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 i.e, critical value is 3.841 2 distrib ution Area to the right of the critica l value Degre 0.995 0.990. 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005 es of freedo m 1 - - 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597 Conclusion: Here the Test Statistic value is greater than the Critical value and it falls in the Rejection region and so there is sufficient evidence to reject the Null Hypothesis thus the subject lies are dependent of the polygraph test detection. Step4: By using TI-83 Calculator we can find test statistics and p value and steps are shown below: 1). To get the MATRIX menu for entering the observed frequencies press . The secreen is displayed below: 2).Select matrix A and enter the observed frequencies The data should be entered as in the original contingency table. 3). Now use the arrow key and select EDIT from the menu item and further press enter , selecting the default MATRIX[A]. The screen is displayed below: 4).The dimensions of the MATRIX[A] is entered now in the format that is where r is the Number of rows and c is the Number of columns. Input the value 2 for r and then click enter and further input value 2 for c and then click enter. The secreen is displayed below. 5).Now enter the observed frequencies in the MATRIX[A], click on enter after entering each frequency. The secreen is displayed below. 6).Now press STAT and select TESTS as shown below. 7). Now use the arrow key to select the -Test as shown below and press enter. 8). Now use the arrow key to select the -Test as shown below and press enter. 9).The value of Test Statistic and P-value with degrees of freedom (df) is obtained as displayed in the below screen. 10).To get the graph select Draw by using the arrow keys and press enter. 11).The graph showing the test statistic and P-value is obtained below. Conclusion: We have = 25.5712 and p value is p = 0.000 comparing p value with level of significance that is p value(p = 0.000) is less than the level of significance = 0.05 Therefore, there is sufficient evidence to reject the Null Hypothesis and also sufficient reason to reject the claim that the sample is drawn from the population where the subject lies are independent of the polygraph test detection. The Test of Independence indicates that the the Test statistic value is greater than the Critical value fetched from the tables and thus it falls in the rejection region. The Null Hypothesis is therefore rejected. The polygraph test is effective in test detection but there are discrepancies resuting in false negatives and false positive answers and thereby they are not much reliable.