Quantitative Methods for Business chapter 16 doc

CHAPTER
Test driving – sampling
theory, estimation and
hypothesis testing
16
Chapter objectives
This chapter will help you to:
■ understand the theory behind the use of sample results for
prediction
■ make use of the t distribution and appreciate its importance
■ construct and interpret interval estimates of population
means and population proportions
■ work out necessary sample sizes for interval estimation
■ carry out tests of hypotheses about population means, pro-
portions and medians, and draw appropriate conclusions
from them
■ use the technology; the t distribution, estimation and hypoth-
esis testing in EXCEL, MINITAB and SPSS
■ become acquainted with the business origins of the t
distribution
In the previous chapter we reviewed the methods that can be used to
select samples from populations in order to gain some understanding of
those populations. In this chapter we will consider how sample results
can be used to provide estimates of key features, or parameters, of the
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 483
populations from which they were selected. It is important to note that
the techniques described in this chapter, and the theory on which they
are based, should only be used with results of samples selected using
probabilistic or random sampling methods. The techniques are based
on knowing, or at least having a reliable estimate of, the sampling error
and this is not possible with non-random sampling methods.

In Chapter 13 we looked at the normal distribution, an important
statistical distribution that enables you to investigate the very many
continuous variables that occur in business and many other fields,
whose values are distributed in a normal pattern. What makes the nor-
mal distribution especially important is that it enables us to anticipate
how sample results vary. This is because many sampling distributions
have a normal pattern.
16.1 Sampling distributions
Sampling distributions are distributions that show how sample results
vary. They depict the ‘populations’ of sample results. Such distribu-
tions play a crucial role in quantitative work because they enable us to
use data from a sample to make statistically valid predictions or judge-
ments about a population. There are considerable advantages in using
sample results in this way, especially when the population is too large
to be accessible, or if investigating the population is too expensive or
time-consuming.
A sample is a subset of a population, that is, it consists of some obser-
vations taken from the population. A random sample is a sample that
consists of values taken at random from the population.
You can take many different random samples from the same popula-
tion, even samples that consist of the same number of observations.
Unless the population is very small the number of samples that you could
take from it is to all intents and purposes infinite. A ‘parent’ population
can produce an effectively infinite number of ‘offspring’ samples.
These samples will have different means, standard deviations and so
on. So if we want to use, say, a sample mean to predict the value of the
population mean we will be using something that varies from sample to
sample, the sample mean (x
–
), to predict something that is fixed, the

population mean.
To do this successfully we need to know how the sample means vary
from one sample to another. We need to think of sample means as
observations, x
–
s, of a variable, X
––
, and consider how they are distributed.
What is more we need to relate the distribution of sample means to the
484 Quantitative methods for business Chapter 16
parameters of the population the samples come from so that we can
use sample statistics to predict population measures. The distribution
of X
––
, the sample means, is a sampling distribution.
We will begin by considering the simplest case, in which we assume
that the parent population is normally distributed. If this is the case,
what will the sampling distributions of means of samples taken from
the population be like?
If you were to take all possible random samples consisting of n obser-
vations from a population that is normal, with a mean ␮ and a standard
deviation ␴, and analyse them you would find that the sample means of
all these samples will themselves be normally distributed.
You would find that the mean of the distribution of all these differ-
ent sample means is exactly the same as the population mean, ␮. You
would also find that the standard deviation of all these sample means is
the population standard deviation divided by the square root of the
size of the samples, ␴/√n.
So the sample means of all the samples size n that can be taken from
a normal population with a mean ␮ and a standard deviation ␴ are dis-

tributed normally with a mean of ␮ and a standard deviation of ␴/√n.
In other words, the sample means are distributed around the same
mean as the population itself but with a smaller spread.
We know that the sample means will be less spread out than the popu-
lation because n will be more than one, so ␴/√n will be less than ␴. For
instance, if there are four observations in each sample, ␴/√n will be
␴/2, that is the sampling distribution of means of samples which have
four observations in them will have half the spread of the population
distribution.
The larger the size of the samples, the less the spread in the values of
their means, for instance if each sample consists of 100 observations the
standard deviation of the sampling distribution will be ␴/10, a tenth
of the population distribution.
This is an important logical point. In taking samples we are ‘aver-
aging out’ the differences between the individual values in the popula-
tion. The larger the samples, the more this happens. For this reason it
is better to use larger samples to make predictions about a population.
Next time you see an opinion poll look for the number of people
that the pollsters have canvassed. It will probably be at least one thou-
sand. The results of an opinion poll are a product that the polling
organization wants to sell to media companies. In order to do this they
have to persuade them that their poll results are likely to be reliable.
They won’t be able to do this if they only ask a very few people for their
opinions!
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 485
The standard deviation of a sampling distribution, ␴/√n, is also
known as the standard error of the sampling distribution because it
helps us anticipate the error we will have to deal with if we use a sam-
ple mean to predict the value of the population mean.
If we know the mean and standard deviation of the parent population

distribution we can find the probabilities of ranges different sample
means as we can do for any other normal distribution, by using the
Standard Normal Distribution.
Example 16.1
Reebar Frozen Foods produces packs of four fish portions. On the packaging they claim
that the average weight of the portions is 120 g. If the mean weight of the fish portions
they buy is 124 g with a standard deviation of 4 g, what is the probability that the mean
weight of a pack of four portions will be less than 120 g?
We will assume that the selection of the four portions to put in a pack is random.
Imagine we took every possible sample of four portions from the population of fish por-
tions purchased by Reebar (which we will assume for practical purposes to be infinite)
and calculated the mean weight of each sample. We would find that the sampling distri-
bution of all these means has a mean of 124 g and a standard error of 4/√4, which is 2.
The probability that a sample of four portions has a mean of less than 120 g is the
probability that a normal variable with a mean of 124 and a standard deviation of 2 is
less than 120.
The z-equivalent of the value 120 in the sampling distribution is
From Table 5 on pages 621–622 in Appendix 1 you will find that the probability that
z is less than Ϫ2.00 is 0.0228 or 2.28%.
We can conclude that there is a less than one in forty chance that four portions in a
pack chosen at random have a mean weight of less than 120 g. You might like to
compare this with the probability of one fish portion selected at random weighing less
than 120 g:
Using Table 5 you will find that the probability that Z is less than Ϫ1.00 is 0.1587 or
15.87%, approximately a one in six chance. This is rather greater than the chance of
getting a pack of four whose mean weight is less than 120 g (2.28%); in general there is
less variation among sample means than there is among single points of data.
z
x 120 124
1.00ϭ

Ϫ
ϭ
Ϫ
ϭϪ
␮
␴ 4
z
x
n

120 124
ϭ
Ϫ
ϭ
Ϫ
ϭϪ
␮
␴ √√44
200.
486 Quantitative methods for business Chapter 16
The procedure we used in Example 16.1 can be applied whether we
are dealing with small samples or with very much larger samples. As
long as the population the samples come from is normal we can be
sure that the sampling distribution will be distributed normally with a
mean of ␮ and a standard deviation of ␴/√n.
But what if the population is not normal? There are many distribu-
tions that are not normal, such as distributions of wealth of individuals
or distributions of waiting times.
Fortunately, according to a mathematical finding known as the
Central Limit Theorem, as long as n is large (which is usually interpreted

to mean 30 or more) the sampling distribution of sample means will be
normal in shape and have a mean of ␮ and a standard deviation of ␴/√n.
This is true whatever the shape of the population distribution.
Example 16.2
The times that passengers at a busy railway station have to wait to buy tickets during the
rush hour follow a skewed distribution with a mean of 2 minutes 46 seconds and a stand-
ard deviation of 1 minute 20 seconds. What is the probability that a random sample of
100 passengers will, on average, have to wait more than 3 minutes?
The sample size, 100, is larger than 30 so the sampling distribution of the sample
means will have a normal shape. It will have a mean of 2 minutes 46 seconds, or
166 seconds, and a standard error of 80/√100 seconds.
From Table 5 the probability that Z is more than 1.75 is 0.0401. So the probability
that a random sample of 100 passengers will have to wait on average more than 3 minutes
is 4.01%, or a little more than a one in twenty-five chance.
PX P
Z
PZ PZ
() 180 seconds
180 166
( 14 8) ( 1.75)
Ͼϭ
ϾϪ
ϭϾ ϭϾ
80 100√
⎛
⎝
⎜
⎞
⎠
⎟

If the samples taken from a population that is not normal consist of
fewer than 30 observations then the Central Limit Theorem does not
apply. The sampling distributions of means of small samples taken
from such populations do not have a normal pattern.
At this point you may find it useful to try Review Questions 16.1 to
16.6 at the end of the chapter.
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 487
16.1.1 Estimating the standard error
The main reason for being interested in sampling distributions is to
help us use samples to assess populations because studying the whole
population is not possible or practicable. Typically we will be using a
sample, which we do know about, to investigate a population, which we
don’t know about. We will have a sample mean and we will want to use
it to assess the likely value of the population mean.
So far we have measured sampling distributions using the mean and
the standard deviation of the population, ␮ and ␴. But if we need to
find out about the population using a sample, how can we possibly
know the values of ␮ and ␴?
The answer is that in practice we don’t. In the case of the population
mean, ␮, this doesn’t matter because typically it is something we are
trying to assess. But without the population standard deviation, ␴, we
do need an alternative approach to measuring the spread of a sam-
pling distribution.
Because we will have a sample, the obvious answer is to use the stand-
ard deviation, s, in place of the population standard deviation, ␴. So
instead of using the real standard error, ␴/√n, we estimate the stand-
ard error of the sampling distribution with s/√n.
Using the estimated standard error, s/√n, is fine as long as the sam-
ple concerned is large (in practice, that n, the sample size, is at least
30). If we are dealing with a large sample we can use s/√n as an approx-

imation of ␴/√n. The means of samples consisting of n observations
will be normally distributed with a mean of ␮ and an estimated stan-
dard error of s/√n. The Central Limit Theorem allows us to do this
even if the population the sample comes from is not itself normal in
shape.
Example 16.3
The mean volume of draught beer served in pint glasses in the Nerry Ash Leavy Arms
pub is known to be 0.538 litres. A consumer organization takes a random sample of
36 pints of draught beer and finds that the standard deviation of this sample is 0.066 litres.
What is the probability that the mean volume of the sample will be less than a pint
(0.568 litres)?
The population mean, ␮, in this case is 0.538 and the sample standard deviation,
s, is 0.042. We want the probability that x
–
is less than 0.568, P (X
––
Ͻ 0.568). The
488 Quantitative methods for business Chapter 16
It is important to remember that s/√n is not the real standard error,
it is the estimated standard error, but because the standard deviation of a
large sample will be reasonably close to the population standard devi-
ation the estimated standard error will be close to the actual standard
error.
At this point you may find it useful to try Review Question 16.7 at the
end of the chapter.
16.1.2 The t distribution
In section 16.1.1 we looked at how you can analyse sampling distribu-
tions using the sample standard deviation, s, when you do not know the
population standard deviation, ␴. As long as the sample size, n, is 30
or more the estimated standard error will be a sufficiently consistent

measure of the spread of the sampling distribution, whatever the shape
of the parent population.
If, however, the sample size, n, is less than 30 the estimated standard
error, s/√n, is generally not so close to the actual standard error, ␴/√n,
and the smaller the sample size, the greater will be the difference
between the two. In this situation it is possible to model the sampling
distribution using the estimated standard error, as long as the popula-
tion the sample comes from is normal, but we have to use a modified
normal distribution in order to do it.
This modified normal distribution is known as the t distribution. The
development of the distribution was a real breakthrough because it
made it possible to investigate populations using small sample results.
Small samples are generally much cheaper and quicker to gather than
a large sample so the t distribution broadened the scope for analysis
based on sample data.
z-equivalent of 0.568 is:
If you look at Table 5 you will find that the probability that Z is less than 2.73 is 0.9968,
so the probability that the sample mean is more than a pint is 0.9968 or 99.68%.
z
x
sn
PX PZ
0.568 0.538
2.73 to 2 decimal places
So 0.568 2.73
ϭ
Ϫ
ϭ
Ϫ
ϭ

ϽϭϽ
␮
√√0 066 36.
()()
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 489
The t distribution is a more spread out version of the normal distri-
bution. The difference between the two is illustrated in Figure 16.1.
The greater spread is to compensate for the greater variation in sample
standard deviations between small samples than between large samples.
The smaller the sample size, the more compensation is needed, so
there are a number of versions of the t distribution. The one that
should be used in a particular context depends on the number of
degrees of freedom, represented by the symbol ␯ (nu, the Greek letter n),
which is the sample size minus one, n Ϫ1.
To work out the probability that the mean of a small sample taken
from a normal population is more, or less, than a certain amount we
first need to find its t-equivalent, or t value. The procedure is very
similar to the way we work out a z-equivalent.
t
x
sn
ϭ
Ϫ
␮
√
Ϫ3 Ϫ20123Ϫ1
0.0
0.1
0.2
0.3

0.4
Figure 16.1
The Standard
Normal Distribution
(solid line) and the
t distribution
(dotted line)
Example 16.4
A customer visiting the pub in Example 16.3 purchases nine pints of draught beer. The
precise volumes of the pints served are known to be normally distributed with a mean
of 0.538 litres and the standard deviation of the volumes of the nine pints bought by
the customer is 0.048 litres. What is the probability that the mean volume of each of
the nine pints is less than a pint (0.568 litres)?
490 Quantitative methods for business Chapter 16
The t value that we used in Example 16.4 could be written as t
0.05,8
because it is the value of t that cuts off a tail area of 5% in the t distri-
bution that has 8 degrees of freedom. In the same way, t
0.01,15
represent
the t value that cuts off a tail area of 1% in the t distribution that has
15 degrees of freedom.
You will find that the way the t distribution is used in further work
depends on tail areas. For this reason, and also because the t distribution
varies depending on the number of degrees of freedom, printed tables
do not provide full details of the t distribution in the same way that
Standard Normal Distribution tables give full details of the Standard
Normal Distribution. Table 6 on page 623 gives selected values of t from
the t distribution with different degrees of freedom for the most com-
monly used tail areas. If you need t distribution values that are not in

Table 6 you can obtain them using computer software, as shown in
section 16.4 later in this chapter.
The population mean, ␮, is 0.538 and the sample standard deviation, s, is 0.048. We
want the probability that X
––
is less than 0.568, P (X
––
Ͻ 0.568). The t value equivalent
to 0.568 is:
You will find some details of the t distribution in Table 6 on page 623 in Appendix 1.
Look down the column headed ␯ on the left hand side until you see the figure 8, the
number of degrees of freedom in this case (the sample size is 9). Look across the row to
the right and you will see five figures that relate to the t distribution with eight degrees
of freedom. The nearest of these figures to 1.875 is the 1.86 that is in the column
headed 0.05. This means that 5% of the t distribution with eight degrees of freedom is
above 1.86. In other words, the probability that t is more than 1.86 is 0.05. This means
that the probability that the mean volume of nine pints will be less than 0.568 litres will
be approximately 0.95.
t
x
sn
PX Pt
0.568 0.538
1.875
So ( 0.568) ( 1.875)
ϭ
Ϫ
ϭ
Ϫ
ϭ

Ͻ
ϭ
Ͻ
␮
√√0 048 9.
Example 16.5
Use Table 6 to find:
(a) t with 4 degrees of freedom that cuts off a tail area of 0.10, t
0.10,4
(b t with 10 degrees of freedom that cuts off a tail area of 0.01, t
0.01,10
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 491
At this point you may find it useful to try Review Questions 16.8 and
16.9 at the end of the chapter.
16.1.3 Choosing the right model for a sampling
distribution
The normal distribution and the t distribution are both models that
you can use to model sampling distributions, but how can you be sure
that you use the correct one? This section is intended to provide a brief
guide to making the choice.
The first question to ask is, are the samples whose results make up
the sampling distribution drawn from a population that is distributed
normally? In other words, is the parent population normal? If the
answer is yes then it is always possible to model the sampling distribu-
tion. If the answer is no then it is only possible to model the sampling
distribution if the sample size, n, is 30 or more.
The second question is whether the population standard deviation,
␴, is known. If the answer to this is yes then as long as the parent popu-
lation is normal the sampling distribution can be modelled using the
normal distribution whatever the sample size. If the answer is no the

sampling distribution can be modelled using the normal distribution
only if the sample size is 30 or more. In the absence of the population
standard deviation, you have to use the sample standard deviation to
approximate the standard error.
Finally, what if the parent population is normal, the population stand-
ard deviation is not known and the sample size is less than 30? In
these circumstances you should use the t distribution and approximate
(c) t with 17 degrees of freedom that cuts off a tail area of 0.025, t
0.025,17
(d) t with 100 degrees of freedom that cuts off a tail area of 0.005, t
0.005,100
.
From Table 6:
(a) t
0.10,4
is in the row for 4 degrees of freedom and the column headed 0.10,
1.533. This means that the probability that t, with 4 degrees of freedom, is
greater than 1.533 is 0.10 or 10%.
(b) t
0.01,10
is the figure in the row for 10 degrees of freedom and the column
headed 0.01, 2.764.
(c) t
0.025,17
is in the row for 17 degrees of freedom and the 0.025 column, 2.110.
(d) t
0.005,100
is in the row for 100 degrees of freedom and the 0.005 column, 2.626.
492 Quantitative methods for business Chapter 16
the standard error using the sample standard deviation. Note that if

the parent population is not normal and the sample size is less than 30
neither the normal distribution nor the t distribution can be used to
model the sampling distribution, and this is true whether or not the
population standard deviation is known.
16.2 Statistical inference: estimation
Businesses use statistical analysis to help them study and solve prob-
lems. In many cases the data they use in their analysis will be sample
data. Usually it is too expensive, or too time-consuming or simply
impossible to obtain population data.
So if there is a problem of customer dissatisfaction they will study data
from a sample of customers, not all customers. If there is a problem
with product quality they will study a sample of the products, not all of
them. If a large organization has a problem with staff training they will
study the experiences of a sample of their staff rather than all their staff.
Of course, they will want to analyse the sample data in order to draw
general conclusions about the population. As long as the samples they
use are random samples, in other words they consist of observed values
chosen at random from the population, it is quite possible to do this.
The use of sample data in drawing conclusions, or making deduc-
tions, about populations is known as statistical inference from the word
infer which means to deduce or conclude. Statistical inference that
involves testing claims about population parameters is known as statis-
tical decision-making because it can be used to help organizations and
individuals take decisions.
In the last section we looked at sampling distributions. These distri-
butions are the theoretical foundations on which statistical inference is
built because they connect the behaviour of sample results to the dis-
tribution of the population the samples came from. Now we will con-
sider the procedures involved in statistical inference.
There are two types of statistical inference technique that you will

encounter in this chapter. The one we shall look at in this section is
estimation, the using of sample data to predict population measures
like means and proportions. The other is hypothesis testing, using sample
data to verify or refute claims made about population measures, the
subject of Section 16.3.
Collecting sample data can be time-consuming and expensive so in
practice organizations don’t like to gather more data than they need,
but on the other hand they don’t want to end up with too little in case
they haven’t enough for the sort of conclusive results they want. You
will find a discussion of this aspect of planning statistical research in
this section.
16.2.1 Statistical estimation
Statistical estimation is the use of sample measures such as means or pro-
portions to estimate the values of their population counterparts. The easi-
est way of doing this is to simply take the sample measure and use it as
it stands as a prediction of the population equivalent. So, we could take
the mean of a sample and use it as our estimate of the population mean.
This type of prediction is called point estimation. It is used to get a ‘feel’
for the population value and is a perfectly valid use of the sample result.
The main shortcoming of point estimation is given away by its name;
it is a single point, a single shot at estimating one number using another.
It is a crude way of estimating a population measure because not only
is it uncertain whether it will be a good estimate, in other words close to
the measure we want it to estimate, but we have no idea of the probability
that it is a good estimate.
The best way of using sample information to predict population
measures is to use what is known as interval estimation, which involves
constructing a range or interval as the estimate. The aim is to be able to
say how likely it is that the interval we construct is accurate, in other
words, how confident we are that the interval does include within it the

population measure. Because the probability that the interval includes
the population measure, or the confidence we should have in the
interval estimate, is an important issue, interval estimates are often
called confidence intervals.
Before we look at how interval estimates are constructed and why
they work, it will be helpful if we reiterate some key points about sam-
pling distributions. For convenience we will concentrate on sample
means for the time being.
■ A sampling distribution of sample means shows how all
the means of the different sample of a particular size, n, are
distributed.
■ Sampling distributions that describe the behaviour of means
of samples of 30 or more are always approximately normal.
■ The mean of the sampling distribution of sample means is the
population mean, ␮.
■ The standard deviation of the sampling distribution of sample
means, called the standard error, is the population standard
deviation divided by the square root of the sample size, ␴/√n.
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 493
494 Quantitative methods for business Chapter 16
The sampling distributions that are normal in shape, the ones that
show how sample means of big samples vary, have the features of the
normal distribution. One of these features is that if we take a point two
standard deviations to the left of the mean and another point two stand-
ard deviations to the right of the mean, the area between the two
points is roughly 95% of the distribution.
To be more precise, if these points were 1.96 standard deviations
below and above the mean of the distribution the area would be exactly
95% of the distribution. In other words, 95% of the observations in the
distribution are within 1.96 standard deviations from the mean.

This is also true for normal sampling distributions; 95% of the sam-
ple means in a sampling distribution that is normal will be between
1.96 standard errors below and 1.96 standard errors above the mean.
You can see this illustrated in Figure 16.2.
The limits of this range or interval are:
␮ Ϫ 1.96 ␴/√n on the left-hand side
and ␮ ϩ 1.96 ␴/√n on the right-hand side.
The greatest difference between any of the middle 95% of sample
means and the population mean, ␮, is 1.96 standard errors, 1.96 ␴√n.
The probability that any one sample mean is within 1.96 standard
errors of the mean is:
The sampling distribution allows us to predict values of sample means
using the population mean. But in practice we wouldn’t be interested in
PnX n()␮␴ ␮␴ 1.96 1.96 0.95ϪϽϽϩ ϭ√√
Standard errors from the mean
Ϫ3Ϫ4 Ϫ201234Ϫ1
0.0
0.1
0.2
0.3
0.4
P(X ϭ x )
Figure 16.2
The middle 95% of
the area of a
sampling distribution
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 495
doing this because we don’t know the population mean. Indeed, typ-
ically the population mean is the thing we want to find out about using
a sample mean rather than the other way round. What makes sampling

distributions so important is that we can use them to do this.
As we have seen, adding and subtracting 1.96 standard errors to and
from the population mean creates an interval that contains 95% of the
sample means in the distribution. But what if, instead of adding this
amount to and subtracting it from the population mean, we add it to
and subtract it from every sample mean in the distribution?
We would create an interval around every sample mean. In 95% of
cases, the intervals based on the 95% of sample means closest to the
population mean in the middle of the distribution, the interval would
contain the population mean itself. In the other 5% of cases, those
means furthest away from the population mean, the interval would not
contain the population mean.
So, suppose we take the mean of a large sample and create a range
around it by adding 1.96 standard errors to get an upper figure, and
subtracting 1.96 standard errors to get a lower figure. There is a 95%
chance that the range between the upper and lower figures will encom-
pass the mean of the population. Such a range is called a 95% interval
estimate or a 95% confidence interval because it is an interval that we are
95% confident, or certain, contains the population mean.
Example 16.6
The total bill sizes of shoppers at a supermarket have a mean of £50 and a standard devi-
ation of £12.75. A group of researchers, who do not know that the population mean bill
size is £50, finds the bill size of a random sample of 100 shoppers.
The sampling distribution that the mean of their sample belongs to is shown in
Figure 16.3. The standard error of this distribution is 12.75/√100 ϭ1.275.
Ninety-five per cent of the sample means in this distribution will be between 1.96
standard errors below the mean, which is:
50 Ϫ (1.96 * 1.275) ϭ 47.50
and 1.96 standard errors above the mean, which is:
50 ϩ (1.96 * 1.275) ϭ 52.50.

This is shown in Figure 16.4.
Suppose the researchers calculate the mean of their sample and it is £49.25, a fig-
ure inside the interval 47.50 to 52.50 that contains the 95% of sample means within
496 Quantitative methods for business Chapter 16
1.96 standard errors of the population mean. If they add and subtract the same 1.96
standard errors to and from their sample mean:
49.25 Ϯ (1.96 * 1.275) ϭ 49.25 Ϯ 2.499 ϭ £46.751 to £51.749
The interval they create does contain the population mean, £50.
Notice the symbol ‘Ϯ’ in the expression we have used. It represents the carrying out
of two operations: both adding and subtracting the amount after it. The addition
46.175 47.45 48.725 50.00 51.275 52.55 53.825
0.0
0.1
0.2
0.3
0.4
x
P(X ϭ x )
46.175 47.45 48.725 50.00 51.275 52.55 53.825
0.0
0.1
0.2
0.3
0.4
P(X ϭ x )
x
Figure 16.3
The sampling distribution in Example 16.6
Figure 16.4
The middle 95% of the sampling distribution in Example 8.1

Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 497
If the researchers in Example 16.6 took many samples and created
an interval based on each one by adding and subtracting 1.96 standard
errors they would find that only occasionally would the interval not
include the population mean.
How often will the researchers in Example 16.6 produce an interval
that does not include the population mean? The answer is every time
they have a sample mean that is among the lowest 2
1
⁄2% or the highest
2
1
⁄2% of sample means. If the sample mean is among the lowest 2
1
⁄2%
the interval they produce will be too low, as in Example 16.7. If the
sample mean is among the highest 2
1
⁄2% the interval will be too high.
As long as the sample mean is among the 95% of the distribution
between the lowest 2
1
⁄2% and the highest 2
1
⁄2%, the interval they pro-
duce will include the population mean, in other words it will be an
accurate estimate of the population mean.
Of course, usually when we carry out this sort of research we don’t
actually know what the population mean is, so we don’t know whether
the sample mean we have is among the 95% that will give us accurate

interval estimates or whether it is among the 5% that will give us inac-
curate interval estimates. The important point is that if we have a sam-
ple mean and we create an interval in this way there is a 95% chance
that the interval will be accurate. To put it another way, on average
produces the higher figure, in this case 51.749, and the subtraction produces the lower
figure, 46.751.
Imagine they take another random sample of 100 shoppers and find that the mean
expenditure of this second sample is a little higher, but still within the central 95% of
the sampling distribution, say £51.87. If they add and subtract 1.96 standard errors to
and from this second mean:
51.87 Ϯ (1.96 * 1.275) ϭ 51.87 Ϯ 2.499 ϭ £49.371 to £54.369
This interval also includes the population mean.
Example 16.7
The researchers in Example 16.6 take a random sample that yields a mean of £47.13.
Calculate a 95% confidence interval using this sample mean.
47.13 Ϯ (1.96 * 1.275) ϭ 47.13 Ϯ 2.499 ϭ £44.631 to £49.629
This interval does not include the population mean of £50.
498 Quantitative methods for business Chapter 16
19 out of every 20 samples will produce an accurate estimate, and
1 out of 20 will not. That is why the interval is called a 95% interval
estimate or a 95% confidence interval.
We can express the procedure for finding an interval estimate for a
population measure as taking a sample result and adding and sub-
tracting an error. This error reflects the uncertainties involved in using
sample information to predict population measures.
Population measure estimate
ϭ
sample result Ϯ error
The error is made up of two parts, the standard error and the number
of standard errors. The number of standard errors depends on how

confident we want to be in our estimation.
Suppose you want to estimate the population mean. If you know the
population standard deviation, ␴, and you want to be (100 Ϫ ␣)% con-
fident that your interval is accurate, then you can obtain your estimate
of ␮ using:
The letter ‘z’ appears because we are dealing with sampling distribu-
tions that are normal, so we can use the Standard Normal Distribution,
the z distribution, to model them. You have to choose which z value to
use on the basis of how sure you want or need to be that your estimate
is accurate.
If you want to be 95% confident in your estimate, that is
(100 Ϫ ␣)% ϭ 95%, then ␣ is 5% and ␣/2 is 2
1
⁄2% or 0.025. To produce
your estimate you would use z
0.025
, 1.96, the z value that cuts off a
2
1
⁄2% tail in the Standard Normal Distribution. This means that a point
1.96 standard errors away from the mean of a normal sampling distri-
bution, the population mean, will cut off a tail area of 2
1
⁄2% of the dis-
tribution. So:
95% interval estimate of ␮ ϭ x
–
Ϯ (1.96 * ␴/√n)
This is the procedure we used in Example 16.6.
The most commonly used level of confidence interval is probably

95%, but what if you wanted to construct an interval based on a higher
level of confidence, say 99%? A 99% level of confidence means we want
99% of the sample means in the sampling distribution to provide accur-
ate interval estimates.
To obtain a 99% confidence interval the only adjustment we make is
the z value that we use. If (100 Ϫ ␣)% ϭ 99%, then ␣ is 1% and ␣/2 is
1
⁄2% or 0.005. To produce your estimate use z
0.005
, 2.576, the z value
xz n (* )Ϯ
␣
␴
2
√
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 499
that cuts off a
1
⁄2% tail in the Standard Normal Distribution:
99% interval estimate of ␮ ϭ x
–
Ϯ (2.576 * ␴/√n)
The most commonly used confidence levels and the z values you
need to construct them are given in Table 16.1.
Notice that the confidence interval in Example 16.8 includes the
population mean, £50, unlike the 95% interval estimate produced in
Example 16.7 using the same sample mean, £47.13. This is because this
sample mean, £47.13, is not amongst the 95% closest to the population
mean, but it is amongst the 99% closest to the population mean.
Changing the level of confidence to 99% has meant the interval is

accurate, but it is also wider. The 95% interval estimate was £44.631 to
£49.629, a width of £4.998. The 99% interval estimate is £43.846 to
£50.414, a width of £6.568.
You can obtain the z values necessary for other levels of confidence by
looking for the appropriate values of ␣/2 in the body of Table 5 on pages
621–622 in Appendix 1 and finding the z values associated with them.
Table 16.1
Selected levels of confidence and associated z values
Level of confidence (100 Ϫ
␣
)%
␣
/2 z
␣
/2
90% 0.050 1.645
95% 0.025 1.960
99% 0.005 2.576
Example 16.8
Use the sample result in Example 16.7, £47.13, to produce a 99% confidence interval
for the population mean.
From Table 16.1 the z value that cuts off a 0.005 tail area is 2.576, so the 99% confi-
dence interval is:
47.13 Ϯ (2.576 * 1.275) ϭ 47.13 Ϯ 3.284 ϭ £43.846 to £50.414
Example 16.9
Use the sample result in Example 16.7, £47.13, to produce a 98% confidence interval
for the population mean.
500 Quantitative methods for business Chapter 16
At this point you may find it useful to try Review Questions 16.10 and
16.11 at the end of the chapter.

16.2.2 Determining the sample size for estimating
a population mean
All other things being equal, if we want to be more confident that our
interval is accurate we have to accept that the interval will be wider, in
other words less precise. If we want to be more confident and retain
the same degree of precision, the only thing we can do is to take a
larger sample.
In the examples we have looked at so far the size of the sample was
already decided. But what if, before starting a sample investigation, you
wanted to ensure that you had a big enough sample to enable you to
produce a precise enough estimate with a certain level of confidence?
To see how, we need to start with the expression we have used for the
error of a confidence interval:
Until now we have assumed that we know these three elements so we
can work out the error. But what if we wanted to set the error and find
the necessary sample size, n? We can change the expression for the
error around so that it provides a definition of the sample size:
This means that as long as you know the degree of precision you
need (the error), the level of confidence (to find z
␣/2
), and the popu-
lation standard deviation (␴), you can find out what sample size you
need to use.
error *
Exchange error and * error
So ( * error)
2
2
2
2

ϭ
ϭ
ϭ
zn
nnz
nz
␣
␣
␣
␴
␴
␴
√
√√:
zn
␣
␴
2
* √
From Table 5 the z value that cuts off a tail area of 0.01 is 2.33, so the 98% confidence
interval is:
47.13 Ϯ (2.33 * 1.275) ϭ 47.13 Ϯ 2.971 ϭ £44.159 to £ 50.101
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 501
At this point you may find it useful to try Review Questions 16.12
and 16.13 at the end of the chapter.
16.2.3 Estimating without
␴
Example 16.6 is artificial because we assumed that we knew the popu-
lation mean, ␮. This helped to explain how and why interval estima-
tion works. In practice we wouldn’t know the population mean, and we

would probably not know the population standard deviation, ␴, either.
Practical interval estimation is based on sample results alone, but
it is very similar to the procedure we explored in Example 16.6. The
main difference is that we have to use a sample standard deviation, s, to
produce an estimate for the standard error of the sampling distribution
the sample belongs to. Apart from this, as long as the sample we have is
quite large, which we can define as containing 30 or more observations,
we can follow exactly the same procedure as before.
That is, instead of
estimate of ␮ ϭ x
–
Ϯ (z
␣/2
*
␴
/√n)
we use
estimate of ␮ ϭ x
–
Ϯ (z
␣/2
* s/√n).
Example 16.10
If the researchers in Example 16.6 want to construct 99% confidence intervals that are
£5 wide, what sample size should they use?
If the estimates are to be £5 wide that means they will have to be produced by adding
and subtracting £2.50 to and from the sample mean. In other words the error will be
2.50. If the level of confidence is to be 99% then the error will be 2.576 standard errors.
We know that the population standard deviation, ␴, is 12.75, so:
n ϭ (2.576 * 12.75/2.50)

2
n ϭ (13.1376)
2
ϭ 172.6 to one decimal place.
Since the size of a sample must be a whole number we should round this up to 173.
When you are working out the necessary sample size you must round the calculated
sample size up to the next whole number to achieve the specified confidence level.
Here we would round up to 173 even if the result of the calculation was 172.01.
We should conclude that if the researchers want 99% interval estimates that are £5
wide they would have to take a sample of 173 shoppers.
502 Quantitative methods for business Chapter 16
In Example 16.11 we are not told whether the population that the
sample comes from is normal or not. This doesn’t matter because the
sample size is over 30. In fact, given that airlines tend to restrict cabin
baggage to 5 kg per passenger the population distribution in this case
would probably be skewed.
16.2.4 Estimating with small samples
If we want to produce an interval estimate based on a smaller sample,
one with less than 30 observations in it, we have to be much more care-
ful. First, for the procedures we will consider in this section to be valid,
the population that the sample comes from must be normal. Second,
because the sample standard deviation of a small sample is not a reli-
able enough estimate of the population standard deviation to enable
us to use the z distribution, we must use the t distribution to find how
many standard errors are to be added and subtracted to produce an
interval estimate with a given level of confidence.
Instead of
estimate of ␮ ϭ x Ϯ (z
␣/2
*

␴
/√n)
we use
estimate of ␮ ϭ x Ϯ (t
␣
/2,␯
* s/√n).
The form of the t distribution you use depends on ␯, the number of
degrees of freedom, which is the sample size less one (n Ϫ 1). You can
find the values you need to produce interval estimates in Table 6 on
page 623 of Appendix 1.
Example 16.11
The mean weight of the cabin baggage checked in by a random sample of 40 passengers
at an international airport departure terminal was 3.47 kg. The sample standard devi-
ation was 0.82 kg. Construct a 90% confidence interval for the mean weight of cabin bag-
gage checked in by passengers at the terminal.
In this case ␣ is 10%, so ␣/2 is 5% or 0.05 and according to Table 16.1 z
0.05
is 1.645.
90% interval estimate of (1.645 * )
3.47 (1.645 * 0.82 40)
3.47 0.213
3.257 to 3.683 kg
␮ ϭϮ
ϭϮ
ϭϮ
ϭ
xsn√
√
You may recall from section 16.1.2 that the t distribution is a modi-

fied form of the z distribution. If you compare the figures in the bot-
tom row of the 0.05, 0.025 and 0.005 columns of Table 6 with the z
values in Table 16.1, that is 1.645, 1.960 and 2.576, you can see that
they are same. If, however, you compare these z values with the equiva-
lent t values in the top row of Table 6, the ones for the t distribution
with just one degree of freedom, which we would have to use for sam-
ples of only 2, you can see that the differences are substantial.
At this point you may find it useful to try Review Questions 16.14
and 16.15 at the end of the chapter.
16.2.5 Estimating population proportions
Although so far we have concentrated on how to estimate population
means, these are not the only population parameters that can be esti-
mated. You will also come across estimates of population proportions,
indeed almost certainly you already have.
If you have seen an opinion poll of voting intentions, you have seen
an estimate of a population proportion. To produce the opinion poll
result that you read in a newspaper pollsters have interviewed a sample
of people and used the sample results to predict the voting intentions
of the entire population.
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 503
Example 16.12
A random sample of 15 employees of a call centre was taken and each employee took a
competency test. The mean of the scores achieved by these employees was 56.3% with
a standard deviation of 7.1%. Results of this test have been found to be normally dis-
tributed in the past. Construct a 95% confidence interval for the mean test score of all
the employees of the call centre.
Here ␣ is 5% so ␣/2 is 2.5% or 0.025 and the number of degrees of freedom, ␯, is
n Ϫ 1, 14.
95% estimate of ␮ ϭ
–

x Ϯ (t
0.025,14
* s/√n)
From Table 6, t
0.025,14
is 2.145, so:
95% estimate of ␮ ϭ 56.3 Ϯ (2.145 * 7.1/√15)
ϭ 56.3 Ϯ 3.932 ϭ 52.378% to 60.232%
In many ways estimating a population proportion is very similar to
the estimation we have already considered. To start with you need a
sample: you calculate a sample result around which your estimate will
be constructed; you add and subtract an error based on the standard
error of the relevant sampling distribution, and how confident you
want to be that your estimate is accurate.
We have to adjust our approach because of the different nature of
the data. When we estimate proportions we are usually dealing with
qualitative variables. The values of these variables are characteristics,
for instance people voting for party A or party B. If there are only two
possible characteristics, or we decide to use only two categories in our
analysis, the variable will have a binomial distribution.
As you will see, this is convenient as it means we only have to deal
with one sample result, the sample proportion, but it also means that
we cannot produce reliable estimates from small samples, those con-
sisting of less than 30 observations. This is because the distribution of
the population that the sample comes from must be normal if we are
to use the t distribution, the device we have previously used to over-
come the extra uncertainty involved in small sample estimation.
The sampling distribution of sample proportions is approximately
normal in shape if the samples involved are large, that is, more than 30,
as long as the sample proportion is not too small or too large. In prac-

tice, because we do not know the sample proportion before taking the
sample it is best to use a sample size of over 100. If the samples are small,
the sampling distribution of sample proportions is not normal.
Provided that you have a large sample, you can construct an interval
estimate for the population proportion, ␲ (pi, the Greek letter p), by
taking the sample proportion, p, and adding and subtracting an error.
The sample proportion is the number of items in the sample that pos-
sess the characteristic of interest, x, divided by the total number of
items in the sample, n.
Sample proportion, p ϭ x/n
The error that you add to and subtract from the sample proportion
is the z value appropriate for the level of confidence you want to use
multiplied by the estimated standard error of the sampling distribu-
tion of the sample proportion. This estimated standard error is based
on the sample proportion:
estimated standard error ϭ
pp
n
()1 Ϫ
504 Quantitative methods for business Chapter 16
Chapter 16 Test driving – sampling theory, estimation and hypothesis testing 505
So the
interval estimate of ␲ ϭ
At this point you may find it useful to try Review Questions 16.16
and 16.17 at the end of the chapter.
16.2.6 Determining the sample size for estimating
a population proportion
If you know how confident you want to be that your interval estimate is
accurate and you need your estimate to have a certain precision, in
other words the error has to be a particular amount, you can work out

the sample size you will need to use.
The precision of the test depends on the estimated standard error of
the sample proportions,
√
p(1 – p)/n
. The value of this depends on p,
the sample proportion. Clearly you won’t know this until the sample
data have been collected, but you can’t collect the sample data until you
have decided what sample size to use. You therefore need to make a
prior assumption about the value of the sample proportion.
To be on the safe side we will assume the worst-case scenario, which
is that the value of p will be the one that produces the highest value of
p
pp
n
z *
(1 )
2
Ϯ
Ϫ
␣
Example 16.13
A study of a sample of 110 supermarkets reveals that 31 offer trolleys suitable for shop-
pers with limited mobility. Construct a 95% interval estimate of the proportion of all
supermarkets that have these trolleys.
These results suggest that we can be 95% confident that the proportion of supermar-
kets with suitable trolleys for shoppers with limited mobility will be between 19.6% and
36.4%.
p
z

31 110 0.28
95%, so 1.96
interval estimate of 0.28 1.96 *
0.28 (1 0.28)
0.28 1.96 * 0.043 0.28 0.084
0.196 to 0.364
ϭϭ
Ϫϭ ϭ
ϭϮ
Ϫ
ϭϮ ϭϮ
ϭ
()%100
110
2
␣
␲
␣
506 Quantitative methods for business Chapter 16
p (1Ϫp). The higher the value of p (1Ϫp) the wider the interval will be,
for a given sample size. We need to avoid the situation where p (1Ϫp)
turns out to be larger than we have assumed it is.
What is the largest value of p (1Ϫp)? If you work out p (1Ϫp) when p
is 0.1, you will get 0.09. If p is 0.2, p (1Ϫp) rises to 0.16. As you increase
the value of p you will find that it keeps going up until p is 0.5, when
p (1Ϫp) is 0.25, then it goes down again.
The error in an interval estimate of a population proportion is:
If p is 0.5, in other words we assume the largest value of p (1 Ϫ p):
This last expression can be re-arranged to obtain an expression for n:
error *

0.5
so
* 0.5
error
and

* error
2
2
2
ϭ
ϭ
ϭ
z
n
n
z
n
z
␣
␣
␣
√
√
2
2
⎡
⎣
⎢
⎤

⎦
⎥
error *
0.5(1 0.5)
*
0.5 * 0.5
*
0.5
2
2
2
ϭ
Ϫ
ϭ
ϭ
␣
␣
␣
z
n
z
n
z
n
error *
(1 )
2
ϭ
Ϫ
z

pp
n
␣
Example 16.14
How many supermarkets would have to be included in the sample in Example 16.13 if
the confidence interval of the proportion of establishments with trolleys suitable for
shoppers with limited mobility has to be within 5% of the actual population proportion
with a 95% degree of confidence.
For the error to be 5%:
This has to be rounded up to 385 to meet the confidence requirement so a random
sample of 385 supermarkets would have to be used.
n
1.96
2 * 0.05
19.6 384.16
2
ϭϭϭ
⎡
⎣
⎢
⎤
⎦
⎥
2