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Question 1170272: In a study of brain waves during sleep, a sample of 29 college students were randomly
separated into two groups. The first group had 15 people and each was given ½ liter of
red wine before sleeping. The second group had 14 people and were given no alcohol
before sleeping. All participants when to sleep at 11 PM and their brainwave activity was
measured from 4-6 AM. The group drinking alcohol had a mean brainwave activity of
19.65 hertz and a standard deviation of 1.86 hertz. The group not drinking alcohol had a
mean of 6.59 hertz and standard deviation of 1.91 hertz. Compute a 90% confidence
interval for the difference in population means of groups drinking alcohol before
sleeping and those not drinking alcohol before sleeping. Explain the meaning of the
confidence intervaL
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 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.)
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