Statistics scramble my brain...

Since $\displaystyle var(x_0 B_0) = x_0^2 var(B_0)$, (this was done in a previous question) it follows that a 90% confidence interval for the expected ressponse at $\displaystyle x = x_0$ has endpoints $\displaystyle B_0 x_0 \pm s_0 x_0 t_{0.05} \frac {(n - 1)}{\sqrt{S_{xx}^{'}}} $ where $\displaystyle b_0 = \frac{S_{xy}^{'}}{S_{xx}^{'}}$ is the relised value of the slope and $\displaystyle s_0^2$ is the residual mean square.

Similarly, since $\displaystyle var(x_0 B_0 + e) = {\sigma}^2 + var(x_0 B_0)$, it follows that a 90% prediction interval for a new observation at $\displaystyle x = x_0$ has endpoints $\displaystyle b_0 x_0 \pm s_0 t_{0.05} (n - 1) \sqrt{1 + {\frac{x_0^2}{S_{xx}^{'}}}}$.

For the fuel consumtion data of an earlier problem we had $\displaystyle b_0 = 10.083 miles/litre, S_{xx}^{'} = 5766.49$ and $\displaystyle s_0^2 = 35.645$ on 5 degrees of freedom.

i) Find a 90% confidence interval for the expectation of distance travelled on a 30 litres of petrol.

ii) Construct a 90% prediction interval for the distance travelled on a typical 30 litres of petrol.

I have so much work to do right now and this is boggling me tho its probably relatively simple. Please help!