Did the Subject Actually Lie?No (Did Not Lie)Yes (Lied)Polygraph test indicated thatthe subject lied.Polygraph test indicated thatthe subject did not lie.20442812 The table below includes results from polygraph (lie detector) experiments conducted by researchers. In each case, it was known if the subjected 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 polygraphtest indication. Do the results suggest that polygraphs are effective in distinguishing between truth and lies?E Click the icon to view the table.Determine the null and alternative hypotheses.O A. Ho: Whether a subject lies is independent of the polygraph test indication.H1: Whether a subject lies is not independent of the polygraph test indication.O B. Ho: Polygraph testing is not accurate.H1: Polygraph testing is accurate.O C. Ho: Polygraph testing is accurate.H1: Polygraph testing is not accurate.O D. Ho: Whether a subject lies is not independent of the polygraph test indication.H1: Whether a subject lies is independent of the polygraph test indication.Determine the test statistic.x = (Round to three decimal places as needed.)Determine the P-value of the test statistic.P-value =(Round to four decimal places as needed.)Do the results suggest that polygraphs are effective in distinguishing between truth and lies?O A. There is not sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication.O B. There is sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication.O C. There is not sufficient evidence to warrant rejection of the claim that polygraph testing is 95% accurate.O D. There is sufficient evidence to warrant rejection of the claim that polygraph testing is 95% accurate.

Name: Did the Subject Actually Lie?No (Did Not Lie)Yes (Lied)Polygraph test
Uploaded: 2024-03-20T07:07:50.000Z
Channel: Bartleby
Description: Did the Subject Actually Lie?No (Did Not Lie)Yes (Lied)Polygraph test indicated thatthe subject lied.Polygraph test indicated thatthe subject did not lie.20442812 The table below includes results from polygraph (lie detector) experiments conducted b

Solution: You have correctly chosen the null and alternative hypotheses. Well done! The chi-square test statistic needs to be determined. For that, first we need to find the expected count or expected frequencies. The formula of expected count is as follows: The table of observed (given) counts is as follows. (These are the Observed Frequencies) Reality No (Did not lie) Yes (Lied) Row Total Polygraph test result Subject lied 20 44 64 Subject did not lie 28 12 40 Column Total 48 56 Grand Total = 104 The table of expected counts is as follows: No(Did not lie) Yes (Lied) Subject lied 29.538462 34.461538 Subject did not lie 18.461538 21.538462 Chi-square test statistic formula is as follows: Rounding 14.872619 to three decimal places, it becomes 14.873. χ2=14.873 P-value: It can be found using MS-Excel. Degrees of freedom = (r-1)*(c-1) Where, r is the number of rows and c is the number of columns. r =2 and c =2 Degrees of freedom = (2-1)*(2-1) = 1*1 = 1 To find the p-value, you need to use excel formula CHIDIST(x, deg_freedom) where, x is the value of test statistic and deg_freedom is degrees of freedom. In any empty cell of an excel sheet, type =CHIDIST(14.873, 1) and hit enter, it gives 0.000115. Rounding it to 4 decimal places, it becomes 0.0001. Thus, P-value = 0.0001 Level of significance is given to be 0.05. Decision: Here, p value is less than the level of significance, so we reject the null hypothesis. Conclusion: There is sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of polygraph test indication.