Examples of tests
Brno 2007
1 Tests on parameters of normal distribution
Let Xi,..., Xn be a sample from W(, G2
) with the density
(2^)-^eXp ( - g ) exp ( - ^ *? + Ji **) ˇ
e = --^, $ = % U(x) = Y,xlT(x)=x.
1.1 Tests on a2
Denote
f) = -
2(7
Then the statistic
V = J2(Xi
-X)2
= U - nT2
is ancillary for 9 and is increasing in U. Then Hi : a < Go is equivalent to 6 < do, and the UMP
unbiased test rejects Hi if J2(Xi -- X)2
> Ca, where
oo
xl-i(y)dy = a.
Consider the hypothesis H2 : a = GQ. Because V is linear in U, the UMP unbiased test rejects H2
provided
EG** - %)2
<Ci or >Co
where the constants are determined so that
C2 ^ eGi
xl-i(y)dy = -- 7 / yxl-i(y)dy = i - a.
c-i n -- L j C l
On the other hand,
1 2 1 \ 1 "~1
--* 1 "+1
1 --Ä 2 1 \
--<K
-M
= <n-i)2"?r(*ř)v
"e =
FFF75I)9 e x
"+l(yh
hence the constants Ci, C2 are determined by the system of equations
xl-i(y)dy = / xl+i(y)dy = i - a.
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1.2 Tests on 
Consider the reparametrization
0 = 4 , # = -7T2, U(x)=x,T(x) = ^2xl
Consider the hypothesis H3 : { = o and without loss of generality assume that {0 = 0 (otherwise we
put X[ = Xi -- o)- Then 0 < 0 is equivalent to  < 0 and
V = * = u
^(x, - xy VT - nlP
is under  = 0 independent of T, because under  = 0 its multiple has tn-i distribution, hence
independent of the nuisance parameter a. Thus the UMP unbiased test for H3 rejects provided
t(x) > Co, where
^Jnx
\J^iE(^-xy
Consider the hypothesis H4 :  = 0 = 0 and the statistic
Then W is also independent of T under  = 0 and it is linear in U = X. Under  = 0, the distribution
of W is symmetric around 0. Thus, the UMP unbiased test of H4 would reject it for \W\ > C, where
P(,=o{W > C) = a. Because
= y/(n-l)nW{x)
^l-nW2
{x) '
then \t(x)\ is an increasing function of |VF(a;)|, and the rejection region of the UMP unbiased test is
equivalent to
\t(x)\>C
and
"~ a f°°
tn-i(y)dy = - , / tn-i(y)dy = a.
C z
JC0
The tests of more general hypotheses with 0 / 0 would have the criterion
/n(x - 0)
t{x) = ^
= T E ( Z * - Ž ) 2
2 Comparing the variances of two normal distributions
Let X = (Xi,..., Xm) and Y = (Yi,..., Yn) be samples from the normal distributions W(, c2
) and
Af(r], r2
). Then the joint density of (X, Y) is
^ A N ( 1 v ^ 2 1 v ^ 2 m
í - nri \
C(,r],a,r)exp ( ^ - ^ ^ x . - -- ^yj + --x + --yj .
Consider the reparametrization
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and the sufficient statistics
We want to test the hypothesis H5 : r2
< Ao<r2
. It is convenient still to reparametrize
0* = - \ + w ^ , V*i=Vi, í = 1,2,3
and to consider the statistics
The test is based on the statistic
v EM - Yf/^o EM - Y)2
/T2
 ( X , - X ) 2
E ( x , - X ) 2 / r 2
Under
r2
= Ao<r2
, the distribution of V does not depend on ,rj,T,a, hence V is stochastically
independent of T-j*, T2*, T3*. The rejection region of the UMP unbiased test of H5 can be written as
^_E(^-^)7Ao(n-l)
(X,-X)2
/(m-l) - °When
r2
= Ao<72
, then T has the F-distribution with n -- 1 and m -- 1 degrees of freedom, thus
Fn_i>m_i(y)dy = a.
0 0
IC0
If we want to test the hypothesis He : T2
= Ao<r2
, we should consider the statistic
w = T.M-*)2
/**
Under r2
= A0<72
, it is also independent of T-j*, T2*, T3* and it is linear in U*. Under r2
= A0<72
, the
distribution of W is the beta-distribution with the density
-p r m-\-n-- 21
,,_! m_x (W) = M
1
2 J
. w ("-3
)/2
(l - W)(TM-3
)/2
, 0 < w < 1
and ÜW = TO"~j_2- We reject H6 provided W < C\ or W > C2 and Ci, C2 are determined through
the relations
f C 2 rC2
/ ďn-l m-l(w)dW = / Bn+1 m-1 (w)ďW = 1 -- (X
7Ci 2
' 2
7Ci 2
' 2
The relation between the beta and F-distribution is such that W = j^y, where ~_l has the
distribution Fra_i;TO_i.
3 Comparing the means of two normal distributions
If we want to compare  and r\ and the variances a2
are r2
are unknown and unequal, then it is the
Behrens-Fisher problem, that cannot be treated in similar way. Thus suppose that a = r. Then the
joint density of (X, Y) is
C(C, rj, <r)exp -2"(I>*2
+ E^) + ~2 E ^ + ^ E ^
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Reparamet rization:
<7Z
<7Z
2<7Z
the sufficient statistics:
Consider the hypothesis H7 : r\ --  < 0. The it is better to reparametrize
i)-j 0*= rnj + nr]
+ l)a2
' l
(m + n)<72!
with the sufficient statistics
U*=Y-X, T?=mX + nY, T2* = J^x
? + E F
/ When
rj = , then we have the statistic
E(x, - x)2
+ (i- - ^)2
VT
2 - ^T
i*2
- Sf7
*2
that does not depend on  = r? and on <r, hence it is independent of T-j^T^. Thus, we reject H7
provided t(X,Y) > Co, where
t(X,Y) =
Y-X/\/ + 
' w
m n
[UXi - Xy + UYj - Y)2
]/(m + n - 2)
and J~ tm+n-2(y)dy = a.
For the hypothesis Hg : rj --  = 0, we should consider the statistics (linear in Č7* with coefficients
dependent only on T*, with distribution symmetric around 0)
w= Y
-x
Y,*? + Y.Y,2
- s ; ( E i + Y,Yj)2
that is related to V through
v = w
ran
m-\-n w2
Hence, we reject Hg provided \t(X, Y)\ > C, and J^ tm+n-2(y)dy = f.
3.1 Test of independence in bivariate normal distribution
Let (Xi, Yi),..., (Xn, Yn) be a sample from the bivariate normal distribution with density
Eo* -^-^ Eo* -*)& -^) + i E(w - ^
1
r l / l ^ , .,2 2p ^ . ... , l ^ . ,2
2TIUT\J\ -- p'
1exp 2(f-p2
)V<T2
^v
' sy
(ir^v
' syvtft
" r2
Consider the hypotheses Hg : p < 0 and H10 : p = 0.
Reparamet rization:
6= , f 9X, i?i = 9 / 1
X
9X, i ? 2 = X
(ir(f-p2
)' L
2a2
(l-p2
), z
2 r 2
( f - p 2\
1 -- p2
\(j2
UTJ ' f -- p2
\tau2
ar
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and the sufficient statistics
Hg is equivalent to 9 < 0. The ancillary statististic is the sample corellation coefficient
Z(Xi-X)(Yi-Y)
R =
VWÜ^WÜY^W1
because it is invariant to transformations Xi i--> l
~^ and Yi i--> -J
--^ and hence under p = 9 = 0 its
distribution is independent of #i,... ,#4. It is nondecreasing in U. Thus the UMP unbiased test of
Hg rejects when R > Co or when
R
^i^W
y/(n -2)>K0
where KQ is determined so that J^° tn-2(y)dy = a. The statistic R is linear in U, hence the UMP
test of H10 rejects when
I fí\ ľ°° (T
^lA--V(n-2)>Kh where J tn-2(y)dy =-.
3.2 Regression
We have observations (Yi,x{),..., (Yn,xn) and consider the simple regression _E[F|a;] = a + ßx.
Start with the transformations
Vi
V52(x
j - x
)2
and denote 7 + övi = a + ßxi, thus Yl v
% = 0, S v
ľ =
1- Then
x ó
a = 7-S .^, ß =^(xj-x)2
' V52(X
J - x
)2
The joint density of Yi,..., Yn is
(2vr)ra
/2
ď
Reparamet rization:
1
exp 2<T2
sufficient statistics
u = Y,viYi, T1 = Y,Y?, T2 = Y,Y.
The UMP test of hypothesis H u : ß = ö = 0 is based on the criterion
\ZviYi\
A =
^EW-^-iE^i)2
!
and rejects when A>C, where fTM tn-2(y)dy = f.
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4 Comparison of two Poisson distributions
Let X and Y be two independent variables with Poisson distributions V{\) and V{ß). Joint distribution
of X, Y is
e-(x+y)
P{X = x,Y = y} = ---- exp
x\y\
ylog- + (x + y) log A
Reparametrization: 9 = log ^, # = log A, sufficient statistics U = Y, T = X + Y. Consider the
hypotheses H12 : ß < A ~ 6 < 0 and H43 : /x = A ~ 6 = 0. The test is conditional for given T = t.
The conditional distribution of Y given X + Y = t is
P I F = y|X + r = í) =
í /x
A + /J.J \X + /J
A í-2/
the binomial B(t, p = ^ r ^ j - The hypothesis H12 is equivalent to p < \ and is rejected when
Y > C (t) and rejected with probability 7(c) when Y = C (t), where
s C H U = 2*a.
i=C(t)+l
Similarly, the hypothesis H43 is equivalent to p = \ and is rejected when Y < C\{t) and Y > C2(i)
and rejected with probability 7J(Č) when F = Cj(í); due to the symmetry it holds C\{t) -- | = |--C2(i)
and 71 (í) =72(í).
5 Comparison of two binomial distributions
Let X, Y be two independent variables with binomial distributions B(m,p\) and B{n,p2), respectively.
Their joint distribution is
>(x = x,Y = y) PÍQÍ
y
P2Q2
m \ n T Tb-- ti TTI Ti
P2Q2 Qi 92exp V ( log -- - log -- ) + (x + y) log --
Q2 <Zi J q\
 = i o g | e / a ) , #= l o g ?'
Reparamet rization:
.Q2' q\j q\
sufficient statistics U = Y, T = X + Y. Under 6 = 0, the conditional distribution of Y given
X + Y = t is the hypergeometric distribution
P(Y = y X + Y = t)
m
t-y
n
y
m + n
t
Hence, we reject the hypothesis H44 : P2 < Pi ~ 0 < 0 when F > C(í) and reject with probability
7(i) when Y = C (t) where
P (Y > C (ť) X + Y = t)+ 7(í)F(r = C (ť) X + Y = t) = a.
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6 Test for independence in a 2 x 2 table
Consider the population of n individuals; for each individual we check whether it has properties A
and B. The results we write in the table
A A sums
B X X' M
B Y Y' N
sums Q Q' s
where X is the number of individuals which have simultaneously A and B, etc. Denote pAB the probability
that an individual has both A and B, and similarly we denote the probabilities PAB, VAB, VÄBMoreover,
let pA and ps be the probabilities that an individual has property A, B, respectively. Then
PA = PAB + PABi PB = PAB + VÄBThe
joint distribution of variables X, Y, X', Y' is multinomial and is given by
T)' i i
P{X = x,X> = x>, Y = y,Y> = y') = - j - ^ ^ p ^ ^ ^ _ (6.1)
nl
, m / , VAB . /, PAB , , PAB
, ,, , n(PÄB) e x
P x
log + x log + y log
xlx'lyly'l V VAB VAB VAB
We wish to test the hypothesis, that properties A, B appear independently in the population. More
precisely, we wish to test the hypothesis of independence
Hi : pAB =PA-PB against Ki : pAB / pA ˇ PB,
or the hypothesis of positive dependence
H2 : PAB >PA-PB against K2 : pAB <PA'PBRewrite
the distribution (6.1) as follows:
n\------(PABTCMÖITI + e2T2 + \theta3T3}
xlx'lyly'l
where
a , VÄB , VAB , , VAB
9\ = -- log -^s
- -- log -^iS
- + log
VÄB VÄB VÄB
Ö2 = l 0 g ^ , Ö 3 = l 0 g ^ ,
VAB VAB
T1=X, T2 = X + X', T3 = X + Y.
The hypotheses Hi and H2 can be then rewritten as
Hi : 0i = 0, Ki : 9 / 0,
H2 : 9i < 0, K2 : 9 > 0.
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The tests are conditional, based on the conditional distribution of T\ given T2, T3 under 9\ = 0.
First we get
Pe1=0 (X = x,Y = y\X + Xl
= m) = ( m
x \ ( n
~ y
m
\ px
/y
(l - PAT^
and this under condition X + Y = q gives
3
öl=o {X = x \ X + X' = m, X + Y = q) =
m \ I n -- m
x J \ q -- x
~7ľT
The conditional test of H2 is called the Fisher-Irwin test; it has the form
1 if x < C\{m,q)
&(x,m,q) = ˇ[ 7(m,(?) if x = C(m,q),
0 if C\{m,q) < x
where Ci,7i,2 (Ci, integer) are determined so that
Y^ I m \ I n -- m \ I m \ I n -- m \ In
i \ Q-i +7l
\Ci [ q-d j =a
[ qí<C1-l