1. As part of a study on the bluntnose minnow (Pimephales notatus) you want to d

Question

1. As part of a study on the bluntnose minnow (Pimephales notatus) you want to determine which factors contribute to how many of these fish live in parts of a stream. So you sampled 80 different sections of a stream and counted the number of minnows per cubic meter, the water temperature (°C), the water depth (meters), and whether or not there was a predator present (yes/no). After confirming all relevant assumptions, you ran a general linear model predicting the density of bluntnose minnows:

b SE

Intercept -4.61 0.45

Water Temp 0.51 0.37

Water Depth 12.33 5.13

Predator (0=No, 1=Yes) -8.71 1.56

a. Calculate the test statistics for each explanatory variable in the model. Which predictors explain a significant amount of variation in the response variable and why? (3pt)

b. Write out the full regression equation for predicting the density of bluntnose minnows in this stream. (1pt)

c. Interpret the estimate for Predator in context. (2pt)

d. Predict the bluntnose minnow density in a part of the stream that is 12°C, 2.5 meters deep, and has a predator present. (1pt)

Explanation / Answer

(A)

Degree of freedom: df=n-4 = 80-4 = 76

We will find the p-value using excel function "ROUND(TDIST(ABS(H3),76,2),4)".

Following table shows the test statistics and p-values:

P-value for Water-temp is greater than 0.05 so we reject the null hypothesis.

b)

The regression model:

y' = -4.61 + 0.51*Water-temp + 12.33 * Water-depth -8.71 * Predator

c)

When Predator is 1 then predicting the density of bluntnose minnows is decreased by 8.71, keeping other things constant.

d)

Water-temp = 12

Water-depth = 2.5

Predator =1

The predicted value is

y' = -4.61 + 0.51*12 + 12.33 *2.5-8.71 * 1 = 23.625

b SE t = b /se P-value Intercept -4.61 0.45 -10.244 0 Water Temp 0.51 0.37 1.378 0.1722 Water Depth 12.33 5.13 2.404 0.0187 Predator(0=No, 1=Yes) -8.71 1.56 -5.583 0

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