Rajesh Singh School of Statistics, D. A.V.V., Indore (M.P.), India
Jayant Singh Department of Statistics Rajasthan University, Jaipur, India
Florentin Smarandache Department of Mathematics, University of New Mexico, Gallup, USA
A Note on Testing of Hypothesis
Published in: Rajesh Singh, Jayant Singh, F. Smarandache (Editors) STUDIES IN STATISTICAL INFERENCE, SAMPLING TECHNIQUES AND DEMOGRAPHY InfoLearnQuest, Ann Arbor, USA, 2009 (ISBN-10): 1-59973-087-1 (ISBN-13): 978-1-59973-087-5 pp. 21 - 25
Abstract :
In testing of hypothesis situation if the null hypothesis is rejected will it automatically imply alternative hypothesis will be accepted. This problem has been discussed by taking examples from normal distribution. Keywords : Hypothesis, level of significance, Baye’s rule.
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1. Introduction
Let the random variable (r.v.) X have a normal distribution N( T , V 2 ), V 2 is assumed to be known. The hypothesis H0 : T = T 0 against H1 : T = T1 , T1 > T 0 is to be tested. Let X1, X2, …, Xn be a random sample from N( T , V 2 ) population. Let X (=
1 n
n
¦X
i
) be the sample
i 1
mean. By Neyman – Pearson lemma the most powerful test rejects H0 at D % level of significance, n X T o
t d D , where d D is such that V
if
f
³
dD
1 2S
e
Z2 2 dZ
=D
If the sample is such that H0 is rejected then will it imply that H1 will be accepted? In general this will not be true for all values of T1 , but will be true for some specific value of T1 i.e., when T1 is at a specific distance from T 0 .
H0 is rejected if
i.e. X t T 0 + d D
n X T o
t dD V
V n
(1)
Similarly the Most Powerful Test will accept H1 against H0
if
n X T1
t - dD V
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V
X t T1 - d D
i.e.
(2)
n
Rejecting H0 will mean accepting H1 (1) Â&#x; (2)
if
i.e.
X t T0 + d D
i.e.
T1 - d D
V
V n
V n
V
d T0 + d D
n
X t T1 - d D
Â&#x;
(3)
n
Similarly accepting H1 will mean rejecting H0 (2) Â&#x; (1)
if
V
T0 + d D
i.e.
V
T1 - d D
d
n
(4)
n
From (3) and (4) we have
T0 + d D
V n
= T1 - d D
i.e. T1 - T 0 = 2 d D
Thus d D
V n
=
V n
V
(5)
n
V T1 T 0 and T1 = T 0 +2 d D . 2 n
Reject H0 if
X >
T0 +
and from (2) Accept H1 if
X >
T1 -
From (1)
T1 T 0 T T = 0 1 2 2 T1 T 0 = 2
T 0 T1 2
Thus rejecting H0 will mean accepting H1
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when
T 0 T1 . 2
X >
From (5) this will be true only when T1 = T 0 + 2 d D
T1 z T 0 + 2 d D
V n
V n
. For other values of
rejecting H0 will not mean accepting H1.
It is therefore, recommended that instead of testing H0 : T = T 0 against H1 : T = T1 , T1 > T 0 , it is more appropriate to test H0 : T = T 0 against H1 : T > T 0 . In this situation rejecting H0 will mean T > T 0 and is not equal to some given value T1 . But in Baye’s setup rejecting H0 means accepting H1 whatever may be T 0 and T1 . In this set up the level of significance is not a preassigned constant, but depends on T 0 , T1 , V 2 and n. Consider (0,1) loss function and equal prior probabilities ½ for T 0 and T1 . The Baye’s test rejects H0 (accepts H1)
if
X >
T 0 T1 2
and accepts H0 (rejects H1)
if
X <
T 0 T1 . 2
[See Rohatagi, p.463, Example 2.] The level of significance is given by
PH0 [ X >
(T T ) n (X T0 ) n T 0 T1 ] = PH0 [ > 1 0 ] 2 V 2V
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§ n (T1 T 0 2V ©
= 1 - ) ¨¨
t
where )(t ) =
³
f
1 2S
e
Z2 2 dZ
· ¸ ¸ ¹
.
Thus the level of significance depends on T 0 , T1 , V 2 and n. Acknowledgement : Author’s are thankful to Prof. Jokhan Singh for suggesting this
problem.
References
Lehmann, E.L. (1976) : Testing Statistical Hypotheses, Wiley Eastern Ltd., New Delhi. Rohatagi, V.K. (1985) : An introduction to probability theory and mathematical statistics, Wiley Eastern Ltd., New Delhi.
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