Question

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 polygraph test indication. Do the results suggest that polygraphs are effective in distinguishing between truth and lies? Click the icon to view the table. Determine the null and alternative hypotheses. Х More Info O A. Ho: Polygraph testing is not accurate. Hy: Polygraph testing is accurate, OB. Ho: Polygraph testing is accurate. H: Polygraph testing is not accurate, O C. 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. OD. Ho: Whether a subject lies is independent of the polygraph test indication. H: Whether a subject lies is not independent of the polygraph test indication. Did the Subject Actually Lie? No (Did Not Lie) Yes (Lied) 40 Polygraph test indicated that the subject lied. Polygraph test indicated that the subject did not lie. 30 12 Determine the test statistic Print Done x2 = (Round to three decimal places as needed.) Determine the P-value of the test statistic. Click to select your answer(s).

Answer 1

Solution:

The given contingency table is

Did	Not lie	Lied	Total
Polygraph indicated lie	7	40	47
Polygraph indicated no lie	30	12	42
Total	37	52	89

Expected frequency = (row sum * column sum)/Grand sum

E1 = 47*37/89 = 19.539

E2 = 47*52/89 = 27.461

E3 = 42*37/89 = 17.461

E4 = 42*52/89 = 24.539

Chi square goodness of fit test :

Category	Observed frequency, Oi	Expected frequency, Ei	(Oi-Ei)^2/Ei
1	7	19.539	8.0468
2	40	27.461	5.7254
3	30	17.461	9.0044
4	12	24.539	6.4072
Total	89	89	29.184

Hypothesis test :

The null and alternative hypothesis is

H0: Whether a subject lies is independent of the polygraph test indication.

H1: Whether a subject lies is not independent of the polygraph test indication.

Answer - option D

Test statistic :

$\chi^2$ = (Oi-Ei)^2/Ei = 29.184

Degree of freedom = df = (r-1)*(c-1) = (2-1)*(2-1) = 1*1 =1

P-value :

P-value corresponding to $\chi^2$ = 29.184 and df = 1 is 0.0000