Question

The table below includes results from polygraph (ie detector) experiments conducted by researchers. In each case, it was known if the subjected ied 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 truth and lies? Click the lcon to view the table. Determine the null and altemative hypotheses. OA. Ho: Whether a subject lies is not independent of the polygraph test indication H: Whether a subject lies is independent of the polygraph test indication OB. Ho Polygraph testing is accurate. H,: Polygraph testing is not accurate. O C. H: Polygraph testing is not accurate. H: Polygraph testing is accurate OD. H: Whether a subject lies is independent of the polygraph test indication. H: Whether a subject lies is not independent of the polygraph test indication. Click to select your answeris)

Answer 1

Solution: Here concept of chi square test for independence of attributes is used.

Solution - The null as follows alternative hypothesis and : are Hoi Whether a subject lies in independent of the polygraph test indication H: Whether a subject lies is not independent of the polygraph test indication. To test the hypothesis we use chi-square test statistic for independence of attributes. The test - statistic is given as a x² = Ě Ē I fij - eij i=1 J=1 Cij [(x-1)(1-1)] where fij: Observed frequency ei: expected frequency Now we shall calculate expected frequencies as follows: Observed frequencies No (Did not lie) Yes (lied) total Polygraph test indicate that subject lied 20 28 23 37 Polygraph test indicate that subject did not lie total 31 34 65

Expected frequencies to observed Expected frequency corresponding frequency & is, a E(8) - 28 X 31 - 13.35385 65 Similarly E(20) = 28*34 = 14.64615 65 E (23) = 37831 = 17.64615 65 E (14) = 34x37 - 19.35385 65 . + .. ? = [18 - 13.35 385)2 + (20-1464615) 13.35385 14.64615 (23-17.64615)² (14- 19.35 385)² 17.64615 19.35385 x2 = 7. 2089 ~ 7.209 Now, for P-value of the test statistic we shall look the x² table and find the probability for which the area is close to 7.209 at (5-1).(0-1) = (2-1) (2-1) = 1 degree of freedom. P-value = P(x² 2 7.209] = 0.0072 • 3 P-value = 0.0072

Given that x = 0.05 Since p-valve is less ie. P-valve < 0.05 reject Ho at 0.05 than level of significance therefore, we shall significance level. Conclusion - There is sufficient warrant rejection of the claim a subject lies is independent polygraph test indication. evidence to that whether of the