Sample Size Determination:
(Using Power Analysis of Two-Sample and Paired T-Test)
One method for assessing the bioavailability of a drug is to note its
concentration in blood and/or urine samples at certain periods of time
after the drug is given. Suppose we want to compare the concentrations
of two types of aspirin (types A and B) in urine specimens taken from
the same person 1 hour after he/ she has taken the drug. In this study
protocol, a specific dosage of either type A or type B aspirin is given
and the urine concentration after 1-hour is measured. One week later,
after the first aspirin has presumably been cleared from the system, the
same dosage of the other aspirin is given to the same person and the
urine concentration after 1-hour is noted. Because the order of giving
the drugs may affect the results, a table of random numbers is used to
decide which of the two types of aspirin to give first. The
concentration will be measured as a percentage, and it is presumed that
the standard deviation of the difference between subjects is
approximately 3%.
Suppose we would like to test whether the mean urine concentration for
Aspirin B is less than Aspirin A at an alpha=0.05 level of
significance. How many samples would we need to collect such that there
would be a 75% chance of correctly rejecting the null hypothesis if the
urine concentration of Aspirin B is 1.5% less than that of Aspirin
A?
We need to find no of samples
Given problem pertains to “Power Analysis of Paired T-Test” t/f load relevant library first:
library(pwr)Data:
Alpha=0.05
Difference in mean=1.5%
Standard Deviation=3%
Therefore d=0.5 (Difference in Means / Standard deviation)
Power=0.75
Writing the Hypotheses:
Null: H0: μA = μB
Alternate: H1: μA > μB
Where μA & μB = mean concentrations of Aspirin A & B in a subject’s blood stream after 1 hour
Finding no of required samples:
pwr.t.test(n=NULL, d= 0.5, sig.level = 0.05,power = 0.75, type="paired", alternative="greater")##
## Paired t test power calculation
##
## n = 22.92958
## d = 0.5
## sig.level = 0.05
## power = 0.75
## alternative = greater
##
## NOTE: n is number of *pairs*Conclusion:
We need to collect 46 samples for total or 23 pairs for Aspirin A and B to have a 75% chance of correctly rejecting the null hypothesis.
Can active exercise shorten the time that it takes an infant to learn how to walk alone? Researchers would at a major university would like to design an experiment to test this hypothesis. Specifically, they would like to use one-week old male infants from white middle-class white families as test subjects and allocate into to one of two treatment groups. Those in the active exercise group will receive stimulation of the walking reflexes for four 3-minute sessions each day from the beginning of the second week through the end of the eighth week following birth; those in the other group will receive no such stimulation. Following this, the time (in months) that it takes subjects from each group to walk independently will then be measured.
Since this is a preliminary project, the researchers would like to use an alpha=0.10 level of significance for this test, but would like to achieve a power of .85 at detecting a mean difference that is half of the (pooled) standard deviation of walking times. How many infants do they need to enroll in each group? How many total??
We need to find no of samples
Given problem pertains to “Power Analysis of Two Sample T-Test”
DATA:
Alpha: 0.10
Power=0.85
d=0.5 (Difference in Means / Standard deviation)
Writing the Hypotheses:
Null: H0: μ1 = μ2
Alternate: H1: μ1 < μ2
Where μ1 = mean time to walk in months for the ‘active exercise group’ and μ2 = mean time to walk in months for the ‘non-stimulated group’
Finding no of required samples:
pwr.t.test(n=NULL, d=-0.5,sig.level=0.1,power=0.85,type=c("two.sample"),alternative = c("less"))##
## Two-sample t test power calculation
##
## n = 43.40568
## d = -0.5
## sig.level = 0.1
## power = 0.85
## alternative = less
##
## NOTE: n is number in *each* groupConclusion:
It will take 44 infants enrolling in each group or 88 total to detect.
getwd()
library(pwr)
#Question # 1:
#We need to find the no of samples
?pwr.t.test
#Question 1:
#Paired T-Test:
#DATA:
#Alpha=0.05
#Difference in mean=1.5%
#Standard Deviation=3%
#Therefore d=0.5 (Difference in Means / Standard deviation)
#Power=0.75
#Now finding no of required samples:
pwr.t.test(n=NULL, d= 0.5, sig.level = 0.05,power = 0.75, type="paired", alternative="greater")
#Question 2:
#2-Sample T-Test
#DATA:
#Alpha: 0.10
#Power=0.85
#d=0.5 (Difference in Means / Standard deviation)
#Now finding no of required samples:
pwr.t.test(n=NULL, d=-0.5,sig.level=0.1,power=0.85,type=c("two.sample"),alternative = c("less"))