Transcript Lecture 5

Sample Size Determination
Text, Section 3-7, pg. 101
• FAQ in designed experiments (what’s the number
of replicates to run?)
• Answer depends on lots of things; including what
type of experiment is being contemplated, how it
will be conducted, resources, and desired
sensitivity
• Sensitivity refers to the difference in means that
the experimenter wishes to detect
• Generally, increasing the number of replications
increases the sensitivity or it makes it easier to
detect small differences in means
• Choice of sample size is closely related to the
probability of type II error b.
• Hypotheses
Ho: m1 = m2
H1: m1  m2
• Type I error – reject H0 when it is true (a)
• Type II error – fail to reject H0 when it is false (b)
• Power = 1 – b = P(Reject HoHo is false)
= P(Fo > Fa,a-1,N-a Ho is false)
• The probability of type II error depends on the true
difference in means d = m1 - m2
Operating Characteristic Curves for
Fixed Effects/Equal Sample Sizes per Treatment Case
 Operating characteristic curves plot b
parameter F where
against a
a
F =
2
n 
i =1
a
2
i
2
 F is related to d. It depends on a (0.01 and 0.05)
and degrees of freedom for numerator (a-1) and
denominator (N-a).
 Assumptions and trials are needed to use the curves
 2 can be estimated through prior experience/
previous experiment/preliminary test/judgment, or
assuming a range of likely values of 2.

=
m

m
i
i
1 a
m =  mi
• ith treatment effect
where
a i =1
Assumed mi’s can be used for which we would like
to reject the null hypothesis with high probability
 Example (etch rate experiment)
If the experimenter is interested in rejecting the null
hypothesis with a probability of at least 0.90 if the
five treatment means are
m1 = 575
m2 = 600
m3 = 650
m4 = 675
 a = 0.01 is planned
 a = 4, N = an = 4n, a – 1 = 3, N – a = 4(n-1)
2

i=1 = 6250 is calculated using assumed mi’s

5
i
  is assumed no larger than 25

ni =1 i2
5
F =
2
a
2
n(6250)
=
= 2.5n
2
4(25 )
 Find the right plot:
 a – 1 = 3 (= v1) determines the use of the upper
plot on page 614 (Appendix Chart V)
 Because a = 0.01 the curves on the right side are
used
Chart V. Operating characteristic curves for the fixed effects model
analysis of variance (Page 614) the upper graph with v1=3 should be
used.
 The objective is to find a b to see if the power is
satisfied
 It needs v2 (or n) to determine the particular curve,
and a value of F to determine b
2.74
a(n-1)
8
0.25
Power (1- b)
0.75
10.0
3.16
12
0.04
0.96
12.5
3.54
16
<0.01
>0.99
n
3
F2
F
7.5
4
5
 Therefore, at least n = 4
b
• It is often difficult to select a set of treatment
means for choosing the sample size
• A very common way to use these charts is to
define a difference in two means D of interest,
then the minimum value of F2 is
2
nD
F2 =
2a 2
• Typically work in term of the ratio of D/ and try
values of n until the desired power is achieved
Other Methods of Determining Sample Sizes
Specifying a Standard Deviation Increase
• As the difference between means increase, the
standard deviation increases
a


2
2
      i / a 
 i =1

• Choose a percentage P for the increase in  of an
observation beyond which the null hypothesis is
rejected, equivalently
a


2
2
    i / a 
 i =1


= 1  P / 100
Specifying a Standard Deviation Increase (continued)
• Rearrange it,
a

i =1
2
i
/a
=

1  P / 100
1
2
• Therefore, F can be expressed as
a
F=

i =1
2
i
/a
/ n
=
1  P / 100
2
 1( n )
• Specify P, a, and the probability to reject the null
hypothesis, then determine n.
Confidence Interval Estimation Method
• Specify in advance how wide the confidence
intervals should be by specifying the accuracy of the
confidence interval
 ta / 2, N a
2MS E
n
• No OC curves are needed. Example: the etch rate
experiment.
• Need a and N (an) to determine ta/2,N-a, and 
to estimate MSE
• Specify the level of confidence (95%, or a
=0.05), difference in mean to be determined
(30Å/min), and (prior) estimate 2 (252 =625)
Confidence Interval Estimation Method (continued)
• Procedure: compare the accuracy with an assumed n,
with the specified accuracy (30Å/min)
• When n = 5, ta/2,N-a =2.120,
ta / 2, N a
2MS E
2(625)
= 2.120
= 33.52  30.00
n
5
• When n = 6, ta/2,N-a =2.086,
2(625)
2.086
= 30.11  30.00
6
• When n = 7, ta/2,N-a =2.064,
2(625)
2.064
= 27.58  30.00
7
Dispersion Effects
• Focus is location effects so far using ANOVA: factor
level means and their differences
• It needs constant variances. If not, using
transformations to stabilize the variances.
• Sometime the dispersion effects are of interest:
whether the different factor levels affect variability
• In such analysis, standard deviation, variance, or
other measures of variability are used as response
variables
An Example – Al Smelting Experiment
• A reaction cell: Alumina and other ingredients (with
a certain ratio) under electric resistance heating
• Four different ratio control algorithms
• Cell voltage is recorded (thousands of voltage
measurements during each run)
• A run consists of one ratio control algorithm
• Average cell voltage (affecting cell temperature), and
the standard deviation of cell voltage over a run
(affecting overall cell efficiency) are response
variables
Ratio
Control
Algorithm
1
2
3
4
5
6
1
4.93(0.05)
4.86(0.04)
4.75(0.05)
4.95(0.06)
4.79(0.03)
4.88(0.05)
2
4.85(0.04)
4.91(0.02)
4.79(0.03)
4.85(0.05)
4.75(0.03)
4.85(0.02)
3
4.83(0.09)
4.88(0.13)
4.90(0.11)
4.75(0.15)
4.82(0.08)
4.90(0.12)
4
4.89(0.03)
4.77(0.04)
4.94(0.05)
4.86(0.05)
4.79(0.03)
4.76(0.02)
Observations
• An ANOVA determines that the ratio control
algorithm had no location effects (the ratio control
algorithm does not change the average cell voltage)
• A transformation is used to study the dispersion
effects
y = -ln(s)
• A standard ANOVA can be done on y, the natural
logarithm of standard deviation -> algorithm 3
produces different standard deviation than others