91影视

The data for the seat belt users and nonusers are provided.

Theexpected frequency formulais computed as,

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

The table with row and column total is represented as,

Theexpected frequency tableis represented as,

The expected values are larger than 5. Assume the sampling is done randomly. Thus, the requirements of the test are satisfied.

To test the independence of the two variables, the hypothesis is formulated as follows;

\({H_0}:\)The amount of smoking is independent of seat belt use.

\({H_1}:\)The amount of smoking is dependent on seat belt use.

The value of the test statistic is computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {175 - 171.5294} \right)}^2}}}{{171.5294}} + \frac{{{{\left( {20 - 19.5882} \right)}^2}}}{{19.5882}} + ... + \frac{{{{\left( {9 - 7.0588} \right)}^2}}}{{7.0588}}\\ = 1.3582\end{aligned}\]

Therefore, the value of the test statistic is 1.3582.

The degrees of freedom are computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {4 - 1} \right)\\ = 3\end{aligned}\)

Therefore, the degrees of freedom are 3.

From the chi-square table, the critical value corresponding to 3 degrees of freedom and at 0.05 level of significance 7.815.

Therefore, the critical value is 7.815.

The P-value is obtained as 0.7154.

Since the critical (7.815) is greater than the value of the test statistic (1.3582). In this case, the null hypothesis fails to be rejected.

Therefore, the decision is that we fail to reject the null hypothesis.

There issufficient evidence to support the claimthatthecounts of cigarettes smoked in a day are independent of seat belt use.

Since the two behaviors are independent of each other, there is no evidence to support the theory that more cigarettes smoked by a person makes him or her less concerned of seatbelt use.