The simple samples are independently taken from normal distribution. The two functions of the sample means and sample variances are considered. The density functions of these two statistics have been derived. These statistics can be applied for verifying the hypothesis on stability of expected value and variance of normal distribution considered, e.g., in statistical process control. The critical values for these statistics have been found using numerical integration. The tables with approximated critical values of these statistics have been presented.density function, sample variance, test statistic, numerical integration, statistical process control

Grzegorz Konczak

Janusz Wywial

English

Research Papers in Economics

BADANIA OPERACYJNE I DECYZJE
Nr 1 2006
Grzegorz KOŃCZAK*
Janusz WYWIAŁ*
TESTING HYPOTHESIS ON STABILITY
OF EXPECTED VALUE AND VARIANCE
The simple samples are independently taken from normal distribution. The two functions of the
sample means and sample variances are considered. The density functions of these two statistics have
been derived. These statistics can be applied for verifying the hypothesis on stability of expected
value and variance of normal distribution considered, e.g., in statistical process control. The critical
values for these statistics have been found using numerical integration. The tables with approximated
critical values of these statistics have been presented.
Keywords: density function, sample variance, test statistic, numerical integration, statistical process
control
1. Introduction
One of the problems of statistical process control is considered. It is a procedure (so
called control charts) for monitoring stability of the expected value and the variance of
diagnostic variables. We assume that during the first  2 ≥ k  periods the mean values of
a diagnostic variable are the same but unknown. The same situation is with the variance
of the variable. The unbiased estimators of the expected value µ and the variance σ
2
evaluated on the basis of data observed in the first k periods and the k + 1 period are  X ,
S ˆ and  1 + k X  and  1 ˆ
+ k S , respectively. Our problem is the following: is the process (char-
acterized by the diagnostic variable) stable in all of the periods? If yes, the distances:
| | 1 + − k X X  or  | ˆ ˆ | 1 + − k S S  should not be significant. Such a problem is considered, e.g.,
in [4]. More formally, we have the problem of testing the hypothesis
                                                     
* Department of Statistics, Katowice University of Economics, ul. Bogucicka 14, 40-226 Katowice.
koncz@ae.katowice.pl, wywial@ae.katowice.plG. KOŃCZAK, J. WYWIAŁ 74
) ˆ ( ) ˆ ( : 1 0 X E X E H k = +    and    ) ˆ ( ) ˆ (
2 2
1 S E S E k = + .
We are going to construct a test statistic for this hypothesis in the next paragraphs.
2. Basic definitions and properties
Let Ja be a column vector consisting of a element equal one and let Ia be a unit
matrix of degree a. Moreover, let  ] [ 1 2 1 X ,..., X ,..., X , X X + = k i , where
[]
i ii i i ji n = X X ,X ,...,X ,...,X 12  (i = 1, 2, …, k + 1, k ≥ 2, j = 1, 2, …, ni).
We will consider the following statistics:
i
i
n i
i
n
j
ij
i
i n
X
n
X J X
1 1
1
= = ∑
=
,
for i = 1, 2, …, k + 1,
n
k
j
i i n
n X
n
X XJ
1 1
1
= = ∑
=
,   where    ∑
=
=
k
i
i n n
1
,
1
) (
1
1 ˆ
1
2 2
−
= −
−
= ∑
= i
T
i i i
n
j
i ij
i
i n
X X
n
S
i X M X
,   
T
n
i
n i i i i n
J J I M n
1
− = ,
T XMX
k n
S k n
k n
S
k
j
i −
= −
−
= ∑
=
1 ˆ ) (
1 ˆ
1
2 2 ,    [ ] i M M = ,
where Mi is the block-diagonal matrix of degree n,
1
) (
1
1 ~
1
2 2
−
= −
−
= ∑
= k
n X X
k
S
T k
i
i i
XMX
,   
T
n n n n
J J M I N
1
− − = ,
where 
2 ˆ
i S  is the sample variance within the i-th group, 
2 ˆ S  is the mean of group vari-
ances and 
2 ~
S  is the variance between groups. Moreover, let us note that  i i M M
2 = ,
M M
2 =  and  N N
2 = .
The particular case of the theorem on independence of the quadratic or linear
forms of normal vectors corresponding to our problem is as follows (see [3], pp. 224
and next).Testing hypothesis on stability of ... 75
Theorem 1. Let X have a non-singular normal distribution N(µJn,  σ
2In) and
T
A Q XAX = , 
T
B Q XBX = , L = Xa where a is a column non-random vector, A, B are
symmetric and non-random matrices of degree n each. Then, the set of necessary and
sufficient conditions for QA and QB to be independently distributed is a) AB = O or b)
ABJn = O. The quadratic form QA and the linear form L are to be independently dis-
tributed if and only if Aa = O.
On the basis of this theorem we can show that the statistics in the following pairs
are independently distributed  ),
~
, (
2 S X   ), ˆ , (
2 S X   ).
~
, ˆ (
2 2 S S  Moreover, the statistics
2
1 1 ˆ , + + k k S X  are independent and they are independent of each of the statistics 
2 ˆ ,S X
and 
2 ~
S .
The obtained result and the well known definitions let us derive the following dis-
tributions:
) 1 , 0 ( :
1 1
1
1 N
n n
X X
Z
k
k
+
−
=
+
+
σ
.( 1 )
Moreover,
2
1 2
2
1 :
~
) 1 (
−
−
= k
S k
U χ
σ
,( 2 )
2
2
2
2 :
ˆ ) (
k n
S k n
U −
−
= χ
σ
(3)
and
2
1 2
2
, 0 :
ˆ ) 1 (
−
−
=
i n
i i
i
S n
U χ
σ
   i = 1, 2, …, k + 1. (4)
On the basis of these expressions we have
) 1 , 1 ( : ~2
2 2
1 − = k F
S
Z
F
σ
,( 5 )
) , 1 ( : ˆ2
2 2
2 k n F
S
Z
F − =
σ
,( 6 )
) , 1 ( : ˆ
ˆ
1 2
2
1
3 k n n F
S
S
F k
k − − = +
+ ,( 7 )
) 1 , 1 ( : ~
ˆ
1 2
2
1
4 − − = +
+ k n F
S
S
F k
k ,( 8 )G. KOŃCZAK, J. WYWIAŁ 76
where F(r, m) denotes the well known F distribution with r and m degrees of freedom
and the following density function:
) (
1 2 2
2 2 ) ( ) , 0 (
) (
2
1
1
2 2
g I
g
m
r
g
m
r
m r
m r
g f
m r
r r
∞
+
−





 +











 ⋅ 










 +
=
Γ Γ
Γ
.( 9 )
3. The statistics Q1 and Q2
Let us consider two statistics Q1 and Q2 given as follows
2
2
2
1
2
2
2
3 1 1 1 ˆ
ˆ
~ ) 1 ( 







− + = − + = +
S
S
S
Z
F F Q k , (10)
2
2
2
1
2
2
2
4 2 2 1 ~
ˆ
ˆ ) 1 ( 







− + = − + = +
S
S
S
Z
F F Q k , (11)
where F1 – F4 are defined by expressions (5)–(8), respectively.
On the basis of the previous results the distributions of the random variables Z, U1,
U2 and U0, k + 1 are independent. Finally, this and the fact the samples are independent
lead to the conclusion that the distributions of the statistics F1 and F3 as well as the
distributions of F2 and F4 are independent.
The density function of random variable F1 is as follows:
) (
1
1
1
) 1 (
1
2
1
2
1
2 ) ( ) , 0 (
2
1 g I
k
g g k k
Γ Γ
k
Γ
g f k ∞






−
+
− 




 −
⋅ 











= , (12)
where



∉
∈
=
A x
A x
x IA 0
1
) ( .Testing hypothesis on stability of ... 77
The density function of random variable F2 is:
) (
1
1
) (
1
2 2
1
2
1
) ( ) , 0 (
2
1 2 g I
k n
g g k n k n
Γ Γ
k n
Γ
g f k n ∞ + −






−
+
− 




 −
⋅ 










 + −
= . (13)
The density functions of F3 and F4 are as follows:
) (
1
) ( ) , 0 (
2
1
2
3 3 g I
g
k n
r
g
c g f k n r
r
∞ − +
−






−
+
= , (14)
) (
1
1
) ( ) , 0 (
2
1
1
2
4 4 g I
g
k
r
g
c g f k r
r
∞ − +
−






−
+
= . (15)
where r = nk+1 – 1 and
2
3
2 2
2
r
k n
r
k n
Γ
r
Γ
k n r
Γ
c 





−





 −
⋅ 










 − +
= ,   
2
4 1
2
1
2
2
1 r
k
r
k
Γ
r
Γ
k r
Γ
c 





−





 −
⋅ 










 − +
= .
Let us derive the distribution of random variable (F3 – 1)
2. If b = g – 1 then we
have g = b + 1 and dg = db and
) (
) 1 (
) ( ) , 0 (
2
1
2
3 3 g I
b
k n
r
k n
r k n
b
c b f k n r
r
a ∞ − +
−






−
+
−
+ −
+
= . (16)
When v = b
2, then we have  v b ± =  and 
v
db
2
1
| | = . If  ] 0 , 1 (− ∈ b , then  v b − =
and for  ) 1 , 0 ( ∈ b , we have  v b = .G. KOŃCZAK, J. WYWIAŁ 78
If  ) 1 , 0 [ ∈ v
  
 






 












−
+
−
+ −
+
+






−
−
−
+ −
−
= − +
−
− +
−
2
1
2
2
1
2
3 3
) 1 ( ) 1 (
2
1
) ( k n r
r
k n r
r
b
v
k n
r
k n
r k n
v
v
k n
r
k n
r k n
v
v
c v f . (17)
If  ) , 1 [ ∞ ∈ v  then
2
1
2
3 3
) 1 (
2
1
) ( k n r
r
b
v
k n
r
k n
r k n
v
v
c v f − +
−






−
+
−
+ −
+
= . (18)
We can write the density function of (F3 – 1)
2 in the following way
+













−
−
−
+ −
−
= − +
−
) (
) 1 (
2
1
) ( ) 1 , 0 (
2
1
2
3 3 v I
v
k n
r
k n
r k n
v
v
c v f k n r
r
b













−
+
−
+ −
+
+ ∞ − +
−
) (
) 1 (
) , 0 (
2
1
2
v I
v
k n
r
k n
r k n
v
k n r
r
. (19)
If W: F(1, k – 1), then the density function of W is given by expression (12). Let V
be the random variable of F distribution with r and n-k degrees of freedom. The den-
sity function of V is given by (19). Now, we are going to evaluate the density func-
tions of the random variable: Q1 = W + V. The density function of the statistic Q1 is as
follows
∫
∞
− =
0
1 3 1 ) ( ) ( ) ( dv v q f v f q h b (20)Testing hypothesis on stability of ... 79
∫
∞
− +
−













−
−
−
+ −
−
=
0
) 1 , 0 (
2
1
2
1 1 ) (
) 1 (
) ( v I
v
k n
r
k n
r k n
v
c q h k n r
r
2
) , 0 (
2
1
2
1
1 ) ( ) 1 (
1
) (
) 1 (
k k n r
r
k
v g
v q v k
v I
v
k n
r
k n
r k n
v






−
−
+ − −
⋅













−
+
−
+ −
+
+ ∞ − +
−
, (21)
where
2
1
2 2
2
2
1
2
1
2
2
r
k n
r
k n
Γ
r
Γ
k n r
Γ
k
Γ Γ
k
Γ
c 





−





 −
⋅ 










 − +





 −
⋅ 











= .
Similarly, the density function of Q2 is derived in the following way.
∫
∞
− =
0
2 4 2 ) ( ) ( ) ( dv v q f v f q h b , (22)
where f4b is the density function of (F4 – 1)
2













−
−
−
+ −
−
= − +
−
) (
1 1
1
) 1 (
2
1
) ( ) 1 , 0 [
2
1
1
2
4 4 v I
v
k
r
k
r k
v
v
c v f k r
r
b
⋅













−
+
−
+ −
+
+ ∞ − +
−
) (
1 1
1
) 1 (
) , 0 (
2
1
1
2
v I
v
k
r
k
r k
v
k r
r
(23)G. KOŃCZAK, J. WYWIAŁ 80
Finally, the density function h2(q) of the statistic Q2 is as follows
   ∫
∞
− +
−













−
−
−
+ −
−
=
0
) 1 , 0 [
2
1
1
2
2 2 ) (
1 1
1
) 1 (
) ( v I
v
k
r
k
r k
v
c q h k r
r
2
1 ) , 0 (
2
1
1
2
1
1
) ( ) (
1
) (
1 1
1
) 1 (
+ − ∞ − +
−






−
−
+
− −
⋅













−
+
−
+ −
+
+ k n k r
r
k n
v g v q v k n
v I
v
k
r
k
r k
v
, (24)
where
2
2 1
2
1
2
2
1
2 2
1
2
2
1 r
k
r
k r
k r
k n
k n
c 





−





 −
⋅ 










 − +





 −
⋅ 










 + −
=
Γ Γ
Γ
Γ Γ
Γ
.
The distribution functions of Q1 or Q2 are evaluated by means of the following in-
tegral
∫ = < =
q
i i i ds s h q Q P q H
0
) ( ) ( ) ( ,   I = 1, 2. (25)
For the given significance level α the quantil q1–α is determined on the basis of the
integral:
α
α
= ∫
∞
−
ds s h
q
i
1
) ( . (26)
4. Numerical computations
An exact solution of equation (26) is very difficult. In this situation, the quantiles
of Q1 and Q2 can be found using numerical integration (see, e.g., [2], [1]). The quan-Testing hypothesis on stability of ... 81
tiles were found for three significance levels (α = 0.01, 0.05, 0.1). Table 1 presents
quantiles of the statistic Q1 and Table 2 presents quantiles of the statistic Q2. These
quantiles were evaluated for the case n1 = n2 = … = nk = nk + 1 = 5. These quantiles have
been evaluated for the numbers of groups k from 4 to 10 and for 15, 20, 25 and 30.
Table 1
Quantiles of statistic Q1
Significance level α Number of
groups k 0.10 0.05 0.01
4 5.85 10.38 31.19
5 4.86 8.08 21.43
6 4.36 6.96 16.75
7 4.06 6.31 14.21
8 3.86 5.90 12.67
9 3.72 5.61 11.64
10 3.62 5.40 10.92
15 3.34 4.85 9.15
20 3.23 4.62 8.45
25 3.16 4.50 8.08
30 3.12 4.42 7.85
0
5
10
15
20
25
30
35
40
2 4 6 8 10 20 30
0.1
0.05
0.01
k
q
Fig. 1. The graphic display of quantiles of statistic Q1 for significance level α  = 0.1, 0.05 and 0.01
The Figure 1 presents quantiles of the statistic Q1 for the significance levels 0.1,
0.05 and 0.01. These quantiles are presented for the same cases as in Table 1. Figure 2
presents the same results as Figure 1 but for the statistic Q2.G. KOŃCZAK, J. WYWIAŁ 82
Table 2
Quantiles of statistic Q2
Significance level α Number of
groups k 0.10 0.05 0.01
4 3.42 5.00 10.71
5 3.30 4.76 9.41
6 3.23 4.62 9.21
7 3.18 4.53 8.34
8 3.15 4.46 8.07
9 3.12 4.41 7.88
10 3.10 4.37 7.74
15 3.04 4.26 7.39
20 3.01 4.20 7.23
25 2.99 4.17 7.21
30 2.98 4.15 7.10
0
1
2
3
4
5
6
7
8
9
10
2468 1 0 2 0
0.1
0.05
0.01
k
q
Fig. 2. The graphic display of quantiles of statistic Q2 for significance level α = 0.1, 0.05 and 0.01
References
[1] BRANDT S., Statistical and Computational Methods in Data Analysis, Springer Verlag, New York
1997.
[2] DAHLQUIST G., BJORCK A., Numerical Methods (in Polish), PWN, Warszawa 1992.
[3] MATHAI A.M., PROVOST S.B., Quadratic forms in random variables, Marcel Dekker, Inc., New York,
Basel, Hong Kong 1992.
[4] THOMPSON J.R., KORONACKI J., Statistical Process Control: The Deming Paradigm and Beyond,
Chapman and Hall/CRC, New York, London 2001.Testing hypothesis on stability of ... 83
Testowanie hipotezy o stabilności wartości oczekiwanej i wariancji
W pracy jest rozważane zagadnienie jednoczesnej stabilności wartości przeciętnej i wariancji. Próby
proste są pobierane niezależnie z populacji o rozkładzie normalnym. Rozważa się dwie funkcje średniej
i wariancji z próby. Dla rozważanych statystyk zostały wyprowadzone funkcje gęstości. Proponowane
statystyki mogą być wykorzystane do weryfikacji hipotezy o stabilności wartości oczekiwanej i wariancji
dla rozkładu normalnego. Hipoteza taka może być rozważana np. w statystycznym sterowaniu procesem
przy konstrukcji kart kontrolnych. Bardzo trudne jest dokładne wyznaczenie kwantyli rozważanych staty-
styk. Dlatego wartości krytyczne dla tych statystyk zostały wyznaczone dla trzech zwykle używanych
poziomów istotności (α = 0,01, 0,05 i 0,1) dla prób o liczebnościach od 4 do 30 z wykorzystaniem cał-
kowania numerycznego. Zaprezentowano tablice wartości krytycznych dla tych statystyk. Zaproponowa-
ne statystyki i wyznaczone wartości krytyczne mogą być również przydatne do wykrywania zmian w proce-
sach produkcyjnych.
Słowa kluczowe: funkcja gęstości, wariancja z próby, test, numeryczne całkowanie, statystyczna kontrola
procesu

Testing hypothesis on stability of expected value and variance

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