Data Analysis Statistics An Introduction to Statistical Inference and Data Analysis

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icance probabilities that correspond to different ways of counting
the xi = M0 as larger and smaller than M0. Actually, it will
suffice to compute the p-value that corresponds to counting all of
the xi = M0 as larger than M0 and the p-value corresponds to
9.2. THE GENERAL 1-SAMPLE LOCATION PROBLEM 191
counting all of the xi = M0 as smaller than M0. If both of these
p-values are less than (or equal to) the significance level �, then
clearly we will reject H0. If neither is, then clearly we will not.
If one is and one is not, then we will declare the evidence to be
equivocal.
" Example 2.3 from Gibbons
Suppose that we want to test H0 : M = 10 vs. H1 : M = 10 at
significance level � = .05.
Suppose that we observe the following sample:
9.8 10.1 9.7 9.9 10.0 10.0 9.8 9.7 9.8 9.9
Note the presence of ties in the data, suggesting that the mea-
surements should have been made (or recorded) more precisely.
In particular, there are two instances in which xi = M0.
If we discard the two xi = 10, then n = 8, s+ = 1, and
p = P (|S+ - 4| e" |1 - 4| = 3)
= P (S+ d" 1 or S+ e" 7)
= 2P (S+ d" 1)
= 2 � .0352 = .0704,
from Table F in Gibbons (see handout).
Since p = .0704 > .05 = �, we decline to reject H0.
" Example 2.4 from Gibbons
Suppose that we want to test H0 : M d" 625 vs. H1 : M > 625 at
significance level � = .05.
Suppose that we observe the following sample:
612 619 628 631 640 643 649 655 663 670
Here, n = 10, s+ = 8, and
p = P (S+ e" 8) = P (S+ d" 2) = .0547,
from Table F in Gibbons (see handout).
Since p = .0547 > .05 = �, we decline to reject H0.
192 CHAPTER 9. 1-SAMPLE LOCATION PROBLEMS
" If n > 20, then we use the normal approximation to the binomial
distribution. Since S+
"
and standard deviation .5 n. The normal approximation is
k - .5 - .5n
.
P (S+ e" k) = P Z e" " ,
.5 n
where Z
" Example 2.4 (continued):
8 - .5 - 5
. . .
P (S+ e" 8) = P Z e" " = 1.58 = .0571.
.5 10
" Notice that the sign test will produce a maximal significance proba-
bility of p = 1 when S+ = S- = .5n. This means that the sign test
is least likely to reject H0 : M = M0 when M0 is a median of the
sample. Thus, using the sign test for testing hypotheses about popu-
lation medians corresponds to using the sample median for estimating
population medians, just as using Student s t-test for testing hypothe-
ses about population means corresponds to using the sample mean for
estimating population means.
" One consequence of the previous remark is that, when the population
mean and median are identical, the Pitman efficiency of the sign
test to Student s t-test equals the asymptotic relative efficiency of the
sample median to the sample median. For example, using the sign test
on normal data is asymptotically equivalent to randomly discarding
36% of the observations, then using Student s t-test on the remaining
64%.
9.2.3 Interval Estimation
" We want to construct a (1 - �)-level confidence interval for the pop-
ulation median M. We will do so by determining for which M0 the
level-� sign test of H0 : M = M0 vs. H1 : M = M0 will accept H0.
" Suppose that we have ordered the data:
x(1)
9.2. THE GENERAL 1-SAMPLE LOCATION PROBLEM 193
" The sign test rejects H0 : M = M0 if |S+ - .5n| is large, i.e. H0 will
be accepted if M0 is such that the numbers of observations above and
below M0 are roughly equal.
" Suppose that
P (S+ d" k) = P (S+ e" n - k) = �/2.
For n d" 20, we can use Table F to determine pairs of (�, k) that satisfy
this equation. Notice that only certain � are possible, so that we may
not be able to exactly achieve the desired level of confidence.
" Having determined an acceptable (�, k), the sign test would accept
H0 : M = M0 at level � if and only if
x(k+1)
hence, a (1 - �)-level confidence interval for M is
(x(k+1), x(n-k)).
" Remark: Since there is no fixed M0 when constructing a confidence
interval, we always use all of the data.
" Example 2.4 in Gibbons (continued): From Table F,
P (S+ d" 2) = P (S+ e" 8) = .0547;
hence, a (1 - 2 � .0547) = .8906-level confidence interval for M is
(628, 655). [ Pobierz całość w formacie PDF ]

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