Convert the interest rate $j_{2}=9\%$ to $j_{12}$ equivalent
$j_{m}:$ nominal (yearly) interest rate which is compounded (payable, convertible) $m$ times per year
$\$1$ at $j_{12}=12i$ will accumulate to $(1+i)^{12}$; $\$1$ at $j_2=9\%$ will accumulate to $\left(1+\frac{\frac{9}{2}}{100}\right)^2=1.092025$
\begin{equation} \label{eq1} \begin{split} (1+i)^{12} & = 1.092025 \\ i & = 1.00736-1 \\ i & = 0.00736 \end{split} \end{equation}
Hence, $j_{12}=12\times 0.00736 = 0.08832\%$
But the solution is $j_{12}=1.5\%$. Where I did wrong? Any help will be appreciated.
I made mistake when wrote $j_{12}=0.08832\%$, which should be $0.08832$ only or $8.83\%$. Moreover, I consider that exercise as a compound interest which is incorrect. It should be considered as a simple interest.