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I've got this hypothesis "People are better at detecting a lie when they are able to see a persons expression, compared to people that are not". I've tested 2 groups of people: 1) one group of 20 people saw a video of a person stating 3 statements of which one of them is a lie and they had to tell which one of the statements they think is a lie 2) other 20 people did the same but not seeing a person, just listening to him in an audio recording

the results were that 3 out 20 people detected a lie correctly from a video (15%) and 5 people out of other 20 detected a lie correctly from an audio recording (25%), which kind of disproves the hypothesis. Now the question is, what kind of test should i use to compare these results to see if the actual results are significant and not just "accidental"? The closest I could find was the "Two proportions" test, but it is based on having different sizes of samples, whereas i have the same size (20 in each medium)

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    $\begingroup$ I might go here link and look for overlap in the 95% confidence intervals. I would make sure to use the right breed of confidence interval - because they are governed by sample size. $\endgroup$ – EngrStudent Apr 21 '17 at 11:16
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    $\begingroup$ Equal sample size is not a problem when comparing two proportions. $\endgroup$ – Michael Chernick Apr 21 '17 at 12:47
  • $\begingroup$ @Velionis Your proposal to the hypothesis testing is not bad. Michael Chermick has pointed out correctly that sample sizes do not cast a shadow when we are using proportions for testing. $\endgroup$ – Subhash C. Davar Apr 21 '17 at 14:41
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Easiest thing to start with would be be to make a 2x2 contingency table with video or audio assignment in the rows, and correctly & incorrectly detecting the lie in the columns. Put in the numbers (not the percentages), and you'll be able to perform a chi-square test.

the table will look something like this:

correct incorrect Total video 3 17 20 audio 5 15 20 Total 8 32 40

As I believe in not giving the answer away completely (also you'll understand the test statistic better if you do the next part yourself), look at the following wikipage to find how to perform the chisquare test and get a p-value for performing the test you've asked for.

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    $\begingroup$ Don't know if I got this wrong, but Chi-square test requires you to have an expected value, which is used to see how well the observed data fits it. However, I don't have an expected value. All i expect from hypothesis that there is a bigger number of people who detect lies from a video rather than just an audio recording. $\endgroup$ – Velionis Apr 21 '17 at 9:39
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    $\begingroup$ Actually, you do have an expected number, and it is the reasons I had added the row and columntotals to the table: $expected value cell_{i,j} = rowtotal_i*columntotal_j/totalN$. You could (1) calculate this expected value for every cell, (2) calculate the difference between observed and expected per cell, (3) square these numbers and then (4) divide by the expected number per cell, (5) sum the values of all four cells, and you'll have calculated the chi-square value. The final step would be to check a chi-square distribution with the appropriate $df$ to obtain a p-value. $\endgroup$ – IWS Apr 21 '17 at 9:49
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    $\begingroup$ I got a bit confused about the steps you provided: are these for generally calculating the chi square value? Also, what does the i,j stand for? (I'm sorry to bother with these stupid questions, I'm just getting confused all over) $\endgroup$ – Velionis Apr 21 '17 at 10:08
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    $\begingroup$ I hate to be rude, but it's all there on the wiki page I mentioned in my answer if you prefer other notation. $\endgroup$ – IWS Apr 21 '17 at 10:16

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