Re: An interesting experiment
You wish to test the hypothesis that the population mean of both tests are actually equal. In other words, we wish to know if there is enough statistical evidence to say that the results of the two tests are not different.
To recall the results of the test 1, Hero has squat and bluffs the river. Do respondents think Villain will call?
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Less than 25% of the time: <font color="red"> 6 (15% of responders) </font>
Between 25% and 50% of the time: <font color="red">11 (27.5% of responders) </font>
Between 50% and 75% of the time: <font color="red">18 (45% of responders) </font>
More than 75% of the time: <font color="red">5 (12.5% of responders) </font>
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In test 2, the action and the board are identical, but Hero has the nuts and bets the river. Do respondents think Villain will call?
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Less than 25% of the time: <font color="red">7 (29.2% of responders) </font>
Between 25% and 50% of the time: <font color="red">12 (50% of responders) </font>
Between 50% and 75% of the time: <font color="red">4 (16.7% of responders) </font>
More than 75% of the time: <font color="red">1 (4.2% of responders) </font>
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The results are reported as a frequency distribution at equal intervals. To calculate the sample mean and variance of each test, I use the average of each interval.
For example:
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Between 25% and 50% of the time
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is considered 38% on average. This does not present a significant problem; statistical tests can be done on interval frequencies.
This is a hypothesis test concerning the equality of the population means of two normally distributed populations based on independent random samples. The test assumes population variances are equal. The test uses Student’s t-Distribution to find the critical value based on a level of significance and degrees of freedom. A test statistic is computed using the sample means, sample variances, and sample sizes. The test statistic is then compared to the critical value to determine whether the null hypothesis can be rejected. I omitted the calculation in this post because (a) they are lengthy and (b) the format is difficult to transcribe in 2+2 forums. I can’t use subscript or superscript, and the formula would look like one big line of parentheses. Also I can't use symbols like mu, or 'x bar' to differentiate sample from population. However I did the calculations and will somehow show them if requested.
Null Hypothesis: Population 1 Mean – Population 2 Mean = 0
Alternative Hypothesis: Population 1 Mean – Population 2 Mean != 0
Inputs used to find the test statistic:
Sample 1 (test 1) Mean: 0.51675 (51.675% chance of Villain call)
Sample 1 Variance: 0.05172 (5.172%)
Sample 1 Size: 40
Sample 2 Mean: 0.36813
Sample 2 Variance: 0.04138
Sample 2 Size: 24
Degrees of Freedom = Sample 1 Size + Sample 2 Size – 2 = 62
The test statistic is calculated to be 3.03368.
The critical value using 62 degrees of freedom at a 1% level of significance is approximately +/- 2.66. Since the test statistic lays above/below the critical value, the null hypothesis can be rejected at a 1% level of significance.
In other words, there is enough statistical evidence to claim that the population mean of test 1 and test 2 are not equal. Based on the level of significance, there is a 1% chance that population mean of test 1 and test 2 are equal.
Using the samples, there is 99% confidence that uNL are weak-tight!
Clark
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