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Course Manual: PG Certificate in Statistics, TCD, Base Module 1

Chapter 1: Statistical Variation and the Normal Curve

1.0 Introduction

Perhaps the most striking feature of observation or real data values from almost any

discipline is the variation they exhibit. If a nurse measures physical dimensions

(heights, weights..) or blood parameters (Serum albumin or cholesterol..) for a sample of

patients, they will (nearly) all be different. If a educational psychologist administers a

psychological test to a sample of children or if impairment is measured using a response

time test, the tests scores in each case will vary greatly. Indeed a manufacturing

engineer who measures the physical dimensions (e.g., outer diameter or length) of

components manufactured to the same specifications will find they exhibit precisely the

same type of variation as the psychological test scores. Similarly, a cosmetic scientist

who measures the depth of wrinkles on women’s faces, or environmental scientists

measuring acidity levels (pH values) in samples from different lakes or streams, or from

one site sampled at different times, or indeed if they measure birds’ eggs, they will find

very much the same kind of variation as did the engineer, psychologist or nurse referred

to above.

This variation is the reason why statistical methods are used across a wide range of

disciplines. Statistics is the science of describing and analysing chance variation, and

statistical methods help us describe the nature of chance or random variation and ‘tame’

it.

This chapter is concerned with describing such variation using:

Pictures

Normal or Gaussian curve

Additive variation and averages – the implications of adding chance quantities

(central to this course)

Measurement systems and control charts.

1.1 Statistical Variation

We begin with a series of examples to illustrate statistical variation in a number of

disciplines. The histogram in Figure 1.1.1 is based on a dataset (Grades) that comes

with the Minitab software [1] that will be used extensively in this course manual.

Investigators in the USA collected data to investigate the relationship between SAT

scores (Scholastic Aptitude Test, often used as a college admission or course placement


criterion) and GPA grades (grade point average – university performance score). Figure

1.1.1 gives the Verbal SAT scores of 200 students. This picture is typical of what we

might expect from psychological/educational test results – the students cluster around

the central value (about 600, here) and the number of students who get scores far from

the centre is quite small – the numbers decrease in a more or less symmetrical manner

in each direction.

800700600500400300

Figure 1.1.1: SAT Verbal scores for 200 university students

The histogram in Figure 1.1.2 shows the distribution of capping torques (in lbs-inches)

of 78 container caps; the capping torque is a measure of the energy expended in

breaking the seal of the screw-capped containers. The data come from a process

capability study of a newly installed filling line in a company that manufactures

chemical reagents. It is interesting to note that the physical measurements display

similar characteristics to the psychological scores – they vary in a roughly symmetrical

way around a central value (about 10.8 for the torques).

13.512.511.510.59.58.57.5

Figure 1.1.2: Capping torques (lb-in) for 78 screw-capped containers


Figure 1.1.3 shows a histogram of the distribution of petal lengths of a sample 50 irises

from the variety Setosa. The data were originally published in the Bulletin of the

American Iris Society [2], but their frequent citation in the statistical literature is due

to their use in the pioneering paper on multivariate analysis by R.A. Fisher [3]. This

botanical/environmental science example exhibits the same symmetry characteristics

noted above for the psychological and industrial data.

1.81.61.41.21.0

Figure 1.1.3: Petal lengths (mm) for 50 irises (Setosa)

In Figure 1.1.4 the histogram displays the distribution of the heights (cm) of pregnant

women (website of Prof. Martin Bland a well-known medical statistician, [4]). Because

the sample size is so large (1794 values) the histogram is much smoother and the

symmetry is much better defined, than for the smaller sample sizes of the first three

examples. In Section 1.3 we will see that all four datasets are consistent with

underlying Normal or Gaussian distributions - smooth symmetrical bell-shaped curves

(see Figure 1.1.6) that will be discussed in some detail in Section 1.2. Consequently,

any departures from symmetry in our histograms are considered due to the variability

necessarily associated with relatively small sample sizes.

Importantly observe that both the women’s heights and petal lengths data have many

repeated values, this occurs due to rounding of a lack of precision in the measurement

systems. For data values arising from the mathematical (i.e. idealised) Normal

Distribution this typically does not occur.


184176168160152144

Figure 1.1.4: Heights (cm) of 1794 pregnant women

While this symmetrical shape is very common, it is by no means universal, as

illustrated by our next example. Figure 1.1.5. -

-OHCS) values in nmol/24h “obtained on a random sample of

100 obese adult females drawn from a fully specified target population”; the data come

from the book on laboratory medicine by Strike [5]. The steroid data show a long tail on

the right, in other words they are strongly skewed (to the right); the characteristic

described by the graph is very different from those of our first four examples. Most of

the steroid measurements are less than 900 but individual values range up to around

1400. While it is now possible to deal directly with skewed data, traditionally, analysts

sought transformations which would allow the data to be analysed using methods based

on the Normal distribution; this is still a common practice.

140012001000800600400200

Figure 1.1.5: Urinary β-hydroxycorticosteroid (nmol/24h) values for 100 obese

females


Figure 1.1.6 shows a histogram (with a superimposed Normal curve) of the transformed

data after the original measurements were subjected to transformation by taking

logarithms (to base 10). As we shall see in Section 1.3 the transformation has been

successful – the transformed data are well modelled by a Normal curve. Where

transformed data follow a Normal curve, the original skewed distribution is said to be

‘log-Normal’.

3.23.02.82.62.4

Figure 1.1.6: Transformed (logarithms) steroid data

There are of course very many different types of shape that may be encountered for

histograms of real data (more will be presented during the lectures). However, Figures

1.1.1-1.1.6 were selected for presentation, both to illustrate the wide applicability of the

Normal curve (in terms of discipline type) and the possibility of transforming data so

that they follow this curve, which is, perhaps, the easiest of all statistical distributions

to use – hence its popularity. It is also of fundamental importance because averages of

data sampled from skewed distributions will, in many cases, follow a Normal curve (see

Section 1.5) – this means that methods developed for Normal data can be applied to

such averages, even though the underlying data are skewed.

Aside: Many important distributions are skewed to the right including the income distribution

and the distribution of city sizes (Zift’s law) – these are example a very important applied

distribution known as the Pareto Distribution; search Wikipedia for details of these distributions

and for example the famous 80-20 Rule derived from the Pareto Distribution.

Note: The fact that a distribution is skewed does not mean that statistical analysis fails, in

particular analysis of the means of many differing samples taken from any one of these

distributions yields to the type of straightforward statistical analysis covered in this course.


1.2 The Normal (Gaussian) curve

If the more or less symmetrical histograms of Sections 1.1 were based on several

thousand observations, the bars of the histograms could be made very narrow and the

overall shapes might then be expected to approximate the smooth curve shown below as

Figure 1.2.1.

Figure 1.2.1: The Normal curve

This is a mathematical curve described using two parameters, the mean μ and standard

deviation σ. It is symmetric about its centre, the mean, and its width is measured by the

standard deviation. The standard deviation is formally defined but can usefully be

thought of as one sixth of the width of the curve, for most practical purposes. In

principle, the curve stretches from minus infinity to plus infinity but in practice (as can

be seen from Figure 1.2.1) the area in the tails becomes infinitesimally small very

quickly.

Interpreting the Normal Curve

Figure 1.2.2 shows three Normal curves with the mean μ and standard deviation σ. The

curves are drawn so that the total area enclosed by each curve is one; when this is done

we call the Normal Curve the Normal (Gaussian) Distribution. Then, approximately

68% of the area lies within 1 standard deviation of the mean; the corresponding values

for two and three standard deviations are 95% and 99.7%, respectively. These areas

will be the same for all Normal curves irrespective of the values of the means and

standard deviations.


Figure 1.2.2: Areas within one, two and three standard deviations of the

mean

The implications of this figure are that if the variation between measurements on

individuals/objects (e.g. blood pressure, SAT scores, torques, etc.) can be described by a

Normal curve, then 68% of all values will lie within one standard deviation of the mean,

95% within two, and 99.7% within three standard deviations of the mean. Similarly, if

we have a measurement error distribution that follows a Normal curve, then in 32% of

cases the chance error will result in a measured value being more than one standard

deviation above or below the mean (the ‘true value’ if the measurement system is

unbiased). In only 5% (or 1-in-20) of cases will the observation be further than two

standard deviations from the mean.

Figure 1.2.3: Distribution of the masses of filled containers

Suppose that the Normal curve shown in Figure 1.2.3 represents the masses of filled

containers sampled from a filling line. The curve is interpreted as an idealised

histogram for which the total area (or probability) is 1.0. The area between any two

points (say x1 and x2) represents the relative frequency (probability) of container

masses between x1 and x2. If this area is 0.2 it means that 20% of all containers filled

by this line have masses between x1 and x2 units. The assumption is that, provided the


filling process remains stable, 20% of all future fillings will result in containers with

masses between x1 and x2 units. Since relative frequencies are generally of interest,

e.g. what fraction of values is less than or greater than a given value, or what fraction

lies between two given values, then, where a Normal curve describes the variability in

question, it can be used to calculate the required relative frequency. Note the area to the

left of x1 is the probability that any observation x selected from this distribution at

random will be less than x1, this is formally written as

𝑃(𝑥 < 𝑥1)

The use of probability as a way of expressing uncertainty has become common in

everyday speech. Our concept of probability is intimately bound up with the idea of

relative frequency and people move easily between the two ways of describing

variability. Thus, it seems quite natural to say that if a filled container is selected at

random, the probability is 0.2 that its mass will lie between x1 and x2. We typically

write this as

𝑃(𝑥1 < 𝑥 < 𝑥2) = 0.2

Both of these ways of describing variability, viz., relative frequency and probability, are

equivalent, and they will be used interchangeably hereafter.

Obtaining Areas under Normal Curves

Figure 1.2.4: A standard Normal curve (z) and an arbitrary Normal curve

(x)


All Normal curves have essentially the same shape once allowance is made for different

standard deviations. This allows for easy calculation of areas, i.e., relative frequencies

or probabilities. If, for any arbitrary curve with mean µ and standard deviation σ, the

area between two values, say x1 and x2, is required, all that is necessary is to find the

corresponding values z1 and z2 on the standard Normal curve, determine the area

between z1 and z2 and this will also be the area between x1 and x2 (as shown

(schematically) in Figure 1.2.4). A standard Normal curve has mean zero and standard

deviation one.

Standard Normal μ=0, σ=1

The z value is the answer to the question “by how many standard deviations does x

differ from the mean μ?”

Once the z value is calculated, the area to its left (i.e. the area from minus infinity up to

this value of z) can be read from a standard Normal table (Table ST-1), this is formally

written as

𝑃(𝑧 < 𝑧1).

Thus, a z value of 2 has an area to its left of 0.9772, or

𝑃(𝑧 < 2) = 0.9772.

while a z value of 1.5 has an area to its left of 0.9332, or

𝑃(𝑧 < 1.5) = 0.9332.

Therefore, the area or probability between z=1.5 and z=2 on the standard Normal curve

is 0.9772 - 0.9332 or 0.0440, or

𝑃(𝑧1 < 𝑧 < 𝑧2) = 0.9772 − 0.9332 = 0.0440.

Similarly, the area between two points which are 1.5 and 2 standard deviations from

the mean of any Normal curve will also be 0.0440.

The relationship or transformation formula between the x and z values is simple:

xz .

And the inverse relationship is

𝑥 = 𝜇 + 𝜎𝑧


Example 1

Figure 1.1.4 showed a distribution of heights for a population of women. The mean of

the 1794 values was 162.4 and the standard deviation was 6.284. If we take the mean

and standard deviation as those of the corresponding population, we can address

questions such as those below.

(i) What fraction of the population has a height greater than 170 cm?

(ii) What fraction of the population has a height less than 145 cm?

(iii) How tall would I be if I was taller than 75% of the population??

(iv) What bounds should we use if we want to enclose the central 95% of the

population within them?

(i) What fraction of the population has a height greater than 170 cm?

To calculate the required proportion of the population, the area to the right of 170 on

the height curve must be found. So we require

𝑃(𝑥 > 170)

Figure 1.2.5: The standard Normal curve (z) and the women’s heights curve (x)

Looking at Figure 1.2.5 we see on the right the Normal Curve for the observed values

centred at μ=164.2. To get the equivalent value on the Standard Normal scale (Normal

curve on the left) we apply the formula to go from (x) values to (z) values. So the

transformation from the height (x) to the standard Normal scale (z) is given by:

21.1284.6

5.162170

xz


Thus, a height of 170 corresponds to the point 1.21 on the standard Normal scale. The

Standard Normal tables give the area to the left of z = 1.21 as

𝑃(𝑧 < 1.21) = 0.8869.

But, the sign is > so we want the area on right, i.e.

𝑃(𝑥 > 170) = 𝑃(𝑧 > 1.21)

Importantly, since the total area is 1, the area to the right is

Area to the right = 1 − 𝐴𝑟𝑒𝑎 𝑡𝑜 𝐿𝑒𝑓𝑡

We formally write this as

𝑃(𝑧 > 1.21) = 1 − 𝑃(𝑧 < 1.21) = 1 − 0.8869 = 0.1131.

Accordingly, approximately 11% of the women will have a height greater than 170 cm.

The calculations are usually summarised in one expression as follows:

𝑃(𝑥 > 170 = 𝑃 (𝑧 >170 − 164.2

6.284) = 𝑃(𝑧 > 1.21) = 1 − 𝑃(𝑧 < 1.21) = 1 − 0.8869 = 0.1131.

We can read P(x >170) as “the proportion of x values greater than 170” or, equivalently,

as “the probability that a single x value, selected at random, will be greater than 170”.

(ii) What fraction of the population has a height less than 145 cm?

The value of z that corresponds to 145 is given by:

z = 284.6

4.162145

x= –2.77.

Statistical Table-1 (ST-1) only gives areas for positive values of z. However, we can use

the symmetry of the curve to find areas corresponding to negative values. Thus,

P(z<2.77) is the area below 2.77; 1–P(z<2.77) is the area above +2.77. By symmetry this

is also the area below –2.77, as shown in Figure 1.2.6


-2.77 2.770

Figure 1.2.6: The symmetry property of the Normal curve (area is 0.0028 in each tail)

Accordingly, we have:

P(X < 145) =P(z < –2.77) = P(z >2.77) = 1 – P(z < 2.77) = 1 – 0.9972 = 0.0028.

Approximately 0.3% will have heights less than 145 cm.

(iii) How tall would I be if I was taller than 75% of the population?

In this case 75% of the population are smaller than I am. So, the area to the left on the

Standard Normal is 0.75, or

𝑃(𝑧 < 𝑧1) = 0.75.

0

0.2

0.4

0.6

0.8

1

-2.5

-2.2

-1.9

-1.6

-1.3

-1.0

-0.7

-0.4

-0.1

0.2

0.5

0.8

1.1

1.4

1.7

2.0

2.3

0.75


So, what is the z-value corresponding to this probability on the Standard Normal. To get

this read the tables is reverse; in other words find the cell value in the body of the table

that is closest to 0.75 and read the associated cell value from the left most column and

top row (add these values), we get

𝑃(𝑧 < 𝑧1) = 0.75 => 𝑧1 = 0.675 .

To convert this back to height (x) scale apply the inverse transformation

𝑥 = 𝜇 + 𝜎𝑧 = 162.4 + 6.284 × 0.675 = 166.6

Thus, if I was 167 cm’s I would be taller than 75% of the women in the population.

(iv) What bounds (A, B) should we use if we want to enclose the central 95% of the

population within them?

Figure 1.2.7: Finding the central 95% limits

To find the values that enclose the central 95% of the standard Normal curve (i.e., an

area of 0.95) we proceed as follows. Enclosing the central 0.95 means we leave 0.025 in

each tail. The upper bound, therefore, has an area of 0.975 below it (as shown in the

leftmost curve in Figure 1.2.7). If we inspect the body of Table ST-1, we find that 0.975

corresponds to a z value of 1.96. By symmetry, the lower bound is –1.96 (the central

curve in Figure 1.2.7). Thus, a standard Normal curve has 95 % of its area between z=–

1.96 and z=+1.96. From the relation:

xz


Using the reverse transformation we get:

B96.1 => B = μ + 1.96 σ

and:

–

A96.1 => A = μ – 1.96 σ

It is clear, then, that the values that enclose the central 95% interval of the population

for an arbitrary Normal curve (x) with mean μ and standard deviation σ are: μ - 1.96 σ

and μ + 1.96 σ; this is generally called the 2 standard deviation interval about the mean

and it is ubiquitous throughout statistical practice in many disciplines -.

In the present case this gives us: 162.4±1.96(6.284), i.e., 162.4±12.3. While this gives

(150.1, 174.7), for practical purposes the bounds would probably be described as 150 to

175 cm.

Example 2

Suppose the distribution of capping torques for a certain product container is Normal

with mean μ=10.6 lb-in and standard deviation σ=1.4 lb-in (these were the values

obtained from the data on which Figure 1.1.2 is based). To protect against leaks, a

minimum value for the torque must be achieved in the manufacturing process. If the

lower specification limit on torque is set at 7 lb-in, what percentage of capping torques

will be outside the lower specification limit?

To calculate the required percentage, the area to the left of 7 on the capping process

curve must be found. To do this, the equivalent value on the standard Normal curve is

calculated; the area to the left of this can be found easily and this gives the required

proportion (as illustrated schematically in Figure 1.2.8).

Figure 1.2.8: The standard Normal curve and the capping process curve


The translation from the torque scale (x) to the standard Normal scale (z) is given by:

xz

z = 4.1

6.107 = – 2.57.

Thus, a torque of 7 on the capping process curve corresponds to the point – 2.57 on the

standard Normal curve. The area to the left of 2.57 under the standard Normal curve is

0.995 (or 0.99491 if your prefer!), so the area to the right is 0.005, since the total area is

1. By symmetry, the area to the left of –2.57 is also 0.005 and, therefore, five caps per

thousand will be expected to be below the capping torque lower specification limit in

future production, provided the process remains stable. A simple statistical tool (a

‘control chart’) may be used to monitor the stability of manufacturing processes. Section

1.7 provides an introduction to and gives several examples of such charts.

Exercises

1.2.1. Weight uniformity regulations usually refer to weights of tablets lying within a certain

range of the label claim. Suppose the weights (masses) of individual tablets are

Normally distributed with mean 100 mg (label claim) and standard deviation 7 mg.

(a) If a tablet is selected at random, what is the probability that its weight is:

(i) less than 85 mg ?

(ii) more than 115 mg ?

(iii) either less than 85 mg or more than 115 mg ?

(b) (i) What percentage of tablets lie between 90 mg and 110 mg ?

(ii) What percentage are heavier than 105 mg ?

(iii) What percentage are lighter than 98 mg ?

(c) (i) What weight value K could be quoted such that 99% of all tablets are heavier than

K ?

(ii) What weight values A and B can be quoted such that 99% of all tablets lie

between A and B ?

1.2.2 Figures 1.1.5 and 1.1.6 are based on data cited by Strike [5]. The data are urinary β-

hydroxycorticosteroid (β-OHCS) values in nmol/24h, obtained on a random sample of

100 obese adult females drawn from a fully specified target population. According to

1 Try re-calculating using standard deviations of 1.38 or 1.41 instead of 1.40 and note the effect on

the resulting probabilities. How often (never?) do engineers know standard deviations to this level of precision?


Strike it is of interest to define a 95% reference range for the β-OHCS measurements (a

reference interval encloses the central 95% of the population of measured values) that

can be used to assess the clinical status of future subjects drawn from the same

population. The distribution of β-OHCS values is quite skewed; accordingly, in order to

construct the reference interval we first transform the data to see if they can be

modelled by a Normal distribution in a transformed scale. Figure 1.1.6 suggests that

this is a reasonable assumption (this will be discussed further in Section 1.3). In the

log10 scale the mean is 2.7423 and the standard deviation is 0.1604.

(i) Calculate the bounds (a, b) for the 95% reference interval in the log scale.

(ii) Use your results to obtain the corresponding reference interval in the original

measurement scale: the bounds will be (10a, 10b).

(iii) What value, U, should be quoted such that only 1% of women from this

population would be expected to have a β-OHCS measurement which exceeds U?

Calculating the standard deviation

Our calculations using the Normal curve assumed the mean and standard deviation

were already known. In practice, of course, they have to be calculated from observations

on the system under study. Suppose that a sample of n objects has been selected and

measured (e.g., SAT scores, people’s heights etc.) and the results are x1, x2, …xn.. The

average of these, x , gives an estimate of the corresponding population value, μ. The

standard deviation, s, of the set of results gives an estimate of the corresponding

population standard deviation, σ. This is calculated as:

1

)(1

2

n

xx

s

n

i

i

i.e. the mean is subtracted from each value, the deviation is squared and these squared

deviations are summed. To get the average squared deviation, the sum is divided by n–

1, which is called the 'degrees of freedom'2. The square root of the average squared

deviation is the standard deviation. Note that it has the same units as the original

measurements.

2 (n–1) is used instead of the sample size n for a mathematical reason; the deviations (before they are

squared) sum to zero, so once the sum of (n–1) is found, the last deviation is determined by the constraint, thus, only (n–1) deviations are free to vary at random – hence, ‘degrees of freedom’. Obviously, unless n is quite small, effectively the same answer is obtained if we get the average by dividing by n.


It is important always to distinguish between system parameters, such as the

population standard deviation, σ, and estimates of these parameters, calculated from

the sample data. These latter quantities are called ‘sample statistics’ (any number

calculated from the data is a ‘statistic’) and are themselves subject to chance variation.

System parameters, such as μ and σ, are considered fixed, though unknown, quantities.

In practice, if the sample size n is very large, the result calculated using the definition

given above will often be labelled σ, since a very large set of measurements will give the

'true' standard deviation. If the sample size is small, the result is labelled s (or

sometimes ̂ ); this makes clear to the user that the calculated value is itself subject to

measurement/sampling error, i.e. if the study were repeated a different value would be

found for the sample standard deviation.

Aside: The Coefficient of Variation (CV) is defined as the standard deviation divided by the

mean. It is often useful to compare standard deviations computed from different distributions.

The assumption that the parameters of the Normal distribution are known without

error is reasonable for the height data of Example 1 (where n=1794), but clearly the

sample size for the capping torque data (n=78) is such that we cannot expect that the

‘true’ process mean or standard deviation is exactly equal to the sample value. Methods

have been developed (tolerance intervals) to allow for the uncertainty involved in having

only sample data when carrying out calculations like those of Examples 1 and 2, though

these are rarely discussed in elementary textbooks. Hahn and Meeker [6] discuss such

intervals in an engineering context.

1.3 Assessing Normality

The statistical methods that will be discussed in later chapters assume that the data

come from a Normal distribution. This assumption can be investigated using a simple

graphical technique, which is based on the type of calculation discussed in the previous

section. Here, for simplicity, the logic of the method will be illustrated using a sample of

10 values, but such a sample size would be much too small in practice, as it would

contain little information on the underlying shape, particularly on the shapes of the

tails of the curve.

The relation between any arbitrary Normal value x and the corresponding standard

Normal value z is given by:

xz

which using the inverse transformation can be re-written as:


x = μ + σ × z.

If x is plotted on the vertical axis and z on the horizontal axis, this is the equation of a

straight line with intercept μ and slope σ, as shown in Figure 1.3.1. Accordingly, if we

make a correspondence between the x values in which we are interested and z values

from the standard Normal table and obtain a straight line, at least approximately, when

they are plotted against each other, we have evidence to support the Normality

assumption.

Figure 1.3.1: The theoretical relationship between x and z when x is Normal

x

Ranks

i

Fraction x

i/n z 4/1

8/3

n

i

N-

scores

12.6 9 0.9 1.282 0.841 1.000

13.1 10 1.0 0.939 1.547

9.8 2 0.2 -0.842 0.159 -1.000

11.7 7 0.7 0.524 0.646 0.375

10.6 4 0.4 -0.253 0.354 -0.375

11.2 6 0.6 0.253 0.549 0.123

12.2 8 0.8 0.842 0.744 0.655

11.1 5 0.5 0.000 0.451 -0.123

8.7 1 0.1 -1.282 0.061 -1.547

10.5 3 0.3 -0.524 0.256 -0.655

Table 1.3.1 Sample data for calculating Normal Scores

X

Z0


The correspondence is established by finding the fraction of the x values equal to or less

than each value in the dataset, and then finding the z value that has the same fraction

of the standard Normal curve below it. In Table 1.3.1, the ten x values are listed in

column 1, their ranks, i, (smallest 1 to largest n=10) in column 2, the fraction equal to or

less than x in column 3 (i/n). Column 4 gives the z value with the area given by column

3 below it; for example, in the first row, 90% of the sample values are equal to or less

than x=12.6 and column 4 indicates that 90% of the area under a standard Normal

curve is below 1.282 (Table ST-1 gives z values to two decimal places, but software such

as Minitab or Excel will give more, if required).

This method for establishing the correspondence between x and z will always result in

the largest x value corresponding to z= (which presents a challenge when plotting!).

To avoid the infinity, some small perturbation of the calculation of the fraction i/n is

required, such that the result is never i/n=1. Simple examples of possible perturbations

would be (i–0.5)/n or i/(n+0.5). Minitab has several options; (i–3/8)/(n+1/4), which is

based on theoretical considerations, is used in Table 1.3.1. The areas which result from

this are shown in column 5 and the corresponding z values (Normal scores) in column 6.

The choice of perturbation is not important: all that is required is to draw a graph to

determine if the values form an approximately straight line. We do not expect a

perfectly straight line even for truly Normal data, due to the effect of sampling

variability. Thus, 12.6 is the ninth largest value in our dataset, but any value between

13.1 and 12.2 would also have been the ninth largest value. Consequently, 12.6 is not

uniquely related to 0.841.

Figure 1.3.2 shows the Normal plot (or Normal probability plot, as it is also called) for

our dataset – a ‘least squares’ line is superimposed to guide the eye (we will discuss

such lines later). The data are close to the line (hardly surprising for an illustrative

example!).

15

14

13

12

11

10

9

8

210-1-2

X

Score

Mean 11.15

StDev 1.328

N 10

AD 0.131

P-Value 0.971

Figure 1.3.2: Normal plot for the sample data


Departures from Normality

The schematic diagrams in Figure 1.3.3 indicate some patterns which may be expected

when the underlying distribution does not conform to the Normal curve. Plots similar

to Figure 1.3.3(a) will be encountered commonly, as outliers are often seen in real

datasets. Figure 1.3.3(c) might, for example, arise where the data represent blood

measurements on patients - many blood parameters have skewed distributions. Figures

1.3.3(b) and (d) are less common but (b) may be seen where the data are generated by

two similar but different sources, e.g. if historical product yields were being studied and

raw materials from two different suppliers had been used in production. The term

‘heavy tails’ refers to distributions which have more area in the tails than does the

Normal distribution (the t-distribution is an example). Heavy tails mean that data

values are observed further from the mean, in each direction, more frequently than

would be expected from a Normal distribution.

Figure 1.3.3: Idealised patterns of departure from a straight line

Real Data

In Section 1.1 we examined histograms for a number of datasets; here, we draw Normal

plots for several of those examples. Figure 1.3.4 is the Normal plot for the SAT scores of

Figure 1.1.1, while Figure 1.3.5 is the Normal plot for the capping torques of Figure

1.1.2. In both cases the lines formed by the plotted points are essentially straight. The

plots also contain statistical significance tests (the Anderson-Darling test, hence AD)

which test for Normality. We will discuss statistical tests in the next chapter. Here, we

use a rule of thumb which says that a p-value (p-values vary between 0 and 1) greater

than 0.05 supports Normality, while one less than 0.05 rejects it. For both Normal plots

the p-values are large and there is no reason to question the assumption of Normality.

Single outlier Mixture Skewness Heavy tails

(a) (b) (c) (d)


800

700

600

500

400

300

3210-1-2-3

Ve

rba

l

Score

Mean 595.6

StDev 73.21

N 200

AD 0.259

P-Value 0.712

Figure 1.3.4: Normal plot for SAT Verbal scores for 200 university students

15

14

13

12

11

10

9

8

7

6

3210-1-2-3

To

rqu

e

Score

Mean 10.84

StDev 1.351

N 78

AD 0.206

P-Value 0.865

Figure 1.3.5: Normal plot for Capping torques for 78 screw-capped containers

Figure 1.3.6 refers to the Urinary β-OHCS measurements, which we first encountered

in Figure 1.1.5. The graph is far from straight and the AD p-value is very much less

than 0.05; the graph displays the characteristic curvature of a right-skewed

distribution. Figure 1.3.7 is the Normal plot of the log-transformed data and it is quite

consistent with an underlying Normal distribution.


1600

1400

1200

1000

800

600

400

200

0

3210-1-2-3

b-O

HC

S

Score

Mean 590.8

StDev 224.0

N 100

AD 1.729

P-Value <0.005

Figure 1.3.6: Normal plot for Urinary β-OHCS measurements

3.2

3.0

2.8

2.6

2.4

2.2

3210-1-2-3

Lo

g1

0(b

-OH

CS

)

Score

Mean 2.742

StDev 0.1604

N 100

AD 0.250

P-Value 0.738

Figure 1.3.7: Normal plot for log-transformed -OHCS measurements

Figure 1.3.8 is a Normal plot of the women’s heights measurements of Figure 1.1.4.

Although it is very close to a straight line, it has a curiously clumped, almost stair-like,

appearance. The p-value is very small indeed. The reason for this is that the

measurements are rounded to integer values (centimetres) and, because the dataset is

so large, there are many repeated values. A Normal distribution is continuous (the

sampled values can have many decimal places) and few, if any, repeated values would

be expected to occur; the AD test detects this unusual occurrence. To illustrate the

effect of rounding on the plots, 1794 random numbers were generated from a Normal

distribution with the same mean and standard deviation as the real data (the mean and

SD of the sample generated are slightly different from those specified, due to sampling

variability).


190

180

170

160

150

140

3210-1-2-3

He

igh

ts (

cm

)

Score

Mean 162.4

StDev 6.284

N 1794

AD 2.985

P-Value <0.005

Figure 1.3.8: Normal plot for women’s heights measurements

190

180

170

160

150

140

3210-1-2-3

Sim

ula

te-1

Score

Mean 162.2

StDev 6.409

N 1794

AD 0.417

P-Value 0.330

Figure 1.3.9: Normal plot for simulated women’s heights measurements

Figure 1.3.9 shows a Normal plot for the data generated; as expected from random

numbers, the sample values give a very good Normal plot. The data were then rounded

to whole numbers (centimetres) and Figure 1.3.10 was drawn – its characteristics are

virtually identical to those of the real data, showing that the cause of the odd line is the

rounding; note the change in the p-value in moving from Figure 1.3.9 to Figure 1.3.10.

Figure 1.3.11 shows a Normal plot for the Iris data of Figure 1.1.3 – the effect of the

measurements being rounded to only one decimal place, resulting in multiple petals

having the same length, is quite striking.


190

180

170

160

150

140

3210-1-2-3

Ro

un

d(S

imla

te-1

)

Score

Mean 162.2

StDev 6.413

N 1794

AD 2.133

P-Value <0.005

Figure 1.3.10: Normal plot for rounded simulated women’s heights

measurements

1.9

1.8

1.7

1.6

1.5

1.4

1.3

1.2

1.1

1.0

210-1-2

Pe

tal Le

ng

th

Score

Mean 1.464

StDev 0.1735

N 50

AD 1.011

P-Value 0.011

Figure 1.3.11: Normal plot for Iris petal lengths

The assumption of data Normality underlies the most commonly used statistical

significance tests and the corresponding confidence intervals. We will use Normal plots

extensively in verifying this assumption in later chapters.

Exercises

1.3.1. For the data in Exercise 1 in Chapter 0 (use Excel if you wish):

(a) Combine the values in the first 3 columns together in one dataset, compute the Normal

Score Table and plot these on a Normal Plot.

(b) Repeat part (a) for the 20 mean values of the variables x1 to x20.

(c) Contrast the plots in (a) and (b); in particular, what is the effect of averaging?


1.4 Combining Random Quantities (Important)

Often the quantities in which we are interested are themselves the result of combining

other quantities that are subject to chance variation. For example, a journey is

composed of two parts: X is the time it takes to travel from A to B, and Y is the time to

travel from B to C. We are interested in S, the sum of the two travel times.

If we can assume that X is distributed with mean 𝜇𝑥 = 100 and standard deviation

𝜎𝑥 = 10 and (units are minutes), while Y is distributed with mean mean 𝜇𝑦 = 60 and

standard deviation 𝜎𝑦 = 5, what can we say about the total travel time, S=X+Y?

If we can assume that X and Y are independent (i.e., if X happens, for example, to be

unusually long, this fact carries no implications for Y being either unusually long or

short), then the following rules hold.

Addition Rules (S = X+Y)

Means and Variances add

Mean: 𝝁𝒔 = 𝝁𝒙 + 𝝁𝒚

Variance: 𝝈𝒔𝟐 = 𝝈𝒙

𝟐 + 𝝈𝒚𝟐

The underlying mathematics shows that both the means and the variances (the squares

of the standard deviations) are additive. The standard deviation of the overall travel

time, S, is:

Standard Deviation of a Sum: 𝝈𝑺 = √𝝈𝒙𝟐 + 𝝈𝒚

𝟐

We might also ask, how much longer than the trip from B to C, the trip from A to B is

likely to take (i.e., what can we say about D=X–Y). To answer such questions we need

subtraction rules.

Subtraction Rules (D = X-Y)

Mean: 𝝁𝑫 = 𝝁𝒙 − 𝝁𝒚

Variance: 𝝈𝑫𝟐 = 𝝈𝒙

𝟐 + 𝝈𝒚𝟐


Note that the variances add, even though we are subtracting. This gives:

Standard Deviation of a Difference: 𝝈𝑫 = √𝝈𝒙𝟐 + 𝝈𝒚

𝟐

The independence assumption is required for the rules related to the standard

deviation; it is not required for combining means.

Normality Assumptions

The rules given above hold irrespective of the distributions that describe the chance

variation for the two journeys, X and Y. However, if we can assume that X and Y are

Normally distributed, then any linear combination of X and Y (sums or differences,

possibly multiplied by constants) will also be Normally distributed. In such cases, the

calculations required to answer our questions are simple; they are essentially the same

as for Examples 1 and 2, once we have combined the component parts into the overall

measured quantity, and then found its mean and standard deviation using the above

rules.

Example 3

Given our assumptions above about the means and standard deviations of the two

journey times, X and Y, and assuming also that journey times are Normally distributed,

what is the probability that the overall journey time, S,

(i) will take more than three hours (S > 180 minutes)?

(ii) will take less than 2.5 hours (S < 150 minutes)?

(i) From our addition rules we obtain:

Mean: 𝜇𝑠 = 𝜇𝑥 + 𝜇𝑦 = 100 + 60 = 160

Variance: 𝜎𝑠2 = 𝜎𝑥

2 + 𝜎𝑦2 = 102 + 52 = 125

𝜎𝑆 = √𝜎𝑥2 + 𝜎𝑦

2 = √125 = 11.18

Overall travel times are, therefore, Normally distributed with a mean of 160 and a

standard deviation of 11.18, as shown schematically in Figure 1.4.1.


Figure 1.4.1: The standard Normal curve and the travel times curve

The probability that S is more than 180 minutes is given by:

0367.09633.01)79.1(1

)79.1()18.11

160180()

180()180(

ZP

ZPZPZPSPS

S

The probability that the overall travel time, S, exceeds three hours (180 minutes)

is approximately 0.04, as illustrated schematically in Figure 1.4.1.

(ii)

Figure 1.4.2: The standard Normal curve and the travel times curve


)18.11

160150()

150()150(

ZPZPSP

S

S

19.0)89.0(1)89.0()89.0( ZPZPZP

The probability that the overall journey, S, will take less than 2.5 hours (150

minutes) is 0.19, as illustrated schematically in Figure 1.4.2.

Example 4

Suppose that X (as given in the introduction, i.e. mean 𝜇𝑥 = 100 and standard deviation

𝜎𝑥 = 10 (units are minutes)) is the time it takes a part-time student to travel to college

each evening and she makes ten such trips in a term.

(i) If T is the total travel time to college for the term, what are the mean and

standard deviation for T?

(ii) What is the probability T exceeds 18 hours (1080 minutes)?

Since only one trip type is involved, we do not need the X sub-script. We have:

𝜇𝑥 = 𝜇 = 100

𝜎𝑥 = 𝜎 = 10

The total travel time is:

10

1iiXT

where Xi is the journey time for trip i and i = 1, 2, 3…10.

All trips have the same characteristics, so 𝜇𝑖 = 100 and 𝜎𝑖 = 10.

(i) Using the addition rules

𝜇𝑇 = 𝜇1 + 𝜇1 + ⋯ + 𝜇10 = 10 × 100 = 1,000

𝜎𝑇2 = 𝜎1

2 + 𝜎22 + ⋯ + 𝜎10

2 = 10 × (102) = 1,000

𝜎𝑇 = √𝜎12 + 𝜎2

2 + ⋯ + 𝜎102 = √1,000 = 31.62

Total travel time is Normally distributed with a mean of 1000 and a standard deviation

of 31.62, where the units are minutes.


(ii)

Figure 1.4.3: The standard Normal curve and the total travel time curve

0057.09943.01)53.2(1

)53.2()62.31

10001080()1080(

ZP

ZPZPTP

The chances that her total travel time to college for the term might exceed 18

hours (1080 minutes) is only about 6 in a thousand, as illustrated schematically in

Figure 1.4.3. Note that the travel characteristics for travelling back from college

could be quite different (different times might mean different traffic levels), so this

would need examination if we were interested in her total amount of college travel.

Exercises

1.4.1. A final year psychology student approaches a member of staff seeking information on

doing a PhD. In particular, she is concerned at how long it will take. The staff member

suggests that there are essentially four sequential activities involved: Literature Review

(reading and understanding enough of the literature in a particular area to be able to say

where there are gaps that might be investigated), Problem Formulation (stating precisely

what will be investigated, how this will be done, what measurement instruments will be

required, what kinds of conclusions might be expected), Data Collection (developing the

necessary research tools, acquiring subjects, carrying out the data collection, carrying out

the statistical analysis), and Writing up the results (drawing conclusions, relating them

to the prior literature and producing the final thesis). She points out that the required

times are variable, but that her experience suggests that they can be modelled by a

sequence of independent Normal distributions with the parameters shown in Table 1.4.1,

where the time units are weeks.


Activity

Mean

Standard

Deviation

Literature review (L)

30

8

Problem formulation

(P)

10 3

Data collection (D) 120 12

Write up (W) 16 3

Table 1.4.1: Estimates for the means and SDs of the four activities (weeks)

Assuming that this is a reasonably accurate description of the situation (in practice, of

course, the activities will overlap), what is the probability that the doctorate will take (i)

less than 3 years (156 weeks)?; (ii) more than 3.5 years (182 weeks)?; (iii) more than 4

years (208 weeks)?

1.5 The effects of averaging

It is rarely the case that decisions are based on single measured values – usually,

several values are averaged and the average forms the basis of decision making. We

need, therefore, to consider the effect of averaging on our data. We will discuss this in

the context of measuring the percentage dissolution of tablets after 24 hours in a

dissolution apparatus. The ideas are general though; precisely the same statements

could be made about examination results, prices of second-hand cars or the pH

measurements made on samples of lake water.

Figure 1.5.1 illustrates the averaging effect. Figure 1.5.1(a) is an idealised histogram of

individual tablet measurements; it represents what we might expect to get by drawing a

histogram for many thousands of measurements of the percentage dissolution of

individual tablets. Because of the large number of observations, the bars of the

histogram could be made very narrow and a smooth curve would be expected to fit the

histogram very closely. In practice, of course, we would rarely have such a large

quantity of data available to us, but that does not prevent us from visualising what

would happen under idealised conditions.

The data on which Figure 1.5.1(b) is based are no longer single tablet dissolution values.

Suppose we randomly group the many thousands of tablet measurements into sets of

four, calculate the mean for each set, and draw a histogram of the means. Four values

selected at random from the distribution described by (a) will, when averaged, tend to

give a mean result close to the centre of the distribution. In order for the mean to be

very large (say greater than 92) all four of the randomly selected tablet measurements


would have to be very large, an unlikely occurrence3. Figure 1.5.1(b) represents an

idealised version of this second histogram. Of course we would never undertake such an

exercise in the laboratory, though we might in the classroom (using simulation). The

implications of the exercise are, however, very important. Figure 1.5.1 shows that the

properties of means are very different from those of single measurements in one key

aspect: they are much less variable. This difference is important, since virtually all

data-driven decision making relies on means rather than on single measurements. The

narrower curve is called ‘the sampling distribution of the mean’, as it describes the

variation of means, based on repeated sampling, and its standard deviation is called ‘the

standard error of the mean’ or ‘standard error’ for short.

Figure 1.5.1: Idealised histograms of the distributions of percentage dissolution

of (a) individual tablets; and (b) of the averages of 4 tablets.

Mathematical theory shows that, if the standard deviation of the distribution of

individual values is σ then the standard error of the distribution of sample means is

σ/n, where n is the number of independent values on which each sample mean is based.

Suppose the individual values vary about a long-run average value of 90 with a

standard deviation of 0.5. It follows that averages of 4 values will also vary about a

long-run mean of 90, but their standard error will be 0.25. Recall that approximately

95% of the area under any Normal curve lies within two standard deviations (or

standard errors) of the mean. This implies that while 95% of individual tablet

measurements would be between 89 and 91, 95% of means of 4 tablet measurements

would be between 89.5 and 90.5. Thus, the effect of the averaging process is that a

sample mean result is likely to be closer to the long-run or batch average, than a

3 If the probability that one tablet was greater than 92 was 0.05, the probability that all four would be

greater than 92 would be 0.00000625!

93.592.591.590.589.588.587.586.5

(b)

(a)


randomly selected single tablet measurement would be. This gain in precision is the

reason why we base decisions on means, in practice.

Example 4 revisited

(i) Suppose we take the journey times to be a sample of size n = 10 journey times

and label journey time as 𝑥1 = 10, 𝑥2 = 10, ⋯ 𝑥10 = 10, (i.e each has a journey

is equal – a rather odd sample indeed).

What is the value of �̅� the overall average travel time to college based on this

sample and what is its standard deviation (i.e. standard error) assuming the

standard deviation of a single journey time is 𝜎 = 10 minutes?

(ii) What is the probability �̅� exceeds 18 hours (1080 minutes)?

Answer

(i) Using the addition rules for random variables the overall sample mean travel

time is for the 10 observed random journey times is:

10

1110

11

i

i

n

i

i XXn

x = 100

10= 10

where Xi is the journey time for trip i and i = 1, 2, 3…10.

The standard error is the standard deviation of �̅�

𝜎�̅�2 =

1

𝑛2∑ 𝜎𝑖

2

𝑛

𝑖=1

=1

102∑ 𝜎2

10

𝑖=1

=10 × (102)

100= 10

since each 𝑥𝑖 has the same standard deviation

giving

𝜎�̅� = √10 = 3.16

NB: the 𝟏

𝒏𝟐 in the formula above – this appears because the variance has the

effect of squaring constants so the 𝟏

𝒏 factor in the mean formula changes to

𝟏

𝒏𝟐


(ii) What is the probability the average travel time �̅� exceeds 15 minutes?

057.0943.01)58.1(1

)58.1()16.3

1015()15(

ZP

ZPZPxP

The chances that her average travel time to college for the term might exceed minutes is

only about 6 in a hundred.

The histograms/curves of Figure 1.5.1 are Normal (probably a reasonable assumption

for tablet dissolution). However, what has been described does not depend on the

assumption of an underlying Normal distribution. Even if the curve for individual

tablets had been skewed, the mean of the sampling distribution would be the same as

that of the individual tablets and the standard error would still be σ/n. What is

interesting, though, is that, as the number of measurements on which each mean is

based increases, the shape of the sampling distribution approaches that of a Normal

curve, even when the underlying distribution shape is non-Normal.

We will illustrate this interesting and important result using some data simulated from

a skewed distribution, specifically from a chi-square distribution with 7 degrees of

freedom (we will encounter the chi-square distribution later when we discuss

significance tests; here, it is used simply as a typically skewed distribution curve).

Figure 1.5.2 shows the theoretical curve, while Figure 1.5.3 shows a histogram of 200

values randomly generated from this distribution.

Figure 1.5.2: A chi-square curve with 7 degrees of freedom


24201612840

50

40

30

20

10

0

Fre

qu

en

cy

Figure 1.5.3: Histogram of 200 values from a chi-square curve with 7 degrees of

freedom

Figure 1.5.4 shows the corresponding Normal plot which has the characteristic curved

shape, we saw previously for skewed distributions.

25

20

15

10

5

0

-5

3210-1-2-3

Sim

ula

ted

Da

ta

Score

Mean 7.485

StDev 4.007

N 200

AD 3.856

P-Value <0.005

Figure 1.5.4:Normal plot of 200 values sampled from a chi-square curve with 7 degrees

of freedom

Thirty values were randomly generated from this distribution and the mean was

obtained. This was repeated 200 times, resulting in a set of 200 sample means, each


based on n=30. A histogram of these means is shown in Figure 1.5.5, while Figure 1.5.6

represents a Normal plot of the set of 200 means.

98765

50

40

30

20

10

0

Mean

Fre

qu

en

cy

Figure 1.5.5: Histogram of 200 means each based on 30 values sampled from a chi-

square curve with 7 degrees of freedom

9

8

7

6

5

4

3210-1-2-3

Me

an

Score

Mean 6.939

StDev 0.7326

N 200

AD 0.211

P-Value 0.858

Figure 1.5.6: Normal plot of 200 means each based on 30 values sampled from a chi-

square curve with 7 degrees of freedom

The effect of averaging has been particularly successful here in producing means which

closely follow a Normal distribution (not all samples will be quite so good). The

mathematical theory says that as the sample size tends to infinity, the means become

Normal; it is clear, though, that even with a sample size of 30 in this case, the so-called

Normal approximation is very good. In statistical theory the Normal approximation is

called the Central Limit Theorem (CLT); these will be used interchangeably. The

importance of the Normal approximation/CLT cannot be over-emphasised.


In pathological cases (e.g., extremely skewed distributions)4 the sampling distribution

curve will not be Normal for the moderate sample sizes that are likely to be used in

practice. In most cases, however, where the underlying curve is not too far from a

Normal curve, the sampling distribution will be reasonably close to a Normal curve, and

this allows us use statistical methods based on the assumption of Normality. Apart

from its common occurrence in practice (as seen in several examples in Section 1.1) this

property of sampling distributions is the reason why the Normal curve is so important

in statistics.

1.6 Measurement Systems

The same symmetrical characteristics as seen for the data of Section 1.1 are displayed

in Figure 1.6.1, but it is different in one important respect – the picture is based on 119

repeated measurements of the same pharmaceutical material, whereas all our earlier

examples were based on single measurements on different people or objects. The

material measured was a control material used in monitoring the measurement process

in a quality control analytical laboratory in a pharmaceutical manufacturing company

[See Mullins [7] Chapters 1, 2]. The methods discussed in Section 1.3 show that a

Normal curve is a good model for these data, also.

97.897.296.696.095.4

Figure 1.6.1: Repeated potency (%) measurements on a control material

In using the Normal curve in a measurement context we have a different conceptual

model for our data than that for the previously cited examples. Where the numbers

referred to different physical entities (whether these were people, industrial components

or flowers) the curves refer to variation between those units – some entities give values

much higher or lower than others, but most cluster in the centre. For measurement

systems we do not have a physical distribution; we have an hypothetical distribution of

the many thousands of measurements that might be made on the single physical entity

4 In particular, theoretically the Normal approximation fails for the Cauchy Distribution.


that is to be measured. In practice, of course, the vast majority of these measurements

will never be made, so our questions about the ‘system’ of measurements (or the

measurement process) are somewhat abstract in nature, but we are so accustomed to

dealing with measurements that this abstraction often goes un-discussed. Two aspects

of measurement quality are illustrated by Figure 1.6.2, namely bias and precision.

11511010510095

A

B

Figure 1.6.2: Measurements on the same entity by two measurement

systems

Suppose this picture represents smoothed histograms, based on thousands of

measurements made on the same object/material by two different measurement systems

A, B. It is clear that System A gives lower values, on average, i.e., the long-run mean

for A is smaller than that for B5, μA < μB. System A is said to be biased downwards

relative to System B. Which system gives the ‘right’ results, on average, (i.e., is

unbiased) we cannot know unless there is a measurement ‘gold standard’. When a

benchmark measurement exists we can define how accurate the system is relative to the

benchmark.

The spread of measurements for System B is greater than that for A – A is said to have

better precision. The width of each curve is measured by the standard deviation, σ; the

shapes of the curves in Figure 1.6.2 mean that σA < σB. There is no necessary

relationship between the two parameters. There is no reason why σA could not be bigger

than σB.

Figure 1.6.3 [7, 8] shows serum ALT6 concentrations for 129 adults. It demonstrates

that there is no necessary relationship between the shape of the curve describing

5 Long-run averages only make sense when the system or process is stable, accordingly care should be taken to

ensure stability before a long-run average is computed. 6 Alanine aminotransferase (ALT) is an enzyme found in most human tissue.


variation in the physical system under study and the shape of the curve that describes

variation in the measurement system used to measure the physical entities. The ALT

distribution (Figure 1.6.3(a)) is right-skewed, that is, it has a long tail on the right-hand

side, which means that some individuals have much higher ALT levels than the

majority of the population.

Figure 1.6.3: Distributions of ALT results (a) 129 adults (b) 109 assays of one

specimen

Figure 1.6.3(b) shows the distribution of results obtained when one of the 129 serum

specimens was re-analysed 109 times. The shape of the distribution that describes the

chance analytical variability is very different from that which describes the overall

variation in ALT measurements. It suggests that a Normal distribution might well

describe the analytical variability. If, therefore, we were interested in studying the

analytical procedure – for, example, to quantify the precision, which is obviously poor –

we might work directly in the arithmetic scale (i.e., the scale in which the

measurements were made. On the other hand, if we wished to make statements about

the distribution of ALT concentrations in the population from which the subjects derive

(by obtaining a reference interval, for example), then we might choose to transform the

data before proceeding with our analysis (in order to get a Normal curve which is easier

to analyse).

Figures 1.6.1 and 1.6.3 underline the importance of the Normal curve for the analysis of

measurement systems, as the graphs of Section 1.1 showed its importance for describing

chance variation in different physical populations.


Exercises

1.6.1. (i) An unbiased analytical system is used to measure an analyte whose true value

for the parameter measured is 100 units. If the standard deviation of repeated

measurements is 0.80 units, what fraction of measurements will be further than

one unit from the true value? Note that this fraction will also be the probability

that a single measurement will produce a test result which is either less than 99

or greater than 101, when the material is measured. What fraction of

measurements will be further than 1.5 units from the true value? Give the

answers to two decimal places.

(ii) Suppose now that the analytical protocol requires making either 2 or 3

measurements and reporting the mean of these as the measured value.

Recalculate the probabilities for an error of one unit and compare them to those

obtained for a single measurement.

1.6.2. (i) A trawl through the website HorsesForum.com suggests that a nailed-on steel

horseshoe will last about 5 to 8 weeks before it needs to be re-set. Assuming the

lifetime of a horseshoe is normally distributed with population mean μ = 7 weeks

and standard deviation σ = 1.25 weeks what is the probability that a nailed-on

horseshoe picked at random from the population will last 9 or more weeks before

needing re-set

(ii) A second website ‘The Chronicle of the Horse’ advises so-called ‘glued-on’

horseshoes, that is shoes that are glued rather than nailed-on, may be used in

cases of laminitis (aka white line disease in USA) where the hoof is unable to

accommodate nails. Gleaning the posts on the website seems to indicate that the

population mean lifetime of glued-on shoes is μ = 5 weeks with standard

deviation σ = 2.0 weeks. Assuming the lifetimes are normally distributed what is

the probability that a glued-on horseshoe picked at random from the population

will last 9 or more weeks before needing re-set.

(iii) Comparing the information from the two websites: What is the probability that a

nailed-on horseshoe will last longer than a glued-on horseshoe, i.e the difference

between the lifetimes of the two types of horseshoe is greater than zero.

(iv) Comparing a sample mean to a population value μ (e.g population mean):

The website data used in part (i) was in fact based on a sample of n=5 with

�̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛 = 7. On the basis of this sample data what is the probability that the

mean lifetime μ for (the population of) nailed-on horseshoes is at least (μ =) 9

weeks.


(v) Exam style question:

Comparing the information based on samples from the two websites: The website

data used in part (i) was in fact based on a sample of n=5 with �̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛 = 7

while in part (ii) the sample size was n=2 and �̅�𝐺𝑙𝑢𝑒𝑑−𝑜𝑛 = 5. On the basis of these

sample sizes what is the probability that a nailed-on horseshoe will last longer

than a glued-on horseshoe, i.e the difference between the means of the lifetimes of

the two types of horseshoe is greater than zero.

1.6.3. Combining data - Meta-analysis study

(i) In a 3-rd world country in Latin America a study shows that infants born to

families who cook on an open Wood stove have an average birth weight of μ = 2.9

kg with standard deviation 𝜎 = 0.5kg.

What is the probability that an infant born to one of these families is over 3.6kg?

(ii) Cooking in the country is also done an open fires and electric cookers in this

country. The respective average birth weight statistics are μ = 2.8kg and 3.1kg

while the standard deviations are 0.4kg and 1.3kg respectively.

Repeat part (i) for an electric stove.

(iii) Combining the information on cooking types: What is the probability that an

infant born in this country will be less than 2.3kg.

(iv) Combining the information on cooking types based on samples from the three

cooking type studies: The study data used in parts (i) and (ii) was in fact based on

3 samples each of size n=36 (108 in all). Taking the sample mean in each case to

be the population mean (i.e. with �̅� = 𝜇) repeat part (iii) above and comment.


1.7 Appendix: Case Study Statistical Process Control Charts

In this chapter we have discussed various aspects of the Normal curve as a model for

statistical variation. This final section illustrates the use of a graphical tool for

monitoring the stability of a system. This tool, the control chart, is based on the

properties of the Normal curve and is an example of a powerful practical tool based on

simple underlying ideas.

Statistical process control (SPC) charts were introduced for monitoring production

systems but, as we shall see, they have much wider applicability. Conceptually they are

very simple: a small sample of product is taken regularly from the process and some

product parameter is measured. The values of a summary statistic (such as the sample

mean) are plotted in time order on a chart; if the chart displays other than random

variation around the expected result it suggests that something has changed in the

process. To help decide if this has happened control limits are plotted on the chart: the

responses are expected to remain inside these limits. Rules are decided upon which will

define non-random behaviour.

The most commonly used charts were developed by Shewhart [9] in Bell Laboratories in

the1920’s. He described variation as being due either to ‘chance causes’ or ‘assignable

causes’. Chance causes are the myriad of small influences that are responsible for the

characteristic Normal shape, which we saw in a variety of settings in Section 1.1.

Shewhart’s crucial insight was that stability must be understood in the context of this

inevitable chance variation, not in the commonplace notion of ‘unchange’. Assignable

causes are larger: they tend to change the mean or the standard deviation of a process

distribution. If a process is stable such that it exhibits only random variation around a

given reference value and the size of that variation (as measured by the standard

deviation) remains constant, Shewhart described it as being ‘in statistical control’ or

simply ‘in control’. The objective in using control charts is to achieve and maintain this

state of statistical control. Control charts were developed to control engineering

production processes, but, as Charts 2, 3 and 4 will show, they can be applied just as

effectively to measurement systems. Any apparent differences between the two

situations disappear when we think of a measurement system as a production process

whose output is measurements. Chart 5 illustrates the use of a control chart in a retail

sales setting.

For an extensive discussion of control charts in a manufacturing environment see

Montgomery [10]. For an introduction to the use of control charts for monitoring

measurement systems see Mullins [7]. Stuart [11] gives a less technical introduction to


control charts in industry than does Montgomery; he also discusses their application in

a retail business context (see Chart 5, which is ‘borrowed’ from Stuart!).

Chart 1: Monitoring Production Systems

Figure 1.7.1 shows a control chart for a critical dimension (mm) of moulded plastic

medical device components, each of which contained a metal insert purchased from an

outside supplier.

Twenty five samples of size n=5 were taken from the stream of parts manufactured over

a period of several weeks, with a view to setting up control charts to monitor the

process. The five sample results were averaged and it is the average that is plotted in

the chart. Sample means are usually denoted x , where x denotes a single measured

value, for this reason the chart is called an ‘X-bar chart’. For now, the centre line (CL)

and control limits (upper and lower control limits, UCL and LCL) will be taken as given;

later, the rationale for the limits will be discussed.

252321191715131197531

3.70

3.65

3.60

3.55

3.50

3.45

Sample

Sa

mp

le M

ea

n

__X=3.5537

UCL=3.6713

LCL=3.4361

Figure 1.7.1: An X-bar chart for the moulded plastic components

In using control charts different companies/users apply different rules to identify

instability. For example, the SPC manual of the Automotive Industry Action Group [12]

(AIAG: a task force set up by the Ford, Chrysler and General Motors companies to unify

their SPC procedures) suggests that action is required if any of the following occur:

• any point outside of the control limits;

• a run of seven points all above or all below the central line;

• a run of seven points in a row that are consistently increasing (equal to or greater

than the preceding points) or consistently decreasing;


• any other obviously non-random pattern.

Runs of nine points, rather than seven, are also commonly recommended. The same

basic principle underlies all the rules: a system that is in statistical control should

exhibit purely random behaviour - these rules correspond to improbable events on such

an assumption. Accordingly, violation of one of the rules suggests that a problem has

developed in the process and that action is required. The rationale for the rules is

discussed below.

The manufacturing process to which Figure 1.7.1 refers appears, by these rules, to be in

statistical control. No points are outside the control limits and there are no long runs of

points upwards or downwards or at either side of the central line.

Chart 2: Monitoring a HPLC Assay

To monitor the stability of a measurement system, a control sample may be measured

repeatedly over time, together with the test samples that are of primary interest. The

control sample results are then plotted in time order on a chart. The centre line for this

chart is the accepted correct value for the control material (typically the mean of recent

measurements on the material), and newly measured values for the control material are

expected to vary at random around the centre line. Sometimes the centre line might be

the certified value for a bought-in or a ‘house’ reference material.

454137332925211713951

95.5

95.0

94.5

94.0

93.5

93.0

Observation

%P

ote

ncy

_X=94.150

UCL=95.519

LCL=92.781

Figure 1.7.2 A control chart for the HPLC potency assay


Figure 1.7.2 shows a control chart for a HPLC (high performance liquid

chromatography) potency assay of a pharmaceutical product [7]. The data displayed in

the chart were collected over a period of several months. At each time point two

replicate measurements were made on a control material, which was just a quantity of

material from one batch of the production material routinely manufactured and then

measured in the laboratory. These results were averaged and it is the average that is

plotted in the chart.

The average level of the measurement system to which Figure 1.7.2 refers appears, by

the AIAG rules, to be in statistical control. No points are outside the control limits and

there are no long runs of points upwards or downwards or on either side of the central

line. Accordingly, we can feel confident that the analytical system is stable and

producing trustworthy results.

Chart 3: Monitoring both Production and Analytical Systems

Figure 1.7.3 illustrates the benefits of having control charts in use for both production

results and analytical control data, when product reviews are carried out.

Figure 1.7.3(a) shows potencies for a series of batches of a drug with a clear downwards

shift in results. Such a shift suggests problems in the production system, but would

often lead to production management questioning the quality of the analytical results.

It has to be someone else’s fault!


Figure 1.7.3: Simultaneous control charts for a production system (a) and the

analytical system (b) used to measure the potency of its product.

As shown in Figure 1.7.3(b), a stable control chart for the analytical system for the same

period is the best possible defence of the stability of the analytical results. If, on the

other hand, there had been a downward shift in the analytical system it would have

been reflected in Figure 1.7.3(b). Recognising such a shift would save much time that

would otherwise be wasted searching for a non-existent production problem. Thus, the

availability of control charts for both the production and analytical systems allows data-

driven decision-making and can reduce inter-departmental friction when problems arise

at the interface7.

Chart 4: An Unstable Analytical System

7 In fact, the problem here could be due to changes in an in-coming raw material; so a third control

chart might well be worthwhile!


Figure 1.7.4, represents 49 measurements on control samples made up to contain 50

parts per billion (ppb) of iron in pure water [7]. The sample was analysed by an

Inductively Coupled Plasma (ICP) Spectrometer over a period of months.

Figure 1.7.4: A control chart for the iron (Fe) content of a water sample

(ppb)

The diagram shows an analytical system which is clearly out of control; in fact, three

gross outliers were removed before this chart was drawn. The data were collected

retrospectively from the laboratory records; the laboratory did not use control charts

routinely. They are presented here because they exhibit the classic features of an out-

of-control system: several points outside the control limits, a run of points above the

centre line and a run of points downwards. There is an obvious need to stabilise this

analytical system.

There are two ways in which control charts are used, viz., for assessing retrospectively

the performance of a system (as in Figure 1.7.4) and for maintaining the stability of a

system, which is their routine use once stability has been established. Where a chart is

being used to monitor a system we would not expect to see a pattern such as is exhibited

between observations 17 and 32 of Figure 1.7.4, where 16 points are all either on or

above the centre line. Use of a control chart should lead to the upward shift suggested

here being corrected before such a long sequence of out-of-control points could develop.

50403020100

55

50

45

40

Run Number

Fe

Res

ult

Mean=48.2

UCL=54.02

LCL=42.38


Note that if the chart were being used for on-going control of the measurement process,

the centre line would be set at the reference value of 50 ppb. Here the interest was in

assessing the historical performance of the system, and so the data were allowed

determine the centre line of the chart.

The Theory Underlying the Control Limits

The centre line (CL) for the chart should be the mean value around which the

measurements are expected to vary at random. In a manufacturing context, this could

be the target value for the process parameter being measured, where there is a target,

but in most cases it is the mean of the most recent observations considered to be ‘in-

control’. Similarly, in a measurement context, the centre line will either be the mean of

recent measurements on the control material, or the accepted ‘true value’ for an in-

house standard or a certified reference material.

The control limits are usually placed three standard deviations above and below the

centre line, Figure 1.7.5(b); the standard deviation of individual measurements is used

if the points plotted are individual measurements, as in Chart 4, and the standard error

of the mean, if the plotted points are means of several measured values, as in Chart 1.

This choice may be based on the assumption that the frequency distribution of chance

causes will follow a Normal curve, or it may be regarded simply as a sensible rule of

thumb, without such an assumption. As we have seen a distribution curve can be

thought of as an idealised histogram: the area under the curve between any two values

on the horizontal axis gives the relative frequency with which observations occur

between these two values. Thus, as shown in Figure 1.7.5(a), 99.74% of the area under

any Normal curve lies within three standard deviations (3σ) of the long-run mean (μ)

and so, while the system remains in control, 99.7% of all plotted points would be

expected to fall within the control limits.

Where only single values are measured (as frequently happens in process industries, for

example) the control limits are:

Upper Control Limit (UCL) = CL + 3σ

Lower Control Limit (LCL) = CL – 3σ

where σ is the standard deviation of individual values.

If, as is more commonly the case in manufacturing of components, the point to be

plotted is the average of a number of values, say n, then an adjustment has to be made

to take account of the fact that means, rather than individual values, are being plotted.

If individual values give a frequency distribution, as shown in Figure 1.7.5(a), with


mean μ and standard deviation σ, then a very large number of averages, each based on

n values from the same process, would give a similar frequency distribution centred on

μ but the width would be defined by σ/ n (the ‘standard error’). The distribution would

be tighter and more of the values would be close to μ. This simply quantifies the extent

to which the averaging process reduces variability, as discussed in Section 1.5: see

Figure 1.5.1, in particular.

Figure 1.7.5: Three sigma limits and a process shift

To obtain the correct control limits when plotting averages, we simply replace σ by

σ/ n in the expressions given above and obtain:

Upper Control Limit (UCL) = CL + 3 σ/ n

Lower Control Limit (LCL) = CL – 3 σ/ n

where σ is (as before) the standard deviation of individual values. The chart based on

single measurements is sometimes called an ‘Individuals or X-chart’ while the chart

based on means is called an ‘X-bar chart’; the two charts are, however, essentially the

same. Note that in a laboratory measurement context, where n is typically 2 or 3, the


simplest and safest approach is to average the replicates and treat the resulting means

as individual values (see Mullins [7], Chapter 2).

The limits described above are known as ‘three sigma limits’, for obvious reasons; they

are also called ‘action limits’. They were first proposed by Shewhart in the 1920’s and

are almost universally used with this type of control chart. Their principal justification

is that if the process is in control then the probability of a point going outside the control

limits is very small: about three in a thousand. Accordingly, they give very few false

alarm signals. If a point does go outside the control limits it seems much more likely

that a problem has arisen (e.g., the process average has shifted upwards or downwards;

see Figure 1.7.5(c)) than that the process is stable, as in Figure 1.7.5(b), but the chance

causes have combined to give a highly improbable result. This is the rationale for the

first AIAG rule, quoted above.

The basis for the second AIAG rule (a run of seven points all above or all below the

central line) is that if the system is in control the probability that any one value is above

or below the central line is 1/2. Accordingly, the probability that seven in a row will be

at one side of the central line is (1/2)7=1/128; again, such an occurrence would suggest

that the process has shifted upwards or downwards. The third rule (a run of seven

points in a row that are consistently increasing (equal to or greater than the preceding

points) or consistently decreasing) has a similar rationale: if successive values are

varying at random about the centre line, we would not expect long runs in any one

direction.

The last catch-all rule (any other obviously non-random pattern) is one to be careful of:

the human eye is adept at finding patterns, even in random data. The advantage of

having clear-cut rules that do not allow for subjective judgement is that the same

decisions will be made, irrespective of who is using the chart. Having said this, if there

really is ‘obviously non-random’ behaviour in the chart (e.g., cycling of results between

day and night shift) it would be foolish to ignore it.

The assumption that the data follow a Normal frequency distribution is more critical for

charts based on individual values than for those based on averages. Averages, as we

saw in Section 1.5, tend to follow the Normal distribution unless the distribution of the

values on which they are based is highly skewed. In principle, this tendency holds when

the averages are based on very large numbers of observations, but in practice samples of

even four or five will often be well behaved in this regard. If there is any doubt

concerning the distribution of the measurements a Normal probability plot (see Section

1.3) may be used to check the assumption.


Chart 5: A retail sales control chart

Stuart [11] considered the problem of monitoring cash variances in the till of a retail

outlet – a sports club bar. At the end of each day's trading, the cash in the till is

counted. The till is then set to compute the total cash entered through the keys, which

is printed on the till roll. The cash variance is the difference between cash and till roll

totals (note that ‘variance’ as used here is an accounting term, which is not the same as

the statistical use of the same word, i.e., the square of the standard deviation). Small

errors in giving change or in entering cash amounts on the till keys are inevitable in a

busy bar. Ideally, the cash variance will be zero. In the presence of inevitable chance

variation, the best that can be hoped for is that the cash variances will centre on zero.

In other words, the cash variance process is ideally expected to be a pure chance process

with a process mean of zero.

-20

-12

-4

4

12

20

10 20 30 40 50

Week

V

a

r

i

a

n

c

e

UC L

CL

LC L

Sundays

Figure 1.7.6: Sunday cash variances control chart

A control chart covering Sunday cash variances for a year is shown in Figure 1.7.6. The

variances are recorded in pounds (the data pre-date the Euro). The centre line, is

placed at zero, the desired process mean. The control limits are placed at ±£12; the

limits were established using data from the previous year.

As may be seen from the chart, this process was well behaved, apart from three points

outside the control limits. For the first point, week 12, value £14.83, the large positive

cash variance indicates either too much cash having been put in the till or not enough

entered through the till keys. When the bar manager queried this, one of the bar

assistants admitted a vague recollection of possibly having given change out of £5

instead of out of £10 to a particular customer. The customer involved, when asked,

indicated that she had suspected an error, but had not been sure. The £5 was

reimbursed.


In the second case, week 34, value –£22.08, an assistant bar manager who was on duty

that night paid a casual bar assistant £20 wages out of the till but did not follow the

standard procedure of crediting the till with the relevant amount.

The last case, week 51, value -£18.63, was unexplained at first. However, other out of

control negative cash variances on other nights of the week were observed, and these

continued into the following year. Suspicion rested on a recently employed casual bar

assistant. An investigation revealed that these cash variances occurred only when that

bar assistant was on duty and such a coincidence applied to no other member of staff.

The services of that bar assistant were ‘no longer required’ and the cash variances

returned to normal.

1.8 Conclusion

This chapter has introduced the concept of statistical variation – the fact that objects or

people selected from the same population or process give different numerical results

when measured on the same characteristic. It also showed that, where the same

object/person is measured repeatedly, different numerical results are obtained, due to

chance measurement error. In many cases, the variation between numerical results can

be described, at least approximately, by a Normal curve.

This curve will often describe the variation between means of repeated samples, also,

even where the raw data on which the means are based come from distributions that

are not themselves Normal. This fact is of great importance for what follows: the most

commonly used statistical procedures are based on means; therefore, statistical methods

based on Normal distributions will very often be applicable in their analysis. For this

reason, the properties of the Normal curve and its use in some simple situations (e.g.,

developing reference intervals in medicine, or use of control charts in industrial

manufacturing or in monitoring measurement systems) were discussed in the chapter.

References

[1] Minitab Statistical Software, ( www.minitab.com )

[2] Anderson, E., The irises of the Gaspé peninsula, Bulletin of the American Iris

Society 59, 2-5, 1935.

[3] Fisher, R.A., The use of multiple measurements in taxonomic problems, Annals of

Eugenics 7, 179-188, 1936.

[4] Bland, M., An Introduction to Medical Statistics, Oxford University Press, 2000.

http://www.minitab.com/


[5] Strike, P.W. Measurement in Laboratory Medicine, Butterworth-Heinemann,

1996.

[6] Hahn, G.J.and Meeker, W.Q., Statistical Intervals: a guide for practitioners, Wiley,

1991.

[7] Mullins, E., Statistics for the Quality Control Chemistry Laboratory, Royal Society

of Chemistry, Cambridge, 2003.

[8] Samuels, M.L., Statistics for the Life Sciences, Dellen Publishing Company, 1989.

[9] Shewhart, W. A., Economic Control of the Quality of Manufactured Products,

Macmillan, London, 1931.

[10] Montgomery, D.C., Introduction to Statistical Quality Control, Wiley, 2005

[11] Stuart, M., An Introduction to Statistical Analysis for Business and Industry,

Hodder, 2003.

[12] Chrysler Corporation, Ford Motor Company, and General Motors Corporation,

Statistical Process Control: Reference Manual, 1995.


Outline Solutions

1.2.1.

(a)

A weight of 85 mg corresponds to a Z value of –2.14 as shown below – the standard Normal

table gives an area of 0.0162 to the left of –2.14, so the area to the left of 85 is also 0.0162.

14.27

10085

xz

P(X < 85) = P(Z < -2.14) = P(Z>2.14) = 1 – P(Z < 2.14) = 1 – 0.9838 = 0.0162.

By symmetry, the area above 115 is also 0.0162.

Approximately, 1.6% of tablets will have weights either greater than 115 mg or less than

85 mg, which means that about 3.2% of tablets are outside these limits.

(b) The calculations for part (b) are carried out in exactly the same way as those for (a).


(c) (i)

If we inspect the standard Normal table to find the value that has an area of 0.99 below it,

we find Z=2.33; therefore, -2.33 will have an area of 0.01 below it.

P(Z < -2.33) = 0.01

7

10033.2

KZ .

This means that K = 100 – 2.33(7) = 100 – 16.31 = 83.69

99% of tablets would be expected to weigh more than 83.69 mg.

(ii)

P(Z < 2.575) = 0.995 => P(Z >2.575) = 0.005

and, by symmetry, P(Z < –2.575) = 0.005

7

100575.2

BZ .

This means that B = 100 + 2.575(7) = 100 +18.025 = 118.025 mg


By symmetry A = 100 – 2.575(7) = 100 –18.025 = 81.975 mg

1.2.2. In the log10 scale we have a Normal distribution with mean 2.7423 and standard

deviation 0.1604.

(i) Calculate the bounds (a, b) for the 95% reference interval in the log scale.

The reference bounds are: μ–1.96σ and μ+1.96σ which gives:

2.7423 ± 1.96(0.1604) or 2.4279 to 3.0567.

(ii) Use your results to obtain the corresponding reference interval in the original

measurement scale: the bounds will be (10a, 10b).

102.4279=267.9 and 103.0567=1139.5, so our reference interval in the original

measurement scale is 267.9 to 1139.5 (i.e., 268 to 1140 in practice). Note that

although the mean 2.7423 is the centre of the interval in the log scale, when we

back transform [102.7423=552.5], the corresponding value in the arithmetic scale is

not centred between the two bounds – the right-hand bound is much further

above 552.5 than the lower bound is below it. This reflects the skewed shape of

the original histogram. In fact, 552.5 is an estimate of the median (the steroid

value below which and above which 50% of the women lie). For a symmetrical

distribution the mean and the median coincide, but for right-skewed distributions

the mean will be higher; thus, the mean of the 100 original steroid values is 590.8

– the long tail pulls the mean upwards. Income distributions, for example, are

typically quite skewed, and the relatively small number of very high income

individuals increases the average income – the median value is, therefore, a

better one-number summary of a ‘typical’ member of the population.

Note also that the calculations above will be quite sensitive to the number of

decimal places used.

(iii) What value, U, should be quoted such that only 1% of women from this

population would be expected to have a β-OHCS measurement which exceeds U?


(note: curves are schematic and not to scale)

The critical value on the standard Normal curve which has 0.99 to the left and 0.01 to

the right is 2.33. From the relation:

xz

it is clear that the corresponding value for a Normal curve with mean μ 2.7423 and

standard deviation σ = 0.1604 is μ+2.33σ=3.1160. Only 1% of the X=Log10(β-OHCS)

values would be expected to exceed X= Log10(U)=3.1160. The corresponding value in the

original measurement scale is U=103.1160= 1306.2.

1.3.1. Follow the steps in the log10 scale we have a Normal distribution with mean 2.7423 and

standard deviation 0.1604.

The real data plot done in Excel (Minitab will be much better) displays that it is right skewed

while the line from means look cannot be differentiated from the straight line that arises with

Normal Data.

The effect of averaging is to produce a set of mean values whose distribution (their sampling

distribution) is approximately Normal – the application of the theoretical Central Limit Theorem

in practice on real data.


Real

Data

Mean

Values

x Ranks Fraction zNormal

Linex Ranks Fraction z

Normal

Line

-4 1 0.04 -1.77 -2.20 3 1 0.05 -1.64 3.20

1 2 0.08 -1.43 -1.22 3.5 2 0.1 -1.28 3.91

1 3 0.12 -1.20 -0.28 3.8 3 0.15 -1.04 4.25

1 4 0.15 -1.02 0.46 4 4 0.2 -0.84 4.53

1 5 0.19 -0.87 1.09 4.5 5 0.25 -0.67 4.76

1 6 0.23 -0.74 1.64 4.8 6 0.3 -0.52 4.97

2 7 0.27 -0.62 2.14 5 7 0.35 -0.39 5.17

2 8 0.31 -0.50 2.61 5.5 8 0.4 -0.25 5.35

2 9 0.35 -0.40 3.05 5.7 9 0.45 -0.13 5.53

3 10 0.38 -0.29 3.48 6 10 0.5 0.00 5.71

3 11 0.42 -0.19 3.89 6.1 11 0.55 0.13 5.88

3 12 0.46 -0.10 4.29 6.2 12 0.6 0.25 6.06

4 13 0.50 0.00 4.69 6.4 13 0.65 0.39 6.24

4 14 0.54 0.10 5.09 6.6 14 0.7 0.52 6.44

4 15 0.58 0.19 5.50 6.7 15 0.75 0.67 6.65

5 16 0.62 0.29 5.91 6.8 16 0.8 0.84 6.88

5 17 0.65 0.40 6.33 6.9 17 0.85 1.04 7.16

6 18 0.69 0.50 6.77 7.2 18 0.9 1.28 7.50

7 19 0.73 0.62 7.24 7.5 19 0.95 1.64 8.01

7 20 0.77 0.74 7.74 7.9 20 0.99 2.33 8.20

8 21 0.81 0.87 8.30

8 22 0.85 1.02 8.92

11 23 0.88 1.20 9.66

11 24 0.92 1.43 10.50

12 25 0.96 1.77 11.00

14 26 1.00 3.72 12.00


Overlaid probability plot for random variable x1 and Mean values of x1-x20

1.4.1. We begin by finding the parameters for the doctorate completion time, T

μT = μL + μP + μD + μW = 30 + 10 + 120 + 16 = 176

𝜎𝑇2 = 𝜎𝐿

2 + 𝜎𝑃2 + 𝜎𝐷

2 + 𝜎𝑊2 = 82+32+122+32=226

03.152222 WDPLT .

Since all four components of the total time are assumed to be Normally distributed, the

total time, T, will also have a Normal distribution.


(i) P(T < 3 years) = P(T < 156 weeks)

09.0)33.1(1

)33.1()33.1()03.15

176156()()156(

TP

TPTPZPT

ZPTPT

T

The probability that the doctorate will take less than 3 years (156 weeks) is 0.09.

(ii) P(T > 3.5 years) = P(T > 182 weeks)

345.0655.01)40.0(1

)3992.0()03.15

176182()()182(

TP

TPZPT

ZPTPT

T

The chances of the postgraduate work taking more than three and a half years (182

weeks) are about 0.35.


(iii)

Similarly, the probability that the work will take more than 4 years (208 weeks) is

about 0.02, as shown below.

0166.09834.01)13.2(1

)13.2()03.15

176208()()208(

TP

TPZPT

ZPTPT

T

This example illustrates the analysis of project planning networks. Of course, the

current set of tasks is the simplest possible ‘network’ – a linear sequence of activities.

Real projects (consider for example the construction of an electricity generating

station) may involve thousands of activities, some of which may proceed

simultaneously (‘in parallel’), while others must be fully completed before others may

begin (e.g., foundations must be in place before walls can be built – these activities

are ‘in series’).

1.6.1.

(i) A measurement of 101 corresponds to a Z value of 1.25 as shown below – the

standard Normal table gives an area of 0.8944 to the left of 1.25, so the area to the

right of 1.25 is 0.1056. The probability that a single measurement will exceed 101 is,

accordingly, 0.1056.

25.18.0

100101

xz

P(X > 101) = P(Z>1.25) = 1 – P(Z < 1.25) = = 1 – 0.8944 = 0.1056

By symmetry, the probability that a single measurement will be less than 99 is also

0.1056. Hence, the probability that a single measurement will be more than one

unit from the true value is 2(0.1056) or 0.21.

The corresponding calculation for a divergence of 1.5 units shows a probability of

0.06.

(ii) We saw in Section 1.5 that means are less variable than single measurements and,

specifically, that the variability of means is given by the standard error (the standard

deviation for single values divided by the square root of the number of values on

which the mean is based). To allow for the averaging in our calculations, therefore,

we simply replace the standard deviation by the standard error.


The probability that a mean of two measurements will exceed 101 is given by:

77.12/8.0

100101

/

n

xz

Where 𝑥 now represents the mean of two measurements.

P(�̅�> 101) = P(Z> 1.77)= 1 – P(Z < 1.77) = = 1 – 0.9616 = 0.0384

By symmetry, the probability that the mean of two measurement will be less than 99

is also 0.0384. Hence, the probability that the mean of two measurements will be

more than one unit from the true value is 2(0.0384) or 0.08.

The value of making even two measurements instead of one is clear – the probability

of an error of one unit has been reduced from 0.21 to 0.08.

The corresponding probability for the mean of three measurements is 0.03.

1.6.2. (i) A trawl through the website HorsesForum.com suggests that a nailed-on steel

horseshoe will last about 5 to 8 weeks before it needs to be re-set. Assuming the

lifetime of a horseshoe is normally distributed with population mean μ = 7 weeks

and standard deviation σ = 1.25 weeks what is the probability that a nailed-on

horseshoe picked at random from the population will last 9 or more weeks before

needing re-set.

Compute z-value based on x = 9:

𝑧 = 𝑥−𝜇

𝜎=

9−7

1.25= 1.6

Compute probability

𝑃(𝑥 ≥ 9) = 1 − 𝑃(𝑥 < 9) = 1 − 𝑃 ( 𝑧 <𝑥−𝜇

𝜎)

= 1 − 𝑃 (9−7

1.25) = 1 − 𝑃(1.6) = 1 − 0.945 =0.055

The probability the nailed-on horseshoe last 9 or more weeks is 0.055 or

just over a 1-in-20 chance.

(ii) A second website ‘The Chronicle of the Horse’ advises so-called ‘glued-on’

horseshoes, that is shoes that are glued rather than nailed-on, may be used in

cases of laminitis (aka white line disease in USA) where the hoof is unable to

accommodate nails. Gleaning the posts on the website seems to indicate that the

population mean lifetime of glued-on shoes is μ = 5 weeks with standard

deviation σ = 2.0 weeks. Assuming the lifetimes are normally distributed what is

the probability that a glued-on horseshoe picked at random from the population

will last 9 or more weeks before needing re-set.


Compute z-value based on x = 9:

𝑧 = 𝑥−𝜇

𝜎=

9−5

2.0= 2

Compute probability

𝑃(𝑥 ≥ 9) = 1 − 𝑃(𝑥 < 9) = 1 − 𝑃 ( 𝑧 <𝑥−𝜇

𝜎)

= 1 − 𝑃 (9−5

2.0) = 1 − 𝑃(2) = 1 − 0.977 =0.023

The probability the glued-on horseshoe last 9 or more weeks is 0.023 or

just under a 1-in-40 chance.

(iii) Comparing the information from the two websites: What is the probability that a

nailed-on horseshoe will last longer than a glued-on horseshoe, i.e the difference

between the lifetimes of the two types of horseshoe is greater than zero.

We want to know whether the difference between nailed-on and glued-on

horseshoe lifetimes is greater than zero based on population means obtained

by ‘received wisdom’ only.

Compute difference in population means 𝝁

𝑑 = 𝜇𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛 − 𝜇𝐺𝑙𝑢𝑒𝑑−𝑜𝑛 = 7 − 5 = 2.

Compute the std err of difference

𝑆𝐸(𝑑) = 𝑆𝐸(𝑥𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛 − 𝑥𝐺𝑙𝑢𝑒𝑑−𝑜𝑛) = √ 𝜎𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛2 + 𝜎𝐺𝑙𝑢𝑒𝑑−𝑜𝑛

2 = √1.252 + 22 = √5.5625

= 2.36

𝑃(𝑑 > 0) = 1 − 𝑃(𝑑 ≤ 0) = 1 − 𝑃 ( 𝑧 <0−𝑑

𝑆𝐸(𝑑))

= 1 − 𝑃 (0−2

2.36) = 1 − 𝑃(−0.847) = 1 − 0.2 =0.8

The probability that a single random nailed-on horseshoe will last longer

than another single random glued-on horseshoe is 0.2 or a 4-in-5 chance.

(iv) Comparing a sample mean to a population value μ (e.g population mean):

The website data used in part (i) was in fact based on a sample of n=5 with

�̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛 = 7. On the basis of this sample data what is the probability that the

mean lifetime μ for (the population of) nailed-on horseshoes is at least (μ =) 9

weeks.

We want to know how far the ‘actual’ sample mean computed from the

observed data values only is from the value of μ = 9 weeks.


Compute the std err of the mean

Note: because we are dealing with samples and the variance of a sample mean is 𝜎2 /𝑛

𝑆𝐸(�̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛) = √𝜎𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛

2

𝑛𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛

= √1.252

5= 0.559

𝑃(𝜇 > �̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛) = 1 − 𝑃(�̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛 ≤ 𝜇) = 1 − 𝑃 ( 𝑧 <7−9

𝑆𝐸(�̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛))

= 1 − 𝑃 ( 𝑧 <�̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛−𝜇

𝑆𝐸(�̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛)) = 1 − 𝑃 ( 𝑧 <

7−9

0.559)

= 1 − 𝑃 (−2

0.559) = 1 − 𝑃(−3.58) = 1 − 0.998 = 0002

So, the chances that the unknown population mean lifetime of nailed-on

horseshoes lasting 9 or more weeks is about 1-in-5000 based on the

sample.

Note: this is a statement about mean values and not about the sample

values themselves. In this case μ = 9 weeks can be thought of as a

hypothesised value for the unknown population mean.

Chapter 2 will take up this idea of hypothesis testing in more detail.

(v) Exam style question:

Comparing the information based on samples from the two websites: The website

data used in part (i) was in fact based on a sample of n=5 with �̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛 = 7

while in part (ii) the sample size was n=2 and �̅�𝐺𝑙𝑢𝑒𝑑−𝑜𝑛 = 5. On the basis of these

sample sizes what is the probability that a nailed-on horseshoe will last longer

than a glued-on horseshoe, i.e the difference between the means of the lifetimes of

the two types of horseshoe is greater than zero.

We want to know whether the difference between nailed-on and glued-on

horseshoe lifetimes is greater than zero based on ‘actual’ sample means

computed from the observed data values only.

Compute difference in sample means 𝒙

�̅� = �̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛 − �̅�𝐺𝑙𝑢𝑒𝑑−𝑜𝑛 = 7 − 5 = 2.

Compute the std err of this difference


Note: because we are dealing with samples and the variance of a sample mean is 𝜎2 /𝑛 we have to use this quantity in our usual formula for adding

two or more variances as follows

𝑆𝐸(�̅�) = 𝑆𝐸(�̅�𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛 − �̅�𝐺𝑙𝑢𝑒𝑑−𝑜𝑛) = √𝜎𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛

2

𝑛𝑁𝑎𝑖𝑙𝑒𝑑−𝑜𝑛

+𝜎𝐺𝑙𝑢𝑒𝑑−𝑜𝑛

2

𝑛𝐺𝑙𝑢𝑒𝑑−𝑜𝑛

= √1.252

5+

22

2= √0.3125 + 2 = 1.52

𝑃(𝑑 > 0) = 1 − 𝑃(𝑑 ≤ 0) = 1 − 𝑃 ( 𝑧 <𝑑−0

𝑆𝐸(�̅�))

= 1 − 𝑃 (2−0

1.52) = 1 − 𝑃(1.32) = 1 − 0.907 =0.093

The probability that an average nailed-on horseshoe will last longer than

an average glued-on horseshoe is 0.093 or just under a chance of 1-in-10.

So, based on our sample information the chance of an average nailed on

horseshoe lasting longer than an average glued-on shoe compared to a

single random nailed on horseshoe lasting longer than a single random

glued-on shoe has halved.

The distinction between making comparison between individual cases

and comparison based on sample mean cannot be overemphasised.

We gain precision using sample information.

1.6.3. Combining data - Meta-analysis study

(i) In a 3-rd world country in Latin America a study shows that infants born to

families who cook on an open Wood stove have an average birth weight of μ = 2.9

kg with standard deviation 𝜎 = 0.5kg.

What is the probability that an infant born to one of these families is over 3.6kg?

Compute z-value based on x = 3.6:

𝑧 = 𝑥−𝜇

𝜎=

3.6−2.9

0.5= 1.6

Compute probability

𝑃(𝑥 ≥ 3.6) = 1 − 𝑃(𝑥 < 3.6) = 1 − 𝑃 ( 𝑧 <𝑥−𝜇

𝜎)

= 1 − 𝑃 (3.6−2.9

0.5) = 1 − 𝑃(1.4) = 1 − 0.919 =0.081

The probability that the weight of an infant in a family that cooks on a

wooden stove is greater 3.6kg is 0.08 or just under a 1-in-12 chance.


(ii) Cooking in the country is also done an open fires and electric cookers in this

country. The respective average birth weight statistics are μ = 2.8kg and 3.1kg

while the standard deviations are 0.4kg and 1.3kg respectively.

Repeat part (i) for an electric stove.

Compute z-value based on x = 3.6:

𝑧 = 𝑥−𝜇

𝜎=

3.6−3.1

1.3= 0.38

Compute probability

𝑃(𝑥 ≥ 3.6) = 1 − 𝑃(𝑥 < 3.6) = 1 − 𝑃 ( 𝑧 <𝑥−𝜇

𝜎)

= 1 − 𝑃 (3.6−3.1

1.3) = 1 − 𝑃(0.38) = 1 − 0.648 =0.352

The probability that the weight of an infant in a family that cooks on an

electric stove is greater 3.6kg is 0.35 or just over a 1-in-3 chance.

(iii) Combining the information on cooking types: What is the probability that an

infant born in this country will be less than 2.3kg.

So we want to combine the information from 3 studies to see whether a child

born in the country is likely to have a small birth weight based on population

means obtained by ‘received wisdom’ only.

Compute the overall population mean 𝝁

𝜇 = ( 𝜇𝑤𝑜𝑜𝑑 + 𝜇𝑜𝑝𝑒𝑛 + 𝜇𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐)/3 = (2.9 + 2.8 + 3.1)/3 = 2.93

Compute the std err of the overall population mean

𝑆𝐸(𝜇) = √ (𝜎𝑤𝑜𝑜𝑑2 + 𝜎𝑜𝑝𝑒𝑛

2 + 𝜎𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐2 )/3 = √(0.52 + 0.42 + 1.32/3) = √2.1/3 = 0.84

𝑃(𝑥 < 2.3) = 𝑃 ( 𝑧 <𝑥−𝜇

𝑆𝐸(𝜇))

= 𝑃 (2.3−2.93

0.84) = 𝑃(−0.75) =0.227

The probability that the weight of a random infant in the population in

this country is less than 2.3kg is 0.23 or a 1-in-4 chance.

(iv) Combining the information on cooking types based on samples from the three

cooking type studies: The study data used in parts (i) and (ii) was in fact based on

3 samples each of size n=36 (108 in all). Taking the sample mean in each case to

be the population mean (i.e. with �̅� = 𝜇) repeat part (iii) above and comment.

So we want to combine the information from 3 studies to see whether the

average birth weight of child born in the country is likely to be under 2.3kg.

Note the important distinction, here we are looking at the average birth

weight of infant in the population not the weight of any random infant in the

population as in part (iii).


Compute overall sample means 𝒙

This is exactly same as in part (iii) but with different notation

�̅� = ( �̅�𝑤𝑜𝑜𝑑 + �̅�𝑜𝑝𝑒𝑛 + �̅�𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐)/3 = (2.9 + 2.8 + 3.1)/3 = 2.93

Compute the std err of the overall sample mean

As in previous question, because we are dealing with samples and the variance of a sample mean is 𝜎2 /𝑛 we have to use this quantity in our usual

formula for adding two or more variances as follows

𝑆𝐸(�̅�) = √(𝜎𝑤𝑜𝑜𝑑

2

𝑛𝑤𝑜𝑜𝑑

+𝜎𝑜𝑝𝑒𝑛

2

𝑛𝑜𝑝𝑒𝑛

+𝜎𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐

2

𝑛𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐

) /3

= √(0.52

36+

0.42

36+

1.32

36) /3 = √(0.007 + 0.004 + 0.045)/3 = 0.137

𝑃(�̅� < 2.3) = 𝑃 ( 𝑧 <�̅�−𝜇

𝑆𝐸(�̅�))

= 𝑃 (2.3−2.93

0.137) = 𝑃(−4.59) =0.00001

The probability that the average birth weight of an infant in a family in this country is less than 2.3kg is 0.004 or a 1-in-100,000 chance.

So, based on our sample information the chance of a child being born with

an ‘average’ weight below 2.3kg is over 25,000 times less than that of any

random infant in the population being born below 2.3kg.

Clearly, statements about the ‘mean’ are very different to similar

sounding statements about particular cases.

who measures the depth of wrinkles on women’s faces they ... · figure 1.2.1. figure 1.2.1: the...

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