Research Methodology - Dr. Krishan K. Pandey

212 Research Methodology
=
57 − 61
453 b . g + 648 b . g 1
× +
5 + 7 − 2 5
1
7
=−3.
053
Degrees of freedom = (n 1 + n 2 – 2) = 5 + 7 – 2 = 10
As H a is two-sided, we shall apply a two-tailed test for determining the rejection regions at 5 per
cent level which come to as under, using table of t-distribution for 10 degrees of freedom:
R : | t | > 2.228
The observed value of t is – 3.053 which falls in the rejection region and thus, we reject H 0 and
conclude that the difference in sales in the two towns is significant at 5 per cent level.
Illustration 10
A group of seven-week old chickens reared on a high protein diet weigh 12, 15, 11, 16, 14, 14, and 16
ounces; a second group of five chickens, similarly treated except that they receive a low protein diet,
weigh 8, 10, 14, 10 and 13 ounces. Test at 5 per cent level whether there is significant evidence that
additional protein has increased the weight of the chickens. Use assumed mean (or A 1 ) = 10 for the
sample of 7 and assumed mean (or A 2 ) = 8 for the sample of 5 chickens in your calculations.
Solution: Taking the null hypothesis that additional protein has not increased the weight of the chickens
we can write:
H : μ μ
0 1= 2
H : μ μ
a 1> 2 (as we want to conclude that additional protein has increased the weight of
chickens)
Since in the given question variances of the populations are not known and the size of samples is
small, we shall use t-test for difference in means, assuming the populations to be normal and thus
work out the test statistic t as under:
with d.f. = (n 1 + n 2 – 2)
t =
X − X
1 2
2
2 b g b g
n1− 1 σs + n2
− 1 σs
1 1
1 2 × +
n + n − 2 n n
1 2 1 2
2 2
From the sample data we work out X1, X2, σ s and σ (taking high protein diet sample as
1 s2
sample one and low protein diet sample as sample two) as shown below: