Page 369 - DMTH404_STATISTICS

2
Unit 26:  - Test Hypothesis

Notes
Example 2: If the correlation of 10 pairs of observations (X, Y) is 0.96, test the hypothesis
that correlation in population is 0.99.
Solution.
We have to test H :  = 0.99 against H :  0.99
0 a
Further,

+
1 0.96
´
=
Z = 1.1513log = 1.1513log 49 1.1513 1.6902 = 1.9459 and
10 10
1 0.96
-
1 0.99
+
=
´
x = 1.1513log = 1.1513log 199 1.1513 2.2989 = 2.6464
10 10
1 0.99
-
-
The test statistic is z = Z x n - 3 = 1.9459 2.6464 7 = 1.8563.
-
Since this value is less than 1.96, there is no evidence against H at 5% level of significance.
0
26.1.3 Test Concerning Equality of Correlations in two Populations
Let there be two independent random samples of sizes n and n from two normal populations
1 2
with correlations  and  respectively. Let r and r be the correlations computed from the
1 2 1 2
respective samples.
If Z , Z , x and x denote Fisher's transformation of r , r , r and  respectively, then
1 2 1 2 1 2 1 2
æ 1 ö æ 1 ö
Z 1 ~ N x 1 , ÷ and Z 2 ~ N x 2 , ÷
ç
ç
è n - 3 ø è n - 3 ø
1
2
æ 1 1 ö
 Z - Z 2 ~ N x - x 2 , + ÷
ç
1
1
è n - 3 n - 3ø
2
1
Z - Z
)
or 1 2 ~ N (0,1 under H :  = 
1 1 0 1 2
+
n - 3 n - 3
1 2
Example 3: The correlation coefficients 0.89 and 0.85 were computed from two independent
samples of sizes 12 and 16 respectively. Test whether they can be regarded to have come from
two bivariate populations with different correlation coefficients?
Solution.
We shall test H :  =  against H :    .
0 1 2 a 1 2
1.89
´
Now Z = 1.1513log = 1.1513log 17.18 = 1.1513 1.2350 = 1.42
1 10 10
0.11
1.85
=
´
and Z = 1.1513log = 1.1513log 12.33 = 1.1513 1.0911 1.26
2 10 10
0.15

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