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The data forthe counts of subjects whotext while driving and who drive while drunk is provided.

The level of significance is 0.05.

Theformula forexpected frequencyis,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The table for observed frequencies with row and column total is represented as,

The table for expected frequency is represented as,

Each of the expected value is greater than 5. It is assumed that the subjects are selected randomly.

\({H_0}:\)Texting while driving and driving while drunk are independent.

\({H_1}:\)Texting while driving and driving while drunk are dependent.

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {731 - 394.744} \right)}^2}}}{{394.744}} + \frac{{{{\left( {3054 - 3390.256} \right)}^2}}}{{3390.256}} + ... + \frac{{{{\left( {4564 - 4227.744} \right)}^2}}}{{4227.744}}\\ = 576.224\end{aligned}\]

Therefore, the value of the test statistic is 576.224.

The degrees of freedomare computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {2 - 1} \right)\\ = 1\end{aligned}\)

Therefore, the degrees of freedom are 1.

From chi-square table, the critical value for row corresponding to 1 degree of freedom at 0.05 level of significance is 3.841.

Therefore, the critical value is 3.841.

Also, the p-value is computed as 0.000.

Since the critical (3.841) is less than the value of test statistic (576.224). In this case, the null hypothesis is rejected.

Therefore, the decision is to reject the null hypothesis.

There is not enough evidenceto support the claim that the two behaviours; textingwhile driving and driving while drunk with alcohol are independent.

Thus, it can be concluded that two risky behaviours are dependent on each other.