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Confidence Interval for Population Quantile

Recall that for a continuous random variable $X$ with cdf F, the median $\nu$, satisfies $P(X \leq \nu)=F(\nu)=\frac{1}{2}$. More generally, let $\xi_p$ denote the population quantile of order $p$. That is $\xi_p$ satisfies

\begin{displaymath}
P(X \leq \xi_p)=F(\xi_p)=p.
\end{displaymath} (8.17)

[Clearly, the median is a special case of this where $\nu=\xi_{.5}$.]

A confidence interval for a population quantile can be obtained using order statistics as follows. Suppose we take a sample of size $n$ from the given population and write the ordered sample as $Y_1, \ldots, Y_n$. Then for integers $r$, $s$, ( $1 \leq r < s \leq n$) the probability $P(Y_r < \xi_p < Y_s)$ can be determined. It is equal to

\begin{eqnarray*}
\mbox{P(\underline{at least} r but \underline{not more than} s...
...}^{s-1}{n \choose x} p^x(1-p)^{n-x}\\
&=& \gamma \mbox {, say}.
\end{eqnarray*}



Alternatively, for a given value of $\gamma$, we'd like to be able to determine $r$ and $s$. Substituting observed values $y_r$ and $y_s$ for $Y_r$ and $Y_s$, the $100\gamma$% CI for $\xi_p$ is ($y_r, \, y_s$). Note that we can't usually choose $r$, $s$ to give the conventional $.9$, $.95$, $.99$, values for $\gamma$. Note also that the above has assumed no particular form for $F$, and is, in that sense, a distribution-free property.



Example 8..10
Let $Y_1 < Y_2 < Y_3 < Y_4 < Y_5$ denote the order statistics of a random sample of size $5$ from a distribution of the continuous type. Compute

\begin{displaymath}(a) \, P(Y_1 < \xi_{.5} < Y_5);
\quad (b) \, P(Y_1 < \xi_{.25}< Y_3).\end{displaymath}

(a)
 

\begin{eqnarray*}
P(Y_1 < \xi_{.5} < Y_5) &=& \sum_{x=1}^4 {5 \choose x}
(.5)^x...
...\choose 5}(.5)^5(.5)^0\right]\\
&=&1 - \frac{2}{32}\\
&=&.9375
\end{eqnarray*}




(b)
 

\begin{eqnarray*}
P(Y_1 < \xi_{.25} < Y_3) &=& \sum_{x=1}^2{5 \choose x}
(.25^x...
... \choose 1}(.25)(.75^4 + {5 \choose 2}(.25)^2(.75)^3\\
&=& .66.
\end{eqnarray*}




next up previous contents
Next: Hypothesis Testing Up: Interval Estimates Previous: Pivotal Method   Contents
Bob Murison 2000-10-31