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\begin{document}

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%\date{January 23, 2003}

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\title{Stat 312: Lecture 11\\
Testing on Population Means}

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\author{Moo K. Chung\\
mchung@stat.wisc.edu}
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\section*{Concepts}
\begin{enumerate}

\item $P(\mbox{Type I error}) = P(\mbox{reject } H_0|H_0 \mbox{
true}) =\alpha.$ The resulting $\alpha$ is called the {\em
significance level} of the test and the corresponding test is
called a {\em level $\alpha$ test}. We will use test procedures
that give $\alpha$ less than a specified level (0.05 or 0.01
usually).

\item Testing mean $\mu$ with known variance $\sigma^2$.
$$H_0: \mu=\mu_0 \mbox{ vs. } H_1: \mu < \mu_0$$
Test statistic: $z=\frac{\bar x -\mu_0}{\sigma/\sqrt{n}}.$
Rejection region for level $\alpha$ test: $z \leq -z_{\alpha}.$

\item Testing mean $\mu$ with unknown variance $\sigma^2$.
$$H_0: \mu=\mu_0 \mbox{ vs. } H_1: \mu < \mu_0$$ Test statistic:
$t=\frac{\bar x -\mu_0}{s/\sqrt{n}}.$ Rejection region for level
$\alpha$ test: $z \leq -t_{\alpha,n-1}.$

\item Testing mean $\mu$ when the sample size is large.
$$H_0: \mu=\mu_0 \mbox{ vs. } H_1: \mu < \mu_0$$ Test statistic:
$z=\frac{\bar x -\mu_0}{s/\sqrt{n}}.$ Rejection region for level
$\alpha$ test: $z \leq -z_{\alpha}.$
\end{enumerate}
\section*{In-class problems}
Assume the IQ of a dog follows $X_i \sim N(\mu, 10^2)$. The IQ of
10 dogs are measured: 30, 25, 70, 110, 40, 80, 50, 60, 100, 60. We
want to test if dogs are as smart as people by testing
$$H_0: \mu = 100 \mbox{ vs. } H_1: \mu < 100$$
at level $\alpha = 0.05$.
\begin{verbatim}
> x<-c(30, 25, 70, 110, 40, 80, 50, 60,
100, 60)
> mean(x)
[1] 62.5
> z<-(mean(x)-100)/(10/sqrt(10))
> z
[1] -11.85854
> qnorm(0.05)
[1] -1.644854
\end{verbatim}
Assume the IQ of a dog follows $X_i \sim N(\mu, \sigma^2),$ where
$\sigma$ is unknown. Test
$$H_0: \mu = 100 \mbox{ vs. } H_1: \mu < 100$$
at level $\alpha = 0.05$.
\begin{verbatim}
> t=(mean(x)-100)/(sd(x)/sqrt(10))
> t
[1] -4.205156
> qt(0.05,9)
[1] -1.833113
> 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}
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%\caption{$t_{1000}}
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\section*{Self-study problems}
Example 8.6., 8.8., 8.9.
\section*{Assignment III.}
Due March 13. 11:00am. 8.24., 8.26, 8.28., 8.36. (a) and (b) only,
8.38.
\end{document}