Question

In a study of high school students at least 16 years of age,

researchers obtained survey results summarized in the accompanying table (based on data from “Texting While Driving and Other Risky Motor Vehicle Behaviors Among U.S. High School Students,” by O’Malley, Shults, and Eaton, Pediatrics,Vol. 131, No. 6). Use a 0.05 significance level to test the claim of independence between texting while driving and irregular seat belt use. Are those two risky behaviors independent of each other?

	Irregular Seat Belt Use?
	Yes	No
Texted while driving	1737	2048
No Texting while driving	1945	2775

Short answer

Texting while driving and irregular seat belt use are dependent on each other.

Step-by-step solution

Given information

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

The level of significance is 0.05.

Compute the expected frequencies and check the requirements

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,

	Irregular seat belt use?
	Yes	No	Row total
Texted while driving	1737	2048	3785
No Texting while driving	1945	2775	4720
Column total	3682	4823	8505

Theexpected frequency tableis represented as,

	Irregular seat belt use?
	Yes	No
Texting while driving	1638.609	2146.391
No Texting while driving	2043.391	2676.609

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

State the null and alternate hypothesis

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.

Compute the test statistic

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.

Compute the degrees of freedom

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.

Compute the critical value

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.

State the decision

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.

State the conclusion

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.