Question

A.

x 11.2.5-T Is Question Help The table below includes results from polygraph (lie detector) experiments conducted by researche

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.

x 11.2.6-T Question Help Results from a civil servant exam are shown in the table to the right. Is there sufficient evidence

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.

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Answer #1

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:

\chi ^{2} = ΣΙΟ-ΕΕ

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:

\chi ^{2} =  ΣΙΟ-ΕΕ = 0.693268+0.713658+1.015143+1.044999 = 3.467068

Chi-Square = \chi ^{2} = 3.467 (Round to three decimal places as needed)

From Chi-Square to p

Calculate Degrees of Freedom

= (rows − 1) \times (columns − 1)

DF = (2 − 1)(2 − 1) = 1×1 = 1

Critical value :

Critical value at  \alpha = 0.05 with   DF = 1

Хо.05- = 3.841

P- value

P- value at  \chi ^{2} = 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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