The table below includes results from polygraph (lie...

The table below includes results from polygraph (lie detector) experiments conducted by researchers. In each case, it was known if the subjected 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 truth and lies? Did the Subject Actually Lie? No (Did Not Lie) | Yes (Lied) Polygraph test indicated that the subject lied. | 19 | 24 Polygraph test indicated that the subject did not lie. | 17 | 8 Please select ALL correct statement from those given below. Answer 1. H0: Whether a subject lies is independent of the polygraph test indication. H1: Whether a subject lies is not independent of the polygraph test indication. 2. X2 test statistic = 3.599 3. P-value = 0.0578 4. There is not sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication.

Transcript

00:01 Hi, i'm david and i'm here to have you answer the question.

00:03 In the question here, we are going to discuss about the high -plus testing.

00:07 Here we want to test whether the subject line will be independent with the polygraph test result.

00:16 So we will have the known hypothesis.

00:20 It will be the subject line is independent of the polygraph test.

00:45 Result and versus the own connective hypothesis the h1 that the subject line is not independent of the body broadcast result now in your question we're given a data set the data frequency from the two variables so let me go back and copy the portion in your question and i put it here now this will be called the observe value i have to create a non the table and this table here will be the expected count so this one will be did not lie will be the no and this one will be the yes and this one will be the test indicated and the subject line and the one not line and to find expected count we would have to compute the total first the total for the column from the 19 plus 17 equal to the 36 and 24 plus 8 equal to the 32 and the total here as well equal to the 19 plus 24 equal to the 43 this only be the 17 plus 8 equal to the 25 and we have the everything here it will equal to the 68 now from here we will be able to find the expected count the first one will turn the total for the column and the right multiply it together divided by the 68 and if we compute it when the 36 times 43 divided by the 68 equal to the 22 form 76 5 higher to the 3 decimal places the next one we will do with the 32 so 32 times window 43 divided by the 68 32 times when the 43 divided by the 68 32 times when the 43 divided by the 68 equal to the 20 point 2 35 so similarly the next one we use under 25 so 36 times 25 the 10 minus 68 equal to the 13 point 2 3 5 and the last 1 32 times 25 the 15 minus 68 equal to the 11 point 765 therefore we can find the task statistics denoted by the k square and this one will be called the expected count denoted by the e so the formula will be the summation of the absorption of the absorption amount expected square divided by exported the first cell will be the 19 and this one when the 19 will manage the 22765 square divided by the 2 2 .765 we keep to lay this until the last cell will be the 8 and this one so 8 minus 11 .765 square divided by the 11 .765.

04:10 Every compute we will have the 19 minus 22.

04:21 765 square divided by 22 .765 plus the 24 minus 203.

04:31 23 .3 .25 square divided by 20.

04:36 1225 plus 17 minus 13 .235 square divided by 13 .25.

04:47 Then we get equal to the 3 .5.

04:51 I route to the 3 decimal prices.

04:55 And that will be the task statistics.

04:57 Then we want to find the p value...

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