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The number of automobile crashes is given for the corresponding lengths of the zodiac signs.

The null hypothesis of this test is as follows:

The number of crashes is proportional to the lengths of the zodiac signs.

The alternative hypothesis is as follows:

The number of crashes is not proportional to the lengths of the zodiac signs.

Let\({O_i}\)denote the observed frequencies of the crashes.

The sums of all the observed frequencies is computed below:

\(\begin{aligned}{c}n = 128005 + 106878 + .... + 154477\\ = 1701704\end{aligned}\)

Let E denote the expected frequencies.

The expected frequencies are computed as shown:

\(E = \frac{{{\rm{Length}}\;{\rm{of the}}\;{\rm{zodiac}}\;{\rm{sign}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{days}}\;{\rm{in}}\;{\rm{a}}\;{\rm{year}}}} \times {\rm{Total}}\;{\rm{crashes}}\)

Therefore, the expected frequencies are computed below:

\(\begin{aligned}{c}{E_1} = \frac{{29}}{{365}} \times 1701704\\ = 135203.87\end{aligned}\)

\(\begin{aligned}{c}{E_2} = \frac{{24}}{{365}} \times 1701704\\ = 111892.87\end{aligned}\)

\(\begin{aligned}{c}{E_3} = \frac{{36}}{{365}} \times 1701704\\ = 167839.30\end{aligned}\)

\(\begin{aligned}{c}{E_4} = \frac{{27}}{{365}} \times 1701704\\ = 125879.47\end{aligned}\)

\(\begin{aligned}{c}{E_5} = \frac{{37}}{{365}} \times 1701704\\ = 172501.50\end{aligned}\)

\(\begin{aligned}{c}{E_6} = \frac{{31}}{{365}} \times 1701704\\ = 144528.28\end{aligned}\)

\(\begin{aligned}{c}{E_7} = \frac{{20}}{{365}} \times 1701704\\ = 93244.05\end{aligned}\)

\(\begin{aligned}{c}{E_8} = \frac{{37}}{{365}} \times 1701704\\ = 172501.50\end{aligned}\)

\(\begin{aligned}{c}{E_9} = \frac{{45}}{{365}} \times 1701704\\ = 209799.12\end{aligned}\)

\(\begin{aligned}{c}{E_{10}} = \frac{{23}}{{365}} \times 1701704\\ = 107230.66\end{aligned}\)

\(\begin{aligned}{c}{E_{11}} = \frac{6}{{365}} \times 1701704\\ = 27973.22\end{aligned}\)

\(\begin{aligned}{c}{E_{12}} = \frac{{19}}{{365}} \times 1701704\\ = 88581.85\end{aligned}\)

\(\begin{aligned}{c}{E_{13}} = \frac{{31}}{{365}} \times 1701704\\ = 144528.28\end{aligned}\)

The table below shows the necessary calculations:

The value of the test statistic is equal to:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{\left( {O - {E^2}} \right)}}{E}} \;\;\;\;\; \sim {\chi ^2}_{\left( {k - 1} \right)}\\ = 4913.487\end{aligned}\)

Thus,\({\chi ^2} = 4913.487\).

Let k be the number of observations.

Here, k=13.

The degrees of freedom for\({\chi ^2}\)is computed below:

\(\begin{aligned}{c}df = k - 1\\ = 13 - 1\\ = 12\end{aligned}\)

The critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with 12 degrees of freedom is equal to 21.0261.

Since the test statistic value is greater than the critical value, the null hypothesis is rejected.

There is enough evidence to conclude that the number of crashes is not proportional to the lengths of the zodiac signs.