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MA181 INTRODUCTION TO STATISTICAL MODELLING

GOODNESS-OF-FIT TEST

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\begin{description}

\item[Example - Mendel's peas]
Mendel's double intercross data for
$(Round,Yellow)\times(Wrinkled, Green)$ peas. The expected
frequencies are in the ratios 9:3:3:1 on the assumption that the
factors segregate independently.

\begin{center}
\begin{tabular}{cccccc}
\hline &amp;$RY$&amp;$WY$&amp;$RG$&amp;$WG$&amp;Total\\ \hline Observed
frequency&amp;315&amp;101&amp;108&amp;32&amp;556\\Expected
frequency&amp;$312\frac{3}{4}$&amp;$104\frac{1}{4}$&amp;$104\frac{1}{4}$&amp;$34\frac{3}{4}$&amp;556\\
\hline
\end{tabular}
\end{center}

$$x^2=\frac{(2\frac{1}{4})^2}{312\frac{3}{4}}+\frac{(3\frac{1}{4})^2}{104\frac{1}{4}}
+\frac{(3\frac{1}{4})^2}{104\frac{1}{4}}+\frac{(2\frac{3}{4})^2}{34\frac{3}{4}}=0.470$$

Critical regions with $\nu=3$ are $x^2&gt;7.815$ for $\alpha=0.05$
and $x^2&gt;11.34$ for $\alpha=0.01$. Therefore accept $H_0$: the two
genes segregate independently.

\item[Example - Pharbitis]
Double intercross data for two genes $A$ and $B$ in Pharbitis. The
expected frequencies are again in the ratios 9:3:3:1, on the
assumption that $A$ and $B$ segregate independently.

\begin{center}
\begin{tabular}{cccccc}
\hline &amp;$AB$&amp;$Ab$&amp;$aB$&amp;$ab$&amp;Total\\ \hline Observed
frequency&amp;187&amp;35&amp;37&amp;31&amp;290\\ Expected
frequency&amp;$163\frac{3}{8}$&amp;$54\frac{3}{8}$&amp;$54\frac{3}{8}$&amp;$18\frac{1}{8}$&amp;290\\
\hline
\end{tabular}
\end{center}

$$x^2=\frac{(23\frac{7}{8})^2}{163\frac{1}{8}}+\frac{(19\frac{3}{8})^2}{54\frac{3}{8}}+\frac{(17\frac{3}{8})^2}{54\frac{3}{8}}+\frac{(12\frac{7}{8})^2}{18\frac{1}{8}}=25.096$$

Critical regions with $\nu=3$

$x^2&gt;7.815$ for $\alpha=0.05$

$x^2&gt;11.34$ for $\alpha=0.01$

$x^2&gt;16.27$ for $\alpha=0.001$

Reject $H_0$ and conclude (very strongly) that the genes are
linked.

\item[Estimating parameters]

\item[Example - Pharbitis revisited]
One theory suggests that the probabilities for the four cells can
be written as $(2+\theta)/4,\ (1-\theta)/4,\ (1-\theta)/4$ and
$\theta/4$ for some parameter $\theta$. The maximum likelihood
estimate of $\theta$ is $\hat{\theta}=0.4835$, which leads to the
expected frequencies given in the following.

\begin{center}
\begin{tabular}{cccccc}
\hline &amp;$AB$&amp;$Ab$&amp;$aB$&amp;$ab$&amp;Total\\ \hline Observed
Frequency&amp;187&amp;35&amp;37&amp;31&amp;290\\ Expected
frequency&amp;180.054&amp;37.446&amp;37.446&amp;35.054&amp;290\\ \hline
\end{tabular}
\end{center}

$$x^2=\frac{(187-180.054)^2}{180.054}+\ldots+\frac{(31-35.054)^2}{35.054}=0.902$$

Critical regions with 2 degrees of freedom are

$x^2&gt;5.991$ for $\alpha=0.05$

$x^2&gt;9/210$ for $\alpha=0.01$

Accept $H_0$: model given as above.

\item[Example Peas in pods]
The table below gives, in its second column, the frequency
distribution of the number $Y$Ł of peas found in the pod of a
four-seeded line of pea. A total of 269 pods were inspected.

$\hat{\pi}=0.5530$

\begin{center}
\begin{tabular}{ccccccc}
\hline Number of peas in pod&amp;0&amp;1&amp;2&amp;3&amp;4&amp;Total\\ \hline Observed
frequency&amp;16&amp;45&amp;100&amp;82&amp;26&amp;269\\ Expected
frequency&amp;10.74&amp;53.15&amp;98.62&amp;81.33&amp;25.15&amp;268.99\\ \hline
\end{tabular}
\end{center}

$$x^2=\frac{(16-10.74)^2}{10.74}+\ldots+\frac{(26-25.15)^2}{25.15}=3.88.$$

Critical regions with three degrees of freedom as on page 1. Do
not reject $H_0$: model given by binomial distribution.

\item[Small expected frequencies]
No expected frequency should be smaller than one and no more than
20\% should be less than five. Otherwise it is necessary to pool
cells.

\item[Example - Poisson distribution]
The number $Y$, of $\alpha$-particles emitted by a film of
Polonium in 2608 intervals of $\frac{1}{8}$ minute was given on
the Poisson distribution handout. The end of the table is as
follows:

\begin{center}
\begin{tabular}{|ccc|}
\hline &amp;Frequency of Intervals&amp;\\ $y$&amp;Observed&amp;Poisson, $E_y$\\
\hline 10&amp;10&amp;11.3\\ 11&amp;4&amp;4.0\\ 12&amp;0&amp;1.3\\ 13&amp;1&amp;0.4\\ 14&amp;1&amp;0.1\\
$\geq15$&amp;0&amp;0.0\\ \hline \end{tabular}
\end{center}

The last four cells may be pooled to give the following complete
table.

$\hat{\mu}=3.8715$

\begin{tabular}{ccccccccccc}
\hline $y$&amp;0&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9\\ \hline
$O_y$&amp;57&amp;203&amp;383&amp;525&amp;532&amp;408&amp;273&amp;139&amp;45&amp;27\\
$E_y$&amp;54.3&amp;210.3&amp;407.1&amp;525.3&amp;508.4&amp;393.7&amp;254.0&amp;140.5&amp;68.0&amp;29.2\\
\hline
\end{tabular}

\begin{tabular}{cccc}
\hline 10&amp;11&amp;$\geq12$&amp;Total\\ \hline 10&amp;4&amp;2&amp;2608\\
11.3&amp;4.0&amp;1.8&amp;2607.9\\ \hline
\end{tabular}


$x^2=13.0$

Critical regions with 11 degrees of freedom

$x^2&gt;19.68$ for $\alpha=0.05$

$x^2&gt;24.72$ for $\alpha=0.01$

$x^2&gt;31.26$ for $\alpha=0.001$

Accept $H_0$: model is given by Poisson distribution.

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