My estimated model is
$$\hat \ln(y_t)=9.873-0.472\ln(x_{t2})-0.01x_{t3}$$
I'm asked to find a predictive CI at 95% confidence for the mean of $y_0$, when $x_{02}=250$, and $x_{03}=8$. We're to assume that $s^2 x_0(X^TX)^{-1}x_0^T=0.000243952$, where $x_0=(250,8)$.
I have a solution from a previous year, that goes like this:
I find the CI of the form $\text{CI}(E[ln(y_0)|x_0])=\left[\hat\ln(y_t)-t_{\alpha/2}s_E,\hat \ln(y_t)+t_{\alpha/2}s_E\right]$, where $t$ is the $\alpha/2$ upper-quantile of distribution $t(n-k)$ and $s_E=\sqrt{0.000243952}$. This gives me $[7.1563,7.2175]$.
Then the author does $\text{CI}(E[y_0|x_0])=[e^{7.1563},e^{7.2175}]=[1282.158,1363.077]$.
I disagree with this last step (by Jensen's inequality we'll underestimate). In Wooldridge's Intro to Econometrics, in page 212, he states that if we're sure the error terms are normal, then a consistent estimator is:
$$\hat E[y_0|x_0]=e^{s^2/2}e^{\hat \ln(y_0)}$$
So, I was thinking of doing
$$\text{CI}(E[y_0|x_0])=\left[e^{s^2/2} 1282.158,e^{s^2/2}1363.077 \right] = \left[ 1282.314,1363.243 \right] $$
Is this correct?
Also, the solution to this exercise states that $\text{CI}(E[y_0|x_0])=[624.020,663.519]$, which is far from either solution I've got.
Any help would be appreciated.
P.S: I've also read that the correction should not be used to the CI but only for the point estimation $\hat E[y_0|x_0]$