Question 1170272
the half-interval is t (0.95, df=27)* sqrt((s1^2/n1+(s2^2/n2))
=1.703*sqrt(0.4912)
=1.19
the difference between the two means is (19.65-6.59)=13.06
so the 90%CI is (11.87, 14.25) hertz
This means that we are highly confident that the difference between the mean EEG frequency with and without alcohol is not 0 but rather positive.  Were the differences not significant, 0 would have been contained in the confidence interval. The true difference is not known, but we are highly confident it lies in this interval (90% confident).  It is not a probability, since the parameter exists, if unknown and not knowable, and therefore either lies in the interval or doesn't.

{alcohol and other sedative drugs produce alpha rhythm, which is a fast rhythm, whereas normal sleep produces low frequency delta waves.)