Statistics 403 Final Exam
December 17, 2010
1. We are planning a study in which our goal will be to accurately estimate a treatment eﬀect, expressed
as the diﬀerence in mean responses between treated and untreated subjects. Our goal is to have
the standard error for our estimate be around 20% of the true value. Based on previous studies, we
anticipate the true value to be around 2. We also expect, based on previous research, that the treated
subjects’ standard deviation is 1, and the untreated subjects’ standard deviation is 2. Our study will
enroll twice as many treated subjects as untreated subjects.
(a) What total sample size should we obtain?
Solution:
The standard error of the estimated treatment eﬀect should be 0
.
2
·
2 = 0
.
4,
thus the variance of the estimated treatment eﬀect should be around 0
.
16. The estimated
treatment eﬀect is
¯
Y
T

¯
Y
U
,
and the variance of the estimated treatment eﬀect is
σ
2
T
/n
T
+
σ
2
U
/n
U
= 1
/
(2
n
) + 4
/n
= 9
/
(2
n
)
,
where
n
is the number of untreated subjects, and thus
n
+ 2
n
= 3
n
is the total sample size.
Thus we need
9
/
(2
n
) = 0
.
16
,
so
n
= 9
/
(2
·
0
.
16)
≈
28, so the total sample size should be around 84.
(b) Consider the power for testing the null hypothesis of zero treatment eﬀect, using the sample size
you found in part (a), and assuming that the eﬀect size is 2, as supposed. Will this power be
greater than 80%? Explain your reasoning, but you do not need to obtain a numerical value for
the power.
Solution:
When
n
= 28, the variance of the estimated treatment eﬀect is 9
/
56. If the actual
eﬀect size is 1
.
5, then
ET
=
2
p
9
/
56
≈
4
.
99
Thus the power is
P
(
T >
2) =
P
(
T

ET >
2

)
=
P
(
Z >
2

4
.
99)
=
P
(
Z >

3)
.
We know from the 689599 rule that 99% of the probability of a standard normal distribution
is between 3 and 3. Thus
P
(
Z >

3) is greater than 0.99. So the power is far greater than
80%.
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 Winter '12
 KerbyShedden
 Statistics, Normal Distribution, Standard Deviation, researcher, treatment effect, total sample size

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