# Statistics for Environmental Engineers

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The hypothesis test is made using:

t =

-1.842

statistic — E(statistic) JV (statistic)

y-n = 7.51-8.0 sy    0.266

The null hypothesis will be rejected if the computed t is less than the value of the lower tail t statistic having probability of a = 0.05. The value of t with a = 0.05 and v = 26 degrees of freedom obtained from tables is tv=26 a=0.05 = -1.706 . The computed value of t = -1.853 is smaller than the table value of -1.706. The decision is to reject the null hypothesis in favor of the alternate hypothesis.

Examples 2.9 and 2.10 are outwardly different, but mathematically and statistically equivalent. In Example

2.9, the experimenter assumes the population parameter to be known and asks whether the sample data can be construed to represent the population. In Example 2.10, the experimenter assumes the sample data are representative and asks whether the assumed population value is reasonably supported by the data. In practice, the experimental context will usually suggest one approach as the more comfortable interpretation.

Example 2.10 illustrated a one-sided hypothesis test. It evaluated the hypothesis that the sample mean was truly to one side of 8.0. This particular example was interested in the mean being below the true value. A two-sided hypothesis test would consider the statistical plausibility of both the positive and negative deviations from the mean.

Use the nitrate data to test the null hypothesis that Ho: n = 8.0 and Ha: n Ф 8.0. Here the alternate hypothesis considers deviations on both the positive and negative sides of the population mean, which makes this a two-sided test. Both the lower and upper tail areas of the t reference distribution must be used. Because of symmetry, these tail areas are equal. For a test at the a =

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