1
2
Means
Test 2.1 Single Mean (One-Sided)
Parameters:
μ = population mean
μ
0
= minimum desired value for μ
σ = population standard deviation
Hypotheses:
H
0
: μ < μ
0
H
1
: μ ≥ μ
0
Data:
=X sample mean
S = sample standard deviation
n = sample size
Critical value(s):
Reject H
0
if:
Xt
S
n
10
+≥
µ
−β
where t
1 − β
= 100*(1 − β) percentile of a central t distribution with n − 1 degrees
of freedom.
2 Equivalence and Noninferiority Tests
Discussion:
If μ = μ
0
, then we would expect the sample mean,
X
, to be greater than:
t
S
n
01
µ−
−β
with probability 1 − β. If
Xt
S
n
01
≥µ −
−β
we would reject the hypothesis that μ < μ
0
. Or, in other words, we reject H
0
if
Xt
S
n
10
+≥
µ
−β
.
If the inequalities in the hypotheses are reversed, that is,
H
0
: μ > μ
0
H
1
: μ ≤ μ
0
then we would reject H
0
if
Xt
S
n
10
−≤
µ
−β
.
The power for the former case is given by:
()
()
+≥µµ=µ
=
−µ
≥− δ=
µ−µ
σ
−β −β
Xt
S
n
n
nX
S
tn
n
Pr |,Pr |,
a
a
10
0
1
0
where
n
a 0
()
δ=
µ−µ
σ
is the noncentrality parameter for the statistic
X
Sn
0
−µ
3Means
which has a noncentral t distribution with n – 1 degrees of freedom (Johnson,
Kotz, and Balakrishnan, 1995).
Example:
Suppose we hypothesize that the average time until failure of a machine is
at least 100 hours. The hypotheses are
H
0
: μ < 100
versus the alternative
H
1
: μ ≥ 100.
We obtain, from a sample of n = 20 times to failure, the data
=X 99.9 hours
S = 3.4 hours.
We choose 1 − β = 0.95, so t
1 − β
≈ 1.729. Thus
Xt
S
n
99.9 1.729
3.4
20
101.2
100
1
+=+≈>
−β
.
Therefore, we reject the null hypothesis in favor of the alternative.
To calculate the power under alternative values of μ = μ
a
, we must calculate
the probability
Xt
S
n
nPr
|,
a10
+≥µµ=µ
−β
.
If μ = μ
0
, then
t
X
Sn
0
=
−µ
has a central t-distribution with n – 1 degrees of freedom, and
Xt
S
n
n
nX
S
tn
Pr |,Pr |,1
100
0
10
()
+≥µµ=µ
=
−µ
≥− µ=µ
=−β
−β −β
.
If μ = μ
a
< μ
0
, then
X
Sn
0
−µ
4 Equivalence and Noninferiority Tests
has a noncentral t-distribution with noncentrality parameter
n
a 0
()
δ=
µ−µ
σ
.
Since, in general, σ is unknown, the power may be plotted against
n
a 0
()
δ
=
µ−µ
σ
or in other words, the difference μ
a
– μ
0
expressed as a proportion of the stan-
dard deviation of the population. Note that if μ
a
= μ
0
, then δ = 0. Figure2.1
shows a power curve for the example.
Condence interval formulation:
The expression:
Xt
S
n
1
+
−β
1
0.9
0.8
0.7
0.6
0.5
Power
0.4
0.3
0.2
0.1
0
–1.2 –1 –0.8 –0.6 –0.4
Delsig
–0.2 0 0.1 0.2 0.3 0.4
FIGURE 2.1
Test 2.1, power curve n = 20.
5Means
is a one-sided upper condence limit for μ. Thus, from the example, the
upper 95 percent condence limit for μ is
Xt
S
n
99.9 1.729
3.4
20
101.2
1
+=
+≈
−β
.
Computational considerations:
Example: n = 25, β = 0.05, μ
0
= 10.00
• SAS code
libname stuff 'H:\Personal Data\Equivalence & Noninferiority\
Programs & Output';
data calc;
set stuff.d20121029_test_2_1_example_data;
run;
proc means data = calc;
var X mu0 beta;
output out = onemean MEAN = xbar popmu betaprob STD = sd N =
n1;
run;
data outcalc;
set onemean;
se = sd/sqrt(n1);
lowlim = xbar + tinv(1-betaprob,n1-1)*se;
run;
proc print data = outcalc;/* has vars xbar popmu betaprob n1
se lowlim */
run;
The SAS System 07:03 Sunday, October 28, 2012 13
Obs _TYPE_ _FREQ_ xbar popmu betaprob sd n1 se lowlim
1 0 25 9.68 10.00 0.05 1.00 25 0.20092 10.0266
• JMP Data Table and formulas (Figure2.2):

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