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I have a hypothesis "People are better at detecting a lie when they are able to see a persons expression, compared to people that are not". The results were that 3 out of 20 people detected a lie correctly from seeing a facial expression and 5 out of other 20 people correctly detected a lie from only hearing it. I've done a chi-square test in R and it gave a p-value of 0.4292, which is higher than 0.05. This basically means, that i should accept the hypothesis. But how can I accept it if I clearly see that there were more people that detected a lie from hearing it rather than seeing a facial expression? In a logical way, my p-value should be less than 0.05, but its not. How do I explain it?

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    $\begingroup$ Intuitively, is 5 out of 20 really "much more" then 3 out of 20? I'd say no. Moreover, your sample is really small so the test is reluctant to reject the null. $\endgroup$ – Tim Apr 23 '17 at 18:34
  • $\begingroup$ What year did you use? $\endgroup$ – Repmat Apr 23 '17 at 18:43
  • $\begingroup$ why is that important? $\endgroup$ – Velionis Apr 23 '17 at 18:44
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The null hypothesis would be "Seeing facial expressions makes no difference to people's ability to detect a lie".

Your p-value of 0.4292 would be the probability of getting 3/20 assuming the null hypothesis is true.

As this is more than 0.05, you fail to reject the null hypothesis, using a significance of 0.05.

In other words there is a fairly high probability of getting 3/20 if there is no difference. Your 3/20 is just down to chance, it's not significant.

So you have no evidence to support the alternative hypothesis "People are better at detecting a lie when they are able to see a persons expression, compared to people that are not"

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  • $\begingroup$ what is the formula for computation of p-value ? and what is the database ? $\endgroup$ – Subhash C. Davar Apr 23 '17 at 22:22
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    $\begingroup$ Your discussion seems to treat the 5/20 as the null case, but the test should be a two-sample test; it's really about whether 3/20 and 5/20 could be consistent with each other than whether 3/20 unusual in itself $\endgroup$ – Glen_b Apr 24 '17 at 1:45
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"Pearson's chi-squared test (χ2) is a statistical test applied to sets of categorical data to evaluate how likely it is that any observed difference between the sets arose by chance."

The null-hypothesis of the test is that both sets do not differ in the frequency. The alpha of p<0.05 is the threshold at which you would reject this null hypothesis (i.e. conclude that sets differ). However, p is far from 0.05, therefore the conclusion is that frequencies of lie detection do not differ significantly between groups. The p value of 0.429 indicates that there is a 42.9% chance of observing a difference this large, or larger, between the two sets assuming the null hypothesis. In other words: The difference between groups is so small that it may likely arise by chance, even if there was 0 difference.

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If you Start with a one sided hypothesis (as you stated in your problem), you should not use the Chi-Square test which is intended for testing a two-sided hypothesis. Also - you should get a p-value greater than 0.5 because the results are opposite to your initial hypothesis - more people detected the lie when hearing. You cannot change your initial hypothesis after you see the results!!

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