Suppose that T is defined as in Definition 7.2. Reference
Chapter 7, Problem 98SE(choose chapter or problem)
The coefficient of variation (CV) for a sample of values 𝑌1, 𝑌2, . . . , 𝑌n is defined by
\(\mathrm{CV}=S / \bar{Y}\)
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 S 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
(CV = 10/1000) might provide acceptable precision, but the second case (CV = 2) would be unacceptable. Let \(Y_{1}, Y_{2}, \ldots, Y_{10}\) 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
\(P\left(-c \leq \frac{S}{\bar{Y}} \leq c\right)=.95\)
Use the result of Exercise 7.33 to find the distribution of
\(\text { (10) } \bar{Y}^{2} / \mathrm{S}^{2}\)
Use the result of Exercise 7.29 to find the distribution of\(S^{2} /\left[(10) \bar{Y}^{2}\right]\)
(c) Use the answer to (b) to find the constant c.
Equation Transcription:
Text Transcription:
Y1, Y2,...,Yn
CV=S / \bar Y
\bar Y
\barY
Y1, Y2,...,Y10
\sigma^2
P\(-c \leq S\bar Y \leq c)=.95
(10) \bar Y^2 /S^2)
S^2[(10) \barY^2\right]
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer