An Introduction to Power Analysis, N = 1
1.
H:   100 H1:  > 100
N = 1,  = 15, Normal Distribution
For  = .10, Zcritical = 1.28, Xcritical = 100 + 1.28(15) = 119.2, therefore: we reject H if X 
119.2. Our chances of rejecting the H are 10% or less if H is true: 10% if  = 100, less than 10% if
 < 100. That is, with our -criterion set at .10, we have no more than a 10% chance of making a
Type I error when H is true.
What if H is false? What are our chances of making a Type II error?
Assume the Truth is that  = 110. We shall fail to reject the false H when X < 119.2.
P(X < 119.2 |  = 110,  = 15) = P(Z < (119.2 - 110) / 15) = P(Z < .61) = approx .73 or 73% = . Our
chances of correctly rejecting this false H, POWER, = P(Z  .61) = 27%. That is, were the H wrong
by 10 points, 2/3 , we would have only a 27% chance of rejecting H.
----------------------------------------------------------------------------------------------------------------------2.
Raise  to .20:
Zcritical = .84
Xcritical = 100 + .84(15) = 112.6
Power = P(Z  (112.6 - 110) / 15) = P(Z  .17) = 43%
Increasing  Raises Power, but at the expense of making Type I errors
more likely.
----------------------------------------------------------------------------------------------------------------------3.
Increase The Difference Between H And The Truth:  = 120
P(Z  (112.6 - 120) / 15) = P(Z  -.49) ==> approximately 69% Power
Big Differences (Between H & Truth) Are Easier To Find Than Small Differences

Copyright 2000, Karl L. Wuensch - All rights reserved.
power1.docx
2
4.
Lower  by using more homogeneous subjects and by holding constant various
extraneous variables:  = 5
Xcritical = 100 + .84(5) = 104.2
P(Z  (104.2 - 120) / 5) = P(Z  -3.16) = 99.9% = Power
Lowering  Increases Power
----------------------------------------------------------------------------------------------------------------------5. Directional vs. Nondirectional Hypotheses
A. Nondirectional
H:  = 100
H1:   100
N=1
 = .10
 = 15
Zcritical =  1.645, that is, reject H if |Z|  1.645
Xcritical
  100 + 1.645(15)   124.68
  100 - 1.645(15)   75. 32
Assume that the true  = 110
P(X  124.68 OR X  75.32) = POWER
P(X  124.68) = P(Z  (124.68 - 110) / 15) = P(Z  .98) =
.1635
P(X  75.32) = P(Z  (75.32 -110) / 15) = P(Z  -2.31) =
-----Power = 17%
.0104
----------------------------------------------------------------------------------------------------------------------B. Directional, correct prediction by H1
See Example 1 on the first page, Power = 27%
3
C. Directional, incorrect prediction by H1
H:   100
H1:  < 100
Zcrit = -1.28 Xcrit = 100 - 1.28(15) = 80.8
Power* = P(X  80.8 | =110) = P(Z  (80.8 - 110) / 15) = P(Z  -1.95) = 3%
Thus, if you can correctly predict the direction of the difference between the truth and the
H, directional hypotheses have a higher probability of rejecting the H than do nondirectional
hypotheses. If you can't, nondirectional hypotheses are more likely to result in rejection of the H .
*One could argue that the probability here represented as “Power” is not Power at all, since the
H (  100) is in fact true ( = 110). Power is the probability of rejecting H given that H is false,
not true.
Copyright 2000, Karl L. Wuensch - All rights reserved.