Now do the next two problems on your own paper. When finished, scan your work and submit.

61. Interest rates on certificates of deposits in recent years are much lower than they were near the end of the twentieth century. For example, in early 1997 one institution offered the following yields:
\begin{tabular}{c|c|c|c|c}
\hline Time & 6 months & 1 yr & 2.5 yr & 5 yr \\
\hline Yield & [tex]$4.75 \%$[/tex] & [tex]$5.03 \%$[/tex] & [tex]$5.25 \%$[/tex] & [tex]$5.54 \%$[/tex] \\
\hline
\end{tabular}

Source: USA Today.

(a) Use logarithmic regression to find a function of the form [tex]$f(x) = a + b \ln x$[/tex] that models the data, where [tex]$x$[/tex] is time in years.

(b) Use [tex]$f$[/tex] to estimate the yield for a 10-year certificate of deposit.



Answer :

Certainly! Let's tackle these problems step-by-step.

### Problem (a)
We are given the following data points:

| Time (years) | Yield (%) |
|--------------|------------|
| 0.5 | 4.75 |
| 1 | 5.03 |
| 2.5 | 5.25 |
| 5 | 5.54 |

We need to use logarithmic regression to find a function of the form [tex]\( f(x) = a + b \ln(x) \)[/tex] that models this data. To do this, we will:

1. Transform the x-values (time in years) using the natural logarithm:
- [tex]\( \ln(0.5) \approx -0.693 \)[/tex]
- [tex]\( \ln(1) = 0 \)[/tex]
- [tex]\( \ln(2.5) \approx 0.916 \)[/tex]
- [tex]\( \ln(5) \approx 1.609 \)[/tex]

2. Fit a linear model to the transformed x-values and the given yields. In this case, the regression line will be:
[tex]\[ y = a + b \ln(x) \][/tex]
where [tex]\( y \)[/tex] are the yields.

The resulting coefficients from the regression are:
- [tex]\( a \approx 0.329 \)[/tex]
- [tex]\( b \approx 4.992 \)[/tex]

Thus, our model function [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = 0.329 + 4.992 \ln(x) \][/tex]

### Problem (b)
Now, we need to estimate the yield for a 10-year certificate of deposit using the function [tex]\( f \)[/tex] we found.

1. Insert [tex]\( x = 10 \)[/tex] into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(10) = 0.329 + 4.992 \ln(10) \][/tex]

2. Calculate the natural logarithm of 10:
[tex]\[ \ln(10) \approx 2.302 \][/tex]

3. Plug this value into the model:
[tex]\[ f(10) = 0.329 + 4.992 \times 2.302 \][/tex]

4. Simplify to find the estimated yield:
[tex]\[ f(10) \approx 0.329 + 4.992 \times 2.302 \approx 0.329 + 11.494 \approx 11.823 \][/tex]

Therefore, the estimated yield for a 10-year certificate of deposit is approximately [tex]\( 11.82\% \)[/tex].

In summary:
- The function that models the given data is [tex]\( f(x) = 0.329 + 4.992 \ln(x) \)[/tex].
- The estimated yield for a 10-year certificate of deposit is approximately [tex]\( 11.82\% \)[/tex].