than or equal to it.
If n is odd, this de nition leads to a unique median which is an
observed value in the data set.
E.g., 9 health insurance claims (dollar amounts):
data: 1100 1900 600 890 690 890000 380 1200 1050
sorted data: 380 600 690 890 1050 1100 1200 1900 890000
) median = 1050 whereas the mean = 99 756:67
If n is even, there are two \middle values" and either middle value or
any number in between would satisfy the de nition. By convention
we take the average of the two middle values.
sorted data: 600 690 890 1050 1100 1200 1900 890000
) median = (1050 + 1100)=2 = 1075
Notice that the median is una ected by the size of the largest claim.
The median is appropriate for ordinal qualitative data as well as
quantitative data. 28 Mode: The mode is simply the most commonly occurring value. This quantity is not unique there may be multiple modes.
In the insurance claims data, all values were distinct, so all values
were modes.
A histogram of the apgar5 data is given below. From this plot it is
easy to see that the mode is 8. The mean is 6.25 and the median is
7. The mode is especially useful for describing qualitative variables or
quantitative variables that take on a small number of possible values.
{ The modal gender in this class is female. The modal academic
program a liation is BHSI.
If two values occur more than others but equally frequently, we say
the data are bimodal, or more generally multimodal.
{ The term bimodal is also sometimes used to describe distributions in which there are two peaks, not necessarily of the same
height. E.g.: 29 Percentiles: Earlier, we noted that the median is the 50th percentile. We
gave the de nition of a percentile on p.20. A procedure for obtaining the
pth percentile of a data set of size n is as follows:
Step 1: Arrange the data in ascending (increasing) order.
Step 2: Compute an index i as follows: i = 100 n.
Step 3:
{ If i is an integer, the pth percentile is the average of the ith and
(i + 1)th smallest data values.
{ If i is not an integer then round i up to the nearest integer and
take the value at that postion.
For example, consider the 9 insurance claims again:
p sorted data: 380 600 690 890 1050 1100 1200 1900 890000
{ For the p = 10th percentile, i = pn=100 = 10(9)=100 = :9.
Round up to 1, so that the 10th percentile is the rst sorted
value, or 380.
{ For the p = 75th percentile, i = pn=100 = 75(9)=100 = 6:75.
Round up to 7, so that the 75th percentile is the seventh sorted
value, or 1200.
Percentiles not only give locate the center of a distribution (e.g., the
median), but also other locations in a distribution. 30 Measures of Dispersion: The two most important aspects of a unimodal distribution are the location
(or central tendency) and the spread (or dispersion).
E.g., consider the time it takes to commute to work by car, train,
and bike. Suppose these are the distributions of commute time by
these modes of transportation. Comparison
Train & Car
Train & Bike
Car & Bike Location
Same
Di erent
Di erent Spread
Di erent
Same
Di erent Measures of Dispersion: Range, Interquartile Range, Variance and Standard Deviation, Coe cient of Variation. 31 Range: The range is simple the maximum value minus the minimum
value in the data set: range = max ; min: The range of the 9 insurance claims was 890 000 ; 380 = 889 620.
Interquartile Range: The range only depends upon the minimum and
maximum, so it is heavily in uenced by the extremes.
That is, the range may not re ect the spread in most of the data.
The interquartile range is the di erence between the third quartile (75th
%'ile) and the rst quartile (25th %'ile). That is,
IQR = Q3 ; Q1 :
For the insurance claim data, we computed the 75th %'ile as Q3 =
1200. To get the 25th percentile, i = pn=100 = 25 9=100 = 2:25.
Rounding up, we take the third smallest value, or Q1 = 690. Thus
IQR = $1200 ; $690 = $510: 32 Variance and Standard Deviation: The most important measures of
dispersion are the variance and its square root, the standard deviation.
Since the variance is just the square of the standard deviation, these
quantities contain essentially the same information, just on di erent
scales.
The range and IQR each take only two data points into account.
How might we measure the spread in the data accounting for the
value of every observation?
Consider the insurance claim data again:
Observation
Number (i)
1
2
3
4
5
6
7
8
9
Sum=
Mean= x
1100
1900
600
890
690
890000
380
1200
1050
897810
99756.67
i x
99756.67
99756.67
99756.67
99756.67
99756.67
99756.67
99756.67
99756.67
99756.67 33 x ;x
98656.7
97856.7
99156.7
98866.7
99066.7
790243.3
99376.7
98556.7
98706.7
0
0
i (x ; x ) 2
9733137878
9575927211
9832044544
9774617778
9814204444
6.24485E+11
9875721878
9713416544
9743006044
7.02547E+11
78060733578
i One way to measure spread is to calculate the mean and then determine
how far each observation is from the mean.
9
1 X x = 1 (1100 + 1900 +
Mean: x = n
9
=1
i + 1050) = 99756:67: i How far an observation is from the mean is quanti ed by the di erence
between that observation and the mean: x ; x. In the entire data set, we
have 9 of these:
x1 ; x x2 ; x : : : x9 ; x:
i One idea is to compute the average of these deviations from the mean.
That is, compute
9
1 X(x
+ (x9 ; x)g = 9
=1 1 f(x ; x) + (x ; x) +
2
91 P i ; x): i Problem: =1 (x ; x) = 0 (always!).
Deviations from the mean always necessarily sum to zero. The positive and negative values cancel each other out.
Solution: Make all of the deviations from the mean positive by squaring
them before averaging.
That is, compute
n
i i (x1 ; x)2 (x2 ; x)2 : : : (x9 ; x)2
and then average. This gives the quantity
9
1 X(x
9 =1 i ; x)2 = 78060733578: i 34 If x1 x2 : : : x9 is the entire population, then x = , the population mean,
and the population size is N = 9. In this case, our formula becomes
1 X(x
N =1
N i ; )2 i which is called the population variance of x1 : : : x , and is usually
denoted by 2 .
Why is this called 2 rather than ? Why the 2 exponent?
Because this is the average squared deviation from the mean (in the claims
data example, the units of this quantity are squared dollars).
To put the variance on the same scale as the original data, we sometimes prefer to work with the population standard deviation
which is denoted as and is just the square root of the population
variance 2 :
N population standard deviation: = p v
uX
u
2=t1
N =1 (x
N i ; )2 : i Suppose now that x1 : : : x are sample values.
How do we compute a sample variance to estimate the population
variance, 2 ?
P
We could simply use 1 =1 (x ; x)2 . However, for reasons we'll discuss
later, it turns out that it is better to de ne the sample variance as
1 X(x ; x)2
2
sample variance: s = n ; 1
=1
n n n i i n i i The sample standard deviation is simply the square root of this quantity: v
u
X
p
u
2=t 1
sample standard deviation: s = s
n ; 1 =1 (x
n i i 35 ; x)2 : The sample variance of the 9 insurance claims is
1X
s = n ; 1 (x
=1
n 2 i ; x)2 i 1
= 9 ; 1 f(x1 ; x)2 + (x9 ; x)2 g = 87818325275 (squared dollars) and the sample standard deviation is
p s = 87818325275 = $296 341:57
A Note on Computation:
The formula for s2 that we just presented,
1X
s = n ; 1 (x
=1
n 2 i ; x)2 i conveys clearly the logic of the standard deviation: it is an average (in
some sense) of the squared deviations from the mean. However, it is not
a good formula to use for computing the SD because it
{ is hard to use and
{ it tends to lead to roundo errors.
For computing, an equivalent but better formula is P s2 = ( n x2) ; nx2 :
n;1 =1 i i When using this formula or any other that requires a series of calculations, keep all intermediate steps in the memory of your calculator
until the end to avoid roundo errors. 36 Coe cient of Variation: In some cases the variance of a variable
changes with its mean.
For example, suppose we are measuring the weights of children of
various ages.
5 year old children (relatively light, on average)
15 year old children (much heaver, on average)
Clearly, there's much more variability in the weights of 15 year olds,
but a valid question to ask is \Do 15 year old children's weights have
more variabilty relative to their average?"
The coe cient of variation allows such comparisons to be made:
population CV = 100% s
sample CV = x 100%: From current CDC data available on the web, I obtained standard
deviations and means for the weights (in kg) of 5 and 15 year old
male children as follows:
Age
5
15 s
2.74
12.05 x
18.39
56.28 CV
0.15
0.21 Thus, 15 year olds' weights are more variable relative to their average
weight than 5 year olds.
Note that the CV is a unitless quantity. 37 Mean and Variance (or SD) for Grouped Data:
Example { Lead Content in Boston Drinking Water
Consider the following data on the lead content (mg/liter) in 12
samples of drinking water in the city of Boston, MA.
data: :035 :060 :055 :035 :031 :039 :038 :049 :073 :047 :031 :016
sorted data: :016 :031 :031 :035 :035 :038 :039 :047 :049 :055 :060 :073
Notice that there are some values here that occur more than once.
Consider how the mean is calculated in such a situation:
+
x = :016 + :031 + :031 + :03512 :035 + :038 + + :073 = :042
031(2) + :035(2)
+
= :016(1) + :(1) + (2) + (2) + + :038(1)+ (1) + :073(1)
(1) + P mf
=1
=P
f
k i i i k =1 i where i k = the number of distinct values in the data
m = the ith distinct value
f = the frequency with which m occurs
i i i Similarly, consider the sample variance: s2 = (:016 ; :042)2 + (:031 ; :042)2 + (:031 ; :042)2 + (:035 ; :042)2 + (:035 ; :042)2 +
+ (:073 ; :042)2 =(12 ; 1) = :015
(:016 ; :042)2 (1) + (:031 ; :042)2 (2) + (:035 ; :042)2 (2) +
=
(1) + (2) + (2) + + (1)] ; 1
P
2
=1 (m ; x) f
=P
=1 f ] ; 1
k i i i k
i i 38 + (:073 ; :042)2 (1) Another Example: Apgar Scores of Low Birthweight Infants Here is a frequency distribution of the Apgar scores for 100 low
birthweight infants in data set lowbwt.
Apgar Score
0
1
2
3
4
5
6
7
8
9
Total= Frequency
6
1
3
4
5
10
11
23
24
13
100 Using the formula for the mean for grouped data we have P
=1
x= P mf
=1 f
k i i i k
i i = 0(6) + 1(1) + 2(3) +
100 + 9(13) = 6:25 which agrees with the value we reported previously for these data.
Similarly, the sample SD is vP
u
u =1 (m
s=t P
k i i ; x)2 fi f ];1
r
2
2
2
;6
= (0 ; 6:25) (6) + (1 ; 6:26) (1) + (2 ; 1 :25) (3) +
100
= 2:43
k =1 i i 39 + (9 ; 6:25)2 (13) zScores and Chebychev's Inequality:
The National Center for Health Statistics at the CDC gives the following
weight
estimate of the body mass index ( height2 ) for 15 year old boys:
x = 19:83
Suppose that a particular 15 year old boy, Fred, has a BMI equal to 25.
How overweight is Fred?
We know he is heavier than average for his age/gender group, but how
much heavier?
Relative to the variability in BMI for 15 year old boys in general,
Fred's BMI may be close to the mean or far away.
Case 1: Suppose s = 10.
This implies that the typical deviation from the mean is about 10.
Fred's deviation from the mean is 5.17, so Fred doesn't seem to be
unusually heavy.
Case 2: Suppose s = 2.
This implies that the typical deviation from the mean is about 2.
Fred's deviation from the mean is 5.17, so Fred does seems to be
unusually heavy.
Thus, the extremeness of Fred's BMI is quanti ed by its distance
from the mean BMI relative to the SD of BMI. 40 The z score gives us this kind of information.
;
z =x s x
where
x = value of the variable of interest for subject i,
x = sample mean
s = sample standard deviation
i i i 19
Case 1: z = 25;10 83 = :517. Fred's BMI is .517 SD's above the mean.
: Case 2: z = 25;19 83 = 2:585. Fred's BMI is 2.585 SD's above the mean.
2
From NCHS data, the true SD for 15 year old boys is s = 3:43. So,
Fred's BMI is z = 25;19 83 = 1:51 SD's above the mean.
3 43
: : : How extreme is a z score of 2? 3? 1.5?
An exact answer to this question depends upon the distribution of the
variable you are interested in.
However, a partial answer that applies to any variable is provided by
Chebychev's inequality. 41 ; Chebychev's Theorem: At least 1 ; 12 100% of the values of any variable must be within k SDs of the mean, for any k > 1.
This results implies (for example):
At least 75% of the observations must be within 2 SDs, since for
k=2
1
1 ; k12
100% = 1 ; 212
100% = 1 ; 4 100% = 75%:
k { For the BMI example, we'd expect at least 75% of 15 year old
males to have BMIs between x ; 2s = 19:83 ; 2(3:43) = 12:97
and x + 2s = 19:83 + 2(3:43) = 26:69.
At least 89% of the observations must be within 3 SDs, since for
k=3
1
1 ; k12
100% = 1 ; 312
100% = 1 ; 9 100% = 89%:
{ For the BMI example, we'd expect at least 89% of 15 year old
males to have BMIs between x ; 3s = 19:83 ; 3(3:43) = 9:54
and x + 3s = 19:83 + 3(3:43) = 30:12.
Note that Chebychev's Thm just gives a lower bound on the percentage falling within k SDs of the mean. At least 75% should fall
within 2 SDs, but perhaps more.
{ Since it only gives a bound and not a more exact statement
about a distribution, Chebychev's Thm is of limited practical
value. 42 We can make a much more precise statement if we know that the distribution of the variable in which we're interest is bellshaped. That is, shaped
roughly like this:
Another bellshaped distribution for Y, say 0.1 0.2 Frequency density of y 0.2
0.1
0.0 Frequency density of x 0.3 0.3 0.4 A bellshaped distribution for X, say 2 0 2 2 x=value of X 0 2 y=value of Y Think of the above pictures as idealized histograms as the sample
size grows toward in nity.
One particular bellshaped distribution is the normal distribution,
which is also known as the Gaussian distribution.
{ The normal distribution is particularly important in statistics,
but it is not the only possible bellshaped distribution. The
distribution above left is normal, the one above right is similar,
but not exactly normal (notice di erence in the tails). 43 For data that follow the normal distribution, the following precise statements can be made:
Excatly 68% of the observations lie within 1 SD of the mean.
Exactly 95% of the observations lie within 2 SDs of the mean.
Exactly 99.7% of the observations lie within 3 SDs of the mean.
In fact, for normally distributed data we can calculate the percentage of
the observations that fall in any range whatsoever.
This is very helpful if we know our data are normally distributed.
However, even if the data aren't known to be exactly normal, but are
known to be bellshaped, then the exact results stated above will be approximately true. This is known as the empirical rule.
Empirical rule: for data following a bellshaped distribution:
Approximately 68% of the observations will fall with 1 SD of the
mean.
Approximately 95% of the observations will fall with 2 SDs of the
mean.
Nearly all of the observations will fall with 3 SDs of the mean. 44 BMIs of 15 Yearold Boys: At age 15, suppose that BMI follows an approximately bellshaped
distribution.
{ Then we would expect approximately 68% of 15 year old boys
to have BMIs falling in the interval (16:40 23:26) = x 1s.
Fred's BMI was 25, so his BMI is more extreme than twothirds
of boys his age.
{ We would expect 95% of 15 yearold boys to have BMIs falling
in the interval (12:97 26:69) = x 2s and nearly all to fall in
the interval (9:54 30:12) = x 3s. (Compare these results with
the Chebychev bounds).
In fact, BMI is probably not quite bellshaped for 15 year olds. It
may be for 5 year olds, but by age 15, there are many obese children
who probably skew the distribution to the right (lots of large values
in the right tail). Therefore, the empirical rule may be somewhat
inaccurate for this variable. 45 Introduction to Probability* Note: we're going to skip ch.s 4 & 5 for now, but we'll come back to
them later.
We all have an intuitive notion of probability.
\There's a 75% chance of rain today."
\The odds of Smarty Jones winning the Kentucky Derby are 2 to 1."
\The chances of winning the Pick5 Lottery game are 1 in 2.3 million."
\The probability of being dealt four of a kind in a 5 card poker hand
is 1=4164."
All of these statements are examples of quantifying the uncertainty in a
random phenomenon. We'll refer to the random phenomenon of interest
as the experiment, but don't confuse this use with an experiment as a type
of research study.
The experiments in the examples above are
{ An observation of today's weather
{ The results of the Kentucky Derby
{ A single play of the Pick5 Lottery game
{ The rank of a 5card poker hand dealt from a shu ed deck of
cards * Read Ch.6 of our text.
46 An experiment generates an outcome through some random process.
Experiment
Weather
Kentucky Derby
Lottery
Poker Hand Outcome
Rains, Does not rain
Smarty Jones wins, places, shows,..., does not nish
Win, Lose
Royal Flush, Straight Flush, Fourofakind, ... Set of outcomes is called the sample space and should consist of
mutually exclusive, exhaustive set of outcomes.
An event is some description of the outcome of an experiment whose
probability is of interest.
A variety of events can be de ned based on the outcome of a given
experiment:
E.g., Events that could be de ned regarding the outcome of the
Kentucky Derby:
{ Smarty Jones nishes
{ Smarty Jones nishes third or better (wins, places, or shows)
{ Smarty Jones wins.
Events of interest need not be mutually exclusive or exhaustive.
The terms \chance(s)", \likelihood", and \probability" are basically
synonymous ways to describe the probability of an event. We denote
the probability of an event A by P ( A)
{ The odds of an event describes probability too, but is a bit
di erent. The odds of an event A are de ned as
PA
odds(A) = P ((A ))
c where A denotes the event that A does not occur, which is
known as the complement of A.
c 47 A number of di erent operations can be de ned on events.
One is the complement: A denotes the event that A does not occur.
The union of events A and B is denoted
c A B:
The union of A and B is the event that A occurs or B occurs (or
both).
{ can be read as \or" (inclusive).
The intersection of events A and B is denoted
A \ B:
The intersection of A and B is the event that A occurs and B occurs.
{ \ can be read as \and".
The following Venn diagrams describe these operations pictorially. 48 There are a number of legitimate ways to assign probabilities to events:
the classical method
the relative frequency method
the subjective method
Whatever method we use, we require
1. The probability assigned to each experimental outcome must be between 0 and 1 (inclusive). That is, if O represents the ith possible
outcome, we must have
i 0 P (O ) 1
i for all i. { Probabilities are between 0 and 1, but they are often expressed
as percentages by multiplying by 100%. That is, to say there
is a .75 chance of rain is the same as saying the chance of rain
is 75%.
2. The sum of the probabilities for all experimental outcomes must
equal 1. That is, for n mutually exclusive, exhaustive outcomes
O1 : : : O , we must have
n P (O1 ) + P (O2) + + P (O ) = 1:
n Classical Method: When all n experimental outcomes are equally likely,
we assign eqach outcome a probability of 1=n.
E.g., when tossing a fair coin, there are n = 2 equally likely outcomes,
each with probability 1=n = 1=2.
E.g., If we pick a card from a wellshu ed deck and observe its suit,
then there are n = 4 possible outcomes, so P (~) = P ( ) = P (}) = P () = 1=n = 1=4: 49 The classical method is really a special case of the more general Relative
Frequency Method. The probability of an event is the relative frequency
with which that event occurs if we were to repeat the experiment a very
large number of times under identical circumstances.
I.e., if the event A occurs m times in n identical replications of an
experiment, then
P (A) = m when n ! 1.
n
Suppose that the gender ratio at birth is 50:50. That is, suppose
that giving birth to a boy and giving birth to a girl are equaly likely
events. Then by the clasical method P (Girl) = 1 :
2
This is also the long run relative frequency. As n ! 1 we should
expect that
number of girls ! 1 :
number of births 2
There are several rules of probability associated with the union, intersection, and complement operations on events.
Addition Rule: For two events A and B P (A B ) = P (A) + P (B ) ; P (A \ B ):
Venn Diagram: 50 Example Consider the experiment of having two children and let A = event that rst child is a girl
B = event that second child is a girl
Assume P (A) = P (B ) = 1=2 (doesn't depend on birth order and
gender of second child not in uenced by gender of rst child).
Then the probability of having at least one girl is
11
P (A B ) = P (A) + P (B ) ; P (A \ B ) = 2 + 2 ; P (A \ B ) But what's P (A \ B ) here?
One way to determine this is by enumerating the set of equally likely
outcomes of the underlying experiment here.
The experiment is observing the genders of two children. It has sample
space (set of possible outcomes):
f(M M ) (M F ) (F M ) (F F )g
which are all equally likely (have probability 1=4 each).
The probability of an event is the sum of the probabilities of the
outcomes satisfying the event's de nition.
Here, the event A \ B corresponds to the outcome (F F ) so
1
P (A \ B ) = 4
and
1
3
P (A B ) = 1 + 2 ; 1 = 4
2
4
Notice that this agrees with the answer we would have obtained by
summing the probabilities of the outcomes corresponding to at least
one girl:
113
P (A B ) = P f(M F )g + P f(F M )g + P f(F F )g = 1 + 4 + 4 = 4 :
4
51 Complement Rule: For an event A and its complement A c P (A ) = 1 ; P (A):
c This is simply a consequence of the addition rule and the facts that P (A A ) = P (entire sample space) = 1
and P (A \ A ) = P (A and not A occur) = 0
c
c Thus, by the addition rule
1 = P (A A ) = P (A) + P (A ) ; P (A{z A } = P (A) + P (A )
\)
c c c c =0 ) P (A) = 1 ; P (A )
c A third rule is the multiplication rule, but for that we need the de nitions
of conditional probability and statistical independence. 52 Conditional Probability: For some events, whether or not one event has occurred is clearly relevant
to the probability that a second event will occur.
We just computed that the probability of having at least one girl in
two births as 3 .
4
Now suppose I know that my rst child was a boy.
Clearly, knowing that I've had a boy a ects the chances of having at
least one girl (it decreases them). Such a probability is known as a
conditional probability.
The conditional proabability of an event A given that another event B has
occurred is denoted P (AjB ) where j is read as \given".
Independence of Events Two events A and B are independent if knowing that B has occurred gives no information relevant to whether or not
A will occur (and vice versa). In symbols A and B are independent if
P (AjB ) = P (A): Multiplication Rule: The joint probability of two events P (A \ B ) is
given by P (A \ B ) = P (AjB )P (B )
or since the A and B can switch places
P (A \ B ) = P (B \ A) = P (B jA)P (A)
Note that this relationship can also be written as
A\
P (AjB ) = P (P (B )B )
as long as P (B ) 6= 0, or
B\
A\
P (B jA) = P (P (A)A) = P (P (A)B )
as long as P (A) 6= 0.
53 Example: Again, the probability of having at least one girl in two births
3 is 4 . Now suppose the rst child is known to be a boy, and the second
child's gender is unknown.
What is the conditional probability of at least one girl given that the
rst child is a boy?
Again, let
A = event that rst child is a girl
B = event that second child is a girl
Then we are interested here in P (A B jA ).
By the multiplication rule
c P (A B jA ) = P f(AP (B ))\ A g :
A
c c c 1
We known that the probability that the rst child is a girl is P (A) = 2 ,
so
P (A ) = 1 ; P (A) = 1 ; 1 = 1 :
22
c In addition, the probability in the numerator, P f(A B ) \ A g, is the
probability that at least one child is a girl (the event A B ) and the rst
child is a boy (the event A ).
Both of these events can happen simultaneously only if the rst child is
a boy and the second child is a girl. That is, only if the outcome of the
experiment is f(M F )g. Thus,
c c P f(A B ) \ A g = P (A
c c \ B ) = P f(M 1
F )g = 4 : Therefore, the conditional probability of at least one girl given that the
rst child is a boy is
41
P (A B jA ) = P f(AP (B ))\ A g = 1=2 = 2 :
A
1=
c c c 54 Another Example: Suppose that among US adults, 1 in 3 obese individuals has high blood
pressure, while 1 in 7 normal weight individuals has high blood pressure.
Suppose also that the percentage of US adults who are obese, or prevalence of obesity, is 20%.
What is the probability that a randomly selected US adult is obese
and has high blood pressure?
Let A = event that a randomly selected US adult is obese
B = event that a randomly selected US adult has high b.p.
Then the information given above is
1
P (A) = 5 P (B jA) = 1
3 1
P (B jA ) = 7 :
c By the multiplication rule, the probability that a randomly selected US
adult is obese and has high blood pressure is P (A \ B ) = P (B \ A) = P (B jA)P (A) = 1
3 55 1 = 1:
5
15 Note that in general given that an event A has occurred, either B occurs, or
B must occur, so the complement rule applies to conditional probabilities
too:
P (B jA) = 1 ; P (B jA):
c c With this insight in hand, we can compute all other joint probabilities
relating to obesity and high blood pressure:
The probability that a randomly selected US adult is obese and does
not have high blood pressure is
2
P (A\B ) = P (B \A) = P (B jA)P (A) = 1;P (B jA)]P (A) = 3
c c c 1 = 2:
5
15 The probability that a randomly selected US adult is not obese and
does have high blood pressure is
1
P (A \B ) = P (B \A ) = P (B jA )P (A ) = P (B jA ) 1;P (A)] = 7
c c c c c 4 = 4:
5
35 The probability that a randomly selected US adult is not obese and
does not have high blood pressure is P (A \B ) = P (B \A ) = P (B jA )P (A ) = 1;P (B jA )] 1;P (A)] = 6
7
c c c c c c c c These results can be summarized in a table of joint probabilities:
Obese
Yes (event A) No (event (A )
1
4
Yes (event B )
High B.P.
15
35
2
24
No (event B )
15
35
28
4
3 =1
15
5
35 = 5 4 = 24 :
5
35 c c 56 19
105
86
105 1 Independence: Two events A and B are said to be independent if
knowing whether or not A has occurred tells us nothing about whether or
not B has or will occur and vice versa.
In symbols, A and B are independent if P (AjB ) = P (A) and P (B jA) = P (B ):
Note that under independence of A and B , the multiplication rule
becomes
P (A \ B ) = P (AjB )P (B ) = P (A)P (B )
and the addition rule becomes P (A B ) = P (A) + P (B ) ; P (A \ B ) = P (A) + P (B ) ; P (A)P (B ):
Note also that the terms mutually exclusive and independent are
often confused, but they mean di erent things.
{ Mutually exclusive events A and B are events that can't happen simultaneously. Therefore, if I know A has occurred, that
tells me something about B namely, that B can't have occurred. So mutually exclusive events are necessarily dependent (not independent).
Obesity and High Blood Pressure Example: The fact that obesity and
high b.p. are not independent can be veri ed by checking that
1 = P (B jA) 6= P (B ) = 19 :
3
105
Alternatively, we can check independence by checking whether P (A \ B ) =
P (A)P (B ). In this example,
1
1
0:0667 = 15 = P (A \ B ) 6= P (A)P (B ) = 5 57 19 = 0:0362
105 Bayes' Theorem:
We have seen that when two events A and B are dependent, then P (AjB ) 6=
P (A).
That is, the information that B has occurred a ects the probability
that A will occur.
Bayes' Theorem provides a way to use new information (event B has occurred) to go from our probability before the new information was available
(P (A), which is called the prior probability) to a probability that takes
the new information into account (P (AjB ), which is called the posterior
probability).
Bayes' Theorem allows us to take the information about P (A) and
P (B jA) and compute P (AjB ). Obesity and High B.P. Example:
Recall A = event that a randomly selected US adult is obese
B = event that a randomly selected US adult has high b.p.
and 1
P (A) = 5 P (B jA) = 1
3 1
P (B jA ) = 7 :
c Suppose that I am a doctor seeing the chart of a patient, and the only
information contained there is that the patient has high b.p.
Assuming this patient is randomly selected from the US adult population, what is the probability that the patient is obese?
That is, what is P (AjB )? 58 By the multiplication rule, we know that A\
P (AjB ) = P (P (B )B ) : () Let's examine the numerator and denominator of this expression and see
if we can use the information available to compute these quantities.
First, notice that the denominator is P (B ), the probability of high blood
pressure. If a random subject has high b.p., then the subject either
a. has high b.p. and is obese, or
b. has high b.p. and is not obese.
That is,
B = (B \ A ) ( B \ A )
so, by the addition rule
c P (B ) = P (B \ A) + P (B \ A ) ; P f(B \ A){z (B \ A )g
\

}
c c =0 Therefore, P (B ) = P (B \ A) + P (B \ A ):
c This relationship is sometimes called the law of total probability, and
is just based on the idea that if B occurs, it has to occur with either
A or A .
So, now (*) becomes
c )
P (AjB ) = P (B \ P ()A \PBB \ A ) :
A+ (
c 59 () Now consider the numerator, P (A \ B ). By the multiplication rule and
using the fact that (A \ B ) = (B \ A), we have P (A \ B ) = P (B \ A) = P (B jA)P (A)
which is useul because we know these quantities.
Applying the same logic to the two joint probabilities in the denominator
of (**), we have that P (B \ A) = P (B jA)P (A) and P (B \ A ) = P (B jA )P (A ):
c c c Therefore, (**) becomes
(A
)
P (AjB ) = P (B jA)PP AB j+ )P((AjA )P (A ) :
( ) PB
c c (y) Equation (y) is known as Bayes' Theorem.
In the example, Bayes' Theorem tells us that the probability that the high
b.p. patient is obese is
(A
)
P (AjB ) = P (B jA)PP AB j+ )P((AjA )P (A )
( ) PB
==
1
= (1=3)(1(15)3)(1(15)7)(4=5) = 19=15 = 0:368
=+=
=105
c 60 c In the example above, we used the law of total probability to compute
P (B ) as P (B ) = P (B \ A) + P (B \ A ) = P (B jA)P (A) + P (B jA )P (A )
c c c where A and A were mutually exclusive, exhaustive events.
Bayes' Theorem generalizes to apply to the situation in which we
have several mutually exclusive, exhaustive events.
Let A1 A2 : : : A be k mutually exclusive, exhaustive events. Then for
any of the events A , i = 1 : : : k, Bayes' Theorem becomes:
c k i P( ) (
P (A jB ) = P (B jA )P (A ) + P (B jB jAP PAA) )
A2 ) ( 2 +
1
1
i i 61 i P (B jA )P (A ) :
k k Another Example  Obesity and Smoking Status:
Let B = event that a randomly selected US adult is obese
A1 = event that a randomly selected US adult has never smoked
A2 = event that a randomly selected US adult is an exsmoker
A3 = event that a randomly selected US adult is a current smoker
and suppose P (B ) = :209
P (B jA1) = :208
P (B jA2) = :239
P (B jA3) = :178
P (A1) = 0:520
P (A2) = 0:250
P (A3) = 0:230:
Given that a randomly selected US adult is obese, what's the probability that he/she is a former smoker?
By Bayes' Theorem P (B A )P (A )
P (A2jB ) = P (B jA )P (A ) + P (B jjA2)P (A2 ) + P (B jA )P (A )
1
1
2
2
3
3
(
= (:208)(:520) + (::239)(::250) + (:178)(:230) = :286
239)( 250)
Note that the denominator is just P (B ), so since we happen to know
P (B ) here, we could have simpli ed our calculations as A)
P (A2jB ) = P (B jP (2BP (A2 )
)
= (:239)(:250) = :286
:209 62 Diagnostic Tests One important application of Bayes' Theorem is to diagnostic or screening
tests.
Screening is the application of a test to individuals who have not
yet exhibited any clinical symptoms in order to classify them with
respect to their probability of having a particular disease.
{ Examples: Mammograms for breast cancer, Pap smears for cervical cancer, ProstateSpeci c Antigen (PSA) Test for prostate
cancer, exercise stress test for coronary heart disease, etc.
Consider the problem of detecting the presence or absence of a particular
disease or condition.
Suppose there is a \gold standard" method that is always correct.
E.g., surgery, biopsy, autopsy, or other expensive, timeconsuming
and/or unpleasant method.
Suppose there is also a quick, inexpensive screening test.
Ideally, the test should correctly classify individuals as positive or
negative for the disease. In practice, however, tests are subject to
misclassi cation errors. 63 De nitions:
A test result is a true positive if it is positive and the individual
has the disease.
A test result is a true negative if it is negative and the individual
does not have the disease.
A test result is a false positive if it is positive and the individual
does not have the disease.
A test result is a false negative if it is negative and the individual
does have the disease.
The sensitivity of a test is the conditional probability that the test
is positive, given that the individual has the disease.
The speci city of a test is the conditional probability that the test
is negative, given that the individual does not have the disease.
The predictive value of a positive test is the conditional probability that an individual has the disease, given that the test is positive.
The predictive value of a negative test is the conditional probability that an individual does not have the disease, given that the
test is negative.
Notation: Let A = event that a random individual's test is positive
B = event that a random individual has the disease
Then
sensitivity =
speci city = predictive value positive =
predicitve value negative = 64 Estimating the Properties of a Screening Test:
Suppose data are obtained to evaluate a screening test where the true
disease status of each patient is known. Such data may be displayed as
follows:
Truth
Diseased (event B ) Not Diseased (event B )
+ (event A)
a
b
Test Result
; (event A )
c
d
n1
n2
What properties of the screening test can be estimated if the data
are obtained:
1. from a random sample of n subjects?
2. from random samples of n 1 diseased and n 2 nondiseased subjects?
3. from random samples of n1 subjects with positive test results and
n2 subjects with negative results?
c c 65 n1
n2
n 1. Suppose a random sample of n subjects is obtained, and each subject
is tested via both the screening test and the gold standard.
In this case,
estimated sensitivity =
estimated speci city =
estimated predictive value positive =
estimated predictive value negative =
2. Suppose that random samples of n 1 diseased and n 2 nondiseased
subjects are obtained, and each subject is tested with the screening
test.
In this case,
estimated sensitivity =
estimated speci city =
but predictive value positive and negative cannot be estimated directly without additional information about the probability (prevalence) of disease.
3. Suppose now that random samples of n1 subjects with positive
screening test results and n2 subjects with negative screening test
results are obtained. Each subject is then tested with the gold standard approach.
In this case,
estimated predictive value positive =
estimated predictive value negative =
but sensitivity and speci city cannot be estimated directly without
additional information about the probability of a positive test result. 66 Notice that only in case 1 is it possible to obtain estimates of all four
quantities from simple proportions in the contingency table.
However, this approach is not particularly quick, easy or e cient because, for a rare disease, it will require a large n to obtain a su cient
sample of truly diseased subjects.
Approach 2 is generally easiest, and predictive values can be computed from this approach using Bayes' Theorem if the prevalence of
the disease is known as well.
Suppose we take approach 2. As before, let A = event that a random individual's test is positive
B = event that a random individual has the disease
Suppose P (B ), the prevalence of disease, is known. In addition, suppose
the sensitivity P (AjB ) and speci city P (A jB ) are known (or have been
estimated as on the previous page).
Then, according to Bayes' Theorem, P (B jA), the predictive value of a
positive test result, is given by
c c (B
)
P (B jA) = P (AjB )PPBAj+ )P((BB )P (B )
( ) P Aj
c c Similarly, P (B jA ), the predictive value of a negative test result, is given
by
(
)B
P (B jA ) = P (A jB PPABjB +P ((A )jB )P (B )
)( ) P
c c c c c c c c c c 67 c Suppose that a new screening test for diabetes has been developed. To
establish its properties, n 1 = 100 known diabetics and n 2 = 100 known
nondiabetics were tested with the screening test. The following data were
obtained:
Truth
Diabetic (event B ) Nondiabetic (event B )
+ (event A)
80
10
Test Result
; (event A )
20
90
100
100
c c Suppose that it is known that the prevalence of diabetes is P (B ) =
:07 (7%).
The sensitivity P (AjB ) here is estimated to be 80=100 = :8.
The speci city P (A jB ) here is estimated to be 90=100 = :9.
From the previous page, the predictive value positive is
c c (B
)
P (B jA) = P (AjB )PPBAj+ )P((BB )P (B ) :
( ) P Aj
c c Therefore, the estimated predictive value positive is
( :07)
estimated P (B jA) = (:8)(:07) +:8)(; :9)(1 ; :07) = :376
(1
This result says that if you've tested positive with this test, then
there's an estimated chance of 37.6% that you have diabetes. 68 ROC Curves:
There is an inherent tradeo between sensitivity and speci city.
Example  CT Scans The following data are ratings of computed
tomography (CT) scans by a single radiologist in a sample of 109 subjects
with possible neurological problems. The true status of these patients is
also known.
True Disease
Status
Normal Abnormal
1
33
3
36
2
6
2
8
Radiologist's Rating
3
6
2
8
4
11
11
22
5
2
33
35
58
51
109
Here, the radiologist's rating is an ordered categorical variable where
1 = de nitely normal
2 = probably normal
3 = questionable
4 = probably abnormal
5 = de nitely abnormal
If the CT scan is to be used as a screening device for detecting
neurological abnormalities, where should the cuto be set for the diagnosis
of abnormality? 69 Suppose we diagnose every patient with a rating 1 as abnormal.
Obviously, we will catch all true abnormals this way  the sensitivity
of this test will be 1.
However, we'll also categorize all normals as abnormal  the specicity will be 0.
Suppose we diagnose every patient with a rating 5 as normal.
Obviously, we won't incorrectly diagnose any normals as abnormal
 the speci city will be 1.
However, we won't detect any true abnormalities  the sensitivity
of this test will be 0.
Clearly, we'd prefer to use some threshold between 1 and 5 to diagnose
abnormality.
We can always increase the sensitivity by setting the threshold high,
but this will decrease the speci city.
Similarly, a low threshold will increase the speci city at the cost of
sensitivity. 70 For each possible threshold value, we can compute the sensitivity and
speci city as follows:
Test Positive Criterion
1
2
3
4
5
>5 Sensitivity
1.00
0.94
0.90
0.86
0.65
0.00 Speci city
0.00
0.57
0.67
0.78
0.97
1.00 A plot of the sensitivity versus (1 ; speci city) is called a receiver operating characteristic curve, or ROC curve. The ROC curve for this
example is as follows: An ROC curve is often used to help determine an appropriate threshold for a screening test. The point closest to the upperleft corner of
the plot has the highest combination of sensitivity and speci city.
{ In this example, the ROC curve suggests that we use a rating
4 to classify patients as abnormal.
The dashed line in this plot shows where there are equal probabilities
of a false positive and a false negative. smallskip
{ A test falling on this line misclassi es normal subjects with the
same frequency with which it misclassi es abnormal subjects.
{ Such a test classi es no better than chance, and thus has no
predicitve value.
71 Estimation of Prevalence from a Screening Test:
Suppose we apply a screening test with known sensitivity and speci city
to a new population for which the prevalence of the disease is unknown.
Without applying the gold standard test, can we estimate the prevalence?
Let's reconsider the diabetes example. Recall how we de ned events: A = event that a random individual's test is positive
B = event that a random individual has the disease
Previously, we obtained
estimated sensitivity = 0:8 = P (dB )
Aj
d
estimated speci city = 0:9 = P (A jB ):
c c (hats indicate that these are estimated quantities).
Recall also that we knew the prevalence of diabetes to be :07.
However, now suppose that this prevalence value was for the US population
and we decide now to apply the screening value in Canada.
Suppose that we screen n = 580 Canadians with our screening test and
we obtain n1 = 105 positive test results:
Truth
Diseased (B ) Not Diseased (B )
+ (A)
?
?
n1 = 105
Test Result
; (A )
?
?
n2 = 475
?
?
n = 580
What is the prevalence of diabetes among Canadians?
c c 72 Using the law of total probability followed by the multiplication rule, we
have
P (A) = P (A \ B ) + P (A \ B )
= P (AjB )P (B ) + P (AjB )P (B )
= P (AjB )P (B ) + P (AjB ) 1 ; P (B )]
With a little algebra, we can solve for P (B ), the prevalence of diabetes as
follows:
( ) ; ; P(
( ) ; P( j
P (B ) = PPAAB ) ; PAABB) ) = PPAAB ) ;11 ; PAAjBB)])]
(j
(j
(j
(j
c c c c c c c c c c P (A), the probability of a positive test result can be estimated as
105
1
Pd) = nn = 580
(A
and the other quantities in this expression, P (AjB ) and P (A jB ), are the
sensitivity and speci city of our test.
Therefore, we can estimate the prevalence of diabetes in Canada as
c d
(A
d ) = Pd) ; 1 ; P (A jB )]
P (B
d
P (dB ) ; 1 ; P (A jB )]
Aj
105 ; 1 ; :9]
= 580; 1 ; :9] = :116
:8
c c c 73 c c Risk Di erence, Relative Risk and Odds Ratio:
Three quantities that are often used to describe the di erence between
the probability (or risk) of disease between two populations are the risk
di erence, risk ratio, and odds ratio.
We will call the two populations the exposed and unexposed populations, but they could be whites and nonwhites, males and females,
or any two populations (i.e., the \exposure" could be being male).
1. Risk di erence: One simple way to quantify the di erence between
two probabilities (risks) is to take their di erence.
Risk di erence = P (diseasejexposed) ; P (diseasejunexposed):
Independence between exposure status and disease status corresponds
to a risk di erence of 0.
Risk di erence ignores the magnitude of risk. E.g., suppose that
among males, the exposed and unexposed groups have disease risks
of .51 and .50, but among females, the exposed and unexposed groups
have risks of .02 and .01.
{ Risk di erence is .01 for males and for females. Risk di erence
does not convey the information that being exposed doubles
the risk for females. 74 2. Relative risk: (also known as risk ratio).
RR = PP (diseasejexposed)
(diseasejunexposed)
Independence between exposure status and disease status corresponds
to a relative risk of 1.
Relative risk especially useful for quantifying exposure e ect for rare
diseases.
{ E.g., the probability that a man over the age of 35 dies of cancer
is 0.002679 for current smokers, and .000154 for nonsmokers.
002679
RR = ::000154 = 17:4 risk di erence = :002679;:000154 = :002525:
Risk di erence and RR convey di erent types of information  both
useful.
3. Odds ratio: RR takes ratio of probabilities. As an alternative to
using probability of disease, can compare odds of disease in exposed
and unexposed group. This leads to the odds ratio (OR):
odds(diseasejexposed)
OR = odds(diseasejunexposed) :
Recall that the odds of an event A are given by P
PA
odds(A) = P ((A )) = 1 ; (A()A)
P
c so the OR is P (diseasejexposed)
OR = P (diseasejunexposed)= 1 ; P (diseasejjexposed)]
= 1 ; P (disease unexposed)] 75 () Independence between exposure status and disease status corresponds
to an odds ratio of 1.
The OR conveys similar information to that of the RR. The main
advantages of the OR are that
a. It has better statistical properties. We'll explain this later, but
for now take my word for it.
b. It can be calculated in cases when the RR cannot.
The latter advantage comes from the fact that using Bayes' Theorem,
it can be shown that
P (exposurejdiseased)
OR = P (exposurejnondiseased)= 1 ; P (exposurejjdiseased)]
= 1 ; P (exposure nondiseased)]
()
{ I.e.. (*) and (**) are mathematically equivalent formulas.
{ This equivalence is useful because in some contexts, the probability of exposure can be estimated among diseased and nondiseased but the probability of disease given exposure status cannot. This occurs in casecontrol studies. 76 Example  Contraceptive Use and Heart Attack A casecontrol study of oral contraceptive use and heart attack. 58 female
heart attack victims were identi ed and each of these \cases" was matched
to one \control" subject of similar age, etc. who had not su ered a heart
attack.
Heart
Attack
Yes No
Contraceptive Use
Yes
23
11
No
35
47
58
58 In this case, the column totals are xed by the study design. Therefore,
the probability of heart attack given whether or not oral contraceptives
have been used cannot be estimated.
Why?
Thus, we cannot estimate the risk of disease in either the exposed
or unexposed group, and therefore cannot estimate the RR or risk
di erence.
However, we can estimate probabilities of contraceptive use given presence
or absence of heart attack:
^
P (contraceptive usejheart attack) = 23=58 = :397
^
P (contraceptive usejno heart attack) = 11=58 = :190:
And from these quantities we can estimate the odds ratio: d
OR = 23
58
11
58 ; 11 ; 23 ; 47
;1 ; 58 = ; 58 ; 58 = 23(47) = 2:808:
23
11
35
11(35)
1 ; 58
58
58 Interpretation: The odds of heart attack are 2.8 times higher for
women who took oral contraceptives than for women who did not.
77 Theoretical Probability Distributions* Probability Distributions:
Some de nitions:
A variable is any characteristic that can be measured or observed
and which may vary (or di er) among the units measured or observed.
A random variable is a variable that takes on di erent numerical
values according to a chance mechanism
{ E.g., any variable measured on the elements of a randomly
selected sample.
{ Discrete random variables are random variables that can take
on a nite or countable number of possible outcomes (e.g.,
number of pregnancies).
{ A continuous random variable can (theoretically, at least) take
on any value in a continuum or interval (BMI).
A probability function is a function which assigns a probability
to each possible value that can be assumed by a discrete random
variable.
The probability function of a discrete random variable (r.v.):
{ de nes all possible values of the r.v.
{ gives the probabilities with which the r.v. takes on each of
those values. * Read Ch.7 of our text.
78 Example Let X =the number of ears a ected by one or more episodes of
otitis media (ear infection) during the rst two years of life. Suppose the
probability distribution function for this random variable is x
0
1
2 P (X = x)
:13
:48
:39 The notation used above is typical. Here, big X is the random
variable, little x is a particular value of the random variable, and we
are giving the probability that the random variable X takes on the
value x for each possible x.
Note that values of x that are not listed are assumed to have probability 0.
Of course, the probability function must assign valid probabilities.
In particular,
{ when summed over all possible values, the probabilities must
sum to 1:
X
P (X = x) = 1:
all x
{ and each probability must be between 0 and 1:
0 P (X = x) 1 for all x.
Probability functions can be given in tables as above, or graphs (e.g.,
a bar graph), or as a mathematical formula.
The probability function allows computation of probabilities for events
de ned in terms of the random variable.
E.g., by the addition rule, the probability of having at least one ear
infection during the rst two years of life is
P (X > 0) = P (X = 1 X = 2)
= P (X = 1) + P (X = 2) ; P (X = 1 \ X = 2) = :48 + :39 = :87

{z
}
=0 79 Expected Value, Variance
The expected value of a random variable is the mean, or average value
of the r.v. over the population of units on which the r.v. is de ned.
For a random variable X , its expected value is usually denoted E(X ),
or , or simply .
The expected value for a discrete r.v. can be computed from its probability
distribution as follows:
X E (X ) = X all x xP (X = x) where this sum is taken over all possible values x of the r.v. X .
E.g., the expected number of ears a ected by ear infection during
the rst two years of life is computed as follows: x
0
1
2 P (X = x)
:13
:48
:39 xP (X = x)
0(:13)
1(:48)
2(:39)
E (X ) = 1:26 { Interpretation: the mean number of ears a ected by otitis media during the rst two years of life is 1.26. 80 The variance of a random variable is the population variance of the r.v.
over the population of units on which the r.v. is de ned.
The variance of X is usually denoted var(X ), or 2 , or simply 2 .
The formula for the variance of a random variable involves taking
expectations:
var(X ) = Ef(X ; )2 g
which, for a discrete r.v. simpli es to
X X var(X ) = X (x ;
all x X )2 P (X = x) where again this sum is taken over all possible values x of the r.v.
X.
E.g., the variance of the number of ears a ected by ear infection
during the rst two years of life is computed as follows: x
0
1
2 P (X = x)
:13
:48
:39 X 1:26
1:26
1:26 (x ; )2 P (X = x)
(0 ; 1:26)2 (:13)
(1 ; 1:26)2 (:48)
(2 ; 1:26)2 (:39)
var(X ) = :452
X The population standard deviation of X is
:673 in our example. 81 x = p 2 X p or :452 = The Binomial Probability Distribution
Many random variables that can be described as event counts where there
is a max number of events that can occur, can be thought of as arising
from a binomial experiment.
A binomial experiment has the following properties:
1. The experiment consists of a sequence of n identical trials.
2. Two outcomes are possible on each trial, one a \success" and the
other a \failure".
3. The probability of success, denoted by p, is the same for each trial.
{ Since the probability of a failure is just 1 ; p, this means that
the failure probability is the same for each trial as well.
4. The trials are independent (what happens on one trial doesn't a ect
what happens on any other trial).
In a binomial experiment we are interested in X , the r.v. de ned to be
the total number of successes that occur over the n trials.
Note that \success" and \failure" are just convenient labels. A success could be identi ed as the birth of a girl, and failure as the birth
of a boy, or vice versa. That is, \success" simply denotes the event
of interest that is being counted.
X in a binomial trial is a discrete random variable with possible
values 0 1 2 : : : n.
For any experiment with the above properties, X will necessarily have a
particular distribution, the binomial probability distribution that is
completely determined by n and p. Examples: A. The number of heads that occur in 4 coin ips
1. Each coin ip is an identical trial.
2. Two outcomes (Heads,Tails) are possible, where \success"=
Heads.
3. Probability of success= P (Heads) = 1=2 on each trial.
4. Coin ips are independent.
82 B. The number of obese subjects out of 3 randomly selected US adults.
1. Observing obesity status of each randomly selected US adult
is an identical trial.
2. Two outcomes are possible (obese, not obese) where \success"
= subject is obese.
3. Probability of success= P (obese) = :209 on each trial.
4. Because selection of subjects is at random, obesity status is
independent from subject to subject. Counter Examples: C. The number of lifetime miscarriages experienced by a randomly selected woman over the age of 50. Suppose the woman had had 5
lifetime pregnancies.
1. The n = 5 pregnancies are the trials, but they are not identical. They occur at di erent ages, under di erent circumstances
(woman's health status di ers, environmental exposures di er,
fathers may di er, etc.).
2. Two outcomes are possible (miscarriage, not miscarriage) where
\success" = miscarriage.
3. Probability of success not constant on each trial. Probability of
miscarriage may be higher when woman is older, may depend
on birth order, etc.
4. Pregnancy outcome may not be independent from one pregnancy to the next (if previous pregnancy was a miscarriage,
that may increase the probability that next pregnancy will be
miscarriage).
D. Out of the n hurricanes that will form in the Atlantic next year, how
many will make landfall in the state of Florida?
1. Each hurricane represents a trial. Not identical.
2. Two outcomes possible (hit FL, not hit FL). \Success" = hit
FL.
3. Probabilities of hitting Florida may not be constant from hurricane to hurricane depending upon when and where they form,
but a priori, it may be reasonable to assume that these probabilities are equal from one hurricane to the next.
4. Hurricane paths are probably not independent. If the previous
hurricane hit FL, that may increase the chances that the next
hurricane will follow the same path and hit FL as well.
83 For any binomial experiment, the probability of any given number of \successes" out of n trials is given by the binomial probability function.
Let the random variable X = the number of successes out of n trials,
where p is the success probability on each trial. Then the probability of x
successes is given by P (X = x) = n p (1 ; p) ; :
x
x ; n x Here, (read \n choose x) is shorthand notation for
a! (\a factorial") is given by n!
!(n;x)! n
x a! = a(a ; 1)(a ; 2) x where and, by convention we de ne0! = 1: (2)(1) For example, to compute the probability of 3 heads out of 4 coin ips,
n = 4, p = 1 , X = number of heads, where we are interested in X = x
2
where x = 3.
Then the binomial probability function says that P (X = 3) = n p (1 ; p) ;
x
4!
13
= 3!(4 ; 3)! 2
4(3)(2)(1)
= f(3)(2)(1)gf(1)g
x n x !
= x!(nn; x)! p (1 ; p) ;
x 1 4;3
1; 2
1 3 1 1=4 1
2
2
2 Where does this formula come from? 84 n 4 x = 0:25 Let's consider example B.
Let X =number of obese subjects out of n = 3 randomly chosen US adults
where p = :209.
Forgetting the formula for a minute, how could we compute P (X = 2),
say?
One way is to list all of the possible outcomes of the experiment of observing 3 subjects and add up the probabilities for the outcomes that
correspond to 2 obese subjects.
Possible outcomes:
Outcome
Number
1
2
3
4
5
6
7
8 First
Subject
O
O
O
O
N
N
N
N Second
Subject
O
O
N
N
O
O
N
N Third
Subject
O
N
O
N
O
N
O
N Probability
of Outcome Outcomes 2, 3, and 5 corresponse to getting a total of X = 2 obese
subjects out of n = 3. What are the probabilitites of these three
outcomes?
Probability of (O O N ):
Recall that for independent events, the joint probability of the events
is the product of the individual probabilities of each event. Here,
whether the subject is obese is independent from subject to subject.
So, the probability of observing (O O N ) is p p (1 ; p) = p2 (1 ; p)1 = p (1 ; p) ;
x where n = 3, x = 2.
85 n x Probability of (O N O): p (1 ; p) p = p2 (1 ; p)1 = p (1 ; p) ;
x n x where n = 3, x = 2.
Probability of (N O O):
(1 ; p) p p = p2 (1 ; p)1 = p (1 ; p) ;
x n x where n = 3, x = 2.
Adding the probabilities of these mutually exclusive events together (addition rule) we get P (X = 2) = p2 (1 ; p)1 + p2(1 ; p)1 + p2 (1 ; p)1 = 3p2 (1 ; p)1
where for n = 3, x = 2 n = 3 = 3! = 3(2)(1) = 3:
x
2
2!(3 ; 2)! f(2)(1)gf(1)g ;3
2 is the number of ways to arrange a sequence with 2 `O's and 1 `N'.
;
More generally, gives the number of ways to choose x objects out
of n to be of one type and n ; x to be of the other type.
So, the probability of 2 obese subjects out of 3 randomly selected
subjects is
n
x n p (1;p) ; = 3 (:209)2 (1;:209)3;2 = 3(:209)2 (1;:209)3;2 = :1037:
x
2
x n x 86 The binomial formula can be used to compute the probability of x successes
out of n trials where the success probability on each trial is p for any value
of n and p.
However, it is convenient to have a table to give the answer for any given
value of n and p, or, even better, a computer function that allows us to
input n and p and outputs the answer.
Table A.1 in Appendix A of our book gives binomial probabilities
for selected values of n and p.
E.g., we computed the probability of x = 3 heads out of n = 4 coin ips
to be .25. Table A.1 uses k instead of x, so we look up n = 4 and k = 3 on
the left side of the table, p = :5 on the top and nd the probability equals
.2500 just as we computed.
Note that the table only gives selected values of p where p :5.
What if we are interested in p = :75, say?
We can handle such a case by considering the number of failures rather
than the number of successes.
That is, if X equals the number of successes out of n trials with success
probability p, then Y = n ; X = number of failures,
where the failure probability is q = 1 ; p. We observe X = x successes
out of n trials if and only if we observe Y = n ; x failures. So,
n
P (X = x) = P (Y = n ; x) = n ; x q ; (1 ; q) ;( ;
n
= n ; x q ; (1 ; q) :
n n n 87 x x x n ) x Example Suppose that 55% of UGA undergraduates are women. In a random sample of 7 UGA undergraduates, what's the probability that 3
of them are women?
Here X = number of women (success) out of n = 7 \trials" where probability of woman on each trial is p = :55. If x = 3 women are observed,
then we necessarily have observed Y = n ; x = 7 ; 3 = 4 men where the
probability of observing a man is q = 1 ; p = 1 ; :55 = :45:
So, the desired probability can be computed based on X :
P (X = 3) = n p (1 ; p) ; = 7 (:55)3 (1 ; :55)7;3
x
3
(7)(6)(5)(4)(3)(2)(1)
= f(3)(2)(1)gf(4)(3)(2)(1)g (:55)3 (1 ; :55)4 = :2388
x n x or, equivalently, based on Y : P (Y = 4) = 7 (:45)4 (1 ; :45)7;4
4
(7)(6)(5)(4)(3)(2)(1)
= f(4)(3)(2)(1)gf(3)(2)(1)g (:45)4 (1 ; :45)3 = :2388
The latter probability is tabulated in Appendix A.1, but the former
is not.
In addition, computer programs give binomial probabilities too. These
have the advantage that they give the result for any value of n and p.
In Minitab, select
Calc > Probability Distributions > Binomial...
Then select \Probability", and enter values for \Number of trials"
and \Probability of success". The value of x desired can be input
under \Input constant" or can be selected from data in a worksheet. 88 The binomial probability function gives the P (X = x) for all possible
values of x: 0 1 2 : : : n. So, the probability function gives the entire
probability distribution of X .
Once we know the probability distribution of a discrete r.v., we can compute its expected value and variance.
Recall: E (X ) = X xP (X = x) all x
= 0P (X = 0) + 1P (X = 1) + and
var(X ) = X all x
= (0 ; (x ;
X X + nP (X = n) = X )2 P (X = x) )2 P (X = 0) + + (n ; X )2 P (X = n) = 2 X Fortunately, these formulas simplify for the binomial distribution so that
we don't have to compute P (X = 0) : : : P (X = n).
In a binomial experiment with n trials, each with success probability p,
the number of successes X has the following expected value and variance: E (X ) = np
var(X ) = np(1 ; p) 89 Example  Obesity Again Suppose I take a random sample of n = 4 US adults. How many obese
subjects should I expect to observe on average?
Here n = 4, p = :209, so I expect to observe E (X ) = np = 4(:209) = 0:836
obese adults out of a sample of n = 4.
In a sample of n = 1000, I'd expect to observe np = 1000(:209) = 209
obese adults. (Make sense?)
The variance of the number of obese adults observed out of n = 1000
would be
var(X ) = np(1 ; p) = 1000(:209)(1 ; :209) = 165:319
p That is, the standard deviation is 165:319 = 12:9.
The interpretation here is that I could select n = 1000 US adults
and count the number of obese subjects over and over again. Over
the long run, the standard deviation of the number of obese subjects
observed when repeating this binomial experiment again and again
is 12:9.
{ That is, I expect to get about 209 out of 1000 obese subjects,
but the actual number obtained is going to vary around 209,
with typical deviation from 209 equal to 12.9. 90 The Poisson Probability Distribution
Another important discrete probability distribution that arisesoften in
practice is the Poisson probability distribution.
The binomial probability function gave the probability for the number of successes out of n trials.
{ Pertains to counts (of the number of successes) that are subject
to an upper bound n.
The Poisson probability function gives the probability for the number of events that occur in a given interval (often a period of time)
assuming that events occur at a constant rate during that interval.
{ pertains to counts that are unbounded. Any number of events
could, theoretically occur during the period of interest.
In the binomial case, we know p = probability of the event (success)
in each trial.
In the Poisson case, we know = the mean (or expected) number of
events that occur in the interval.
{ Or, equivalently, we could know the rate of events per unit of
time. Then , the mean number of events during an interval
of length t would just be t rate. Example  Tra c Accidents: Based on longrun tra c history, suppose that we know that an average
of 7 tra c accidents per month occur at Broad and Lumpkin. That is,
= 7 per month. We assume this value is constant throughout the year.
What's the probability that in a given month we observe exactly 8
accidents?
Such probabilities can be computed by the Poisson probability function.
If X = the number of events that occur according to a Poisson experiment
with mean , then
e; :
P (X = x) = x!
x Here, e denotes the base of the natural logarithm function. This is
a constant (like ) equal to 2:71828 : : :.
91 In the example, the probability of getting exactly 8 accidents in a month
is
;
;7 8
P (X = 8) = e x! = e 8!7 = :130:
x Poisson probabilities are tabulated in Table A.2 of Appendix A of
our text. They also may be computed in computer programs like
Minitab.
Often we are interested in cumulative probabilities.
For example, we may be interested in the probability that we have
no more than 8 accidents in a given month.
A probability P (X x) like this can be computed simply by summing up
P (X = 0) P (X = 1) : : : P (X = x).
In this example, the probability of no more than 8 accidents in a
month is given by P (X 8) = P (X = 0) + + P (X = 8)
e;770 + + e;778 = :000912 + + :130 = :729:
= 0!
8!
Fortunately, computer programs like Minitab usually have functions
for cumulative probabilities like this so that the individual probabilities need not be computed separately and then summed.
Of course, if I were interested in knowing the probability of having
more than x accidents in a month I could get that via
P (X > x) = 1 ; P (X x)
So, for example, the probability of having 9 or more accidents in a
month is 1 ; :729 = :271.
Cumulative binomial probabilities can be computed in the same way
as Poisson cumulative probabilities. That is, the formulas P (X x) = P (X = 0)+ +P (X = x) and P (X > x) = 1;P (X x)
hold for X a binomial outcome as well.
92 The Poisson distribution has the remarkable property that its expected value (mean) and variance are the same. That is, for X
following a Poisson distribution with mean ,
E(X ) = var(X ) =
For binomial experiments involving rare events (small p) and large
values of n, the distribution of X = the number of success out of
n trials is binomial, but it is also well approximated by the Poisson
distribution with mean = np.
E.g., Suppose that the prevalence of testicular cancer among US males is
.000113. Suppose we take a random sample of n = 10 000 male subjects.
Then the probability of observing 2 or fewer males with a lifetime diagnosis
of testicular cancer is given by the binomial cumulative probability: P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)
= 10000 :0001130(1 ; :000113)10000;0 + + 10000 :0001132(1 ; :000113)10000;2
0
2
= :894312
This is the exact answer, but it is pretty well approximated by a Poisson
probability with mean = np = 10000(:000113) = 1:13. Using the Poisson
probability function P (X 2) = P (X = 0) + P (X = 1) + P (X = 2)
e;1 131:130 + e;1 131:131 + e;1 131:132 = :894301
0!
1!
2!
: : 93 : Continuous Probability Distributions
Recall that for a discrete r.v. X , the probability function of X gave P (X =
x) for all possible x, thus describing the entire distribution of the r.v. X .
We'd like to do the same for a continuous r.v.
How do we calculate probabilities for continuous random variables?
For a continuous r.v., the probability that it takes on any particular
value is 0! Therefore, we can't use a probability function to describe
it!
{ E.g., the probability that a randomly selected subject from this
class weighs 146.923578234785079074... lbs is 0.
Instead of a probability function that gives the probability for each particular value of X , we quantify the probability that X falls in some interval
or region of all possible values of X .
This works because while the probability that a random student
weighs 146.923578234785079074... lbs is 0, the probability that he/she
weighs between 145 and 150 lbs, say, is not 0.
So, instead of describing the distribution of a continuous r.v. with a probability function, we use what is called the probability density function.
The probability density function for a continuous r.v. X gives a curve
such that the area under the curve corresponding to some interval
on the horizontal axis gives the probability that X takes a value in
that interval. 94 0.015
0.010
0.005
0.0 Probability density 0.020 0.025 E.g., suppose the probability density function for X =body weight for a
randomly selected student in this class looks like this: 100 150 200 250 Weight (lbs) The dashed vertical lines are at weight=145 lbs and weight=150 lbs.
The area under the curve between these lines gives the probability
that a randomly selected student weighs between 145 and 150 lbs.
In general, the are under the curve between x1 and x2 where x1 < x2
gives
P (x1 < X < x2 )
Note that the curve extends to the left and right, getting closer and
closer to zero.
{ That is, weights greater than x lbs, say, are possible (have
nonzero probability) no matter how big x is, but they are increasingly unlikely as x gets bigger.
{ Similarly, smaller and smaller weights are decreasingly probable.
The entire area under the probability density function is 1, representing the fact that P (;1 < X < 1) = 1
95 Note that for a continuous r.v. X , P (X = x) = 0 for all x. Therefore, P (X x) = P (X = x) + P (X < x) = 0 + P (X < x) = P (X < x):
Similarly, P (X x) = P (X = x) + P (X > x) = 0 + P (X > x) = P (X > x):
{ That is, for X continuous, there's no di erence between < and
probability statements, and also no di erence between >
and probability statements. Not true in the discrete case.
The Normal Distribution
Many continuous random variables have distributions such that
 values close to the mean are most probable, and values further away
from the mean are decreasingly probable (unimodal)
 values c units larger than the mean are just as probable as values
that are c units smaller than the mean (symmetry).
That is, many continuous random variables have probability distributions
that look like this: 0.2
0.0 0.1 f(x)=the p.d.f. of x 0.3 0.4 A normal probability density with mean 0 and variance=1 4 2 0 2 4 x The probability density function or p.d.f. given above is the p.d.f. of
the normal probability distribution (sometimes called the Gaussian
probability distribution).
96 The normal distribution is not the only distribution whose p.d.f.
looks bellshaped, but it is the most important one, and many real
world random variables follow the normal distribution, at least approximately.
The normal distribution, like the binomial and Poisson, is an example
of a parametric probability distribution. It is completely described
by a small number of parameters.
{ In the case of the binomial, there were two parameters, n and
p.
{ In the case of the Poisson, there was just one parameter, , the
mean of the distribution.
{ In the case of the normal, there are two parameters:
= the mean of the distribution, and
2
= the variance of the distribution.
That is, if X is a r.v. that follows the normal distribution, then that
means that we know exactly the shape of the p.d.f. of X except for
= E(X ), the mean of X , and 2 = var(X ), the variance of X .
{ We will use the notation X N( 2 ) to denote that the r.v. 2X folllows a normal distribution with
mean and variance .
{ E.g., X N (3 9) means that X has a normal distribution
with mean 3 and variance 9 (or SD=3).
The normal curve given above has mean 0 and variance 1. I.e., it is
N (0 1), which is called the standard normal distribution. 97 Normal distributions with di erent means have di erent locations.
Normal distributions with di erent variances have di erent degrees
of spread (dispersion).
{ Below are three normal probability distributions with di erent
means and variances.
0.4 Normal probability densities with different means and variances 0.2
0.0 0.1 f(x) 0.3 mean=0,var=1
mean=0,var=4
mean=4,var=1 5 0 5 10 x The normal p.d.f. is a function of x that maps out a bellshaped curve.
That is, it is a function f (x) that gives a probability density (a value along
the vertical axis) for each value of x (a value along the horizontal axis).
For a r.v. X N ( ), the speci c mathematical form of the normal
probability density of X is
1 e; (x2; 2)2
f (x) = p 2
2
where again, e denotes the constant 2.71828... and denotes the constant
3.14159.... 98 Facts about the normal distribution:
1. It is symmetric and unimodal.
{ As a consequence of this, the mean, median and mode are all
equal and occur at the peak of the normal p.d.f.
2. The normal p.d.f. can be located (have mean) anywhere along the
real line between 1 and extends inde nitely away from its mean
in either direction without ever touching the horizontal axis.
{ That is, if X N ( 2 ), then any value of X is possible,
although values far from will not be very probable.
3. As with any p.d.f., the area under the normal curve between any two
numbers x1 x2 where x1 < x2 gives
P (x1 < X < x2 )
and the total area under the p.d.f. is 1.
In particular, here are a few notable normal probabilities:
{ For x1 = ; 1 , x2 = + 1 ,
P ( ; 1 < X < + 1 ) = :6826
That is, 68.26% of the time a normally distributed r.v. falls
within 1 SD of its mean (i.e., has z score between 1 and 1).
{ For x1 = ; 2 , x2 = + 2 ,
P ( ; 2 < X < + 2 ) = :9544
That is, 95.44% of the time a normally distributed r.v. falls
within 2 SDs of its mean (i.e., has z score between 2 and 2).
{ For x1 = ; 3 , x2 = + 3 ,
P ( ; 3 < X < + 3 ) = :9972
That is, 99.72% of the time a normally distributed r.v. falls
within 3 SDs of its mean (i.e., has z score between 3 and 3).
These results are where the \empical rule" comes from.
99 Example  Height Suppose that US adult women have heights that are normally distributed
where the population mean height is 65 inches and the population standard
deviation for women's height is 2.5 inches.
Suppose that US adult men have heights that are normally distributed
with population mean 70 inches and population SD of 3 inches.
Let X = the height of a randomly selected adult US woman, and Y = the
height of a randomly selected adult US man. Then X N( 2 X X ) = N (65 2:52 ) Y N( 2 Y Y ) = N (70 32 ): For women, one SD below the mean is ; 1 = 65 ;1(2:5) = 62:5.
One SD above the mean is + 1 = 65 + 1(2:5) = 67:5.
{ So, the probability that a randomly selected woman has height
between 62.5 and 67.5 inches is
P (62:5 < X < 67:5) = :6826
X X X X (68.26% of women have heights between 62.5 and 67.5 inches).
The height p.d.f.s for men and women are given below.
0.15 Height p.d.f.s for men and women 0.0 0.05 f(height) 0.10 Women
Men 55 60 65 70 75 80 height Clearly, the area under the curve between 62.5 and 67.5 inches for
men is much less than 68.26%.
{ In fact the area under the male height curve between 62.5 and
67.5 inches turns out to be .1961 or 19.61%.
100 ...
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 Summer '08
 Staff
 Statistics, Biostatistics, Standard Deviation, Variance, Probability theory

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