Determine the P-value of the test statistic. Do the results suggest that polygra

Question

Determine the P-value of the test statistic.

Do the results suggest that polygraphs are effective in distinguishing between truth and lies?
A.
There is sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication.
This is the correct answer.B.
There isThere is sufficient evidence to warrant rejection of the claim that polygraph testing is 95% accurate.
Your answer is not correct.C.
There is notThere is not sufficient evidence to warrant rejection of the claim that polygraph testing is 95% accurate.
D.
There is notThere is not sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication.

The table below r cludes results orn p ly gr p e detector experiments conducted y resea chers each casewas kno n the s b eded i d or did not ile so the table indicates when the polygraph test was correct 0 05 significance level to test the claim that whether a subject lias is independent of the polygraph test indication Do the results suggest that polygraphs are efectivg in distinguishing between tuth and les? 22 Click the icon to view the table se-

Explanation / Answer

Solution:

For the given scenario of null and alternative hypothesis, first we have to find chi square test statistic and then we have to find p-value. After finding p-value we have to take decision regarding the null hypothesis.

The test statistic formula is given as below:

Test statistic = Chi square = [(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Expected frequencies = E = row total * column total / grand total

Calculation tables are given as below:

Observed Frequencies (O)

Did the subject actually lie?

Row variable

No

Yes

Total

Polygraph test indicated that the subject lied

8

31

39

Polygraph test indicated that the subject did not lie

14

14

28

Total

22

45

67

Expected Frequencies (E)

Did the subject actually lie?

Row variable

No

Yes

Total

Polygraph test indicated that the subject lied

39*22/67 =12.80597

39*45/67 =26.19403

39

Polygraph test indicated that the subject did not lie

28*22/67 =9.19403

28*45/67 =18.80597

28

Total

22

45

67

( O - E)

-4.80597

4.80597

4.80597

-4.80597

(O - E)^2/E

1.803639

0.881779

2.512212

1.228192

We are given

Level of significance = = 0.05

Number of rows = r = 2

Number of columns = c = 2

Degrees of freedom = (r – 1)*(c – 1) =(2 – 1) * (2 – 1) = 1*1 = 1

Critical value = 3.841459

Test statistic = Chi square = [(O – E)^2/E] = 6.425822

P-value = 0.011247

P-value < = 0.05

So, we reject the null hypothesis that whether a subject lies is not independent of the polygraph test indication.

There is sufficient evidence to conclude that whether a subject lies is not independent of the polygraph test indication.

There is sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication.

Observed Frequencies (O)

Did the subject actually lie?

Row variable

No

Yes

Total

Polygraph test indicated that the subject lied

8

31

39

Polygraph test indicated that the subject did not lie

14

14

28

Total

22

45

67

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