Steve M., Darby D.M., Geostatistics Explained - An Introductory Guide for Earth Scientists

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50 were found in total and there were 5 slabs) would be expected on each

slab. Unfortunately, these data are not suitable for a goodness-of-ﬁt test

because the ﬁve conditions are neither mutually exclusive nor contingent

categories within a sample. This is clear if you consider that any dinosaur

could have walked on multiple types of sediments, or that it could have

chosen to walk on only one. The numbers in each slab are actually single

replicates of ratio scale data.

To avoid the pitfall of confusing categories and samples, you need to ask

yourself “Do I have data for categories that are mutually exclusive and

contingent within each sample, or are my ‘categories’ really separate inde-

pendent samples?”

18.7 Recommended tests for categorical data

Several tests have been developed for data that are frequencies in

mutually exclusive and contingent categories. The following are broad

recommendations.

When comparing the frequencies in two or more categories within a

single sample to their expected proportions, Yates’ corrected chi-square

can be used where no more than 20% of expected frequencies are less

than ﬁve. A randomization test can be used for any sized sample.

For 2 × 2 contingency tables the Fisher Exact Test will give an unbiased

probability and is available in most statistical packages.

For contingency tables with more than two rows and columns, the chi-

square G test or can be used if no more than 20% of expected frequencies are

less than ﬁve. A randomization test will give an unbiased probability for any

sized sample.

18.8 Comparing proportions among two or more related

samples of nominal scale data

If you have measured the same variable more than once on each sampling

or experimental unit, then the samples are not independent and need to

be analyzed using a test for related samples. Table 18.8 gives data for the

accretion and growth of 12 beaches on a glacial spit oﬀ the coast of

Massachusetts. Surveys and photographs were used to document the size

of beaches in two consecutive years. In the ﬁrst year, 11 of the 12 beaches

18.8 Two or more related samples 243

were accreting sand, adding hundreds of yards of sand to the coastline each

year. Over the intervening winter several new jetties were constructed to

control and funnel the ﬂow of water and sand, in an attempt to maintain the

course of the nearby river channel and improve its navigability for shipping.

By the following spring, deposition patterns had dramatically changed: only

ﬁve beaches continued to accrete sand and the remaining seven were

eroding. The null hypothesis was that the proportion of accreting/eroding

beaches would be unaﬀected by the jetties, while the alternate hypothesis

was that the proportion of accreting/eroding beaches would be a ﬀ ected.

These are two related samples so it is not appropriate to analyze them with

a test that compares two independent ones.

The McNemar test for the signiﬁcance of changes compares two related

samples of nominal scale data in two categories. The data in Table 18.8 are

summarized in a 2 × 2 table giving the number of individuals in all four

possible combinations of categories and samples. These are (a) accreting

before and after the jetty construction, (b) accreting before and eroding

after, (c) eroding before and accreting after and (d) eroding before and

eroding after (Table 18.9).

Table 18.8 The status of 12 ocean beaches, as measured in August

of two successive years, preceding and following construction of

jetties at the mouth of a nearby river. These two samples are not

independent because they contain the same 12 beaches.

Beach Year 1 Year 2

1 Accreting Accreting

2 Accreting Accreting

3 Accreting Eroding

4 Accreting Accreting

5 Accreting Eroding

6 Eroding Eroding

7 Accreting Accreting

8 Accreting Eroding

9 Accreting Accreting

10 Accreting Eroding

11 Accreting Eroding

12 Accreting Eroding

244 Non-parametric tests for nominal scale data

The null hypothesis predicts that there will be no diﬀerence in the

proportions of eroding beaches between the two samples, while the alternate

predicts there will be a diﬀerence. Therefore, under the null hypothesis, the

beaches that did change status (combinations (b) and (c)) would be

expected to include equal numbers that changed from eroding to accreting

and from accreting to eroding, so you would expect cells (b) and (c) of

Table 18.9 to contain equal frequencies. If, however, the proportion of

eroding beaches diﬀered before and after construction of the jetties, then

the frequencies in these two cells would be expected to be unequal.

In this example, six beaches changed status, so three would be expected to

change from eroding to accreting and vice versa. The McNemar test ignores

categories (a) and (d) where no change has occurred and compares the

observed and expected frequencies in cells (b) and (c) using a goodness-of-

ﬁt test (e.g. the chi-square, exact or randomization tests for two mutually

exclusive categories discussed earlier in this chapter, or the exact probability

calculated from the binomial distribution discussed in Chapter 6). If there is

a statistically signiﬁcant diﬀerence between the proportions in these two

categories it indicates a change between the two samples.

For three or more related samples of nominal scale data in two categories,

the Cochran Q test is an extension of the McNemar test. These tests are also

included in most statistical packages.

18.9 Questions

(1) The ratio of left-handed to right-handed people in the human pop-

ulation is about 1 : 9. When one of us (SM) was i n ﬁrs t year of

Table 18.9 The status of 12 beaches tested before and after

construction of jetties at the mouth of a nearby river. Two cells

show the beaches that changed; these are (b) from accreting to

eroding and (c) from eroding to accreting. Note that in this

example there are no beaches in the second category.

After

Before Accreting Eroding

Accreting (a) 5 (b) 6

Eroding (c) 0 (d) 1

18.9 Questions 245

university, he was in a tutorial group of 14 that contained 13 left-handers

and one right-hander. Students had been assigned to tutorial groups at

random. Is this departure from the expected proportion of left-handers

signiﬁcant? What could you conclude about this occurrence?

(2) To help understand how the chi-square statistic is related to depar-

tures from expected proportions, it is u seful to use a statistical package

to compare a sample of contrived data to the expected proportions

for a population. Use expected proportions of 1:1 (e.g. left- and right-

hand forams) and a sample size of 100. (a) Initially set the sample

numbers at 50:50 left to right and run a single-sample chi-square test.

What would you expect the value of chi-square to be? (b) Now, change

the sample proportions to 45 and 55 and rerun the test. Repeat this

for increasing departures from the expected ratio (e.g. 40 and 60,

35 and 65, 30 and 70). What happens to the value of chi-square?

What happens to the probability?

(3) A team of paleontologists searched two sandstone outcrops for three

hours each and found 8 trilobites in one outcrop and 22 in the other.

They compared the numbers using a chi-square test and an expected

ratio of 15:15. Was this test appropriate? Please discuss.

(4) Do you think the sampling described in Section 18.8 was well designed?

Might there be other possible reasons for the change in status of

the 12 beaches from Year 1 to Year 2? How might you improve the

experimental design to take other possible inﬂuences into account?

246 Non-parametric tests for nominal scale data

19 Non-parametric tests for ratio,

interval or ordinal scale data

19.1 Introduction

This chapter describes some non-parametric tests for ratio, interval and

ordinal scale univariate data. These tests do not use the predictable distribution

of sample means, which is the basis of most parametric tests, to infer whether

samples are from the same population. Consequently non-parametric tests are

generally not as powerful as their parametric equivalents, but if the data are

grossly non-normal and cannot be satisfactorily improved by transformation,

it is necessary to use one of these tests.

Non-parametric tests are often called “distribution free tests” but most

nevertheless assume that the samples being analyzed are from populations

with the same distribution. Therefore, most non-parametric tests should

not be used where there are gross diﬀerences in distribution (including

the variance) among samples. The general rule that the ratio of the largest

to smallest sample variance should not exceed 4 : 1 discussed in Chapter 13

also applies to non-parametric tests.

Many non-parametric tests for ratio, interval or ordinal data calculate a

statistic from a comparison of two or more samples and work in the

following way.

First, the raw data are converted to ranks. For example, the lowest value is

assigned the rank of “1”, the next highest “2” etc. This transforms the data to

an ordinal scale (see Chapter 3) with the ranks indicating only their relative

order. Under the null hypothesis that the samples are from the same

population you would expect a similar range of ranks within each, with

diﬀerences among samples occurring only by chance.

Second, a statistic that reﬂects any diﬀerences in the ranks among

samples is calculated and its value compared to the expected distribution

of this statistic when samples have been taken from the same population. If

the calculated value falls within the range generated by the most extreme 5%

247

of departures from the null hypothesis the result is considered statistically

signiﬁcant. Most statistical packages give the value of the test statistic,

together with the probability of that outcome. Randomization and exact

tests can also be used to compare two or more samples of ratio, interval or

ordinal scale data and are described in this chapter.

19.2 A non-parametric comparison between one sample

and an expected distribution

The Kolmogorov–Smirnov one-sample test can be used to compare the

distribution of a single sample of continuous data to an expected or known

distribution. Here is an example. Limestone is relatively soft, easily worked,

and has an attractive appearance, so it is often cut into blocks and used as

building stone and tombstones. Unfortunately, it is composed mostly of

calcium carbonate (calcite, or CaCO

) and therefore vulnerable to dissolu-

tion by rainwater, which is naturally slightly acidic but can be extremely so

(hence the term “acid rain”) in polluted areas.

Tombstones in a graveyard east of Detroit, Michigan have a bimodal

distribution of the amount of weathering: some stones start to show signs of

partial dissolution after only 10 years, while others take much longer. All of

these tombstones come from a nearby quarry that has been worked for more

than 200 years. The amount of weathering in the tombstones can be quantiﬁed

by examining the lead that was initially inserted into the carved lettering and

then polished ﬂush with the surface of the stone: acid rain will slowly dissolve

the limestone but not the lead. Thus, a “lead lettering index” (LLI) that is the

absolute diﬀerence in height between the lead letters and the eroded surface of

the stone around them was created to quantify the weathering of a tombstone.

Generations of geology students have visited the town’s graveyards and

amassed LLI data along with age of the limestone tombstones. These data

have been obtained from such a large number of stones over several decades

that they can be assumed to be the distribution for the population.

The owner of the local quarry where all the tombstones originated noticed

that some areas of the quarry walls, containing large quantities of fossils,

showed very little weathering. The owner wondered if stone from these areas

might produce more lasting tombstones, so they commissioned a geologist to

test the hypothesis that tombstones containing the most fossils were more

resistant to dissolution. The geologist visited a local graveyard and measured

248 Further non-parametric tests

the LLIs of 36 tombstones, each of which was 100 years old and contained

prominent fossils. This distribution was compared to the known distribution

of LLIs for the (much larger) population of 100 year old tombstones measured

by generations of students. Both sets of data are summarized in Table 19.1.

If you make a preliminary inspection of these data by drawing a histogram

you will ﬁnd the LLI distribution is bimodal and clearly not appropriate for

analysis using a parametric test such as a single-sample t test. A non-

parametric test is needed.

For a Kolmogorov–Smirnov one-sample test you need to subdivide the

range (here the LLI for the tombstones) into several intervals of equal width,

and count the number of cases within each. This is the same procedure used

for drawing a frequency histogram described in Section 3.3.2 and you

should follow those guidelines.

Table 19.1 A worked example of a Kolmogorov–Smirnov one-sample test. The LLIs

of 36 fossil-rich gravestones that are 100 years old are summarized as frequencies

that are converted to proportions of the sample. These are expressed as cumulative

proportions. First each expected cumulative proportion (from the known or

expected distribution of the population) is subtracted from the observed cumulative

proportion and expressed as the absolute diﬀerence D. The greatest value of the

statistic is identiﬁed. In this case, it is 0.1217 (shown in bold in the far right-hand

column). If the probability of obtaining this or a more extreme value of the D statistic

is less than 5%, then the two distributions are considered signiﬁcantly diﬀerent.

LLI (lead

lettering

index) in

Observed

numbers

in each

category

Expected

cumulative

proportions

(from the

population)

Absolute

diﬀerence D

(observed

minus

expected

D ¼ F



<2.0 5 0.1389 0.1389 0.1045 0.0344

2.0–2.49 10 0.2778 0.4167 0.3943 0.0224

2.5–2.99 4 0.1111 0.5278 0.5623 0.0345

3.0–3.49 2 0.0556 0.5833 0.6236 0.0403

3.5–3.99 8 0.2222 0.8056 0.6839 0.1217

4.0–4.49 5 0.1389 0.9444 0.9325 0.0119

≥4.50 2 0.0556 1.0000 1.0000 0.0000

Total 36 1.0000

19.2 One sample and a distribution 249

A worked example is given in Table 19.1. First, the number of cases in

each interval is converted to relative frequencies (and thus their proportions

of the sample). Second, the relative frequencies are progressively added

together to give the cumulative relative frequency (i.e. the cumulative

proportion). Here too, this procedure is the same as drawing a cumulative

frequency graph (Section 3.3.3).

The cumulative proportions for the sample have to be compared to the

cumulative proportions of the known distribution for the population. To do

this you calculate the absolute value of the diﬀerence between the cumu-

lative observed relative frequency (F

) and cumulative expected relative

frequency (

) for each interval, which will always be positive:

D ¼jF



j (19:1)

If the observed and expected proportions in each interval are the same the

value of D will always be zero. As the discrepancy between the observed and

expected proportions increases, the value of D will increase.

The largest value is found and called the D statistic. For the worked

example in Table 19.1 the D statistic is 0.1217. Statistical packages generate

the cumulative frequency distributions, calculate the D statistic and give the

probability.

The most common use of the Kolmogorov–Smirnov one-sample test is to

compare the distribution of one sample with an expected distribution such

as the normal curve, and most statistical packages include this option.

19.3 Non-parametric comparisons between two

independent samples

19.3.1 The Mann–Whitney test

The Mann–Whitney test is used to compare two independent samples.

Table 19.2 gives data for the length, in cm, of Devonian-age Paracyclas

clams in each of two outcrops. The null hypothesis is that the two samples

have come from the same population.

First, the values are ranked over both samples as shown in Table 19.2.The

smallest value is given the rank of 1, the next largest the rank of 2 etc., so the

largest will have the rank of n

+ n

(which is the sum of the number of cases

in each sample). For the data in Table 19.2, the largest possible rank is 9.

250 Further non-parametric tests

If two or more values are the same (that is, they are tied), each is given the

average of the ranks assigned to that many values. For example, if the data in

Table 19.2 contained two 4 cm long clams and these were the smallest, each

would be given the average of ranks 1 and 2, which is 1.5.

If most of the clams were longer in one outcrop than the other, the ranks

would diﬀer between these two samples. In contrast, if the clams were of a

similar length in both outcrops the ranks within each sample would also be

similar.

The ranks are summed separately for each sample (these are R

and R

Table 19.2) and the two Mann–Whitney statistics U and U′ calculated:

U ¼ n

 n

ðn

þ 1Þ

 R

(19:2)

and

¼ n

 n

ðn

þ 1Þ

 R

(19:3)

where n

and n

are the size of each sample.

These equations may appear complex, but are easily explained by sepa-

rating them into three components as shown for U in Equation (19.2).

U ¼ n

 n

ðn

þ 1Þ

 R

ðcomponent AÞðcomponent BÞðcomponent CÞ

(19:4)

Table 19.2 The length, in centimeters, of Devonian-age Paracyclas clams

in each of two outcrops. Ranks are shown in the two right-hand columns,

together with the rank sums (R

and R

) for each treatment.

Length in

outcrop A

Length in

outcrop B

Rank for

outcrop A

Rank for

outcrop B

24 22 7 6

41 6 9 2

17 11 5 3

38 15 8 4

=4 n

=5 R

=29 R

=16

19.3 Two independent samples 251

Component A will increase with the size of both samples. Component B

will only increase as the size of sample 1 increases. In contrast, component C

will be aﬀected by the way the ranks are distributed between the two samples.

A lot of low ranks in sample 1 will give a relatively small value of R

and vice

versa. Therefore, since U is calculated by taking component C away from the

sum of components A and B, it will be large compared to U′ when sample 1

contains mainly low ranks. In contrast, if sample 1 contains mainly high

ranks, the value of U will be small compared to U′.Finally,ifbothsamples

contain similar ranks then neither U nor U′ will be relatively large or small.

When both samples are from the same population, most values of U and

U′ will be similar but diﬀerences between them will occur by chance and the

most extreme 5% of discrepancies will give values of U or U′ that will exceed

a critical value. For a two-tailed test, if either of the U statistics exceeds the

critical value then the probability the samples are from the same population

is less than 5%.

19.3.2 Randomization tests for two independent samples

Another way of comparing two independent samples, without assuming they

are from a normal distribution, is to use a randomization test. These tests were

ﬁrst discussed in relation to samples of categorical data in Chapter 18.

If two independent samples are taken from the same population, then the

values within each should diﬀer only by chance. A randomization test takes the

combined set of ranks from both sampl es (a group of size n

+ n

), repeatedly

samplesitatrandomandassignstherankstotwogroupsofsizen

and n

The simulated sampling is iterated several thousand times and used to

generate the expected distribution of U and U′ from the data set and

therefore identify the most extreme 5% of departures from the outcome

expected under the null hypothesis. Finally, the U statistics for the actual

outcome are compared to these distributions and if the probability is less

than 5% it is statistically signiﬁcant (Figure 19.1).

19.3.3 Exact tests for two independent samples

Data for two samples can also be analyzed by tests that calculate the exact

probability, and work in a very similar way to the Fisher Exact Probability

Test described in Chapter 18.

252 Further non-parametric tests