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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

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

Is the hypothesis test left-tailed, right-tailed, or two-tailed?

A randomized controlled trial was designed to compare the effectiveness of splinting versus surgery in the treatment of carpal tunnel syndrome. Results are given in the table below (based on data from 鈥淪plinting vs. Surgery in the Treatment of Carpal Tunnel Syndrome,鈥 by Gerritsen et al., Journal of the American Medical Association,Vol. 288,

No. 10). The results are based on evaluations made one year after the treatment. Using a 0.01 significance level, test the claim that success is independent of the type of treatment. What do the results suggest about treating carpal tunnel syndrome?

Successful Treatment

Unsuccessful Treatment

Splint Treatment

60

23

Surgery Treatment

67

6

Motor Vehicle Fatalities The table below lists motor vehicle fatalities by day of the week for a recent year (based on data from the Insurance Institute for Highway Safety). Use a 0.01 significance level to test the claim that auto fatalities occur on the different days of the week with the same frequency. Provide an explanation for the results.

Day

Sun.

Mon.

Tues.

Wed.

Thurs.

Fri.

Sat.

Frequency

5304

4002

4082

4010

4268

5068

5985

Cybersecurity The accompanying Statdisk results shown in the margin are obtained from the data given in Exercise 1. What should be concluded when testing the claim that the leading digits have a distribution that fits well with Benford鈥檚 law?

Questions 6鈥10 refer to the sample data in the following table, which describes the fate of the passengers and crew aboard the Titanic when it sank on April 15, 1912. Assume that the data are a sample from a large population and we want to use a 0.05 significance level to test the claim that surviving is independent of whether the person is a man, woman, boy, or girl.


Men

Women

Boys

Girls

Survived

332

318

29

27

Died

1360

104

35

18

Is the hypothesis test left-tailed, right-tailed, or two-tailed?

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