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 driving when drinking alcohol. Are those two risky behaviors independent of each other?

	Drove when drinking Alcohol?
	Yes	No
Texted while driving	731	3054
No Texting while driving	156	4564

Short answer

Texting while driving and driving while drunk are dependent behaviours of driving.

Step-by-step solution

Given information

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

The level of significance is 0.05.

Compute the expected frequencies

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,

	Drove when drinking Alcohol?
	Yes	No	Row total
Texted while driving	731	3054	3785
No Texting while driving	156	4564	4720
Column total	887	7618	8505

The table for expected frequency is represented as,

	Drove when drinking Alcohol?
	Yes	No
Texted while driving	394.744	3390.256
No Texting while driving	492.256	4227.744

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

State the null and alternate hypothesis

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

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

Compute the test statistic

The value of the test statisticis computed as,

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

Therefore, the value of the test statistic is 576.224.

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 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.

State the decision

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.

State the conclusion

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.