Chapter 7, Problem 95SE

The coefficient of variation (CV) for a sample of values $Y_1,Y_2,\dots ,Y_n$ is defined by

$\text{CV=}S/\bar{Y}$.

This quantity, which gives the standard deviation as a proportion of the mean, is sometimes informative. For example, the value $S=10$ has little meaning unless we can compare it to something else. If 5 is observed to be 10 and $\bar{Y}$ is observed to be 1000, the amount of variation is small relative to the size of the mean. However, if S is observed to be 10 and $\bar{Y}$ is observed to be 5, the variation is quite large relative to the size of the mean. If we were studying the precision (variation in repeated measurements) of a measuring instrument, the first case $\left(\text{CV =}10/1000\right)$ might provide acceptable precision, but the second case (CV = 2) would be unacceptable.

Let $Y_1,Y_2,\dots ,Y_{10}$ denote a random sample of size 10 from a normal distribution with mean 0 and variance ${\sigma }^2$. Use the following steps to find the number c such that

$P\left(-c\le \frac{S}{\bar{Y}}\le c\right)=\mathrm{.9.5}$

1. a Use the result of Exercise 7.33 to find the distribution of $\left(10\right){\bar{Y}}^2/S^2$.
2. b Use the result of Exercise 7.29 to find the distribution of $S^2/\left[\left(10\right){\bar{Y}}^2\right]$.
3. c Use the answer to (b) to find the constant c.

7.33 Use the structures of T and F given in Definitions 7.2 and 7.3, respectively, to argue that if T has a t distribution with v df, then $U=T^2$ has an F distribution with 1 numerator degree of freedom and v denominator degrees of freedom.

7.29 If Y is a random variable that has an F distribution with v₁ numerator and v₂ denominator degrees of freedom, show that $U=1/Y$ has an F distribution with v₂ numerator and v₁ denominator degrees of freedom.