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Suppose we have $100$ people and we obtain the average blood pressure $x$. Now suppose we have another group of $40$ people and they have an average blood pressure $\approx x$. How do we quantify the fact that the second experiment is more "efficient" than the first experiment? Would we do a power analysis? What would be the null hypothesis?

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What do you mean by saying that "the second experiment is more 'efficient' than the first experiment"? The fact that 2 groups have similar means is not what statisticians mean by 'efficiency'. –  gung Jan 14 '13 at 16:43
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Note sure I pefectly understand your questions but here is my contribution. That's just mean that your two means are from samples drawn from the same a population with... a mean of x. If you don't use a referent(population) mean to compare these estimated means then you don't have null hypothesis and no power to calculate. Do you mean you want to test these means vs a referent mean, for example with a one sample t-test? If so, with about the same estimated x values (and so same effect size) in your two samples, then, by definition, you'll have much power with N=100 because of the higher N. –  Alric Jan 14 '13 at 17:26
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By "efficient" I gather you mean something like: "We get roughly the same information back, at 40 per cent of the cost." Yes? If so, then you might phrase your statement in terms of the cost of uncertainty reduction or the value of information. In that case, it would be helpful to understand the larger context in which this information is intended to be used, in order to understand what are the costs of testing versus the costs of error. –  Arthur Small Jan 14 '13 at 17:42
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I can't think of a meaning of "efficient" that would be tested by two groups having similar means. I think you need to think about measures of spread and precision of estimate. –  Peter Flom Jan 14 '13 at 17:51

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You didn't get the same information: you got a confidence interval of a certain width from measuring 40 people, and then after measuring 100 people you got a confidence interval of a (probably) much smaller width- that's information.

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You appear to assume that the CI will shrink as the sample size goes up. That is a general tendency, to be sure--assuming the two groups were measured in the same way and that the CI's were computed in the same way. In discussions of efficiency, precisely these two factors are of concern: efficiency is used to evaluate different ways of measuring the same quantity and to compare different statistical procedures to estimate a given quantity. –  whuber Jan 15 '13 at 17:41
    
@whuber: Of course you're right, but I'm pretty sure that's what the OP is asking- measured from the same populations in the same way. (In fact, if you're estimating the variance from the sample it is theoretically possible that the new sample variance is so much higher than the old sample variance that it makes up for the change in n, but that's so extremely unlikely that it's trivial). –  David Robinson Jan 15 '13 at 18:01
    
(I have upvoted your answer based on your edit.) When you have to guess what the OP is asking, you have two good courses of action: either request clarification in a comment to the question (which several interlocutors have done here) or, if you feel you can interpret the question, at a minimum please make explicit within your answer your interpretation and all assumptions needed for your answer to be correct. (We have had some very good statisticians get into lots of trouble here by not following this advice, because they have no defense when it later turns out they misinterpreted.) –  whuber Jan 15 '13 at 18:06

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