Predicting $y$ when response variable is $\ln(y)$ 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 $\hat E[y_0|x_0]$
You do not find the same answer because of what I suspect to be a typographical error, which would thus be the main reason of your problem: $x_{03}$ would be set to $80$, not $8$. Another possibility, if you keep $x_{03}=8$, is an error in the second estimated coefficient, say, $\hat{\beta}_2 = -0.1$ instead of $-0.01$.

Anyway, one of these modifications solves everything and yields the same result as the solution to this exercise.

Considering this change, with $t_{\alpha/2}=1.96476138969835$, one gets

Method 1

$\text{CI}(E[y_0|x_0])=[e^{6.43618291164626},e^{6.49755798189177}]=[624.020307335178,663.519326788772]$
(the given solution to this exercise)

or Method 2
(as stated In Wooldridge's Intro to Econometrics, in page 212) if we're sure the error terms are normal (and one is extremely lucky)

$$\text{CI}(E[y_0|x_0])=\left[e^{s^2/2}624.0203,e^{s^2/2}663.5193 \right] = \left[ 624.0960,663.6002 \right] $$

however

the method 2 is very unlikely to be correct, since as you mention in your question [...] the (underestimation) correction should not be used to the CI but only for the point estimation.

Why ? I would say because of the dependency betweeen the two terms, knowing the expectations of $e^{s^2/2}$ on the one hand and $\hat{y_0}$ on the other hand does not mean one knows the one of $e^{\frac{s^2}{2} + \hat{\ln(y_0)}}$.