概率论与数理统计(英文) 第九章

9. Nonparametric Statistics
9.1 Sign Test 符号检验
1
The simplest of all nonparametric methods is the sign test, which is usually used to test the significance of the difference between two means in a paired experiment.
最简单的非参数检验是符号检验
检验两个总体均值差的显著程度
It is particularly suitable when the various pairs are observed under different conditions, a case in which the assumption of normality may not hold. However, because of its simplicity, the sign test is often used even though the populations are normally distributed. As is implied by its name in this test only the sign of the difference
between the paired variates is used.
若两个总体的均值相等，那么符号‘＋’、‘－’的概率一样。

D = sign of (X 1－X 2 )
If p denotes the probability of a difference D being positive and
q the probability of its being negative, we have as hypothesis p=1/2. appropriate test statistic is X , X~B (n, p)， X --- N(‘+”)
we will reject 0H
in favor of
1H
only if the proportion of plus
signs is sufficiently less than 1/2, that is , when the value x of our random variable is small. Hence, if the computed P -value
12()P P X x when p =≤=
is less than or equal to the significance level α, we reject 0H
in
favor of
1H .
we reject
0H
in favor
1H
when the proportion of plus signs is
significantly less than or significantly greater than 1/2. This, of course, is equivalent to x being sufficiently small or sufficiently large, respectively. Therefore, if /2x n < and the computed P-value 122()P P X x when p =≤=
is less than or equal to α, or if /2x n > and the computed P-value 122()P P X x when p =≥= is less than or equal to α, we reject 0H
in favor
1H .
Car Radial tires Belted tires D
1 4.
2 4.1 ＋ 2 4.7 4.9 －
3 6.6 6.2 ＋
4 7.0 6.9 ＋
5 6.7 6.8 －
6 4.5 4.4 ＋
7 5.7 5.7
8 6.0 5.8 ＋
9 7.4 6.9 ＋
10 4.9 4.9
11 6.1 6.0 ＋
12 5.2 4.9 ＋
13 5.7 5.3 ＋
14 6.9 6.5 ＋
15 6.8 7.1 －
16 4.9 4.8 ＋
符号检验的利弊
n 必须比较大
因为对于n =5的样本，会出现永远不拒绝“总体均值相等“的假设。

（极端情形2(1/2)5=0.0625，全部为正号情形）
对双边检验，n至少6以上，越大越好。

9-2 Rank-Sum Test 秩和检验
is an appropriate alternative to the two-sample
a sample (x1,x2, …,x n1) from X ’s population a sample (y1,y2, …,y n2) from Y ’s population Hypothesis: 12μμ= 总体均值是否相等 n= n1 +n2 In Wilcoxon ’s test:
These n variates are graded (or ranked) according to in creasing size
n 个变量从小到大排列并编号 x1, y1, x2, x3, y2, … 1 2 3 4 5 w1=1+3+4+… w2=2+5+…. w1<w2
Find P = p{ ∑i
R ≤w1}
If P <α (significant level )
Then reject H 0; (otherwise, we accept H 0 )
Example 检验制造水泥的一种新方法是否提高了抗压强度
New method: 148, 143, 138, 145, 141 Standard method: 139, 136, 142, 133, 140
Does this indicate that the new method has increased the compressive strength(5% level) ? Solution.
H0: u standard = u new H1: u standard <u new
The variaes belong to the sample with smaller mean are underlined
Original data:
133, 136, 138, 139, 140, 141, 142, 143, 145, 148 ranks: 1 2 3 4 5 6 7 8 9 10
w1= 1+2+4+5+7=19
w2=3+6+8+9+10=36
the total of all ranks is
w1+w2= 19+36 =55
(如果新方法没有显著效果，w1应该和w2差不多。

事实上，任意一组5个数的组合，都应相近）there are 12 cases, a sum of rank less than or equal to 19
since N( ∑i
R ≤w1) = 12 )
5
10C = 252
so P= p{ ∑i
R ≤ 19 } =
252
12
=0.0477 <0.05=α
reject H 0. (用古典概型定义)
If there is no significant difference between the two sample means, the total of the ranks corresponding to the first and those corresponding to the second sample should be about the same . If, however, the total of the ranks for one sample is appreciably less than that of the other, we calculate ——under the hypothesis of equal population means ——the probability of obtaining by chance alone sum of ranks less than or equal to that obtained in the given experiment. If this probability is less than the significance level, we reject the hypothesis; otherwise, we accept it.
The procedure can be better understood by an example .
E xample 9.2.1
The following data give the results of tests on two preparations(药剂) of a fly spray, in terms of the percentage of mortality.
Sample A: 68, 68, 59, 72, 64, 67, 70, 74 Sample B: 60, 67, 61, 62, 67, 63, 56, 58 The ranks are determined as follows:
Original data: 56 58 59 60 61 62 63 64 67 67 67 68 69 70 72 74
Ranks: 1 2 3 4 5 6 7 8 10 10 10 12 13 14 15 16
In the case of ties (identical observation 有相同观察数的处理), we replace the observations by the mean of the ranks that the observations would have if they were distinguishable, here the ninth and tenth and eleventh observations are identical, we assign a rank 10 to each of the three observations.
here 128,8n n ==, 1216n n +=,
112456101045w =++++++=
12(1)
2n n w w ++=.
2(16)(17)
45912
w =-=.
In choosing repeated samples of size 1n
and
2n ,
we would
expect
1w , and therefore 2w , to vary. Thus we may think of 1w
and
2w
as values of the random variable
2W
and 2W , respectively. The
null hypothesis 12μμ= will be rejected in favor of the alternative
12μμ< only
if
1w is sufficiently small; the alternative 12μμ> is
accepted if
2w is sufficiently small; and the alternative 12μμ≠ is
accepted if the minimum of 1w
and
2w
is sufficiently small.
of the related statistic
1U or 2U , respectively, or on the value u of
the statistic U , the minimum of 1U
and
2U . These statistics simplify
the construction of tables of critical values. Since both
1U
and 2U
have symmetric sampling distributions and assume values in the interval from 0 to 12n n
such that
1212u u n n +=.
Table F gives critical value of
1
U and
2
U for some level of
significance. If the observed value of 1u
and 2u
or u is less than or
equal to the tabled critical value , the null hypothesis is rejected at the level of significance indicated by the table.
Suppose, for the example above, that we wish to test if there is a significant differences (5% level) between the two preparations. In other words we wish to test the null hypothesis that 12μμ= against the two-tailed alternative that 12μμ≠ at the 0.05 level of significance for random samples of size 18n =
and
28n =
that yield
the value
145w =. It follows that
111111
(1)4589922
u w n n =-+=-⨯⨯=.
Our two-tailed test is base on the statistic
1U . Using Table F,
we reject the null hypothesis of equal means when
113u ≤.
( { u < u 0 } is rejected regions ) Since
19u =
falls in the rejection region , the null hypothesis can be
rejected and we conclude that there is a significant difference between the two preparations.
Under the null hypothesis that two samples come from identical
populations, it can be shown that the mean and the variance of the sampling distribution of 1U are
1
12
2
U n n μ=
and 1
212(1)
2
U
n n n σ+=. If there are ties in rank, these formulas provide only approximations,
but if the number of ties is small, these approximations will generally be good.
Since numerical studies have shown that the sampling
distribution of 1
U can be approximated closely by a normal distribution when
1n
and
2n
are both greater than 8, the test of null
hypothesis that the two samples come from identical populations can be based on
1
1
1U U
U Z μσ-=
which is a random variable having approximately the standard normal distribution.
The use of the wilcoxon rank-sum test is not restricted to
nonnormal populations. It can be used in place of the two-sample t-test when the populations are normal, although the power will be smaller. The Wilcoxon rank-sum test is always superior to the t-test for decidedly nonnormal populations.
9.3 Signed-Rank Test
Wilcoxon signed-rank test .（符号检验，秩和检验结合）
The reader should note that the sign test utilizes only the plus and
minus signs of the differences between the observations in the two sample case, and 0μ in the one-sample case. It does not take into consideration the magnitudes of these differences. A test utilizing
both direction and magnitude was proposed in 1945 by Frank Wilcoxon, called Wilcoxon signed-rank test .
1. 原理
H 0: 0μμ= (0:210=-μμH ) )(0
i i i i i y x D x D -=-=μ
∑+=+)(i R w
∑-=-)(i R w
When H 0 is true,
-+≈w w (should nearly equal )
H 1: 0μμ< , when w + is small and
w - is large
H 1:0μμ>, when
w +
is large and
w -
is small.
H 1: double tail
w =
-+w or w , when w is sufficiently small.
For one-tailed α<<=-}{0w w P value P , reject H 0.
For two-tailed: α<<=-}{20w w P value P , reject H 0
Example (麦种处理效果比较)
Is there a significant difference between the two seed treatments ?
Solution
Let X be the first sample variates
X Y D=X －Y rank
588
327
30 6
5 1
-7 3
6 2
11 5
10 4
n=8,
w-= 3 (负数序数和) w+=n(n+1)/2－W-
There are a total of 256 equally likely possibilities for the sequence of signs.（28）
No negative D-value: 1 case(＋＋＋＋＋＋＋＋) One negative D-value: 3 case
(－＋＋＋＋＋＋＋)
(＋－＋＋＋＋＋＋)
(＋＋－＋＋＋＋＋)
two negative D-value: 1 case(――＋＋＋＋＋＋)
P{∑
R≤w-} = 5/256 = 0.0195
i
P=2 P{∑i R≤w-} =0.0390 < 0.05
(拒绝Ho)
Table 9-16 Signed-Rank Test
2. 查表（w0）
It should be noted that whenever 5
n<and the level of significance does not exceed 0.05 for a one-tailed test or 0.01 for a
two-tailed test, all possible values of w
+,w
-
, or w will lead to the
acceptance of the null hypothesis.
when 530
n
≤≤, Appendix Table G shows approximate critical
values of W
+, W
-
or W for some levels.
Solution 1.
0: 1.8H μ=,
1: 1.8H μ≠
2. 0.05α=
3. Critical region: Since n = 10, after discarding the one measurement that equals 1.8, Table G(Table 1.16) shows the critical region to be 8w ≤.
4. Calcutation: Subtracting 1.8 from each measurement and then ranking the differences by their absolute values, we have
Now
13w +=
and
42w -=
so that 13w =, the small of
w +and w -.
5. Decision: Do not reject 0H
and conclude that the average
operating time is not significantly different form 1.8 hours.
Example 9.3.2
Solution
Now we find that
5 2.57.5w +=+=. 5. Decision: Reject 0:H
and conclude that sample problems do not,
on the average, increase one ’s graduate record score by as much as 50 points.
When 15n ≥, the sampling distribution of
W +
(or
W -
)
approaches the normal distribution with mean (1)
4
W n n μ+
+=
and variance 2
(1)(21)
24
W n n n σ+
++=
. Therefore, when n exceeds the largest value in Table A.16, the statistic W
W
W Z μσ+
+-=
can be used to determine the critical region for our test.
Homework chap 9
9.1, 9.3, 9.4, 9. 10。