Q. 1 You want to study the effect of a new drug for insomnia for one week. The o
ID: 2929991 • Letter: Q
Question
Q. 1 You want to study the effect of a new drug for insomnia for one week. The outcome measure is duration of sleep one week post-intervention. The mean sleep time in the control group is 4 hours. You believe that an increase in sleep time from 4 to 6 hours is a clinically meaningful difference worth detecting. You will use two-sided test, = 0.05, power = 90%, and standard deviation () = 2hr. How many subjects do you need per group? Show all work.
Q. 2 Regarding Question #1: You are interested in detecting an increase in sleep time of 1 hour (and other information is the same as in Question 1). How will this affect your sample size requirement? Show all work.
Explanation / Answer
Here Type I error = 0.05
Probability of type II error = 1 - power = 1 - 0.90 = 0.10
standard deviation () = 2hr.
Let say sample sie per group = n
so, at 5% level. the upper 95% confidence bound = xbar + Z95% ( / sqrt(n))
= 4 + 1.645 * 2/ sqrt(n) = 4 + 3.29/sqrt(n)
Here Pr( Type II error) = Pr(Sleep < 4 + 3.29/ n ; 6 ; 2/n) = 0.01
Z - value for p - value = 0.10 is
Z = -1.28
-1.28 = (4 + 3.29/ n - 6)/ 2/n
-2.56/n = 3.29/n -2
2 = 5.85/n
n = 5.85/2 = 2.925
n = 8.55 or 9
Question 2. Here we wish to detect an increase in sleep time of 1 hour.
so now let say new sample size is n.
As all given information is same.
so, at 5% level. the upper 95% confidence bound = xbar + Z95% ( / sqrt(n))
= 4 + 1.645 * 2/ sqrt(n) = 4 + 3.29/sqrt(n)
Here we have to detect an increase in sleep time of 1 hour that means of 5 hour
Here Pr( Type II error) = Pr(Sleep < 4 + 3.29/ n ; 5 ; 2/n) = 0.01
Z - value for p - value = 0.10 is
Z = -1.28
-1.28 = (4 + 3.29/ n - 5 )/ 2/n
-2.56/n = 3.29/n -1
1 = 5.85/n
n = 5.85/1 = 5.85
n = 34.222 or 35
that makes it increase four time increase in sample size.
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