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% these def are just for making the typing easier
\def\wh{\widehat}
\def\wb{\overline}
\def\wt{\widetilde}
\def\dbc#1#2{{ \left( \matrix{ #1\cr  #2\cr } \right) }}
% Text Binomial Coefficient (tbc)
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\begin{document}

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\begin{slide}

\begin{center}
{\large \bf Confidence Interval for Pop. Mean $\mu$}
\end{center}

\[ \overline{X}-(z_{\alpha/2})\mbox{se}(\overline{X})<\mu<
\overline{X}+(z_{\alpha/2})\mbox{se}(\overline{X}) \]

% \[ \wb{X}-z_{\alpha/2}\frac{s}{\sqrt{n}}<\mu<
% \wb{X}+z_{\alpha/2}\frac{s}{\sqrt{n}} \]
\[ \left(\wb{X}-z_{\alpha/2}\frac{s}{\sqrt{n}}\;,\;
\wb{X}+z_{\alpha/2}\frac{s}{\sqrt{n}}\right) \]

Example from $F_2$ mouse data: $\wb{X}=14.4$, $s=2.1$

\[ \wb{X} - z_{\alpha/2}\frac{s}{\sqrt{n}}=
14.4-1.96\frac{2.1}{\sqrt{28}}=13.6   \]

\[ \wb{X} + z_{\alpha/2}\frac{s}{\sqrt{n}}=
14.4+1.96\frac{2.1}{\sqrt{28}}=15.2   \]

\end{slide}

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\begin{slide}

\begin{center}
{\large \bf General C.I. for Estimator $\wh{\theta}$}
\end{center}

\vspace*{.1in}

\[ \wh{\theta}-(z_{\alpha/2})\mbox{se}(\wh{\theta})<\theta<
\wh{\theta}+(z_{\alpha/2})\mbox{se}(\wh{\theta}) \]

\vspace*{.1in}

se($\wh{\theta}$) = standard error of $\wh{\theta}$ =
est. std. deviation of $\wh{\theta}$
% \hspace*{.2in} = estimated standard deviation of $\wh{\theta}$

\[ \mbox{se}(\wb{X}) = \frac{s}{\sqrt{n}}  \]

\[ \mbox{se}(\wh{p}) = \sqrt{\frac{\wh{p}(1-\wh{p})}{n}} \]

\end{slide}

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\begin{slide}

\begin{center}
{\large \bf C.I. for $p$ of a Binomial($n,p$) Random V.}
\end{center}

\[ \wh{p}-(z_{\alpha/2})\sqrt{\frac{\wh{p}(1-\wh{p})}{n}} <p<
\wh{p}+(z_{\alpha/2})\sqrt{\frac{\wh{p}(1-\wh{p})}{n}} \]

Savant Example: $\wh{p}$ = 28/70 = 0.40

\[ 1.96\sqrt{\frac{(.4)(.6)}{70}}=.11   \]

95\% confidence interval for $p$=P(stay wins) is

(.40 - .11, .40 + .11) = (.29,.51)

\end{slide}
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\begin{slide}

\begin{center}
{\large \bf Bootstrap Standard Errors}
\end{center}

\begin{enumerate}
\item Mimic the true sampling scheme as well as you can, replacing populations by
samples when necessary.  This gives a bootstrap sample

\[ X_1^*, \ldots, X_n^* \]

\item For this bootstrap sample, calculate the estimator for which we want a
standard error, e.g., the sample mean or sample median, in general denoted by
$\widehat{\theta}^*$.

\item Repeat this process a total of $B$ times to get a sample of bootstrap
values

\[ \widehat{\theta}^*_1, \ldots, \widehat{\theta}^*_B  \]
\end{enumerate}

\end{slide}
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\begin{slide}

\begin{enumerate}
\item[4.] Calculate the sample standard deviation of this sample of bootstrap values

\vspace*{.1in}

\[ s_{\widehat{\theta}^*}=\sqrt{\frac{1}{B-1}\sum_{i-1}^{B}\left(\widehat{\theta}^*_i
-\overline{\widehat{\theta}^*}\right)^2} \]

\end{enumerate}

\vspace*{.1in}

$s_{\widehat{\theta}^*}$
is the bootstrap standard error of $\widehat{\theta}$. \\

How large should we choose $B$? 20--200

\end{slide}
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\begin{slide}
{\large \bf Example: Inbreeding Coefficient $f$}

In a sample of 70 individuals, the genotypes for a particular trait were

\[ n_{AA} = 24, n_{Aa} = 26, n_{aa} = 20 \]

An estimate of the inbreeding coeff. (Weir, 1996, p.65) is

\[\wh{f}=1-2n\frac{n_{Aa}}{(2n_{AA}+n_{Aa})(2n_{aa}+n_{Aa})}=0.25 \]

What about a standard error for $\wh{f}$?

\end{slide}
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\begin{slide}
\begin{enumerate}
\item Place in a hat 24 pieces of paper marked $AA$, 26 marked $Aa$, and
20 marked $aa$.  Draw out 70 pieces of paper {\bf with replacement}.
\item Calculate $\wh{f}^*$.
\item Repeat $B=100$ total times yielding $\wh{f}^*_1, \ldots,
\wh{f}^*_{100}.$
\item
\[ s_{\widehat{f}^*}=\sqrt{\frac{1}{B-1}\sum_{i-1}^{B}\left(\widehat{f}^*_i
-\overline{\widehat{f}^*}\right)^2}=0.123 \]
\end{enumerate}
(three more times: 0.122, 0.117, 0.091)

\end{slide}
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\begin{slide}
{\large \bf Confidence Intervals}

Based on normal approximation:

\[ 0.25 \mp (1.96)(0.123) \Rightarrow (0.01,0.49)\]

Percentile Method = Middle $(1-\alpha)$ of Bootstrap Dist.
\[ \left(\wh{\theta}^*_{(k)},
\wh{\theta}^*_{(B+1-k)} \right) \qquad
k=\mbox{max}\left(1,[(B+1)\alpha/2]\right) \]

\[ B=100:\qquad \left(\wh{f}^*_{(2)},\wh{f}^*_{(99)} \right)=(-0.03,0.51)\]

\[ B=2000:\qquad \left(\wh{f}^*_{(50)},\wh{f}^*_{(1951)}
\right)=(0.03,0.48)\]

\end{slide}
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\begin{slide}
\begin{figure}
\psfig{figure=/ncsu/boos/srcis/genetics/boot.100.ps,width=8in}
\vspace*{-9.2in}
\end{figure}

\end{slide}

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\begin{slide}
\begin{figure}
\psfig{figure=/ncsu/boos/srcis/genetics/boot.2000.ps,width=8in}
\vspace*{-9.2in}
\end{figure}

\end{slide}

\end{document}


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