Reference no: EM131052592

Sleep And investigator here at OHSU studies sleep, circadian rhythm, and the health consequences associated with disrupting this natural rhythm. (There seem to be many negative effects.)One of the characteristics he measures is a person's blood pressure, both systolic (SBP) and diastolic (DBP), and how disrupting a person's sleep (from obstructive sleep apnea [OSA]) canalter the "phase" at which SBP or DBP dips to a minimum. ( FYI: Over the course of a night's sleep, a healthy person's blood pressure slowly drops down to a minimum at some point before slowing rising back to normal.) He wants to study two group: those with OSA and a separate control group with no history of apnea. Pilot data suggests the actual standard deviation in 'phase' measurements is around 100 units of phase. A phase shift of 90 units equates to 6 hrs of time, while a phase shift of 60 units is a shift of 4 hrs. He plans to test fora difference between groups with respect to the mean phase using a two-sided test conducted at the 0.05 level.

• Assuming he splits his total sample size N equally between the two groups, how many subjects per group are needed to ensure an 80% chance of detecting a genuine difference of 4 hrs between the groups (60 units of phase)?
• Sleep studies are expensive, so suppose he can afford to use only n1 = n2 = 10 people per group (N = 20 total).

1. Can you tell him what his power (chances) will be to detect a genuine 6 hr difference between the two groups using a two-sided 0.05-level test?

2. What is his power if he is willing to relax α and accept a 10% chance of a 'false discovery'and is also willing to conjecture the effect is such that the mean phase for OSA is higher than the mean phase for control?

3. How does simultaneously lowering stringency of α and specifying a direction for the alternative improve things while keeping N = 20?