In Exercise?? est the given? ?claim. Testing a Lie Detector? 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 Actuall y Lie? N Y o e ( s D ( i L d i N e o t d L ) i e ) Polygraph test indicated 1 4 that the subject lied 5 2 Polygraph test indicated 3 that the subject did not 9 2 lie

Solution 7BSC Step1: From the above given problem we have 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. Did the subject actually LIe No(did not lie) Yes(lied) Polygraph test indicated that 15 42 the subject lied Polygraph test indicated that 32 9 the subject did not lied With = 0.05 level of significance Step2: Our aim is 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 Consider the hypothesis, H 0 The row and column variables are independent that is the subject lies are independent of the polygraph test detection. H 1 The subject lies are dependent of the polygraph test detection. With = 0.05 level of significance Step3: Here, Did the subject actually LIe Row total No(did not lie) Yes(lied) Polygraph test 15 42 57 indicated that the subject lied Polygraph test 32 9 41 indicated that the subject did not lied Column total 47 51 Grand total = 98 Now, we need to find expected frequencies and it can be calculated by using the formula E = (row total)(column total) grand total Observed frequencies Expected frequencies (57)(47) Polygraph test No(did not lie) 15 98 = 27.3367 indicated that the subject lied Polygraph test Yes(lied) 42 (57)(51= 29.6633 98 indicated that the subject lied Polygraph test No(did not lie) 32 (41)(47= 19.6633 98 indicated that the subject did not lied Polygraph test Yes(lied) 9 (41)(51= 21.3367 98 indicated that the subject did not lied Now, The goodness of fit can be calculated by using the formula = (OE)2 E Where, O = observed frequencies in a cell E = expected frequencies in a cell Observe Expected O - E (O E) 2 (O E) E d frequencie frequenc s ies (57)(47) Polygraph test No(did 15 98 = -12.3367 152.1950 5.5674 indicated that the not lie) 27.3367 subject lied Polygraph test Yes(lied) 42 (57)(51= 12.3367 152.1950 5.1308 98 indicated that the 29.6633 subject lied Polygraph test No(did 32 (41)(47= 12.3367 152.1950 7.7401 98 indicated that the not lie) 19.6633 subject did not lied Polygraph test Yes(lied) 9 (41)(51= -12.3367 152.1950 7.1330 98 indicated that the 21.3367 subject did not lied (OE)2 25.5712 E And the degrees of freedom is (r-1)(c-1) = (2-1)(2-1) = 1 Where, Number of rows. Number of columns. Critical value can be obtained by using A-6 table using 1 degrees of freedom with 0.05 level of significance...