Research Methodology - Dr. Krishan K. Pandey

Testing of Hypotheses I 227
and
2
X i X
2 2 2
σ s
Hence, F
2
= ∑ −
d i .
2
s2
2
s
n
2
− 1
314
= = 28 55
12 − 1
2 2 eQσs σ
2 s1j
σ
= >
σ
1
2855 .
= = 214 .
1333 .
Degrees of freedom in sample 1 = (n 1 – 1) = 10 – 1 = 9
Degrees of freedom in sample 2 = (n 2 – 1) = 12 – 1 = 11
As the variance of sample 2 is greater variance, hence
v 1 = 11; v 2 = 9
The table value of F at 5 per cent level of significance for v 1 = 11 and v 2 = 9 is 3.11 and the table
value of F at 1 per cent level of significance for v 1 = 11 and v 2 = 9 is 5.20.
Since the calculated value of F = 2.14 which is less than 3.11 and also less than 5.20, the F ratio
is insignificant at 5 per cent as well as at 1 per cent level of significance and as such we accept the
null hypothesis and conclude that samples have been drawn from two populations having the same
variances.
Illustration 20
Given n 1 = 9; n 2 = 8
d 1i 2
1i
d 2i 2
2i
∑ X − X = 184
∑ X − X = 38
Apply F-test to judge whether this difference is significant at 5 per cent level.
2 2
Solution: We start with the hypothesis that the difference is not significant and hence, H0: σp = σp
.
To test this, we work out the F-ratio as under:
F
2
s1
2
s
d 1i 2
1i
b
2
1 g
2i 2 2
σ ∑ X − X / n − 1
= =
σ ∑ X − X / n − 1
2
d i b g
184/ 8 23
= = = 425 .
38/ 7 543 .
v 1 = 8 being the number of d.f. for greater variance
v 2 = 7 being the number of d.f. for smaller variance.
1 2