NHST: Comparing the effect with a standard. The size.
We are, supposedly, trying to reprove the accepted fact that the observed/measured Normal Population has mean= 3. Following my experiment I had found, with a size 30 random sample, mean=5.917, standard deviation=0.663, we intend to build a 95% Confidence Interval containing D, the excess from established mean=3. Hence
(2.045 are the 97.5% quantile of Student T, 29 degrees of freedom). Therefore we discover that D stays inside the interval [2.67, 3.16]: because it doesn?t contain 0, we can state our find will falls down the common value 3. Recurring to only standard statistical procedures, the problem could be correctly solved. __________
The size an effect is revealed
The above scientist intending to forecast how much data should to use in order to prove H0: mean > 5.8, do the following: sqrt(n) > (0.663 * 2.045)/(5.947 - 5.8) = 11.588, resulting that the size should be at least 135. Of course . . . if it happens, this time, the standard deviation be a little larger than 0.662 and the intent is spoiled: if it was 1.000 the size amounts to194 or more.