Question

A study of seat belt users andnonusers yielded the randomly selected sample data summarized in the given table (based on data from “What Kinds of People Do Not Use Seat Belts?” by Helsing and Comstock, American Journal of Public Health,Vol. 67, No. 11). Test the claim that the amount of smoking is independent of seat belt use. A plausible theory is that people who smoke more are lessconcerned about their health and safety and are therefore less inclined to wear seat belts. Is this theory supported by the sample data?

	Number of Cigarettes Smoked per Day
	0	1-14	15-34	35 and over
Wear Seat Belts	175	20	42	6
Don't Wear Seat Belts	149	17	41	9

Short answer

The amount of smoking is independent of seatbelt use. There is not enough evidence to support the theory.

Step-by-step solution

Given information

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

Compute the expected frequencies

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,

	0	1-14	15-34	35 and over	Row Total
Wear Seat Belts	175	20	42	6	243
Don't Wear Seat Belts	149	17	41	9	216
Column Total	324	37	83	15	459

Theexpected frequency tableis represented as,

	0	1-14	15-34	35 and over
Wear Seat Belts	171.5294	19.5882	43.9412	7.9412
Don't Wear Seat Belts	152.4706	17.4118	39.0588	7.0588

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

State the null and alternate hypothesis

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.

Compute the test statistic

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.

Compute the degrees of freedom

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.

Compute the critical value

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.

State the decision

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.

State the conclusion

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.