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The data fortexting while driving and irregular seat belt use is provided.

The level of significance is 0.05.

The formula for expected frequencies,

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

The table for observed counts along with row and column total is,

Theexpected frequency tableis represented as,

Each expected count is greater than 5. The requirements would be satisfied if it is assumed that the subjects are randomly selected.

As per the claim of the study, the hypotheses are formulated as,

\({H_0}:\)Texting while driving and irregular seat belt use are independent.

\({H_1}:\)Texting while driving and irregular seat belt use are dependent.

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {1737 - 1638.609} \right)}^2}}}{{1638.609}} + \frac{{{{\left( {2048 - 2146.391} \right)}^2}}}{{2146.391}} + ... + \frac{{{{\left( {2775 - 2676.609} \right)}^2}}}{{2676.609}}\\ = 18.773\end{aligned}\]

Therefore, the value of the test statistic is 18.773.

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 degrees of freedom and at 0.05 level of significance 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 (18.773). 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 texting while driving and irregular seat belt use are independent.

Thus, the two risky behaviours--texting while driving and seat belt use-- irregularity are dependent on each other.