Research Methodology - Dr. Krishan K. Pandey

Testing of Hypotheses I 217
Solution: Let the sales before campaign be represented as X and the sales after campaign as Y and
then taking the null hypothesis that campaign does not bring any improvement in sales, we can write:
H 0 : μ μ
1 2
= which is equivalent to test H : D = 0
H : μ μ
a 1< 2 (as we want to conclude that campaign has been a success).
Because of the matched pairs we use paired t-test and work out the test statistic ‘t’ as under:
diff
0
D − 0
t =
σ . / n
To find the value of t, we first work out the mean and standard deviation of differences as under:
Table 9.9
Shops Sales before Sales after Difference Difference
campaign campaign squared
X i Y i (D i = X i – Y i ) D i 2
A 53 58 –5 25
B 28 29 –1 1
C 31 30 1 1
D 48 55 –7 49
E 50 56 –6 36
F 42 45 –3 9
2
n = 6 ∑ Di = −21
∑ Di = 121
∴ D
σ diff
.
Di
=
n
∑
= ∑ − ⋅
21
=− =−35
.
6
di b g
2 2 2
i
.
D D n
=
n − 1
− −
Hence, t = = − 35 . 0 35 .
=−2.
784
308 . / 6 1257 .
121 − − 35 × 6
= 308 .
6 − 1
Degrees of freedom = (n – 1) = 6 – 1 = 5
As H a is one-sided, we shall apply a one-tailed test (in the left tail because H a is of less than type)
for determining the rejection region at 5 per cent level of significance which come to as under, using
table of t-distribution for 5 degrees of freedom:
R : t < – 2.015
The observed value of t is – 2.784 which falls in the rejection region and thus, we reject H 0 at 5
per cent level and conclude that sales promotional campaign has been a success.