The coeffcient of variation (CV) for a sample of values $Y_1,\ldots, Y_n$ is defined by $$ CV = S/ \bar{Y}.$$ Let $Y_1,\ldots, Y_n$ be a random sample of size $10$ from a normal distribution with mean $0$ and variance $\sigma^2$.
Find the distribution of $(10) \bar{Y}^2/S^2$.
What I have so far $$(10) \bar{Y}^2/S^2 = \sqrt{10} \bar{Y}/S = \frac{\bar{Y}}{S/\sqrt{10}} $$
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My approach is to let $Z = \sqrt(10) \bar{Y}/\sigma$, which follows standard normal and $W =(10-1) s^2/\sigma^2$ which has a chi-squared distribution with 9 degrees of freedom.
Then $T = \frac{Z}{\sqrt(W/v)}$...after some calculation which gives me $T = \frac{\bar{Y}}{S/\sqrt{10}}$. By the definition above, since Z and W are independent by construction for T variable, T has a t distribution with 9 degrees of freedom.
Can anyone find faults in this approach?
Let $Z= (root10)\bar{Y})/σ$ will follow Standard normal. Hence $Z^2$ is Chisquare with df 1. Again $T=(10-1)S^2/(σ^2)$ follows Chisquare with df 9. $((Z^2)/1)/(T/9)$ follows F with $df$ 1,9.. I hope u hve gt ur hint.