Prey attracts predators Here is one way in which nature regulates the size of animal populations: high population density attracts predators, which remove a higher proportion of the population than when the density of the prey is low. One study looked at kelp perch and their common predator, the kelp bass. On each of four occasions, the researcher set up four large circular pens on sandy ocean bottoms off the coast of southern California. He randomly assigned young perch to $1$of $4$pens so that one pen had $10$perch, one pen had $20$perch, one pen had $40$perch, and the final pen had $60$perch. Then he dropped the nets protecting the pens, allowing bass to swarm in, and counted the number of perch killed after two hours. A regression analysis was performed on the$16$ data points using x=number of perch in pen and y=proportion of perch killed. Here is a residual plot and a histogram of the residuals. Check whether the conditions for performing inference about the regression model are met.

Here is computer output from the least-squares regression analysis of the perch data.

a. Find the critical value for a $90\%$confidence interval for the slope of the true regression line. Then calculate the confidence interval.

b. Interpret the interval from part (a).

c. Explain the meaning of “$90\%$ confident” in this context.

Step by step solution

01

Part (a) step 1 : Given information

We have to find the critical value and confidence interval for a $90\%$confidence interval.

02

Part (a) Step 2 : Simplification

We will use the following formula for the boundaries of the confidence interval :-

$$\begin{gathered} b-t*\times SEb1 \\ b+t*\times SEb1 \end{gathered}$$

In the row "Perch" and the column "Coefficient" of the computer output, the slope $b1$is mentioned.

$b1=0.008569$

In the row "Perch" and the column "stdev." of the mention output from the computer, the computed standard deviation of the slope $SEb1$is mentioned.

$SEb1=0.002456$

degrees of freedom :-$16-2=14$

In the student's T distribution table $df=14$and column of $c=90$percent, the $t$-value may be found.

$t*=1.761$

Theconfidenceinterval'sbounds

$$\begin{gathered} b-t*\times SEb1=0.008569-1.761\times 0.002456=0.004243984 \\ b+t*\times SEb1=0.008569+1.761\times 0.002456=0.012894016 \end{gathered}$$

03

Part (b) step 1 : Given information

We have to explain the interval from part (a).

04

Part (b) Step 2 : Simplification

From part (a) ,

$$\begin{gathered} b-t*\times SEb1=0.008569-1.761\times 0.002456=0.004243984 \\ b+t*\times SEb1=0.008569+1.761\times 0.002456=0.012894016 \end{gathered}$$

$$\begin{gathered} b-t*\times SEb1=0.008569-1.761\times 0.002456=0.004243984 \\ b+t*\times SEb1=0.008569+1.761\times 0.002456=0.012894016 \end{gathered}$$

There is a $90\%$chance that the true slope of the population regression line is between $0.004243984$and $0.012894016.$

05

Part (c) step 1 : Given information

We have to explain the meaning of $90\%$confident.

06

Part (c) Step 2 : Simplification

The slope of the correct regression line is shown in the $90$percent confidence range. $90$percent confidence also means that it is projected that around $90\%$of all samples will have a $90\%$confidence interval including the true population parameter.

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