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Given a sample of 6 workers, three were randomly selected for working environment A and three were selected for working environment B. The observed response (the number of widgets produced) was for workers under treatment A: 50, 120, 250 and for workers under treatment B: 250, 140, 150.

(a) Tell the null hypothesis where the average number of widgets produced under each treatment is equal versus a two-sided alternate hypothesis.
(b) Test the same hypothesis as in (a) but use a randomization test (using permutations).

So for (a), I have $H_o:\mu =0$ and $H_a:\mu\neq 0$ so I've let $d=A-B$, yielding -200, -20 and 100 respectively. Then $\bar{d}=-40$. I have $5$ d.f. and so my t-statistics is $t_5=(-40-0)/(\sigma/\sqrt{6})$. My estimate of $\sigma$ is $95.5$ using the regular formula, so we have $t_5=-1.026$. Then, p = P{|t_5|\geq 1.026}= 1.65 using R, which seems ridiculous.

My issue here is that this is not a matched-pairs design -- did I do something wrong with the setup?

For (b), I am not sure of the right way to randomize between the 6 samples.

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    (Quick note: the larger-or-equal sign $\leq$ is written `\leq`. I've made quite a few 'greater or equal ... `\geq`?' thinkos, myself.)2012-04-05
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    You seem to be using R in a wrong way. The probability that a $t_5$ is greater than 1.026 in absolute value is computed by 2*(1-pt(1.026,5)). You seem to have calculated 2*pt(1.026,5) instead. Of course a probability cannot be greater than 1.2012-04-05
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    What the hell was I thinking when I wrote my leq/geq comment? `leq` is less than or equal. D'oh! >_<2012-04-05

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