Critical Thinking: Was Allstate wrong? The Allstate insurance company once issued a press release listing zodiac signs along with the corresponding numbers of automobile crashes, as shown in the first and last columns in the table below. In the original press release, Allstate included comments such as one stating that Virgos are worried and shy, and they were involved in 211,650 accidents, making them the worst offenders. Allstate quickly issued an apology and retraction. In a press release, Allstate included this: “Astrological signs have absolutely no role in how we base coverage and set rates. Rating by astrology would not be actuarially sound.”

Analyzing the Results The original Allstate press release did not include the lengths (days) of the different zodiac signs. The preceding table lists those lengths in the third column. A reasonable explanation for the different numbers of crashes is that they should be proportional to the lengths of the zodiac signs. For example, people are born under the Capricorn sign on 29 days out of the 365 days in the year, so they are expected to have 29/365 of the total number of crashes. Use the methods of this chapter to determine whether this appears to explain the results in the table. Write a brief report of your findings.

Zodiac sign	Dates	Length(days)	Crashes
Capricorn	Jan.18-Feb. 15	29	128,005
Aquarius	Feb.16-March 11	24	106,878
Pisces	March 12-April 16	36	172,030
Aries	April 17-May 13	27	112,402
Taurus	May 14-June 19	37	177,503
Gemini	June 20-July 20	31	136,904
Cancer	July21-Aug.9	20	101,539
Leo	Aug.10-Sep.15	37	179,657
Virgo	Sep.16-Oct.30	45	211,650
Libra	Oct.31-Nov 22	23	110,592
Scorpio	Nov. 23-Nov. 28	6	26,833
Ophiuchus	Nov.29-Dec.17	19	83,234
Sagittarius	Dec.18-Jan.17	31	154,477

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.