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The frequencies of digits that appear in the lottery are recorded for 60 days.

As per the requirements of the chi-square test, the samples must be randomly selected, and the expected value must be larger than 5.

Let O denote the observed frequencies of the games.

The following values are obtained for the 10 digits:

\(\begin{aligned}{l}{O_0} = 21\\{O_1} = 30\\{O_2} = 31\\{O_3} = 33\\{O_4} = 19\end{aligned}\)

\(\begin{aligned}{l}{O_5} = 23\\{O_6} = 21\\{O_7} = 16\\{O_8} = 24\end{aligned}\)

\({O_9} = 22\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 21 + 30 + ...... + 22\\ = 240\end{aligned}\)

Let E denote the expected frequencies.

It is given that the digits are expected to be selected in a way that they are equally likely.

The expected frequency for each of the 10 digits is the same and is equal to:

\(\begin{aligned}{c}E = \frac{{240}}{{10}}\\ = 24\end{aligned}\)

Assuming the samples are randomly taken, the requirements of the test are satisfied.

The table below shows the necessary calculations:

The value of the test statistic is equal to:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \;\\ = 0.375 + 1.5 + ....... + 0.1667\\ = 11.583\end{aligned}\)

Thus,\({\chi ^2} = 11.583\).

Let k be the number of digits, which are 10.

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

\(\begin{aligned}{c}df = k - 1\\ = 10 - 1\\ = 9\end{aligned}\)

The critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with 9 degrees of freedom is obtained from chi-square table as 16.919.

The p-value is,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 11.583} \right)\\ = 0.238\end{aligned}\)

Since the test statistic value is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to be rejected.

There is enough evidence to favor the claim that the different digits in the lottery do occur equally likely.