Exercise 2.1
A researcher studying reading habits stands outside a London railway station
and approaches people at random as they leave the station. She asks them
which newspapers they read. What are the inadequacies
...
Exercise 2.1
A researcher studying reading habits stands outside a London railway station
and approaches people at random as they leave the station. She asks them
which newspapers they read. What are the inadequacies of this way of drawing
the sample? What differences might you expect between samples drawn at
9 am on a weekday and at 7 pm on a Saturday?
The point being made in Exercise 2.1 is that in drawing a sample it is essential
that it should represent the population being studied. In a study of road traffic
accidents, you might decide to examine a sample of 50 accidents. However,
unless you wanted only to make inferences about motorway accidents, these
50 should not just be motorway accidents but should be drawn from the entire
population of different types of road accident.
2.1.2 Probability
Example 2.3 Street lamp bulbs
Suppose that a local council suspects that the latest consignment of sodium
tubes for its street lamps is of poor quality, with an unacceptably large proportion of them being faulty and not working. To investigate this, a researcher
examines a sample of bulbs from this consignment waiting in the warehouse.
This example is fairly typical of a particular problem of quality control: the
estimation of 'percentage defectives' in a batch of supplied items. In practice,
quality procedures are much more rigidly prescribed than is suggested by the
sequential testing approach described in what follows, both for the supplier
and end-user. However, we shall be mainly concerned with the essential idea
of testing a random sample from a population (the population being, in this
case, the recent consignment of bulbs).
The researcher enters the warehouse and randomly chooses bulbs. The first
bulb works. Amongst the first ten bulbs he examines, he finds that only
one does not work. Amongst the first fifty, nine do not work. Amongst the
first 100, fifteen do not work. The researcher could go on, examining all the
bulbs in the warehouse, and find the exact proportion of bulbs which do not
work. Alternatively, he may use statistical arguments to support a claim that
the proportion of faulty bulbs within a sample of a certain size provides a
sufficiently accurate estimate of the overall proportion that is faulty.
As the researcher increases his sample size he will observe an interesting
phenomenon. For small samples, quite substantial deviations between the
proportions of faulty bulbs for different sample sizes are observed. Suppose
that the first two bulbs he examined were functioning, then he came across
a faulty one so that the sequence of the observed proportions of defectives
initially was 0/1,0/2 and 113. These may be regarded as successive estimates
of the unknown overall proportion of defectives and the last of these three
estimates is quite different from the first two. As the sample gets larger,
however, the deviations between successive estimates tend to become smaller.
Suppose for the 98th, 99th, and 100th bulbs that he tested the estimates
were 14/98,15/99 and 15/100. These proportions are much closer to one
another. The proportion of bulbs observed to be faulty seems to be settling
down towards some constant value (about 0.15). H
A 'random sample' here is strictly
one where no item in the
population is any more likely than
any other to be incorporated into
the sample. In this simplified
description of events, bulbs less
accessible than others are probably
less likely to be sampled! Hence
the need, in practice, for strict
quality procedures.Chapter 2 Section 2.1
We can simulate this phenomenon ourselves, that is, we can produce a simplified version or model of this situation which behaves in the same way. Later
on you will be encouraged to perform such simulations using a computer, but
for this first exploration the results will be more striking if we carry it out
for some physical system. The simulation tools that we shall use are dice and
coins. In many ways these provide ideal models because they strip away much
of the complexity so that we can focus on the essentials.
Exercise 2.2
Using a die from a family board game, carry out a sequence of 30 rolls with a
view to obtaining either a three or a six. Each time that a three-spot face or
a six-spot face occurs record a 1, to represent success. Each time that some
other face of the die occurs record a 0, to represent failure. In this way you
will generate a sequence of Is and OS. At any stage you will not be able to
predict whether the next number will be a 1 or a 0-so the next number is a
random variable.
By adding the Is and OS as you go along you can calculate the total number
of successes after each roll. At each roll you can also calculate the proportion
of rolls so far that have produced a three- or six-spot face. Letting P denote
this proportion, plot a graph of P, on the vertical axis, against the number
of rolls, on the horizontal axis. Join together successive points to produce a
jagged line running from left to right across the page.
Your sequence in Exercise 2.2 might have begun
0 0 1 1 0 0 0 0 0 0 1 1 ,
which would give you a sequence of points to plot as shown in Table 2.2.
Table 2.2 The first 12 rolls
Not everybody has easy access to a
six-sided die-some people will not
possess a board game. You can get
round the problem this way.
Obtain a six-sided pencil or
ball-point pen and identify
particularly two sides of the six (to
match the 'three' and the 'six' of
the exercise). These could be the
side impressed with the maker's
name and the next side round; or,
in the case of a ball-point'pen, the
side with the maker's name and the
side with the little hole in the
casing.
Roll number 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2
Observedresult 0 0 1 1 0 0 0 0 0 0 1 1
Total so far 0 0 1 2 2 2 2 2 2 2 3 4
Proportion (P) 011 012 113 214 215 216 217 218 219 2/10 3/11 4/12
An example of the kind of graph you might have obtained is given in Figure 2.2.
The thirty rolls which produced this graph gave the following results.
0 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 1 0 1 0 0 1 0 0
Roll
Figure 2.2 Proportion P, thirty rolls of a dieElements of Statistics
The phenomenon described above is apparent in this graph. As the number
of rolls increases, the sequence of estimates given by successive values of P
becomes less variable. The sequence appears to be tending to a particular
constant value, in this case about i. Did you notice something like this
happening for your sequence of estimates?
The sequence of rolls that resulted in the graph in Figure 2.2 was taken
further-up to 500 rolls of the die. The points in Figure 2.3 show the proportions obtained. The phenomenon is even more marked here.
Roll
Figure 2.3 Proportion P, 500 rolls of a die
Exercise 2.3
Take a coin and carry out a sequence of 40 tosses, recording 1for Heads and
0 for Tails. Again, the outcome of each toss is a random variable. Plot the
proportion of tosses producing Heads against the number of tosses. What do
you observe about the proportion of Heads as the number of tosses increases?
In Exercise 2.3, you might have observed behaviour similar to that of
Exercise 2.2, but with the points defining the jagged line converging within a
narrow region close to $ instead of $. However, what is readily apparent from
as many as 500 repetitions of an experiment (Figure 2.3) will not necessarily
be evident in as few as 30 repetitions (Figure 2.2) or even 40.
The four examples described so far (leaf lengths, quality testing of street
lamp bulbs, the die and coin experiments) have some common features. In
each example a number of cases is observed in which recorded measurements
on a random variable may differ from case to case. In the leaf measuring
example, the length of the leaves differed. In the street lamp example, bulbs
may or may not be faulty. In the die-rolling and coin-tossing examples, there
were two possible outcomes and each repetition produced one of these.
The leaf length example is different from the other three because, in this case,
the measured random variable is continuous. For the moment, let us considerChapter 2 Section 2.1
only the other three. Each example consists of a sequence of observations
on a binary random variable (two possible outcomes only). Using OS and Is
to label the different outcomes in each case, the results for any one of the
examples could be recorded as a sequence of OS and Is, such as
The term Bernoulli trial is used to describe a single statistical experiment
resulting in one of only two possible outcomes. Each of these examples therefore consists of a sequence of Bernoulli trials. The outcome of each trial is
uncertain (it was not possible to predict beforehand what it would be). We
have also noticed, for the bulb, die and coin examples at least, a tendency for
the observed proportions to settle down towards some constant value.
The value towards which the proportion is tending as the number of trials
increases is called the probability of the particular outcome of interest.
In the introduction to this chapter it was remarked that probability models
are used to represent the important aspects of random phenomena, and that
it is often necessary in the construction of such models to make simplifying
assumptions. This is true for the cases under discussion now. It is often
assumed that a coin is fair-that there is no reason to suppose that Heads
is more or less likely than Tails. So we might construct a probability model
for the outcome of a coin-tossing experiment by starting with the assumption
that the probability that Heads will appear is the same as the probability that
Tails will appear-both being equal to exactly i.
Similarly, for the die example, we might assume that each face has the same
probability of appearing. That is, we might choose to model the outcomes of
rolls of a real die by those of a perfect die in which the appearance of each
face is equiprobable. If we do this, the probability that a particular one of
the six faces appears is i,and so the probability that one of the two faces
showing three or six spots will appear is exactly 2 , or i.
Two other terms sometimes used in statistics are sample frequency, to
describe the number of times an outcome occurs, and sample relative frequency, to describe the proportion of times an outcome occurs in a particular
experiment. In a coin-tossing experiment in which a coin is tossed 100 times,
coming up Heads 47 times, we would say that the sample frequency (or just
'frequency') of Heads was 47 and the sample relative frequency of Heads was
= 0.47.
In the examples above, the modelling assumption is made that the probability
that a three- or six-spot face will appear in a roll of a perfect die is i;and
that the probability that Heads will appear in a toss of a perfect coin is i,.
The bulb-testing example is different. We cannot deduce, from any kind of
theoretical argument based on symmetry, a numerical value for the probability
that a bulb will be faulty. This probability has to be estimated from data-in
Example 2.3, this probability estimate was the sample relative frequency of
failures, 0.15 (based on 15 observed failures in 100 trials). We might say, on
the basis of this experiment, that we believe the proportion of faulty bulbs in
the latest consignment to be about 0.15.Elements of Statistics
Exercise 2.4
In an experiment to explore the issue of whether people are generally more
helpful to females than to males, eight students approached people and asked
if they could change a 5p coin. Altogether 100 people were approached by the
male students and 105 by the female students. The results of the experiment
are displayed in Table 2.3.
Table 2.3 Helping behaviour
Gender of student Help given Help not given
Male 71 29
Female 89 16
(a) Use these data to estimate the probability that a male is given help under
Sissons, M. (1981) Race, sex and
helping behaviour. British Journal
of Social Psychology, 20, 285-292.
The sum of 5p may seem very
small indeed to the modern reader,
and prompt the question why
change might be required for such
a sum. At the time the experiment
these circumstances.
(b) What would you estimate this probability to be for a female?
(c) Do the results of the experiment support the notion that people are more .
helpful to females?
was carried out, a local telephone
call could be made from public
telephone boxes for as little as 2p.
In Exercise 2.4 you were asked to estimate two probabilities. A large part of
this course is concerned with estimation-methods for obtaining estimates and
assessing how reliable these estimates are. You were also asked to comment on
the meaning of the results of the experiment. Later in the course (Chapter 8)
you will see formal ways of doing this-so that you can actually quantify the
extent to which experimental results support a research hypothesis.
Now let us take a slightly more complicated example. In a Bernoulli trial
there are only two possible outcomes: what if we were to consider, say, each
of the six faces of a die as distinct possible outcomes? Suppose the die is
rolled. What is the probability that the face with a single spot appears, what
is the probability that the face with two spots appears and so on?
We could carry out an experiment analogous to those above-rolling the die a
large number of times and calculating the relative frequency with which each
of the faces came up. This would enable us to directly estimate the probability
that each face occurs. Alternatively, since there is no reason to suppose that
any one of the six faces is favoured more than any of the others, we could argue
that each face is equally likely to appear-that the probabilities are equal for
each face. That is, our model would be that each face has a probability of i
of appearing. This would mean that about i of the time we should expect
the face with one spot to come up, about i of the time we should expect the
face with two spots to come up and so on. As we rolled the die more and
more times, so, under our model, we should expect the proportion of times
each face appears to get closer and closer to 6.
Example 2.4 A simulated die-rollingexperiment
Here is a simulation of an experiment based on the assumption of a perfect
die. The simulation was achieved using a computer to mimic the rolls and also
to record the outcomes. The die was rolled a large number of times, and the
observed frequencies of Is, 2s, ..., 6s (six possible outcomes) recorded after
30, 300 and 3000 rolls. At each stage the sample relative frequency for each
outcome was calculated.
52Chapter 2 Section 2.1
Table 2.4 Simulation results for a perfect die
Outcome 1 2 3 4 5 6
30 rolls Frequency 5 3 6 4 5 7
Relative frequency 0.1667 0.1000 0.2000 0.1333 0.1667 0.2333
300 rolls Frequency 55 54 56 52 46 3 7
Relative frequency 0.1833 0.1800 0.1867 0.1733 0.1533 0.1233
3000 rolls Frequency 506 508 497 525 467 497
Relative frequency 0.1687 0.1693 0.1657 0.1750 0.1557 0.1657
B
It is apparent from the observed frequencies that the computer model is
capable of reflecting the random variation inherent in the physical process.
Although we might 'expect' 50 observations of each of the six equiprobable
outcomes after 300 rolls of the die, we would nevertheless be somewhat surprised to see exactly that: actually, there were only 37 6s. However, after a
large number of rolls, the observed relative frequency for each of the possible
outcomes seems to be showing a tendency to approach the theoretical value
of A = 0.1667. This tendency is shown graphically in Figure 2.4, summarizing
the-findings in Table 2.4.
Relative frequency Relative frequency
1 2 3 4 5 6
Outcome
Rela
0.25
0.20
0.15
o.ia
0.05
0
re frequency
1 2 3 4 5 6
Outcome
Figure 2.4 Relative frequencies from a simulation of a perfect die
(a) 30 rolls (b) 300 rolls (c) 3000 rolls
Assuming a perfect die and applying a symmetry argument, Figure 2.5 illustrates the conclusions of a very extended experiment involving many rolls of
the die.
Relative freq~lency
Outcome
Figure 2.5 The perfect dieElements of Statistics
Put another way, the diagram in Figure 2.5 shows the theoretical probability
distribution over the six possible outcomes (1,2,.. .,6) for a die that is
assumed to be perfect.
It will be useful if at this stage some simple notation is introduced. In this
course, the probability that a statistical experiment results in a particular
outcome will be denoted P(.). For instance, the earlier statement that, in
tossing a perfect coin, Heads and Tails are equally likely so each outcome has
probability i,would be written
The theoretical probability distribution for the perfect die may be written
In particular, referring to the original experiment (Exercise 2.2),
This concept of a probability distribution is a very important and fundamental one in statistics. It tells us how likely it is that different outcomes (or,
in general, events) will occur. For example, you will see later how the idea
can be used to work out for a random sample how likely it is that a particular average will occur. Conversely, we might express our confidence, given a
sample average, that some particular underlying probability distribution might
have generated the data. This serves as a test of our understanding of what
is going on. If it is very unlikely that the data could have arisen from the
model that we believed to be a realistic representation for the phenomenon,
then maybe our beliefs need modification.
The next three sections of this chapter explore the idea of a probability distribution in more detail.
Exercise 2.5
Table 2.5 gives the numbers of yeast cells found in each of 400 very small
squares on a microscope slide when a liquid was spread over it. The first
row gives the number n of yeast cells observed in a square and the second
row gives the number of squares containing n cells. (For instance, 213 of
the 400 squares contained no yeast cells at all; no square contained more
than 5 cells.)
Table 2.5 Yeast cells on a microscope slide
Cells in a square, n 0 1 2 3 4 5
Frequency 213 128 37 18 3 1
Evidently, the number of cells per square is a random variable taking (in
this experiment) observed values between 0 and 5. Would you model this
variation using a discrete or continuous probability model?
Table 2.6 gives the lengths (in mm) of the jawbones of 23 kangaroos of
the Macropus Giganteus species.
Table 2.6 Jawbone lengths (mm)
Would you choose to model jawbone length as a discrete or continuous
random variable?
The following is a list of 20 operational lifetimes (in hours) for components.
Table 2.7 Operational lifetimes (hours)
'Student' (1906) On the error of
counting with a haemocytometer.
Biometrika, 5, 351-360. 'Student1
was the pseudonym of William
Sealy Gosset (1876-1937). Gosset
pursued his research whilst working
for the Guinness brewery company,
which prohibited its employees
from publishing their work-hence
the need to write under a pen
name. The probability distribution
bearing the name 'Student' is a
very important statistical model,
which we shall meet in Chapter 7.
Andrews, D.F. and Herzberg, A.M.
(1985) Data. Springer Verlag, New
York, p. 311.
Angus, J.E. (1982) Goodness-of-fit
tests for exponentiality based on a
loss-of-memory type functional
equation. J. Statistical Planning
and Inference, 6, 241-251.
What sort of model would you adopt for the variation in lifetimes?Chapter 2 Section 2.2
Daily rainfall (in mm) was recorded over a 47-year period in Turramurra,
Sydney, Australia. For each year, the wettest day was identified, that is,
that having the greatest rainfall. Table 2.8 shows the rainfall recorded for
the 47 annual maxima.
Table 2.8 Annual maxima, daily rainfall (mm)
What sort of random variable would you use to model the variation evident here?
Incidentally, the seventh observation of 3830mm seems very exceptional,
even in a location as wet as Turramurra would appear to be! In fact, it
turns out that a fairly standard probability model fits these data quite well
without ignoring that observation as an outlier, or assuming a misprint.
Without local knowledge, it is difficult to say whether or not the units
of measurement (mm) have been incorrectly recorded or whether perhaps
there is confusion over reporting what was actually measured. However,
the fact that something has gone wrong in the recording of these data is
clear: weather records give the annual average rainfall in Sydney as about
47 inches or about 1200mm! The Encyclopedia Britannica (15th edition,
1985) reports that the heaviest daily rainfall then recorded anywhere in
the world was 1870mm at Cilaos, La Reunion in the Indian Ocean, during
15-16 March 1952.
This kind of situation where the data units are wrong, or omitted, or
where it is not entirely clear what was being recorded, occasionally happens in statistics.
Data are given in Table 2.9 for library book usage. A sample of 122 books
was selected and for each of them the number of times they had been
borrowed in the preceding twelve months was counted. (For instance, of
the 122 books sampled, 65 had been borrowed just once in the preceding
year.)
Table 2.9 Library book usage
Loans per year 1 2 3 4 5 6 7
Frequency 65 26 12 10 5 3 1
The number of loans is a random variable. Is the random variable discrete
or continuous?
Here is a further example.
Example 2.7 Memory recall times
In a study of memory recall times, a series of stimulus words was shown to a
subject on a computer screen. For each word, the subject was instructed to
recall either a pleasant or an unpleasant memory associated with that word.
Successful recall of a memory was indicated by the subject pressing a bar
Rayner, J.C.W. and Best, D.J.
(1989) Smooth tests of goodness of
fit. Oxford University Press, p. 146.
Burrell, Q.L. and Cane, V.R.
(1982) The analysis of library data.
J. Royal Statistical Society,
Series A, 145, 439-471. The
authors collected the data from
several libraries. These data are
from one of the sections of the
Wishart Library in Cambridge.
Dunn, G. and Master, D. (1982)
Latency models: the statistical
analysis of response times.
Psychological Medicine, 12, 659-665.Elements of Statistics
on the computer keyboard. Table 2.10 shows the recall times (in seconds)
for twenty pleasant and twenty unpleasant memories. The random variable
'recall time' is an example of a continuous random variable-in principle, any
value within a certain range could arise.
Of key interest in this study was whether pleasant memories could be recalled
more easily and quickly than unpleasant ones. Figure 2.10 shows a comparative boxplot for these two samples and we can see that although the smaller
values of recall time for the unpleasant memories have values similar to the
smaller values for the pleasant memories, the larger values tend to be substantially greater for the unpleasant memories. Both distributions are skewed. We
shall explore more formal techniques for examining problems like this later in
the course.
pleasant,
I I I I I I I I I I I
1 2 3 4 5 6 7 8 9 10 1 1
Time (seconds)
Figure 2.10 Boxplot for memory recall times
Exercise 2.6
m
(a) According to the model suggested in Example 2.8, what proportion of
ODD.
gaps exceeds 5 seconds?
(b) According to the suggested model, can you find a formula for the probability that a gap is not longer than t, for some general value of t?
In your solution to Exercise 2.6(b), you used geometrical arguments to find a
formula for the probability P(T < t) for the triangular density defined over
the interval 0 5 t 5 20. A different approach is to use integration and say
This would give you the same answer. However, most of the results we shall
use in this course are standard and do not need to be worked out from first
principles. If you cannot remember a particular formula, then all you need do
is turn the pages to find it.
This section finishes with a summary of the ideas introduced so far, thus
identifying the important points.
2.2.3 Properties of probability models
We have briefly looked at two probability models for discrete random variables.
The first was proposed in order to describe the variation that would be observed when a six-sided die is rolled. Under the assumption that the die is
perfectly fair, the corresponding probability function is
where the random variable X denotes the number of spots on the face that
appears.
The second probability model was used to describe'the variation in outcome
when a Double-Five is rolled. Denoting the outcome of that experiment by
the random variable Y, the corresponding probability function is
and the range of Y is {l,3,4,5,6).
For continuous random variables, we needed a different mathematical entity, reflecting the fact that the range of observable values is continuous. We
saw that in this case we can represent the variation that might be observed
through a probability density function f (.), and calculate the probability that
a particular value is observed between a lower limit xl and an upper limit x2
asChapter 2 Section 2.2
where the expression on the right-hand side means the area under the curve
f (X)between X = x1 and X = 22. In the traffic example (Example 2.8), we
used the triangular density function
as a continuous probability model for traffic waiting times (in seconds).
Different models will be appropriate in different contexts: for example, data
suggest that a model very different to the triangular density would be appropriate to represent variation in leaf lengths, at least for one particular bush
of one particular species.
The expression 'probability density function' is often abbreviated to p.d.f.
and we may summarize the situation so far as follows.
Probability density function
For a continuous random variable X , observed variation may be
modelled through the probability density function (or simply the
density function) f(x). This is often abbreviated to p.d.f. Then the
probability that an observation on X lies between limits xl and 2 2 may
be calculated as the area under the curve of f(x) between X = xl and
X = 22. This is written in mathematical notation as the integral
The p.d.f. is defined for all values of X in the range of X .
What we have so far loosely called the probability function p(x) describing
the probability distribution of a discrete random variable X is sometimes
called a probability mass function, which clearly distinguishes it from the
probability density function f (X)for a continuous random variable. The term
is often abbreviated to p.m.f.
Probability mass function
The probability function for a discrete random variable is usually called
the probability mass function (6r simply the mass function) of the
random variable. This is often abbreviated to p.m.f. For a discrete
random variable X , the probability mass function gives the probability
distribution of X:
p(.) = P ( X = X).
The p.m.f. is defined for all values of X in the range of X .
During this discussion of probability mass functions and probability density
functions some important properties of probability models have been identified. First, a probability (whether of taking a particular value, or of observing
a value within a particular interval) is a number representing a long-term
proportion for an event; and that event will either never happen, sometimesElements of Statistics
happen, or always happen. Second, of all the possible outcomes (all the values
of X in the range of X ) one will certainly happen. Thus we are led to these
properties of probability functions.
For a discrete random variable X with probability mass function p(x),
0 l P(X) 5 1
for all X in the range of X , and
where the summation is over all X in the range of .X.
For a continuous random variable Y with probability density function
f (y), the p.d.f. cannot be negative,
and
where the integration is over the whole range of possible values of the
random variable Y. (The total area under the curve f (y) over the entire
range of Y is equal to 1.)
Using the appropriate probability function, we can calculate the probability
that a random variable will lie in any given interval by summing probabilities
(or by integration in the continuous case). For the sort of standard probability density function that will be considered in this course, it will not be
necessary to go through the process of integration, since most results are
well-documented.
It is often useful to know the probability that the random variable will not
exceed some specified value. Suppose, for example, that we wish to know the
probability that a random variable X, which only takes non-negative values,
will not exceed some specified value X. That is, we want to find the probability
P ( X 5 X).
If X is discrete, this probability may be obtained from the p.m.f. p(x) by summing appropriate terms: so, for a random variable X with range O,1,2,. . .,
this probability may be written
On the other hand, if X is continuous, then we would have
r x
(the lower limit of the integral being, in general, the lowest observable value
of X, not necessarily zero).
In either case, the function F(x) = P ( X < X)is called the cumulative distribution function of the random variable X .Chapter2 Section 2.2
The cumulative distribution function
The cumulative distribution function F(x) of a random variable X
is the probability that X takes a value less than or equal to X:
F(x) = P ( X 2 X),
where X is in the range of X.
Often the abbreviation c.d.f. is used, or the simpler term distribution
function.
In the following example, the c.d.f. of a discrete random variable is obtained.
Example 2.9 The c.d.f. for a fair die
For a fair die, the probability mass function is given by
P ( X = x ) = p ( x ) = k, x = 1 , 2 , 3 , 4 , 5 , 6 ,
and the cumulative distribution function is defined by F(x) = P ( X 5 X). For
instance,
The p.m.f. p(x) and the c.d.f. F(x) for the random variable X may conveniently be set out in a table as shown in Table 2.12. .
Exercise 2.7
If the random variable Y taking values y = 1,3,4,5,6is used to model the
outcome of a roll of a perfectly engineered Double-Five,and if the c.d.f. of Y
is written F(y), construct a table like Table 2.12 to represent the probability
distribution of the random variable Y.
Exercise 2.8
You have already found in Exercise 2.6(b) the c.d.f. of T, the random variable
describing the waiting time (in seconds) between consecutive vehicles using
the Kwinana Freeway in Perth. It was
Use the c.d.f. to calculate the probabilities
(a) P(T 5 10);
(b) P(T> 5).
(Again, you should notice that the supposition that these calculations are
useful is based on the assumption that F(t) as given above provides a good
model for the probability distribution of the time intervals between cars.)
The notation p ( . ) is standard for
probability mass functions; so is
the notation f (.) for probability
density functions. In either case,
F ( . ) is standard notation for the
cumulative distribution function.
Table 2.12 The probability
distribution for a fair dieElements of Statistics
The probability function for a random variable tells us all there is to know
about the probability that a random variable will take particular values or
fall in a particular interval of the range. If, for a given random variable, we
knew this function, then life would be straightforward-questions about likely
outcomes could be answered simply by reading off values from this function or
by using appropriate formulas. Unfortunately, life is not so straightforward.
In general, the probability function will not be known. This means that to
make headway in answering questions about probabilities we shall have to
estimate the function. Histograms and bar charts based on samples of data
provide one general way of estimating probability functions. In some cases
we may have more information on which to base our model for the process
generating the values of the random variable.
In the street lamp example (Example 2.3) there were only two possible outcomes (faulty or not faulty) so that the probability distribution had only two
peaks (0.15 and 0.85).
In the coin-tossing example (Exercise 2.3) which is very similar, we might
go further and postulate that the coin is fair-it favours neither Heads nor
Tails-so P(Heads) = ~ ( ~ a i l s ) . Since, as we have seen, the sum of the probabilities of all possible outcomes is 1,i.e. ~ ( ~ e a d s ) + P(Tai1s) = 1, we must
have P(Heads) = P(Tai1s) = $. This means that the two non-zero peaks of
the corresponding probability distribution in the coin-tossing example must
have equal heights. If we do not wish to make the assumption that the coin
is fair, then the situation is, from a mathematical viewpoint, identical to the
bulb-testing example.
In both these cases we shall have to estimate the probabilities from a sample,
that is, from the data. In other situations, such as the leaf length example
(Example 2.1) or the blood plasma P endorphin concentration data from
Chapter 1, Example 1.4, we might also have some reason (perhaps based
on theory or on extensive experience with similar situations) for believing
that the probability function will have a particular form. In the rest of this
chapter, and in Chapters 3 and 4, particular forms which occur frequently
and have very widespread application are described.
The probability models associated with the bulb-testing example, the cointossing example and the first die-rolling example (with two outcomes:
either the three- or six-spot face, or one of the other four) all used Bernoulli
trials: this illustrates an important feature of such situations. This is that
the probability function in question will typically be a member of a family of
similar functions. In these three examples, each function consisted of just two
peaks, but the heights of the peaks differed between the examples. We can
index the family by the heights of the peaks. In fact, in each case the height
of only one of the peaks needs to be given since the other can be found by
subtraction from 1. This indexing value is called a parameter of the distribution. The value of the parameter tells us exactly which member of a family
of distributions is being
Exercise 2.9
In a study in the Norwegian county of Sogn and Fjordane of all 6503 women Keiding, N. (1991) Age-specific
aged between 35 and 49, 591 of them were found to have passed menopause. imidence and prevalence: a
statistical perspective. J. Royal
(a) If we were to choose one of these women at random, what is the probability Statistical Society, Series A, 154,
that she would have passed the menopause? 371-412.
(b) Define a suitable random variable X to indicate whether a randomly
chosen woman has passed the menopause, and write down its probability
mass function.
Exercise 2.10
Telephone surveys are surveys conducted over the telephone. These are easier and cheaper than sending out interviewers, but not all households have
telephones so the coverage will be incomplete. In 1986, 78% of British households had telephones. For a randomly chosen British household, write down
the probability distribution of the random variable X which takes the value
1if the household does have a telephone and 0 if it does not
Exercise 2.11
An experiment is performed consisting of a sequence of 15 independent
Bernoulli trials. If a trial is successful (the interpretation of 'success' depends
on the context) a score of 1 is recorded; otherwise, a score of 0 is recorded.
The total 'score' for the experiment, Y, a random variable, is obtained by
adding together all 15 Is and OS.
What is the range of the random variable Y?
The situation where what is of interest is the total number of successful trials
in a set of independent Bernoulli trials is a very common one in statistics.
The probability model for this random variable is one of the standard models
that we shall be using in the course. Its definition follows.
If a set of n independent Bernoulli trials each has identical probability
of success, p, then the random variable Y, defined as the total number of
successes over all the trials, is said to follow a binomial distribution
with parameters n and p. This is written Y B(n,p).
Sometimes the expression 'Y is a binomial random variable' or 'Y is binomially
distributed' or, simply, 'Y is binomial' is used.
Before going on to examine properties of the binomial probability distribution,
we shall first explore different situations where it might or might not provide
an adequate model.Chapter 2 Section 2.3
Example 2.13 Coin tossing
Suppose a (fair) coin is tossed three times. Each time it may come up Heads or
Tails, and each of these outcomes has probability i. It is reasonable to assume
that the trials are independent-that the fact that it came up Heads (or Tails)
the first time does not influence the chance of it doing so at the second (or any
subsequent) trial. These three trials thus satisfy the conditions for the total
number of Heads to follow a binomial distribution with parameters n = 3 and
p = l.
2
This means that the probability distribution for the total number of Heads
(either 0, 1, 2 or 3) is binomial B(3,3). W
Example 2.14 Political attitudes in one household
In a survey of political attitude, we might ask all adults in a household how
they intend to vote in a two-party system. There are two possible outcomes.
Let us define, for each adult in the household, a random variable X, taking
the value 1 if they say they intend to vote RED and 0 if they say BLUE.
Then we could add up these random variables to produce a new one, Y, the
total number of adults who say they intend to vote RED. However, the X
scores would not be independent since adults within a household are likely to
influence the way the others vote, so the random variable Y would not follow
a binomial distribution. W
Exercise 2.12
Half of the married people in the United Kingdom are female and half male.
Thus, if a married person is chosen at random, the probability of picking a
female is i. If 100 married people are randomly picked, the distribution of the
number of females will follow the binomial distribution B(100,i).However,
suppose that 100 married people are picked by choosing 50 different people
at random, along with their spouse. Does the number of females now follow
a binomial distribution?
Example 2.15 Visual perception
In an experiment on visual perception, it was necessary to create a screen of Laner, S., Morris, P. and Oldfield,
small squares ( h inch by inch) and, at random, colour the squares black or (1957) A random pattern
white. The size of the screen was 27 inches by 40 inches, so there were 155520 Screen. Quarterly
Experimental Psychology, 9,
small squares to colour. A computer was used to make the decision according 105-108.
to a very long sequence of Bernoulli trials with predetermined probabilities
After this was done, and before performing the experiment, the screen was
sampled to see whether the colouring algorithm had operated successfully.
A total of 1000 larger squares (iinch by inch) each containing 16 of the
small squares were randomly selected and the number of black small squares
was counted in each case. The results should have been consistent with 1000
observations from the binomial probability distribution B(16,0.29). The results are shown in Table 2.13.Elements of Statistics
If the computer algorithm for the random colouring was correct, then-the data
in Table 2.13 give us our first glimpse of a random sample from a binomial
distribution. Figure 2.17 gives the corresponding bar chart for these data.
You can see that the sample is unimodal and moderately skewed.
Table 2.13 Counts on a
random screen pattern
Number of Frequency
black squares
Frequency
Screen counts
Figure 2.17 Counts on a random screen pattern
The statistical procedures for testing the quality of the fit of a hypothesized
probability model to a particular data set will be discussed in Chapter 9.
Indeed, at this stage we have only discussed circumstances under which the
binomial model might be assumed to hold, and the model itself has not yet
been developed. That is, we have not yet discovered how to calculate the
probability that a particular total will be obtained if-weobserve data assumed
to arise from a binomial distribution.
To illustrate the calculations, let us use the following example. Suppose that
we are interested in how likely people are to visit their doctor. We might be
interested in this as members of a team reporting to a local health authority on
the demand for health treatment. To investigate this, we could draw a random
sample of people and ask each person if they have visited their doctor in the
last year. We shall assume that whether or not one person has visited the
doctor implies nothing about the probability that any other person will have
visited the doctor-the observations are independent. Let us also suppose
that, unknown to us, 1 in 3 people will answer Yes to the question, so the
probability is !j that a randomly chosen person will answer Yes.
We shall go through the calculation in two stages. First, suppose that three
people are asked the question: have you visited your doctor in the past twelve
months? The probability that the first answers Yes is 5 . The probability that
the second answers Yes is also 5 . The probability that the third gives a No
response is $. An assumption of the binomial model is that these responses
are independent. What then is the overall probability that the question elicits
the responses
Yes Yes No
in that order? The easiest way to think about this is to calculate the proportion
of occasions in repeated samples that the first three responses would take
precisely that form. Only in one-third of such samples would the first response
be Yes; and only in one-third of those would the second also be Yes-so only
in one-ninth of repeated samples would the first two responses be Yes:
In reality, estimating this
probability would be the purpose,
or one of the purposes, of such an
investigation.
P(Yes Yes) = P(Yes) X P(Yes) = f X f = B.Chapter2 Section 2.3
Continuing the process, you can see that only in two-thirds of those samples
would the third response be No:
P(Yes Yes No) = P(Yes) X P(Yes) X P(No) = i X $ X -2 - 1 3 - 2 7 ' We can apply the same sort of
argument to any particular
Now we move to the second stage of the calculation. In pursuing this survey, sequence. For example, the
we are not actually interested in recording the order in which the responses rob ability that the sequence
occurred, only in counting the different responses. What, for instance, is the Yes Yes Yes
after asking the question of five
probability that amongst the first three people questioned, two say Yes and
people is
one says No? There are exactly three different ways this could happen and $ x ~ x ? x ' x ' = 4
their probabilities are as follows. 3 3 3 3 243'
P(Yes Yes No) = i X i X i = ;1Z?
P(Yes No Yes) = $ X i X =
P(No Yes Yes) = i X i X -31 - 2 - 27
If we now disregard the order of response, we find that the probability of
achieving two Yes responses and one No after questioning these three people
is
P(Yes Yes No) + P(Yes No Yes) + P(No Yes Yes)
- - 5 + & + & = 3 x & . 2
Fortunately, it is not necessary to list all the possible sequences of responses
in order to establish, say, the probability that in a sample of ten people questioned seven responded Yes; or that more than twelve responded Yes in a
sample of twenty. There is a formula that makes such efforts unnecessary. It
states generally that the number of different ways of ordering X Yeses (and
n - X Noes) in a sequence of n Yes-No responses is given by
for X taking any value from 0 (all Noes) to n (all Yeses).
The notation X!is a shorthand for the number 1 X 2 X 3 X ... X X. You might
also have seen the notation
'The number X!is read 'X factorial'. The number (E) is read 'n C X'.
So, for instance, the number of different ways of obtaining 7 Yeses (and 3
Noes) in a sample of ten people questioned is
7 3
Each one of these 120 ways occurs with probability (i) ( g ) ; for instance,
P(Yes Yes Yes No No Yes Yes Yes No Yes)Elements of Statistics
Overall, then, the probability of obtaining 7 Yes responses and 3 No responses
in a sample of size ten is
This is the probability that a binomial random variable with parameters
n = 10 (the sample size) and p = (the probability of obtaining a Yes response) will take the value 7. That is, if Y- B (10, $),then
or, more generally, if Y- B (10,i),then
The number O! is defined to be 1. Setting y equal to 0 in this formula gives
which is the probability that all ten responses are No.
The binomial probability model may be summarized as follows.
The binomial probability model
The random variable X follows a binomial distribution with parameters
n and p if it has probability mass function
where 0 < p < 1,and
This is written X- B(n,p) and provides a probability model for the
total number of successes in a sequence of n independent Bernoulli trials,
in which the probability of success in a single trial is p.
Exercise 2.13 m
Suppose that a study is undertaken to compare the safety and efficacy of two H
antidepressant drugs. Eighteen patients are each randomly allocated to one of
three groups, six to a group. The first group is treated with Drug A and the
second with Drug B. Patients in the third group are treated with a placebo (a
substance which is inert as far as antidepressive effects go, but which is given
to the patients in the same way as the treatments being studied, so that the
analysis can be controlled for any natural remission).
One of the problems associated with studies of this sort is that patients occasionally 'drop out': they cease treatment before the study is completed.
This might be for reasons unrelated to their course of treatment, or because
78Chapter 2 Section 2.3
they suffer from side-effects, or it might be because they perceive no beneficial
effect from their treatment. The phenomenon is consequently a complicating
feature in a statistical analysis of the results of such studies.
A previous study suggests that the proportion of patients in placebo groups
who drop out might be about 14%.
(a) Using this estimate for the value of the parameter p in a binomial model,
calculate for the placebo group in the present study
(i) the probability that all six patients drop out;
(ii) the probability that none of the six drop out;
(iii) the probability that exactly two drop out.
(b) An assumption of the binomial model is that of 'independence from trial to
trial'. Interpret this assumption in the context of the study, and comment
on whether you believe that, in this case, it is a reasonable assumption.
Example 2.16 Multiple choice examination scores
One of the components of assessment of the statistics students at a certain
British university is a multiple choice examination consisting of twenty questions. . For each question the correct answer is one of five possible options.
Students indicate the one of the five they believe to be correct. Sometimes
some of the students give the impression that they have gone through the
paper guessing answers at random.
Since such a student guesses answers at random, an answer to any particular
question is independent of an answer to any other question. Moreover, since
there are five possible answers to each q6estion and since the selection is made
at random, the probability of picking the correct one is $ for each question.
Thus the answers of a student who guesses at random form a sequence of
twenty independent Bernoulli trials, each with probability of success i , i.e.
0.2. So the total number of correct answers given to the twenty questions is
a random variable following a binomial distribution with parameters n = 20
and p = 0.2. That is, letting the random variable T denote the total number
of correct answers, T B(20,0.2).
In such a situation we might like to know the proportion of students who
would score more than the pass mark if they guessed in the way described
above. For example, if the pass mark was 10, the probability that a student
adopting a random guessing procedure such as that described above would
score less than 10 (and so fail the examination) is given by
Here, we have summed the probabilities of all the scores less than 10.
Table 2.14 gives the probability mass function P ( T = t) = p(t) and the cumulative distribution function P ( T 5 t )= F(t) for the binomial random variable
T B(20,0.2). For example,
P ( T = 3)
Dunbar, G.C., Cohn, J.B., Fabre,
L.F., Feighner, J.P., Fieve, R.R.,
Mendels, J. and Shrivastava, R.K.
(1991) A comparison of paroxetine,
imipramine and placebo in
depressed outpatients. British
Journal of Psychiatry, 159,
394-398.Elements of Statistics
The probabilities are given correct to four decimal places, but notice that, Table 2.14 The probability
because of rounding effects, the numbers in the right-hand column are not all distribution of 2' B(20,0.2)
exactly equal to the 'running sum' of the probabilities in the middle column. P(T= t ) P(T5 t )
For instance, the table gives
0
P(T= 0) = 0.0115, P ( T = 1)= 0.0576;
adding gives
P ( T = 0) + P ( T = 1)= 0.0691;
but
P(T 5 1)= 0.0692 (to 4 decimal places).
This kind of discrepancy occurs occasionally in probability calculations involving rounding and can be ignored.
If the pass mark is set at 10, the probability that a student who randomly
guesses his answers passes may be found as follows:
P(T2 10) = 1- P ( T 5 9) = 1- 0.9974 = 0.0026.
In general, the c.d.f. of a binomial random variable Y B(n,p) is given by
Unfortunately, this does not have a convenient mathematical form that is
quick and easy to evaluate. (If you are doing a statistical analysis using a statistical computer package, then the computer will probably do the calculations
for you and produce a probability value directly. This is the easiest way of
obtaining particular values of p(y) and F(y) when Y is binomial. Otherwise,
you can sum the individual probabilities using a calculator.)
In Figure 2.18 are diagrams of four typical binomial probability distributions.
They show
(a) the number dropping out in the placebo group (Exercise 2.13): B(6,0.14),
(b) the number of correct answers guessed (Example 2.16): B(20,0.2),
(c) the number of faulty bulbs in a box of ten (Example 2.3): B(10,0.15),
(d) the number of 5s in eight rolls of a Double-Five (Example 2.6): B (8,4).
Exercise 2.14
(a) If V W B(8,0.3), find the probability P ( V = 2). 1
(b) If W N B(12,0.5),find the probability P ( W = 8).
(c) If X - B(6,0.8), find the probability P ( X > 4).
(d) If Y B (6,i),find the probability P ( Y 5 2).
(e) If Z B (10,i),find the probability P ( Z 5 7).Chapter 2 Section 2.3
(a) Number of dropoilt.~, g (h) Number of correct guesses, y
0.35 0.30
0.30 0.25
0.25
0.20
0.20
0.15
0.15
0.10 0.10
0.05 0.05
0 0
0 1 2 3 4 5 6 7 8 9 1 0 0 1 2 3 4 5 6 7 8
(C) Number of faillt,y hillhs, (d) Number of Ss, ?/
Figure 2.18 Typical members of the binomial family
Exercise 2.15
The science of genetics includes the study of hereditable physical features R
m.
such as eye and hair colour, bone structure, blood group or unusual skin
conditions. The Augustinian monk Gregor Mendel (1822-1884) investigated,
in some experiments on pea plants, what happened when plants with round
yellow peas of a certain kind were crossed. The offspring were not all round
and yellow-some were wrinkled and others green, according to the following
combinations and in the proportions shown. (These proportions are based on
theory. However, observed experimental frequencies were very close to the
theoretical predictions.)
P(round and yellow) = &
P(round and green) =
P(wrink1ed and yellow) =
P(wrink1ed and green) =
Round yellow peas are crossed and produce eight offspring peas.
(a) What is the distribution of the number of wrinkled yellow peas amongst
the offspring?
(b) What is the probability that all eight offspring are wrinkled yellow peas?
(c) What is the probability that there are no wrinkled green peas amongst
the offspring?
Calculation of binomial probabilities using the formula can involve a surprisingly large number of key strokes on your calculator. This is especially
true when you are calculating more than one probability in order to calculate the probability that a particular random variable takes one of several
81Elements of Statistics
possible values. A set of statistical tables is another option, but these are
not necessarily very easy to read and usually contain only a limited number
of possibilities for the binomial parameter p. (For example, they may list
binomial probabilities for n from 2 to 20 and for p from 0.1 to 0.9 by steps
of 0.1 or possibly 0.05. It may be that the problem of interest to you has
p equal to 0.37.) An option is to use a computer.
Exercise 2.16
Use your computer to calculate directly these probabilities.
(a) P ( X = 4) when X N B(10,0.3);
(b) P ( X = 6) when X N B(20,0.17);
(C) P(X 5 8) when X N B(ll,O.33);
(d) P(X 1 8) when X N B(20, i).
Parts (e) and (f) illustrate two applications of the binomial distribution.
(e) Which is more probable: to obtain at least one 6 in one roll of four dice,
or at least one double-6 in twenty-four rolls of two dice? (This problem
was discussed by Cardano (1501-1576) and arose again at the gambling
table when, in 1654, de M&-&proposed it to Pascal.)
( f ) On any given day of the year (365 days), past experience suggests that
the probability of rain is about 0.3. What is the probability that in a
whole year it rains at least 100 times?
Exercise2.17
At the time of writing this section, one of the authors had a library of 5152 Data supplied by K.J. McConway,
books-too many for his shelves-so many of them were stored in boxes. He The Open University.
intended to move h&se andihoped that in his new house he would be able to
have enough shelving for all his books. He considered that the books currently
on his shelves poss&sed the same distribution of widths as his entire collection
and so took a sample of 100 books from his shelves to estimate the shape of
this distribution. The data are shown in Table 2.20.
Table 2.20 Widths of 100 books (mm)
(a) Plot a histogram of this sample, and calculate its mean and standard
deviation. Using your calculated mean, obtain an estimate in metres of
the total length of shelf space the author needed.
(b) Comment on any similarities or differences between your sample histogram and the bell-shaped distribution of S?ction 2.4.
In Section 2.1 you conducted some simulations using a die and a coin. Simulation can be a very powerful tool to investigate natural phenomena and is
normally done using a computer. For example, suppose we wish to explore
the statistical properties of an engineering production line. We may believe
that the objects produced each have a probability p of being faulty, and that
whether or not an object is faulty is independent of whether any other object
is faulty. We might be interested in how often all six objects in a groupElements of Statistics
of six will be in working order (perhaps the objects .are packed in boxes of
six). We could try to explore this mathematically and indeed the problem
is not very hard. Following the description (in Section 2.3) of the binomial
distribution, we know that the proportion of boxes containing six working
components out of six is (1-p)6. Alternatively, since we have a well-defined
mathematical model, we could program a computer to behave in the same
way as we believe the production line behaves. We could get the computer
to generate random OS and 1s so that a proportion p of them were 1s and
the others OS, and interpret 1as corresponding to a faulty product. Then we
would look through the sequence of OS and 1s generated by the computer and
grouped in sixes, and see how often a set of six is all OS.
This is an example of simulation. It is a rather artificial example in that it is
easy to work out the answer algebraically, as has been done. As the course
progresses, we shall discover more realistic situations and reasons for using
simulation approaches. To get used to the idea of generating data on your
computer, try the following exercises.
Exercise 2.18
(a) Using the Bernoulli random number generator of your computer, generate 10 values from a Bernoulli distribution with parameter p = 0.2, and
calculate the total number of 1s amongst the 10 values.
Now use the binomial random number generator to generate a single value
from a binomial distribution B(10,0.2).
What is the relationship between the populations from which these two
numbers have come?
(b) Again using the binomial random number generator, generate 10 values
from a B(20,0.5) distribution, 100 values from the same B(20,0.5) distribution and then 500 values from the same B(20,0.5) distribution.
In each of the three cases produce a bar chart of the samples.
Comment on the differences between the three bar charts in terms of how
jagged and irregular they are.
(c) Using the binomial random number generator, generate 500 values from
(i) a binomial distribution with parameters n = 10 and p = 0.1;
(ii) a binomial distribution with parameters n = 10 and p = 0.3;
(iii) a binomial distribution with parameters n = 10 and p = 0.5.
In each case, produce a bar chart for the 500 values.
How do you think the parameter p affects the shape of the binomial
distribution?
(d) Using the binomial random number generator, generate 500 values from
(i) a binomial distribution with parameters n = 10 and p = 0.2;
(ii) a binomial distribution with parameters n = 30 and p = 0.2;
(iii) a binomial distribution with parameters n = 50 and p = 0.2.
In each case produce a bar chart of the 500 values.
What effect do you think the parameter n has on the shape of the binomial
distribution?Chapter 2 Section 2.5
The next exercise is about the simulation of an industrial process.
Exercise 2.19
Fuses for use in domestic electrical appliances are packaged in bags of eight
for sale to customers. The probability that any fuse is defective is 0.012.
A retailer buys a box of 100 of these bags direct from the manufacturer. Use
your computer to 'open the bags' and count the number of defective fuses in
each one. How many bags contained no defectives? How many contained just
one? Complete a frequency table like the one in Table 2.21.
Table 2.21 Defective fuses
Number of Frequency
defective fuses
Total 100
The next exercise is about generating the results from simulated rolls of a die.
Exercise 2.20
(a) Use your computer to roll a fair six-sided die 6 times; compare the
sample relative frequencies for each of the six different outcomes with
the theoretical probabilities.
(b) Now use your computer to roll the die 600 times and repeat the calculations.
Exercise 2.21
A researcher is exploring the theory that after early injury to the left half
of the brain, the brain reorganizes itself to compensate for the loss. She
hypothesizes that one manifestation of such reorganization would be that
people who would otherwise have been right-handed would, after such damage
and reorganization, become left-handed. To test this hypothesis she wishes
to compare the proportion of left-handed people in the population of the
United Kingdom with the proportion of left-handed people in a sample of
brain-damaged people.
She knows that in the United Kingdom population, around 10% of the people
are left-handed.
In a sample of 33 people who were brain-damaged at birth, 8 were found to
be left-handed. Eight out of 33 represents a proportion of about 24%.
Data were provided by
Dr S.L. Channon, Middlesex
Hospital, University College
London.Elements of Statistics
This suggests that there may be some basis for the researcher's theory. However, we should explore the random variation that might be observed in
samples from the wider population. If a count as high as 8 turns out to
be not unusual, then, on the evidence of this experiment at least, there may
be no foundation for her belief.
Using the binomial random number generator on your computer, generate 100
samples from a binomial distribution with parameters n = 33 and p = 0.1, i.e.
10%, and see what fraction of these samples produce a count of 8 or more.
This fraction gives you an indication of how often you might expect to obtain
as many as 8 out of 33 left-handed people, if the proportion of left-handed
people in the overall population from which the sample was drawn was only
10%. If this fraction is very small then the researcher can feel confident that
there is some substance to her hypothesis. If the fraction is large then such
an outcome (at least 8 out of 33) could easily have arisen by chance.
This is an example of a statistical hypothesis test; hypothesis testing is described more formally in depth later in the course.
If your computer incorporates functions or procedures for the triangular
density Triangular(B), you can use the following exercise to investigate the
command syntax. Otherwise, use your calculator.
Exercise 2.22
(a) If the continuous random variable V follows a triangular distribution with
parameter 60, i.e. V- Triangular(GO), find
(i) P ( V I 2 0 ) ;
(ii) P ( V > 40); \
(iii) P(20 5 V 5 40).
(h) The following table gives the waiting times, that is, the gap (to the nearest Cox, D.R. and Snell, E.J. (1981)
half-hour) between consecuti've admissions at an intensive care unit. (The ,Applied Statistics-Princ~ples and
data are read across the rows.) Examples. Chapman and Hall,
London, D. 53. The data were
Table 2.22 Waiting times between admissions (hours)
collectedAbyDr A. Barr, Oxford
Regional Hos~ital Board.
(i) Obtain a histogram for these data.
(ii) Comment on the feasibility of fitting a triangular model to these data,
and decide on a sensible value for the indexing .parameter 8.
(iii) Use your model to determine the long-term proportion of waiting
times that would exceed 100 hours. Compare this with the corresponding
estimate of that proportion, based on these data.Chapter 2 Section 2.
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