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Math Help - [SOLVED] Confidence/Prediction intervals

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    Super Member Deadstar's Avatar
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    [SOLVED] Confidence/Prediction intervals

    Statistics scramble my brain...

    Since 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 x = x_0 has endpoints B_0 x_0 \pm s_0 x_0 t_{0.05} \frac {(n - 1)}{\sqrt{S_{xx}^{'}}} where b_0 = \frac{S_{xy}^{'}}{S_{xx}^{'}} is the relised value of the slope and s_0^2 is the residual mean square.

    Similarly, since 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 x = x_0 has endpoints 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 b_0 = 10.083 miles/litre, S_{xx}^{'} = 5766.49 and 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!
    Last edited by mr fantastic; February 5th 2009 at 05:16 AM. Reason: Fixed the latex
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