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The table below includes results from polygraph (lie detector) experiments conducted by researchers Charles R. Honts (Boise State University) and Gordon H. Barland (Department of Defense Polygraph Institute). In each case, it was known if the subject lied or did not lie, so the table indicates when the polygraph test was correct. Use a 0.05 significance level to test the claim that whether a subject lies is independent of the polygraph test indication. Do the results suggest that polygraphs are effective in distinguishing between truths and lies?

Did the subject Actually Lie?


No (Did Not Lie)

Yes (Lied)

Polygraph test indicates that the subject lied.


15

42

Polygraph test indicates that the subject did not lied.


32

9

Short Answer

Expert verified

A polygraph test is effective in distinguishing truth and lies.

Step by step solution

01

Given information

The data forthe polygraph test is provided.

The level of significance is 0.05.

02

Compute the expected frequencies and check the requirements

Theexpected frequency is computed as,

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

The table of observed values with row and column total is represented as,


Did the subject Actually Lie?



No (Did Not Lie)

Yes (Lied)

Row total

Polygraph test indicates that the subject lied.

15

42

57

Polygraph test indicates that the subject did not lie.

32

9

41

Column total

47

51

98

Theexpected frequency tableis represented as,


Did the subject Actually Lie?


No (Did Not Lie)

Yes (Lied)

Polygraph test indicates that the subject lied.

27.3367

29.6633

Polygraph test indicates that the subject did not lie.

19.6633

21.3367

Assume that the subjects are randomly selected and assigned to treatment groups. Also, the expected values are greater than 5.

Thus, all the requirements are satisfied.

03

State the hypotheses

To test if the true results are independent of the results obtained from the polygraph, the hypotheses are formulated as:

\({H_0}:\)Thesubject鈥檚 lie is independent of the polygraph test indication.

\({H_1}:\) The subject鈥檚 lie is not independent of the polygraph test indication.

04

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {15 - 27.3367} \right)}^2}}}{{27.3367}} + \frac{{{{\left( {42 - 29.6633} \right)}^2}}}{{29.6633}} + ... + \frac{{{{\left( {9 - 21.3367} \right)}^2}}}{{21.3367}}\\ = 25.571\end{aligned}\]

Therefore, the value of the test statistic is 25.571.

05

Compute the degrees of freedom

The degrees of freedomwith total number of rows (r) and columns (c)are 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 P-value

From chi-square table, the P-value for row corresponding to 1 degree of freedom and at 0.05 level of significance is 0.000.

Therefore, the P-value is 0.000.

Also, the critical value is obtained at 0.05 level of significance as 3.841.

07

State the decision

Since the P-value (0.000) is less than the level of significance (0.05). In this case, the null hypothesis is rejected.

Therefore, the decision is to reject the null hypothesis.

08

State the conclusion

There is enough evidence to reject the claim that the true results for lies are independent of polygraph test results. Thus, polygraphs appear to detect the lies effectively.

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Most popular questions from this chapter

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 鈥淭exting While Driving and Other Risky Motor Vehicle Behaviors Among U.S. High School Students,鈥 by O鈥橫alley, 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

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No Texting while driving

156

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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

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152

In his book Outliers,author Malcolm Gladwell argues that more

American-born baseball players have birth dates in the months immediately following July 31 because that was the age cutoff date for nonschool baseball leagues. The table below lists months of births for a sample of American-born baseball players and foreign-born baseball players. Using a 0.05 significance level, is there sufficient evidence to warrant rejection of the claim that months of births of baseball players are independent of whether they are born in America? Do the data appear to support Gladwell鈥檚 claim?


Born in America

Foreign Born

Jan.

387

101

Feb.

329

82

March

366

85

April

344

82

May

336

94

June

313

83

July

313

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Aug.

503

91

Sept.

421

70

Oct.

434

100

Nov.

398

103

Dec.

371

82

Mendelian Genetics Experiments are conducted with hybrids of two types of peas. If the offspring follow Mendel鈥檚 theory of inheritance, the seeds that are produced are yellow smooth, green smooth, yellow wrinkled, and green wrinkled, and they should occur in the ratio of 9:3:3:1, respectively. An experiment is designed to test Mendel鈥檚 theory, with the result that the offspring seeds consist of 307 that are yellow smooth, 77 that are green smooth, 98 that are yellow wrinkled, and 18 that are green wrinkled. Use a 0.05 significance level to test the claim that the results contradict Mendel鈥檚 theory.

Benford鈥檚 Law. According to Benford鈥檚 law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. In Exercises 21鈥24, test for goodness-of-fit with the distribution described by Benford鈥檚 law.

Leading Digits

Benford's Law: Distributuon of leading digits

1

30.10%

2

17.60%

3

12.50%

4

9.70%

5

7.90%

6

6.70%

7

5.80%

8

5.10%

9

4.60%

Detecting Fraud When working for the Brooklyn district attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 15, 0, 76, 479, 183, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford鈥檚 law, the check amounts appear to result from fraud. Use a 0.01 significance level to test for goodness-of-fit with Benford鈥檚 law. Does it appear that the checks are the result of fraud?

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