Let \(Y_t\) be the value of \(Y\) in year \(t\). The natural logarithmic value of \(Y_t\) is regressed on \(t\). A simple linear regression yields the statistically significant estimates of intercept and slope coefficient as \(12\) and \(0.1512\), respectively. The calculated compound annual growth rate of \(Y\) is
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For a log-linear trend equation:
\[
\ln Y_t = a + bt
\]
the growth rate is:
\[
(e^b-1)\times100
\]
and not simply \(b\times100\).
Step 1: Write the fitted line.
The log of $Y$ is regressed on time, so
\[ \ln Y_t=12+0.1512\,t \]
The slope $0.1512$ is the key number.
Step 2: Link slope to growth.
When the left side is a natural log, a one period rise in $t$ multiplies $Y$ by $e^{b}$. So the yearly growth rate is
\[ \text{CAGR}=(e^{b}-1)\times100 \]
Step 3: Plug in the slope.
\[ \text{CAGR}=(e^{0.1512}-1)\times100 \]
Step 4: Work out the number.
Since $e^{0.1512}\approx1.1632$,
\[ \text{CAGR}=(1.1632-1)\times100=16.32\% \]
Note that just using $b\times100$ would give a wrong $15.12\%$.
\[ \boxed{16.32\%} \]