Page 377 - DMTH404_STATISTICS

2
Unit 26:  - Test Hypothesis

Solution. Notes

H : The drug is not effective in curing cold
0
Using the Brandt and Snedecor formula, we have

2
160 160 52 2 10 2 18 2 80 ö
æ
´
2
 = ç + + - ÷ = 2.12.
´
80 80 è 96 20 44 160 ø
This value is less than the tabulated value (= 5.99) for 2 d.f. and 5% level of significance. Thus,
there is no evidence against H .
0
26.3 Summary
 Here we have to test whether  is different from zero. Accordingly, H and H are  = 0 and
0 a
  0 respectively.
For small samples, the test statistic can be obtained from the sampling distribution of b.
We note that if r = 0, then b would also be zero.
b S 1 S S n - 2 n - 2
Therefore, = r × Y × = r × Y × X r = will follow t - distribution
-
-
E
E
S . . ( ) b S X S . . ( ) b S X S Y 1 r 2 1 r 2
n - 2
with (n - 2) d.f. Hence, r can be taken as the test statistic. We note that
-
1 r 2
1 r 2
-
E
S . . ( ) r = . Therefore, 100(1 - a)% confidence limits of r can be written as r ± t
n - 2 a/2
S.E.(r).
 Let there be two independent random samples of sizes n and n from two normal
1 2
populations with correlations  and  respectively. Let r and r be the correlations
1 2 1 2
computed from the 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 ~ N x , and Z ~ N x ,
1 ç 1 ÷ 2 ç 2 ÷
è n - 3 ø è n - 3 ø
2
1
æ 1 1 ö
 Z - Z ~ N x - x , +
1 2 ç 1 2 ÷
è n - 3 n - 3ø
1
2
Z - Z
)
or 1 2 ~ N (0,1 under H :  = 
1 1 0 1 2
+
n - 3 n - 3
2
1
26.4 Keywords
Fisher’s test: It is applicable whether n is small or large. If r is correlation in sample, then its
+
1 1 r
Fisher's Z transformation is given by Z = log e .
2 1 r
-
2
 - test: It can be used to test, how far the fitted or the expected frequencies are in agreement
with the observed frequencies.
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