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].
