/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q11-2-10BSC The accompanying table lists res... [FREE SOLUTION] | 91影视

91影视

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

Short Answer

Expert verified

Winning an overtime game and whether playing under the old rule or the new rule are independent. And hence the rule change is not effective.

Step by step solution

01

Given information

The data for overtime football games before and after the overtime rule was changed in the National Football League in 2011 is provided.

02

Compute the expected frequencies

Theexpected frequencyis computed as,

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

The observed frequencies(O), along with row and column totals is tabulated below,


Before Rule Change

After Rule Change

Row total

Overtime Coin Toss Winner Won the Game

252

24

276

Overtime Coin Toss Winner Lost the Game

208

23

231

Column total

460

47

507

Theexpected ( E) frequency table is represented as,


Before Rule Change

After Rule Change

Overtime Coin Toss Winner Won the Game

250.4142

25.5858

Overtime Coin Toss Winner Lost the Game

209.5858

21.4142

Assume the subjects are randomly selected for the study.

Since all the expected values are larger than 5, the requirement for chi-square test are fulfilled.

03

State the null and alternate hypothesis

The claim to test the independence of the two variables; the hypotheses are formulated as follows,

\({H_0}:\)Winning an overtime game and whether playing under the old rule or the new rule are independent.

\({H_a}:\)Winning an overtime game and whether playing under the old rule or the new rule are dependent.

04

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {252 - 250.4142} \right)}^2}}}{{250.4142}} + \frac{{{{\left( {24 - 25.5858} \right)}^2}}}{{25.5858}} + ... + \frac{{{{\left( {23 - 21.4142} \right)}^2}}}{{21.4142}}\\ = 0.2378\end{aligned}\]

Therefore, the value of the test statistic is 0.2378.

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 the row corresponding to 1 degrees of freedom and at 0.05 level of significance is 3.841.

Therefore, the critical value is 3.841.

Also, the p-value is obtained from the table as 0.626.

07

State the decision

Since the critical (3.841) is greater than the value of the test statistic (0.2378), in case the null hypothesis fails to be rejected.

Therefore, the decision is that we fail to reject the null hypothesis.

08

State the conclusion

There issufficient evidence to support the claim that winning an overtime game and whether playing under the old rule or the new rule are independent.

The results suggest that the change in rule is not effective on the winning in over time game.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Cybersecurity When using the data from Exercise 1 to test for goodness-of-fit with the distribution described by Benford鈥檚 law, identify the null and alternative hypotheses.

In a clinical trial of the effectiveness of echinacea for preventing

colds, the results in the table below were obtained (based on data from 鈥淎n Evaluation of Echinacea Angustifoliain Experimental Rhinovirus Infections,鈥 by Turner et al., NewEngland Journal of Medicine,Vol. 353, No. 4). Use a 0.05 significance level to test the claim that getting a cold is independent of the treatment group. What do the results suggest about the

effectiveness of echinacea as a prevention against colds?

Treatment Group


Placebo

Echinacea:

20% Extract

Echinacea:

60% Extract

Got a Cold

88

48

42

Did Not Get a Cold

15

4

10

The table below shows results since 2006 of challenged referee calls in the U.S. Open. Use a 0.05 significance level to test the claim that the gender of the tennis player is independent of whether the call is overturned. Do players of either gender appear to be better at challenging calls?

Was the Challenge to the Call Successful?


Yes

No

Men

161

376

Women

68

152

Exercises 1鈥5 refer to the sample data in the following table, which summarizes the last digits of the heights (cm) of 300 randomly selected subjects (from Data Set 1 鈥淏ody Data鈥 in Appendix B). Assume that we want to use a 0.05 significance level to test the claim that the data are from a population having the property that the last digits are all equally likely.

Last Digit

0

1

2

3

4

5

6

7

8

9

Frequency

30

35

24

25

35

36

37

27

27

24

When testing the claim in Exercise 1, what are the observed and expected frequencies for the last digit of 7?

Car Repair Costs Listed below are repair costs (in dollars) for cars crashed at 6 mi/h in full-front crash tests and the same cars crashed at 6 mi/h in full-rear crash tests (based on data from the Insurance Institute for Highway Safety). The cars are the Toyota Camry, Mazda 6, Volvo S40, Saturn Aura, Subaru Legacy, Hyundai Sonata, and Honda Accord. Is there sufficient evidence to conclude that there is a linear correlation between the repair costs from full-front crashes and full-rear crashes?

Front

936

978

2252

1032

3911

4312

3469

Rear

1480

1202

802

3191

1122

739

2767

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.