Download Power and Sample Size in Testing One Mean

Power and Sample Size in
Testing One Mean
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Type I & Type II Error
• Type I Error: reject the null hypothesis
when it is true. The probability of a Type
I Error is denoted by .
• Type II Error: accept the null
hypothesis when it is false and the
alternative hypothesis is true. The
probability of a Type II Error is denoted
by .
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Null hypothesis
Decision
True
False
Reject H0
Type I Error
Correct
Decision

1
Correct
Decision
1
Type II Error
Do not reject H0

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A Population
The serum cholesterol levels for all
20- to 24-year-old males of a certain
population was claimed to be
normally distributed with
a mean of 180 mg/100ml and
standard deviation is 46 mg/100ml.
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Research Hypothesis
Is the mean cholesterol level of the
population of 20- to 74-year-olds
higher than 180 mg/100ml ?
H0:  = 180 mg/100ml (or   180 mg/100ml)
Ha:  > 180 mg/100ml
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Power of the Test
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Power of the Test
The power of the statistical test is the ability
of the study as designed to distinguish between
the hypothesized value and some specific
alternative value. That is, the power is the
probability of rejecting the null hypothesis if the
null hypothesis is false.
Power = P(reject H0 | H0 is false)
Power = 1  
The power is usually calculated given a simple alternative
hypothesis: Ha:  = 211 mg/100ml
Power = P(reject H0 | Ha is true)
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Assuming that we use a sample of size 25 and
test the hypothesis with a level of significant  =
0.05, what would be the  and the power of the
test, i.e. 1  ?
If we take the critical value approach, the critical
value would be 1.645. At the level of significant
 = 0.05, we would reject H0 if z  1.645.
Rejection region
Right tail area is 0.05
z
0
1.645
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If a sample of size 25 is selected, what would be
the critical value in terms of the mean
cholesterol level?
Rejection region
Right tail area is 0.05
z
0 1.645
Rejection region
x
180 195.1
?
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X  N (  X  180,  X 
)
25
180  1.645 
46
25
 195.1
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If the alternative hypothesis Ha:  = a = 211
mg/100ml, what would be the probability of
NOT rejecting H0, i.e. ?
H0:   0 = 180 v.s. Ha:    = 211
That is to say “how powerful can this test detect
a 31 mg/100ml increase in average cholesterol
level?”
x
180 195.1 211
 = ?0.042
Power of the test
when sample size is 25:
Power = 1 – 
= 1 – 0.042
= 0.958. 10
Assuming that we use a sample of size 100
and test the hypothesis with a level of
significant  = 0.05, what would be the  and the
power of the test, i.e. 1  ?
Rejection region
z
Right tail area is 0.05
0 1.645
Rejection region
x
180 187.6
?
46
X  N (  X  180,  X 
)
100
46
180  1.645 
 187.6
100
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If we assume the actual population mean is 211
mg/100ml for Ha, what is the probability of
accepting H0, i.e.  ?
x
180 187.6 211
 =?0.00
 = .05
Power of the test when sample size is 100:
Power = 1 –   ?1 – 0  1
Recall that the power was .958 when sample size
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was 25.
n = 100
n = 25
x
180
187.6
211
195.1
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Sample Size Estimation
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H0:  = 180 mg/100ml (or   180 mg/100ml) v.s.
HA:  > 180 mg/100ml,
to perform a test at the power level of 0.95, with
the level of significance of 0.01, how large a
sample do we need?
(Power of 0.95 means we want to risk a 5%
chance of failing to reject the null hypothesis if
the true mean is as large as 211 mg/100ml. Or,
to say, we want to have a 95% chance of
rejecting the null hypothesis if the true mean is
as large as 211 mg/100ml.)
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x
180
 = 0.05
211
 = .01
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Let the unknown sample size to be n.
If  = .01, then the critical value in z-score would be
2.32. The critical value in average cholesterol level
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would be
x*  180  2.32 
n
If the true mean is 211 mg/100ml, with a power of
0.95, i.e.  = .05, we would accept the null hypothesis
when the sample average is less than
46
x*  211  1.645 
n
x
118
 = 0.05
x * 211
 = .01
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x
x * 211
Set the two equations equal to each other:
118
46
46
180  2.32 
 211  1.645 
n
n
Solve for n:
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 (2.32  1.645)(46) 
n
 34.6

211  180


So, the sample size needed is 35.
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Formula: The estimated sample size for a onesided test with level of significance  and a power
of 1   to detect a difference of a – 0 is,
 ( z  z  )   
n

 a  0 
2
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For a two-sided test, the formula is
 ( z / 2  z  )   
n

 a  0 
2
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Example:
For testing hypothesis that
H0:  = 0 = 70 v.s.
(Two-sided Test)
Ha:   70
with a level of significance of  = .05. Find
the sample size so that one can have a power
of 1   = .90 to reject the null hypothesis if the
actual mean is a = 80. The standard deviation,
, is approximately equal 15.
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 ( z / 2  z  )   
n

 a  0 
 = .05
1   = .90
  = .10
2
z/2 = z.025 = 1.96
z = z.1 = 1.28

 (1.96  1.28) 15 
n
 23.62  24

80  70


2
The sample size needed is 24.
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Example:
For testing hypothesis that
H0:  = 0 = 70 v.s. Ha:  < 70
(One-sided Test)
with a level of significance of  = .05. Find the
sample size so that one can have a power of
1    .95 to reject the null hypothesis if the
actual mean is a  60 (or the actual
difference in means is 10). The standard
deviation, , is approximately equal 15.
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Example:
For testing hypothesis that
H0:  = 0 = 70 v.s. Ha:   70
(Two-sided Test)
with a level of significance of  = .05. Find the
sample size so that one can have a power of
1    .95 to reject the null hypothesis if the
actual mean is a  75. The standard
deviation, , is approximately equal 15.
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