Problem

A right-tail randomization test has observed statistic \(+5\) and \(5\) of \(500\) shuffled statistics at least as extreme. Compute \(p\), compare it with \(\text{alpha}~=~0.05\), and interpret the treatment effect.

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What this problem is really about

Preserve the signed treatment-minus-control statistic when locating the observation in the randomization distribution. The stated alternative determines which tail counts as at least as extreme, and the empirical p-value is that tail count divided by all shuffles. Compare it with alpha for the null decision, then use the sign and units to describe direction and observed magnitude without claiming an exact true effect.

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Because the observed statistic +5 is positive, ignore p = 5/500 = 0.010 and do not reject the no-effect null; there is no evidence that treatment increases response.

Because 0.010 < 0.05, reject the no-effect null; the result is evidence that treatment increases response by about 5 units.

Compute p by dividing the 5 extreme results by the observed statistic: 5/5 = 1. Because 1 is at least 0.05, do not reject the no-effect null.

Because 0.010 < 0.05, reject the no-effect null; conclude that there is no treatment effect despite the positive observed difference.

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A right-tail randomization test has observed statistic \(+5\) and \(5\) of \(500\) shuffled statistics at least as extreme. Compute \(p\), compare it with \(\text{alpha}~=~0.05\), and interpret the treatment effect.

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