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 Actually Lie?
	No (Did Not Lie)	Yes (Lied)
Polygraph test indicates that the subject lied.	15	42
Polygraph test indicates that the subject did not lied.	32	9

The data forthe polygraph test is provided.

The level of significance is 0.05.

Theexpected frequency is computed as,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The table of observed values with row and column total is represented as,

Theexpected frequency tableis represented as,

Assume that the subjects are randomly selected and assigned to treatment groups. Also, the expected values are greater than 5.

Thus, all the requirements are satisfied.

To test if the true results are independent of the results obtained from the polygraph, the hypotheses are formulated as:

\({H_0}:\)Thesubject’s lie is independent of the polygraph test indication.

\({H_1}:\) The subject’s lie is not independent of the polygraph test indication.

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {15 - 27.3367} \right)}^2}}}{{27.3367}} + \frac{{{{\left( {42 - 29.6633} \right)}^2}}}{{29.6633}} + ... + \frac{{{{\left( {9 - 21.3367} \right)}^2}}}{{21.3367}}\\ = 25.571\end{aligned}\]

Therefore, the value of the test statistic is 25.571.

The degrees of freedomwith total number of rows (r) and columns (c)are computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {2 - 1} \right)\\ = 1\end{aligned}\)

Therefore, the degrees of freedom are 1.

From chi-square table, the P-value for row corresponding to 1 degree of freedom and at 0.05 level of significance is 0.000.

Therefore, the P-value is 0.000.

Also, the critical value is obtained at 0.05 level of significance as 3.841.

Since the P-value (0.000) is less than the level of significance (0.05). In this case, the null hypothesis is rejected.

Therefore, the decision is to reject the null hypothesis.

There is enough evidence to reject the claim that the true results for lies are independent of polygraph test results. Thus, polygraphs appear to detect the lies effectively.