Question

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

Short answer

A polygraph test is effective in distinguishing truth and lies.

Step-by-step solution

Given information

The data forthe polygraph test is provided.

The level of significance is 0.05.

Compute the expected frequencies and check the requirements

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,

	Did the subject Actually Lie?
	No (Did Not Lie)	Yes (Lied)	Row total
Polygraph test indicates that the subject lied.	15	42	57
Polygraph test indicates that the subject did not lie.	32	9	41
Column total	47	51	98

Theexpected frequency tableis represented as,

	Did the subject Actually Lie?
	No (Did Not Lie)	Yes (Lied)
Polygraph test indicates that the subject lied.	27.3367	29.6633
Polygraph test indicates that the subject did not lie.	19.6633	21.3367

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.

State the hypotheses

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.

Compute the test statistic

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.

Compute the degrees of freedom

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.

Compute the P-value

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.

State the decision

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.

State the conclusion

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.