A.
Determine the Test Statistic
x2=______(Round to three decimal places as needed)
Determine the P-Value of the test statistic
P-value=_____(Round to four decimal places as needed)
Do the results suggest that polygraphs are effective in distinguishing between truth and lies?
A.There is not sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication.
B.There is sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication.
C.There is not sufficient evidence to warrant rejection of the claim that polygraph testing is 95% accurate.
D.There is sufficient evidence to warrant rejection of the claim that polygraph testing is 95% accurate.
B.
Determine the Test Statistic
x2=______(Round to three decimal places as needed)
Determine the P-Value of the test statistic
P-value=_____(Round to four decimal places as needed)
Is there sufficient evidence to support the claim that results from the test are discriminatory?
(a) there is not sufficient evidence to support the claim that results are discriminatory. (
b) there is sufficient evidence to support the claim that the results are discriminatory.
(c) there is not sufficient evidence to reject the claim that a white candidate is more likely to pass the test than a minority candidate.
(d) there is sufficient evidence to reject the claim that a white candidate is more likely to pass the test than a minority candidate.
SOLUTION:
From given data,
Add up rows and columns:
(O) | No (Did Not Lie) | Yes(Lied) | Total |
Polygraph test indicated that the subject lied. | 17 | 24 | 41 |
Polygraph test indicated that the subject did not lie. | 18 | 10 | 28 |
Total | 35 | 34 | 69 |
Calculate "Expected Value" for each entry:
Multiply each row total by each column total and divide by the overall total:
E = (Row total)(Column total) / Grand Total
(E) | No (Did Not Lie) | Yes(Lied) |
Polygraph test indicated that the subject lied. | 35*41/ 69 = 20.7971 | 34*41/ 69 = 20.2029 |
Polygraph test indicated that the subject did not lie | 35*28 / 69 = 14.2029 | 34*28/ 69= 13.7971 |
Subtract expected from actual, square it, then divide by expected:
=
No (Did Not Lie) | Yes(Lied) | |
Polygraph test indicated that the subject lied. |
(17-20.7971)2 / 20.7971 =0.693268 |
(24-20.2029)2 / 20.2029 =0.713658 |
Polygraph test indicated that the subject did not lie |
(18-14.2029)2 / 14.2029 =1.015143 |
(10-13.7971)2 / 13.7971 = 1.044999 |
Determine the Test Statistic
Now add up those values:
=
= 0.693268+0.713658+1.015143+1.044999 = 3.467068
Chi-Square = =
3.467 (Round to three decimal places as needed)
From Chi-Square to p
Calculate Degrees of Freedom
= (rows − 1) (columns −
1)
DF = (2 − 1)(2 − 1) = 1×1 = 1
Critical value :
Critical value at = 0.05
with DF = 1
= 3.841
P- value
P- value at =
3.467 with DF = 1
P- value =0.0500136
approx , P- value = 0.0500 (Round to four decimal places as needed)
Since P- value is greater than critical value,
we reject null hypothesis. At 5% level , there is sufficient evidence.
B.There is sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication. (correct)
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