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\begin{document}

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%\date{January 23, 2003}

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\title{Stat 312: Lecture 15\\
$P$-values}

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\author{Moo K. Chung\\
mchung@stat.wisc.edu}
\date{November 2, 2004}
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\begin{enumerate}


\item The $P$-value is the smallest level of significance at which
$H_0$ would be rejected.\\

Example 8.5, 8.15. Nicotine contents $X_1,\cdots X_{32}$ of
cigarettes follow $N(\mu,0.2^2)$. We are interested in testing if
$$H_0: \mu = 1.5 \mbox{ vs. } H_1: \mu > 1.5.$$
Assume $\bar x = 1.57$. Given $\alpha$ level, the corresponding
rejection regions are given by
\begin{verbatim}
> -qnorm(0.05)
[1] 1.644854
> -qnorm(0.025)
[1] 1.959964
> -qnorm(0.01)
[1] 2.326348
> -qnorm(0.005)
[1] 2.575829

> (1.57-1.5)/(0.2/sqrt(32))
[1] 1.979899

> 1-pnorm(1.979899)
[1] 0.02385744
\end{verbatim}


\item $P$-value $\leq \alpha \to$ reject $H_0$ at level $\alpha$.\\
$P$-value $> \alpha \to$ do not reject $H_0$ at level $\alpha$.\\
The smaller the $P$-value, the easier to reject $H_0$.


Example. Dog IQ example.
\begin{verbatim}
> t.test(x,mu=100,alternative="less",
 conf.level=0.95)
        One Sample t-test
 data:  x t = -4.2036, df = 9,
 p-value = 0.001147
 alternative hypothesis: true mean is
 less than 100
 95 percent confidence interval:
    -Inf 78.8531
 sample estimates: mean of x 62.5
\end{verbatim}

\item P-value computation for two sided tests. Let us continue the
nicotine content example and find the P-value for testing
$$H_0:\mu = 1.5 \mbox{ vs. } H_1:\mu \neq 1.5.$$

\begin{verbatim}
> 2*(1-pnorm(1.979899))
[1] 0.04771488
\end{verbatim}
\end{enumerate}

{\em Review Problems.} Example 8.17., 8.18.

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