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\title{Stat 312: Lecture 24\\
Chi-square Goodness-of-fit Test}
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\author{Moo K. Chung\\
mchung@stat.wisc.edu}
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\section*{Concepts}
\begin{enumerate}


\item The null hypothesis of interest is
$$H_0: p_1=c_1, \cdots, p_k=c_k,$$
$$H_1: p_j \neq c_j \mbox{ for some } j.$$

\item The test statistic
$$\chi^2 =\frac{(N_1 -\mathbb{E}N_1)^2}{\mathbb{E}N_1} + \cdots
\frac{(N_k -\mathbb{E}N_k)^2}{\mathbb{E}N_k}.$$ Assuming
$\mathbb{E}N_j \geq 5$ for every $j$, $$\chi^2 \sim
\chi_{k-1}^2.$$ {\em Proof.} When $k=2$ (binomial data), it is
easy to prove. Note that $\chi^2 =
\frac{(N_1-np_1)^2}{np_1(1-p_1)}$. We have shown that for large
$n$, $$\frac{N_1/n-p_1}{\sqrt{p_1(1-p_1)/n}} \sim N(0,1)$$

 \item $\chi^2$ goodness-of-fit test can be used to test whether
 sample data follows a particular distribution.
\end{enumerate}

\section*{In-class problems}
Example 1. There are four blood types (ABO system). It is believed
that 34, 15, 23 and 28 $\%$ of people have blood type A, B, AB and
O respectively. The blood samples of 100 students were collected:
A(12), B(56), AB(2), O(30). Test if the collected blood sample
contradicts the previous belief.
 {\em Solution.}
 The null hypothesis would be
 $$H_0:p_A=0.34, p_B=0.15,p_{AB}=0.23,p_O=0.28.$$
\begin{verbatim}
> chi <-(12-34)^2/34+(56-15)^2/15+
(2-23)^2/23+(30-28)^2/28
[1] 145.6187
> 1-pchisq(chi,3)
[1] 0
\end{verbatim}
Since the $P$-value is almost 0, we reject $H_0$.

Example 2. 90 rats proceed down a ramp to one of three doors. Test
if the rats have no preference concerning the choice of a door
when the observed frequencies for the three doors are $n_1=23,
n_2=36, n_3=31$. {\em Solution.} We compute $\chi^2 = 2.87$. The
$P$-value the goodness-of-fit test is
$1-\tt{pchisq(2.87,2)}=0.24.$ Since this is somewhat large, we do
not reject $H_0$.

Example 3. The number of accidents $X$ per week at an intersection
was checked for 50 weeks: 32weeks with no accident, 12 weeks with
1 accident, 6 weeks with 2 accidents. Test if $X$ follows a
Poission distribution.\\
{\em Solution.} $H_0: P(X=x)=\lambda^x e^{-\lambda}/x!$ for all $
x=0,1,2,\cdots$. Since $\lambda$ unknown, we estimate it with MLE.
$\hat \lambda = \bar x =(12+2\times6)/50=0.48$. Define three
categories by $X=0,X=1$ and $X \geq 2$. The expected probabilities
are estimated to give $\hat p_1 =0.619, \hat p_2=0.297, \hat p_3 =
1-\hat p_1 -\hat p_2=0.084$. Then the test statistic $\chi^2 \sim
\chi^2_1$ (the additional degree of freedom is lost due to the
estimation of $\lambda$). $\chi^2=1.354$ and the $P$-value would
be 0.25. We do not reject $H_0$.
\section*{Self-study problems}
Example 14.2. 14.8, 14.9.



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