Research Methodology - Dr. Krishan K. Pandey

Testing of Hypotheses I 209
nX + nX
X12
. =
n + n
1 1 2 2
1 2
3. Samples happen to be small samples and population variances not known but assumed
to be equal:
In this situation we use t-test for difference in means and work out the test statistic t as
under:
t =
with d.f. = (n 1 + n 2 – 2)
Alternatively, we can also state
t =
with d.f. = (n 1 + n 2 – 2)
X − X
1 2
2
2
d i d i
∑ X1i − X1 + ∑ X2i − X2
1 1
× +
n + n − 2
n n
1 2 1 2
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
Illustration 7
The mean produce of wheat of a sample of 100 fields in 200 lbs. per acre with a standard deviation
of 10 lbs. Another samples of 150 fields gives the mean of 220 lbs. with a standard deviation of
12 lbs. Can the two samples be considered to have been taken from the same population whose
standard deviation is 11 lbs? Use 5 per cent level of significance.
Solution: Taking the null hypothesis that the means of two populations do not differ, we can write
H : μ = μ
0 2
and the given information as n 1 = 100; n 2 = 150;
H : μ μ
a 1≠ 2
X = 200 lbs . ; X = 220 lbs . ;
1 2
σ = 10 lbs . ; σ = 12 lbs . ;
s s
1 2
and
σ p = 11 lbs.
Assuming the population to be normal, we can work out the test statistic z as under:
z =
X1− X2
200 − 220
=
2F
1 1I
2F
1 1
σ p + bg
11 +
n n HG100
150
HG
1 2
KJ
I K J