As part of a research project, you’re responsible for tracking the length of rainbow trout (Oncorhynchus mykiss) over the growing season. At the beginning of the growing season, 400 fish are stocked in a pond with a surface area of ~ 0.10 Ac and a depth of 3 ft. The pond is fed by a nearby cold spring that's also supplemented by surface runoff. We need to determine and report the average length of the fish every two weeks. How many fish should sample?

Is it practical to collect and measure all 400 fish every two weeks? Why or why not?

No, this isn't practical. Begin by thinking through the logistical issues associate with collecting 400 fish that have been dispersed in ~100,000 gallons of water. (Note that in this system, each fish is afforded almost 33 cubic feet of water!) Now, recognize that every time a fish is handled, there is a increased chance for disease to be introduced into the pond in addition to the increase in general stress levels that result from handling. Since these factors can affect fish growth (what we’re supposed to be studying), care should be taken to minimize fish handling.

This means that we need to sample enough fish to draw statistically relevant conclusions from our data while minimizing the potential adverse impacts to the fish as well as the [personnel] cost associated with measuring all 400 fish every two weeks.

We need to set a few parameters including the level of confidence and margin of error. To start, approximately how long are rainbow trout? -millimeters long? -centimeters long? -meters long? It’s safe to say that we would be measuring fish length in centimeters.

Like many other statistical tests, let’s assume we’re 95% confident that the average length of the population will fall within our confidence interval (margin of error, E).

To find “n”, we need to find or estimate “z” and “s”. Since we know our confidence level, we can find z from standard tables (z = 1.960 at 95% confidence) or from the information presented previously in class. But what about “s”?! We could perform a few preliminary studies or make a reasonable estimate. Let’s estimate! Standard deviation is a measure of the amount of variation we expect to see in our measurements. If we’re measuring fish that are 20 to 40 cm long, a standard deviation of 2 cm might be reasonable. Since this is just a calculation, we can also run our estimate of “n” at different standard deviations. Now, what about the margin of error? Again, we’ve got to specify a reasonable value. Perhaps E = 2 cm is worth a try? Using these conditions, we should sample 4 fish.

Let’s see if this works. Consider data on the lengths of 393 fish. Let's select four fish at random.

Using a random number generator (www.random.org), we'll need to select four fish at random and calculate their mean length.

So, for an average fish length of 24.5 cm and E = 2 cm, we predict (with 95% confidence) that the mean length of the population will fall between 22.5 and 26.5 cm. The population mean is 25.8 cm, which falls between 22.5 and 26.5 cm.

We should take a minute to remember that an underlying assumption we've made is that our data are normally distributed. Are they, really? Let's take a look at a frequency histogram of our data.

One way we can test for normality is by using the coefficient of variation (COV). For the population described in the frequency histogram, the population mean and standard deviation are 25.7 cm and 2.5 cm, respectively. This corresponds to a COV of 0.1 (0.097), so our data can be considered to be normally distributed.

It would be good to see how changes to our criteria can impact our prediction of n. Let's consider margins of error of 1 and 2 cm and levels of confidence of 90 and 95%:

Before we're completely finished, we need to look at our table of reduced data and consider the physical significance of our findings. I recommend revising the table to sample whole numbers of fish, after all, it's difficult to sample a fraction of a fish!