Question

1. Two streams (Brimer and Standifer creeks) located in the same watershed are similar in size/shape and most habitat conditions (e.g., temperature, dissolved oxygen, channel substrate); however, they exhibit different pH values. Brimer Creek has a mean pH of 6.1, whereas Standifer Creek has a pH of 5.6. A stream ecologist wishes to determine whether the pH has an influence on the local distributions of benthic macroinvertebrates (annelids, crustaceans, insects) in the two streams. Two study sites were established (one in each stream) that are in close proximity to each other (~50 m apart, separated by a ridge) and located near the mouths of their respective streams. Benthic invertebrates were collected at the two sites using a standardized kick-sampling technique (equal sampling times and areas). Invertebrates were counted in samples from the two sites. The data are summarized below. Number of invertebrates per sample: Brimer Creek, n = 1373 individuals Standifer Creek, n = 955 individuals Perform a G-test for goodness of fit (α = 0.05) to test the ecologist’s hypothesis.

Answer 1

To test whether pH of the creek has an influence on the local distributions of benthic macroinvertebrates,

i.e. To test: H₀ : pH does not influence the local distributions of benthic macroinvertebrates

Vs H_a : pH does influence the local distributions of benthic macroinvertebrates

Using SPSS,

Enter the data as:

Creek Frequency

Brimer 1373

Standifer 955

Since, the frequencies are already obtained for each category, entering data as above, alone won't do it.We need to weight the cases.

Next, to test the goodness of fit,

The output is obtained as:

Since the p value obtained for the test 0.000<0.05, we may reject the null hypothesis at 5% level of significance.

Hence, it may be concluded that the pH does influence the local distributions of benthic macroinvertebrates.

Using excel,

Formula for the test statistic

Here, H_{0 can be expressed as p = 0.5, where, p = proportion of invertebrates.}

_{p value = 0.000< 0.05, and we arrive at the same conclusion as above.}