Chapter 9, Problem 23E

(a)

To determine

To explain how does this model compare to the one in previous exercise.

Answer to Problem 23E

It fits comparably well but slopes are different.

Explanation of Solution

In the previous exercise in the question there was a time plot of the interest rate over the years till $1980$ . Now in the question the time plot is showing the trend that changed dramatically after $1980$ . The dependent variable is interest rate. Also, it is given that:

  $\begin{array}{l}{R}^2=74.5\%\\ s=1.630\\ \text{Intercept}=21.0688\\ \text{Year}-1950=-0.356578\end{array}$

Now, from the above information and by looking at the model we can say that this model compare to the previous exercise model fits comparably well but they have very different slopes. As in the previous exercise it is positive and now it is negative.

(b)

To determine

To explain what does this model estimate the interest rate to have been in $2000$ and how does this compare to the rate you predicted in previous exercise.

Answer to Problem 23E

It is $3.24\%$ .

Explanation of Solution

In the previous exercise in the question there was a time plot of the interest rate over the years till $1980$ . Now in the question the time plot is showing the trend that changed dramatically after $1980$ . The dependent variable is interest rate. Also, it is given that:

  $\begin{array}{l}{R}^2=74.5\%\\ s=1.630\\ \text{Intercept}=21.0688\\ \text{Year}-1950=-0.356578\end{array}$

Now, we predict that the interest rate to have been in $2000$ will be as $3.24\%$ that is much lower than the other model predicts.

(c)

To determine

To explain do you trust this newer predicted value or not.

Answer to Problem 23E

Yes, it can be trusted.

Explanation of Solution

In the previous exercise in the question there was a time plot of the interest rate over the years till $1980$ . Now in the question the time plot is showing the trend that changed dramatically after $1980$ . The dependent variable is interest rate. Also, it is given that:

  $\begin{array}{l}{R}^2=74.5\%\\ s=1.630\\ \text{Intercept}=21.0688\\ \text{Year}-1950=-0.356578\end{array}$

Now, we can trust the new predicted value because it is in the middle of the data used for the regression. Thus, the new model can be useful. And it will give the accurate results.

(d)

To determine

To explain what would you predict the interest rate on three-month treasury bills will be in $2020$ .

Answer to Problem 23E

No, we cannot predict.

Explanation of Solution

In the previous exercise in the question there was a time plot of the interest rate over the years till $1980$ . Now in the question the time plot is showing the trend that changed dramatically after $1980$ . The dependent variable is interest rate. Also, it is given that:

  $\begin{array}{l}{R}^2=74.5\%\\ s=1.630\\ \text{Intercept}=21.0688\\ \text{Year}-1950=-0.356578\end{array}$

We cannot predict the interest rate on three-month treasury bills will be in $2020$ because it can be different for the different years and by following the previous trend than the future years interest rates are very difficult to predict.