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

Expert verified

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

Step by step solution

01

Given information

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

The level of significance is 0.05.

02

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.

03

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.

04

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {1737 - 1638.609} \right)}^2}}}{{1638.609}} + \frac{{{{\left( {2048 - 2146.391} \right)}^2}}}{{2146.391}} + ... + \frac{{{{\left( {2775 - 2676.609} \right)}^2}}}{{2676.609}}\\ = 18.773\end{aligned}\]

Therefore, the value of the test statistic is 18.773.

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 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.

07

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.

08

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.

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Most popular questions from this chapter

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

-36

Flight 19

-5

-32

-13

-9

-19

49

-30

-23

Flight 21

-23

28

103

-19

-5

-46

13

-3

P-Value If we use a 0.05 significance level in analysis of variance with the sample data given in Exercise 1, what is the P-value? What should we conclude? If a passenger abhors late flight arrivals, can that passenger be helped by selecting one of the flights?

Clinical Trial of Lipitor Lipitor is the trade name of the drug atorvastatin, which is used to reduce cholesterol in patients. (Until its patent expired in 2011, this was the largest-selling drug in the world, with annual sales of $13 billion.) Adverse reactions have been studied in clinical trials, and the table below summarizes results for infections in patients from different treatment groups (based on data from Parke-Davis). Use a 0.01 significance level to test the claim that getting an infection is independent of the treatment. Does the atorvastatin (Lipitor) treatment appear to have an effect on infections?


Placebo

Atorvastatin 10 mg

Atorvastatin 40 mg

Atorvastatin 80 mg

Infection

27

89

8

7

No Infection

243

774

71

87

Chocolate and Happiness Use the results from part (b) of Cumulative Review Exercise 2 to construct a 99% confidence interval estimate of the percentage of women who say that chocolate makes them happier. Write a brief statement interpreting the result.

Is the hypothesis test described in Exercise 1 right tailed, left-tailed, or two-tailed? Explain your choice.

A case-control (or retrospective) study was conductedto investigate a relationship between the colors of helmets worn by motorcycle drivers andwhether they are injured or killed in a crash. Results are given in the table below (based on datafrom “Motorcycle Rider Conspicuity and Crash Related Injury: Case-Control Study,” by Wellset al., BMJ USA,Vol. 4). Test the claim that injuries are independent of helmet color. Shouldmotorcycle drivers choose helmets with a particular color? If so, which color appears best?

Color of helmet


Black

White

Yellow/Orange

Red

Blue

Controls (not injured)

491

377

31

170

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213

112

8

70

26

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