Using the $F$ distribution table: I think it should be $ P\{\frac{S_1}{S_2} \geq3\} = 1 - P\{\frac{S_1}{S_2} \leq 3\} = 1 - 0.05 = 0.95$. But my notes say it should be $0.05$?
F distribution: $P\{S_1\geq 3S_2\}$ with $n_1 = 7, n_2 = 13$
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0The mistake is saying that P{S1/S2<=3} = 0.05 as the given answers show it is close to 0.95. – 2012-07-04
1 Answers
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Let $X = \frac{S_1}{S_2}$ be F-distributed random variable. Then $ f_X(x) = \frac{1}{B\left(\frac{n_1}{2}, \frac{n_2}{2}\right)} \frac{n_1}{n_2} \left( \frac{n_1 x}{n_2} \right)^{\frac{n_1}{2} -1} \left( 1 + \frac{n_1 x}{n_2} \right)^{-\frac{n_1+n_2}{2}} $ and $ \mathbb{P}\left( X \geqslant 3 \right) = \int_3^\infty f_X(x) \mathrm{d} x $ The plot of $f_X(x)$ for given parameters looks as follows:
Clearly the probability $\mathbb{P}(X \geqslant 3)$ is small (it is the orange area in the plot). Using quadratures it equals $0.04156$.
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0Thanks, I was interpreting wrongly the $\alpha$ in the graph, now I know it measures the area to the right from $F = 3$. – 2012-07-04