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some relevant statistics theory

Posted By: Bob Koca
Date: Saturday, 20 November 2010, at 7:58 p.m.

In Response To: LOL (higonefive)

Do not agree.

First of all "adding the intervals" is not the right way to say what you mean. The confidence interval is a range. For example if you see "the 95% confidence interval is .72 +- .03" the confidence INTERVAL is [.69, .75]. What I am guessing you are thinking could be said as "the confidence intervals do not overlap".

Secondly, Even if that is true you won't be "sure" of anything. You would have a very very high (much higher than 95%) confidence though. Requiring that the 95% CI levels do not overlap is a very stringent condition.

Roughly, (this works if the +- values are about equal which will usually be the case if the plays were rolled out the same number of times) to find the margin of error for the difference you should add the two +- values and divide by SQRT(2), approx 1.41

As an example:

Play A 95% CI is .70 +- .10

Play B 95% CI is .55 +- .10

The 95% CI for the difference is .15 +- .141 = [.009, .291] This does not include 0 so you have greater than 95% confidence (barely) that play A is better.

The +- values have a term "margin of error" and for a 95% confidence interval are calculated as approximately 1.96 * std error. The 1.96 value comes from the normal distribution and is chosen so that P( -1.96 < Z < 1.96) = .95 where Z is a normally distributed variable with mean 0 and std. dev 1.

Working backwards here we can determine that the std errors for A and B are .10/1.96 = .05102

The std error for the difference is (.05102+.05102)/SQRT(2) = .07215 The difference of .15 is that value times 2.08. Using a normal table (or software) P(-2.08 < Z < 2.08) = .9624 so there is 96.24% confidence in my example that play A is better than play B.

It is left as an exercise what would be the confidence if the plays were actually .2 (the sum of the 95% confidence interval margins of error) apart.

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