Question

Squirrels and their food supply ($3.2$) Animal species produce more offspring when their supply of food goes up. Some animals appear able to anticipate unusual food abundance. Red squirrels eat seeds from pinecones, a food source that sometimes has very large crops. Researchers collected data on an index of the abundance of pinecones and the average number of offspring per female over $16$years.$4$Computer output from a least-squares $4$regression on these data and a residual plot are shown here

a. Is a linear model appropriate for these data? Explain.

b. Give the equation for the least-squares regression line. Define any variables you use.

c. Interpret the values of $r^2$and $s$in context.

Short answer

(a) Yes, the linear model is absolutely appropriate for the data.

(b) The equation for the least square regression line is $\hat{y}=1.4146+0.4399x$.

(c) The$r^2$means variations in variables and$s$means average error.

Step-by-step solution

Part (a) Step 1: Given information

We need to find whether the linear model is appropriate for the given data i.e.

Part (a) Step 2: Explanation

We know that

The residual plot shows no discernible trend, and the residuals appear to be centered around$0$, indicating that the linear model is a good fit for the data.

Part (b) Step 1: Given information 

We need to find out the equation for the least-squares regression line.

Part (b) Step 2: Explanation

We know that

The general least square eq. is given as

$\hat{y}=a+bx$

And here, $a=1.4146,b=0.4399$

Therefore, the equation is

role="math" localid="1654085044915" $\hat{y}=1.4146+0.4399x$

Where $x$is cone index and $\hat{y}$is the average no. of offsprings per female.

Part (c) Step 1: Given information

We need to interpret the values of $r^2$and $s$.

Part (c) Step 2: Explanation

We know that

The value of $r^2=57.2$percent indicates that the linear model can explain $57.2$percent of the variation between the variables.

The value of $s=0.600309$indicates that the average prediction error is around $0.600309$.