Cochran’s Theorem
3.7 Determining Sample Size
■ F I G U R E 3 . 1 4 JMP output from Example 3.1 Response Etch rate
Whole Model
RF power
Least Squares Means Table
Level Least Sq Mean Std Error Mean
160 551.20000 8.1694553 551.200
180 587.40000 8.1694553 587.400
200 625.40000 8.1694553 625.400
220 707.00000 8.1694553 707.000
Residual by Predicted Plot
30 20 10 0 –10 –20 –30
Etch rate Predicted Etch rate Residual
550 600 650 700
Summary of Fit
6 0 6 2 9 . 0 e
r a u q S R
6 9 1 2 1 9 . 0 j
d A e r a u q S R
6 4 7 6 2 . 8 1 r
o r r E e r a u q S n a e M t o o R
5 7 . 7 1 6 e
s n o p s e R f o n a e M
0 2 )
s t g W m u S r o ( s n o i t a v r e s b O
Analysis of Variance
Source DF Sum of Squares Mean Square F Ratio
Model 3 66870.550 22290.2 66.7971
Error 16 5339.200 333.7 Prob F
C. Total 19 72209.750 .0001
Effect Tests
Source Nparm DF Sum of Squares F Ratio Prob > F
RF power 3 3 66870.550 66.7971 .0001
Actual by Predicted Plot
750 700 650 600 550
550
Etch rate Predicted P < .0001 RSq = 0.93 RMSE = 18.267
Etch rate Actual
600 650 700
A significant problem with this approach to using OC curves is that it is usually difficult to select a set of treatment means on which the sample size decision should be based. An alternate approach is to select a sample size such that if the difference between any two treatment means exceeds a specified value, the null hypothesis should be rejected. If the difference between any two treatment means is as large as D, it can be shown that the minimum value of 2is
(3.45) Because this is a minimum value of 2, the corresponding sample size obtained from the operating characteristic curve is a conservative value; that is, it provides a power at least as great as that specified by the experimenter.
To illustrate this approach, suppose that in the plasma etching experiment from Example 3.1, the experimenter wished to reject the null hypothesis with probability at least 0.90 if any two treatment means differed by as much as 75 Å/min and 0.01. Then, assuming that 25 psi, we find the minimum value of 2to be
2 n(75)2
2(4)(252)1.125n 2 nD2
2a2
3.7 Determining Sample Size
107
E X A M P L E 3 . 1 0
Consider the plasma etching experiment described in Example 3.1. Suppose that the experimenter is interested in rejecting the null hypothesis with a probability of at least 0.90 if the four treatment means are
She plans to use 0.01. In this case, because 2500, we have (1/4)2500625 and
Thus, 6250. Suppose the experimenter feels that the standard deviation of etch rate at any particular level of
4i12i
4467562550 3365062525 22600625 25 11575625 50
4i1i 1575 2600 3650 and 4675
power will be no larger than 25 Å/min. Then, by using Equation 3.44, we have
We use the OC curve for a1413 with Na a(n1)4(n1) error degrees of freedom and 0.01 (see Appendix Chart V). As a first guess at the required sample size, try n3 replicates. This yields 22.5n 2.5(3)7.5, 2.74, and 4(2)8 error degrees of free- dom. Consequently, from Chart V, we find that 0.25.
Therefore, the power of the test is approximately 1 10.250.75, which is less than the required 0.90, and so we conclude that n3 replicates are not sufficient.
Proceeding in a similar manner, we can construct the fol- lowing display:
2 ni1
4 2ia2 n(6,250) 4(25)2 2.5n
n 2 a(n1)  Power (1)
3 7.5 2.74 8 0.25 0.75
4 10.0 3.16 12 0.04 0.96
5 12.5 3.54 16 0.01 0.99
Thus, 4 or 5 replicates are sufficient to obtain a test with the required power.
Now we can use the OC curves exactly as in Example 3.10. Suppose we try n4 repli- cates. This results in 21.125(4)4.5, 2.12, and 4(3) 12 degrees of freedom for error. From the OC curve, we find that the power is approximately 0.65. For n5 replicates, we have 25.625, 2.37, and 4(4)16 degrees of freedom for error. From the OC curve, the power is approximately 0.8. For n6 replicates, we have 26.75, 2.60, and 4(5)20 degrees of freedom for error. From the OC curve, the power exceeds 0.90, so n6 replicates are required.
Minitab uses this approach to perform power calculations and find sample sizes for single-factor ANOVAs. Consider the following display:
In the upper portion of the display, we asked Minitab to calculate the power for n5 repli- cates when the maximum difference in treatment means is 75. Notice that the results closely match those obtained from the OC curves. The bottom portion of the display the output when the experimenter requests the sample size to obtain a target power of at least 0.90. Once again, the results agree with those obtained from the OC curve.
3.7.2 Specifying a Standard Deviation Increase
This approach is occasionally helpful in choosing the sample size. If the treatment means do not differ, the standard deviation of an observation chosen at random is . If the treatment means are different, however, the standard deviation of a randomly chosen observation is
If we choose a percentage P for the increase in the standard deviation of an observation beyond which we wish to reject the hypothesis that all treatment means are equal, this is
2i1a 2i/a
Power and Sample Size One-way ANOVA
Alpha 0.01 Assumed standard deviation 25 Number of Levels 4
Sample Maximum
SS Means Size Power Difference
2812.5 5 0.804838 75
The sample size is for each level.
Power and Sample Size One-way ANOVA
Alpha 0.01 Assumed standard deviation 25 Number of Levels 5 4
Sample Target Maximum
SS Means Size Power Actual Power Difference
2812.5 6 0.9 0.915384 75
The sample size is for each level.
equivalent to choosing
or
so that
(3.46) Thus, for a specified value of P, we may compute from Equation 3.46 and then use the operating characteristic curves in Appendix Chart V to determine the required sample size.
For example, in the plasma etching experiment from Example 3.1, suppose that we wish to detect a standard deviation increase of 20 percent with a probability of at least 0.90 and
0.05. Then
Reference to the operating characteristic curves shows that n10 replicates would be required to give the desired sensitivity.
3.7.3 Confidence Interval Estimation Method
This approach assumes that the experimenter wishes to express the final results in terms of confidence intervals and is willing to specify in advance how wide he or she wants these con- fidence intervals to be. For example, suppose that in the plasma etching experiment from Example 3.1, we wanted a 95 percent confidence interval on the difference in mean etch rate for any two power settings to be 30 Å/min and a prior estimate of is 25. Then, using Equation 3.13, we find that the accuracy of the confidence interval is
Suppose that we try n5 replicates. Then, using 2(25)2625 as an estimate of MSE, the accuracy of the confidence interval becomes
which does not meet the requirement. Trying n6 gives
Trying n7 gives
Clearly,n7 is the smallest sample size that will lead to the desired accuracy.
⫾2.064
2(625)7 ⫾27.58⫾2.086
2(625)6 ⫾30.11⫾2.120
2(625)5 ⫾33.52⫾t /2,Na
2MSn E(1.2)21(n)0.66n
i1a 2i/a
/n (10.01P)21(n)
i1a 2i/a
(10.01P)21
2i1a 2i/a
10.01P (Ppercent)
3.7 Determining Sample Size
109
The quoted level of significance in the above illustration applies only to one confi- dence interval. However, the same general approach can be used if the experimenter wishes to prespecify a setof confidence intervals about which a jointorsimultaneous confidence statementis made (see the comments about simultaneous confidence intervals in Section 3.3.3). Furthermore, the confidence intervals could be constructed about more general con- trasts in the treatment means than the pairwise comparison illustrated above.