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\centerline{Mathematics 375 -- Probability and Statistics 2}
\centerline{Clarification of Class Notes from February 8}
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In our first example of using a ``pivotal quantity'' to 
derive confidence intervals this day, I wasn't very clear
about exactly how you derive the formulas for the endpoints.
This sheet should clarify the algebra involved.

We considered the 
case of sampling from a $N(\mu,\sigma^2)$ population
to estimate the target parameter $\mu$ (the population mean). 
From our work in section 7.2 of the text, we know that with 
$n$ independent samples, the quantity
$$Q = {{\overline Y} - \mu \over s/\sqrt{n}},$$
(where $s = {1\over n-1} \sum_{i=1}^n (Y_i - \overline{Y})^2$, the 
sample standard deviation) satisfies
\medskip
\item{$\bullet$} $Q$ contains $\mu$ as the only unknown, and
\item{$\bullet$} $Q$ has a $t(n-1)$ distribution.
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\noindent
Hence we can use $Q$ as a pivotal quantity to derive
confidence intervals.  Here are complete derivations
of each of the three cases (being more careful about
signs than I was in class):
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\noindent
{\it Two-sided, $(1-\alpha)\times 100\%$ confidence interval}
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Let $t_{\alpha/2}$ be the ${\alpha/2}$-percentile point for 
the $t$-distribution: that is $P(T > t_{\alpha/2}) = \alpha/2$.
Then by symmetry:
$$\eqalign{
1 - \alpha &= P\left(-t_{\alpha/2}(n-1) 
< {{\overline Y} - \mu \over s/\sqrt{n}} < t_{\alpha/2}(n-1)\right)\cr
&= P\left(-t_{\alpha/2}(n-1) {s\over \sqrt{n}} < {\overline Y} - \mu < 
t_{\alpha/2}(n-1) {s\over \sqrt{n}}\right)\cr
&=P\left(-\overline{Y}-t_{\alpha/2}(n-1) {s\over \sqrt{n}} < - \mu < 
-\overline{Y}+t_{\alpha/2}(n-1) {s\over \sqrt{n}}\right)\cr
&=P\left(\overline{Y} + t_{\alpha/2}(n-1) {s\over \sqrt{n}} > \mu > 
\overline{Y}-t_{\alpha/2}(n-1) {s\over \sqrt{n}}\right)\cr
&=P\left(\overline{Y} - t_{\alpha/2}(n-1) {s\over \sqrt{n}} < \mu < 
\overline{Y}+t_{\alpha/2}(n-1) {s\over \sqrt{n}}\right)}$$
(Note: at the next-to-last step, we multiply through by $-1$, 
which reverses all inequalities, then we switch back to write the 
terms increasing left to right as usual).  
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\noindent
{\it (turn over)}
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\noindent
{\it One-sided, $(1-\alpha)\times 100\%$ lower confidence bound}
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{\it Note:} The formula I gave for this in class {\it was correct},
but the way I suggested to derive it wasn't (d'oh!).
We start from the statement (with $t_\alpha$ instead of $t_{\alpha/2}$):
$$\eqalign{1 - \alpha &= P\left(
{{\overline Y} - \mu \over s/\sqrt{n}} < t_\alpha(n-1)\right)\cr
&=  P\left({\overline Y} - \mu < t_\alpha(n-1){s\over\sqrt{n}}\right)\cr
&=  P\left(- \mu < - \overline{Y} + t_\alpha(n-1){s\over\sqrt{n}}\right)\cr
&=  P\left(\mu > \overline{Y} - t_\alpha(n-1){s\over\sqrt{n}}\right)\cr}$$
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\noindent
{\it One-sided, $(1-\alpha)\times 100\%$ upper confidence bound}
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Starting from 
$$1 - \alpha = P\left(-t_\alpha(n-1) < {{\overline Y} - \mu \over s/\sqrt{n}}\right)$$
and proceeding as for the lower confidence bound,
$$1 - \alpha = P\left(\mu <  \overline{Y} + t_\alpha(n-1) {s\over \sqrt{n}}\right)$$
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