Question

6. (12 marks) A biologist wishes to examine the collision avoidance behaviour of bullfrogs housed in experimental boxes. He placed an object in the upper visual field of each frog and then moved the object progressively closer until the frog jumped to avoid collision. He is interested in how the angle (in degrees) of a frog's escape direction depends on its angular velocity (degrees/second). The data (for only the 69 frogs whose escape direction was less than 50 degrees) are available in the file frog50.csv. a. b. (2 marks) Give the equation of the fitted regression line. (2 marks) Create a plot of direction vs. velocity, including the fitted regression line. (2 marks) Interpret the slope of the fitted regression line in the language of the problem. (2 marks) State the R2 value and its interpretation (in the language of the problem). (2 marks) What is the best prediction of the velocity of a jump with a direction of 25 degrees? (2 marks) Can you provide an accurate prediction of the velocity of a jump with a direction of 60 degrees? If so, provide this prediction. If not, explain why. e. f.

Answer 1

Here,

Dependent Variable (response) = Direction

Independent Variable ( predictor) = Velocity

a. Linear regression model:

$Direction = 1.0490 + 0.1535*Velocity$

b. Plot of direction v/s velocity, along with regression line:

Scatterplot of Direction vis Velocity 40 30 Direction 20 10 0 0 50 100 150 200 250 300 Velocity

c. Interpreting the slope of the regression line:

Our value of slope obtained = 0.1535

We interpret the slope as the rate of change in the angular escape direction of the frog as its angular velocity changes.

In other words for unit increase in angular velocity , the angular escape velocity will increases by 0.1535 points.

d. Finding R² and interpreting it:

R²=0.9245,

Now,  $R^{2}= \frac{Explained Variation}{Total Variation}$

so , from our finding , we can say 92.45% variability in the angular escape velocity of the bullfrogs has been explained by our Linear model.

e. Prediction of Velocity with Direction =25 degrees

Now, we construct another linear model with response= Velocity and predictor = direction

we get

  $Velocity= -0.3787 + 6.0235 *Direction$

we put Direction = 25

we get Velocity= -0.3787 + (6.0235*25) = 150.2088

f. We can surely, obtain a prediction for direction = 60 degree

by using our fitted equation

   $Velocity= -0.3787 + 6.0235 *Direction$

required direction = 361.0313

But this is not an accurate prediction as we have our fitted regression line taking a limited number of samples into consideration where the direction was "<50 degrees" for all the samples. But our given direction = "60 degrees". Had we taken samples with direction ">50 degrees" into consideration, the regression line would have been different, and our prediction would be much reliable and accurate.

Hence our required prediction will not be accurate based on our given sample.

We have used r programming for the above analysis. The r-code is provided below:

data=read.csv("frog50.csv",sep=",",header=TRUE)
attach(data)

LinMod=lm(Direction~Velocity)#finding regression line for direction = response, velocity= predictor

#plotting scatterplot with regression line
plot(Velocity,Direction,main="Scatterplot of Direction v/s Velocity",pch = 16, cex = 1, col = "blue")
abline(LinMod,col="red",cex=3)

summary(LinMod)#finding R2

LinMod1=lm(Velocity~Direction)#finding regression line for direction = predictor, velocity= response