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

55

Cases (injured or killed)

213

112

8

70

26

Short Answer

Expert verified

Injuries are dependent on helmet color.

The proportion of the subjects that were not injured was least when the subjects wore blue color.

Step by step solution

01

Given information

Data for relationship between the colors of helmets worn by motorcycle drivers and their injuries or deaths in crashes.

02

Check the requirements of the test

Theexpected frequency formulais,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The observation table with row and column total is,


Black

White

Yellow/Orange

Red

Blue

Row total

Controls(not injured)

491

377

31

170

55

1124

Cases(injured or killed)

213

112

8

70

26

429

Column Total

704

489

39

240

81

1553

Theexpected frequency tableis represented as,


Black

White

Yellow/

Orange

Red

Blue

Controls (not injured)

509.5274

353.9189

28.2267

173.7025

58.6246

Cases (injured or killed)

194.4726

135.0811

10.7733

66.2975

22.3754

Assume the experimental units are randomly selected.

The expected frequencies are greater than 5.

Thus, the requirements of the test are satisfied.

03

Formulate the hypotheses

The hypotheses are formulated as follows:

\({H_0}:\)Injuries are independent of helmet color.

\({H_1}:\)Injuries are dependent on helmet color.

04

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {491 - 509.5274} \right)}^2}}}{{509.5274}} + \frac{{{{\left( {377 - 353.9189} \right)}^2}}}{{353.9189}} + ... + \frac{{{{\left( {26 - 22.3754} \right)}^2}}}{{22.3754}}\\ = 9.971\end{aligned}\]

Therefore, the value of the test statistic is 9.971.

05

Compute the degrees of freedom

The degrees of freedomare computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {5 - 1} \right)\\ = 4\end{aligned}\)

Therefore, the degrees of freedom are 4.

06

Compute the critical value

From the chi-square table, the critical value corresponding to 4 degrees of freedom and at 0.05 level of significance 9.488.

Therefore, the critical value is 9.488.

The p-value is obtained as 0.041.

07

State the decision

Since the critical value (9.488) is less than the value of the test statistic (9.971). In this case, the null hypothesis is rejected.

Therefore, the decision is to reject the null hypothesis.

08

State the conclusion

There isinsufficient evidence to support the claim that Injuries are independent of helmet color.

Thus, the helmet color is dependent on the injuries.

Therefore, the motocycle drivers must choose a particular color to avoid injuries. From the sample data, the proportion of least controls (not injured) under each of the color categories is:


Black

White

Yellow/Orange

Red

Blue

Controls(not injured)

0.6974\(\left( {\frac{{491}}{{704}}} \right)\)

0.7710

\(\left( {\frac{{377}}{{489}}} \right)\)

0.7949

\(\left( {\frac{{31}}{{39}}} \right)\)

0.7083

\(\left( {\frac{{170}}{{240}}} \right)\)

0.6790

\(\left( {\frac{{55}}{{81}}} \right)\)

Thus, lowest proportion of no injuries occurred when subjects wore blue helmets.

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Most popular questions from this chapter

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

Critical Thinking: Was Allstate wrong? The Allstate insurance company once issued a press release listing zodiac signs along with the corresponding numbers of automobile crashes, as shown in the first and last columns in the table below. In the original press release, Allstate included comments such as one stating that Virgos are worried and shy, and they were involved in 211,650 accidents, making them the worst offenders. Allstate quickly issued an apology and retraction. In a press release, Allstate included this: “Astrological signs have absolutely no role in how we base coverage and set rates. Rating by astrology would not be actuarially sound.”

Analyzing the Results The original Allstate press release did not include the lengths (days) of the different zodiac signs. The preceding table lists those lengths in the third column. A reasonable explanation for the different numbers of crashes is that they should be proportional to the lengths of the zodiac signs. For example, people are born under the Capricorn sign on 29 days out of the 365 days in the year, so they are expected to have 29/365 of the total number of crashes. Use the methods of this chapter to determine whether this appears to explain the results in the table. Write a brief report of your findings.

Zodiac sign

Dates

Length(days)

Crashes

Capricorn

Jan.18-Feb. 15

29

128,005

Aquarius

Feb.16-March 11

24

106,878

Pisces

March 12-April 16

36

172,030

Aries

April 17-May 13

27

112,402

Taurus

May 14-June 19

37

177,503

Gemini

June 20-July 20

31

136,904

Cancer

July21-Aug.9

20

101,539

Leo

Aug.10-Sep.15

37

179,657

Virgo

Sep.16-Oct.30

45

211,650

Libra

Oct.31-Nov 22

23

110,592

Scorpio

Nov. 23-Nov. 28

6

26,833

Ophiuchus

Nov.29-Dec.17

19

83,234

Sagittarius

Dec.18-Jan.17

31

154,477

Cybersecurity The table below lists leading digits of 317 inter-arrival Internet traffic times for a computer, along with the frequencies of leading digits expected with Benford’s law (from Table 11-1 in the Chapter Problem).

a. Identify the notation used for observed and expected values.

b. Identify the observed and expected values for the leading digit of 2.

c. Use the results from part (b) to find the contribution to the\({\chi ^2}\)test statistic from the category representing the leading digit of 2.

Leading Digit

1

2

3

4

5

6

7

8

9

Benford’s

Law

30.1%

17.6%

12.5%

9.7%

7.9%

6.7%

5.8%

5.1%

4.6%

Leading Digits

of Inter-Arrival

Traffic Times

76

62

29

33

19

27

28

21

22

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.

California Daily 4 Lottery The author recorded all digits selected in California’s Daily 4 Lottery for the 60 days preceding the time that this exercise was created. The frequencies of the digits from 0 through 9 are 21, 30, 31, 33, 19, 23, 21, 16, 24, and 22. Use a 0.05 significance level to test the claim of lottery officials that the digits are selected in a way that they are equally likely.

The accompanying table lists results of overtime football

games before and after the overtime rule was changed in the National Football League in 2011. Use a 0.05 significance level to test the claim of independence between winning an overtime game and whether playing under the old rule or the new rule. What do the results suggest about

the effectiveness of the rule change?

Before Rule Change

After Rule Change

Overtime Coin Toss Winner Won the Game

252

24

Overtime Coin Toss Winner Lost the Game

208

23

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