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266 Statistics for Environmental Science and Management, Second Edition

assumptions but use the data more efficiently. As to whether or not it is reasonable to assume a normal distribution, this depends very much on the circumstances. Some types of data are almost always normal or nearly normal. For example, measurements of lengths of body parts for animals and plants fall in this category. In other cases, a simple data transformation, such as taking logarithms, will change a nonnormal distribution into a normal one.

The following is a brief summary of some of the most commonly used tests of significance. The book by Kanji (1999) covers just about any test that is likely to be used, with example calculations. However, in practice, the easiest way to do these things is by using one of the standard statistical computer packages.

A1.4.1  The One-Sample t-Test

The purpose of this test is to see whether the mean of a random sample of n values from a population is significantly different from some hypothetical value μ. The sample mean and standard deviation (x and s) are used to calculate the statistic

t = (x − μ)/(s/√n)

As noted in Section A1.3, if the population mean is really μ, then this follows the t-distribution with n − 1 df. Hence the significance of the sample result can be tested by seeing whether the observed statistic is a reasonable value from this distribution. This test requires the assumption that the population being sampled is normally distributed.

A1.4.2  The One-Sample Chi-Squared Test

This test is used to see whether a set of sample counts is in reasonable agreement with the frequencies expected on the basis of some hypothesis. The test statistic is

χ2 =∑(O −E)2/E

where O is an observed frequency, E is an expected frequency, and the summation is over all such frequencies. If the differences between the observed and expected frequencies are merely due to sampling errors, then the test statistic will be approximately a random value from the chi-squared distribution with n − 1 df. A significantly large value for the test statistic in comparison with the chi-squared distribution is evidence that the basis for calculating the expected frequencies is not correct. It is important to remember that the chi-squared test is only valid with count data. It is, for example, quite wrong to use this test with percentages. Also, the use of the chi-squared distribution requires that the expected frequencies are not too small, with most being at least five. For more details about this test, see Kanji (1999, test 37).

A1.4.3  The Contingency-Table Chi-Squared Test

A contingency table is obtained when N observations are classified on the basis of two different criteria, and a count is made of the number of observations Oij that fall into the ith class for the first criterion of classification (row i of the table) and the jth class for the second criterion of classification (column j of the table). This is done for all values of i and j. For example, a 2 × 3 contingency table is produced if 100 streams are classified as being either in the north or the south of a region (the two row categories) and as having low, medium, or high pollution (the three column categories).

The contingency-table chi-squared test is designed to see whether there is any evidence that the two classifications are related, i.e., whether the probability of a sample unit being in class i for the first classification varies according to which of the classes the unit is in for the second classification. On the null hypothesis that the classifications are unrelated, it can be shown that the expected count for the cell in row i and column j of the table is Eij = RiCj/N, where Ri is the total observed count in row i and Cj is the total observed count in column j. The test statistic is then

χ2 =∑i ∑j (Oij −Eij )2/Eij

where the double summation means adding over all the rows and columns in the table. If the null hypothesis of independent classifications is true, then this statistic will be approximately a random value from the chi-squared distribution with (r − 1)(c − 1) df, where r is the number of rows and c is the number of columns in the table. A significantly large value for the test statistic in comparison with the chi-squared distribution is evidence against the null hypothesis. As for the one-sample chi-squared test, the expected frequencies should not be too small when this test is used. Preferably, they should all be five or more. For more details about this test, see Kanji (1999, test 44).

A1.4.4  The Paired t-Test

This is used when the data are naturally paired. The null hypothesis being tested is that the differences between paired values have a particular mean value μd, which is often zero. The test statistic is

t = (d − μd)/(sd/√n)

where d is the mean, and sd is the standard deviation of the sample differences. This statistic is compared with the t-distribution with n − 1 df. The test requires the assumption that the paired differences are normally distributed. Essentially, it is just the one-sample t-test calculated on differences, with a null hypothesis mean of zero.

268 Statistics for Environmental Science and Management, Second Edition

A1.4.5  The Two-Sample t-Test

This is designed to test whether the difference between the means of two independent samples are significantly different. The test statistic is

t = (x1 − x2)/[sp√(1/n1 + 1/n2)]

where sample 1, of size n1, has a of mean x1, and sample 2, of size n2, has a mean x2. Also,

sp2 = [(n1 − 1)s12 + (n2 − 1)s22]/(n1 + n2 − 2)

is an estimate of variance based on pooling the variances, s12 and s22, for the two samples. To assess the significance of the sample mean difference, the calculated t-value is compared with the critical values in the t-table with n1 + n2 − 2 df. This test assumes that the two distributions being sampled are normal with the same standard deviation. For more details, see Kanji (1999, test 8). Test 9 in the same book also contains a variation on this test for situations where the distributions do not have the same standard deviation.

A1.4.6  The Wilcoxon Signed-Ranks Test

This is a nonparametric alternative to the paired t-test. It is based on ranking the differences between two measurements and using sums of positive and negative ranks as test statistics. This test does not require a normal distribution for the differences. For more details see Kanji (1999, test 48).

A1.4.7  The Mann-Whitney U-Test

This is a nonparametric alternative to the two-sample independent t-test, which tests whether the two samples come from distributions with the same mean on the assumption that the distributions have the same shape. To use this test, it is only necessary to be able to rank the data in order for the two samples combined, and no assumption of normality is required. For more details see Kanji (1999, test 52).

Example A1.1:  Testing the Mean Level of TcCB at a Site

As an example of a test of significance, consider the data in Table A1.1 on measurements of 1,2,3,4-tetrachlorobenzene (TcCB) from n = 47 different locations at a particular site. Suppose that a mean level of 0.5 is considered to be acceptable, and the question is whether the mean at this site is significantly higher than 0.5. A one-sample t-test is used to answer this question.

Following the framework of steps 1–6 defined in Section A1.4, the following results are obtained:

1.The null hypothesis to be tested is that μ = 0.5, where μ is the mean level of TcCB over the entire site.

2.The alternative of interest to the null hypothesis that needs to be detected is that μ exceeds 0.5.

3.The test statistic used is t = (x − 0.5)/(s/√n) from equation (A1.4) because the data appear to follow a distribution that is reasonably close to normal (Figure A1.2), and this statistic measures the difference between the sample mean and the hypothetical mean of 0.5.

4.The distribution of the test statistic is a t-distribution with 46 df if the null hypothesis is true.

5.For the observed data, the mean is x = 0.599 and the standard deviation is s = 0.284. The test statistic is therefore

t = (0.599 − 0.5)/(0.284/√47) = 2.39

6.The presence of only positive values for the test statistic indicates that the mean of the site is higher than 0.5. Hence it is necessary to find the probability, p, of obtaining a mean as large as, or larger than, 2.39 for the t-distribution with 46 df. To this end, Table A2.2 in Appendix 2 can be consulted. This does not have a row for 46 df, but it does show that, with 40 df, the probability of a value as large as or larger than 2.39 is between 0.025 and 0.01, and that with 60 df the probability is 0.01. It follows that the probability of a value as large as or larger than 2.39, with 46 df, is about p = 0.02.

7.As the p-value is less than 0.05, it is significant at the 5% level, giving some evidence against the null hypothesis. It appears that the mean for the site is higher than 0.5.

These days, probabilities for the t-distribution are provided in spreadsheet programs. It is therefore easy enough to determine the p-value accurately for a test like this, although that is not really necessary. The exact probability of a value from the t-distribution with 46 df equaling or exceeding 2.39 is, in fact, p = 0.0105.

A1.5  Confidence Intervals

Confidence limits for a parameter of a distribution give a range within which the parameter is expected to lie. For example, a 90% confidence limit for a distribution mean defines a range, which is called a confidence interval, within which the mean is expected to lie 90% of the time, in the sense that if many such intervals are calculated, then about 90% of them will contain the true value of the parameter.

As an example of what this means, suppose that a random sample of size 20 is taken from a normal distribution with an unknown mean μ. Then it is known that

270 Statistics for Environmental Science and Management, Second Edition

t = (x − μ)/(s/√n)

will be a random value from the t-distribution with 19 df. For this distribution, Table A2.2 in Appendix 2 shows that

Prob(t > 1.729) = 0.05

and, as the distribution is symmetrical about zero, it is also true that

Prob(t < −1.729) = 0.05.

Hence, if it is asserted that

then this statement will be true with probability 1 − 2(0.05) = 0.9. Rearranging the left-hand side of this equation shows that

−1.729 < (x − μ)/(s/√n)

is equivalent to

μ < x + 1.729(s/√n).

Similarly, rearranging the right-hand side shows that it is equivalent to

x − 1.729(s/√n) < μ.

Thus equation (A1.7) is exactly equivalent to

Since equation (A1.7) is true with probability 0.9, equation (A1.8) must be true with the same probability. In this sense, equation (A1.8) gives a 90% confidence interval for the true population mean.

This argument can easily be generalized for a random sample of size n from a normal distribution to give a 100(1 − α)% confidence interval for the mean of the distribution of the form

where tα/2,n−1 is the value that is exceeded with probability α/2 by a random value from the t-distribution with n − 1 df.

The confidence interval in equation (A1.9) is one of many that can be derived for a parameter of a distribution, using a similar type of argument.

A confidence interval is a range within which a parameter will lie with a certain defined probability.

Example A1.2:  A Confidence Interval for the Mean TcCB at a Site

Considering again the data in Table A1.1, suppose that there is interest in determining a 95% confidence interval for the mean level of 1,2,3,4-tetra- chlorobenzene at the site in question. The mean and standard deviation for the sample of size n = 47 are x = 0.599 and s = 0.284, respectively, and the critical value of the t-distribution with 46 df that is exceeded with

probability α/2 = 0.025 is t0.025,46 = 2.01 from Table A2.2 in Appendix 2. Hence the 95% confidence interval given by equation (A1.9) is

0.599 − 2.01(0.284/√47) < μ < 0.599 + 2.01(0.284/√47)

i.e.,

0.516 < μ < 0.682

A1.6  Covariance and Correlation

Suppose that a random sample of size n from a large population of items is taken, and two variables X and Y are measured on each of the sampled items. Let the values of X and Y on the ith sampled item be xi and yi, respectively.

One sample statistic that is used to measure the relationship between X and Y is the covariance

i=1

where x and y are the sample means for X and Y, respectively. If large values of X tend to occur with large values of Y, then cxy will be positive. Conversely, if large values of X tend to occur with small values of Y, then cxy will be negative.

Covariances as defined by equation (A1.10) are used in many statistical procedures, but they have a disadvantage in that there is no simple way to judge whether a particular value for cxy indicates that X and Y are closely related. Therefore, for many purposes, the covariance is adjusted so that it becomes the covariance between X and Y after both variables have been scaled to have sample standard deviations of exactly 1. The resulting statistic is then referred to as the sample correlation, or sometimes as Pearson’s correlation coefficient, which can be shown to be the same as