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

Expert verified

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

Step by step solution

01

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.

02

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.

03

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.

04

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.

05

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.

06

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.

07

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.

08

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.

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Most popular questions from this chapter

In Exercises 5–20, conduct the hypothesis test and provide the test statistic and the P-value and, or critical value, and state the conclusion.

Police Calls Repeat Exercise 11 using these observed frequencies for police calls received during the month of March: Monday (208); Tuesday (224); Wednesday (246); Thursday (173); Friday (210); Saturday (236); Sunday (154). What is a fundamental error with this analysis?

In Exercises 5–20, conduct the hypothesis test and provide the test statistic and the P-value and , or critical value, and state the conclusion.

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Equivalent Tests A\({\chi ^2}\)test involving a 2\( \times \)2 table is equivalent to the test for the differencebetween two proportions, as described in Section 9-1. Using the claim and table inExercise 9 “Four Quarters the Same as $1?” verify that the\({\chi ^2}\)test statistic and the zteststatistic (found from the test of equality of two proportions) are related as follows:\({z^2}\)=\({\chi ^2}\).

Also show that the critical values have that same relationship.

In Exercises 1–4, use the following listed arrival delay times (minutes) for American Airline flights from New York to Los Angeles. Negative values correspond to flights that arrived early. Also shown are the SPSS results for analysis of variance. Assume that we plan to use a 0.05 significance level to test the claim that the different flights have the same mean arrival delay time.

Flight 1

-32

-25

-26

-6

5

-15

-17

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Flight 19

-5

-32

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-9

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49

-30

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28

103

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13

-3

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