Agriculture Reference
In-Depth Information
s
1
Solution.
Under the given information, we want
to test
H
0
: ρ ¼
0 against
H
1
: ρ 6¼
0
:
The test
statistic is given by
t ¼
r
h
i
X
: d
2
2
S
d
¼
d
i
8
ð
8
1
Þ
s
1
p
1
¼
½
87804
:
0
87780
:
5
n
2
ð
7
Þ
p
with
ðn
2
Þ
d
:
f
:
r
1
7
23
:
5
r
2
p
33
:
571
¼
¼
¼
1
:
832
:
Given that
r ¼
0.850,
n ¼
20, so
85
p
1
p
0
18
d
p
¼
104
:
p
¼
75
104
:
75
0
:
ð
20
2
Þ
0
:
85
=
=
So
t ¼
t ¼
q
¼
p
¼
:
:
6
846
:
sd
0
648
1
:
832
:
2775
2
ð
0
:
85
Þ
¼
:
:
161
651
The table value of
t
0
:
005
;
18
¼
2
:
878. Since
From the table value, we have
t
0
:
95
;
7
¼
jj>
table
t
0
:
005
;
18
, that is, 6.846
>
2.878, so the
1
:
895 and
t
:
99
;
7
¼
2
:
998
:
The
calculated
null hypothesis
H
0
: ρ ¼
0 is rejected, we accept
value of
is less than the table value at both the
level of significance. Hence, the test is significant
at 1% level of significance. So we reject the null
hypothesis
t
H
1
: ρ 6¼
So we can conclude that the age and
lactation period in Indian cows are correlated.
12.
0
:
Test for Equality of Two Population Vari-
ances from a Bivariate Normal Distribution
The null hypothesis for testing the equality
of two variances is
.
That is, there was significant effect of hormone
spray on fruit weight of papaya.
11.
H
0
: μ
x
¼ μ
y
and accept
H
1
: μ
x
<μ
y
2
2
y
.
Let us derive two new variables
H
0
: σ
X
¼ σ
To Test for Significance of Population Cor-
relation Coefficient, That
U
and
V
Is, H
0
: ρ ¼
0
such that
U ¼ X
+
Y
and
V ¼ X Y
.So
Against H
1
: ρ 6¼
0
The test statistic under
the
Cov(
U
,
V
)
¼
Cov(
X
+
Y
,
X Y
)
¼
H
0
will be
t ¼
σ
2
X
σ
2
y
. Under
the
null
hypothesis
p
1
r
p
n
2
=
r
2
at (
n
2) d.f. where
r
is
H
0
: σ
2
X
¼ σ
2
y
, Cov(
U
,
V
)
¼
0, and thus
U
the sample correlation coefficient between
X
and
are two normal variates with correla-
tion coefficient
V
and
, the two variables considered.
If the calculated value of
Y
ρ
UV
¼
0 when
H
0
is true.
jj
is less than the
2
2
y
Hence,
H
0
:
σ
X
¼ σ
is equivalent to test
H
0
:
tabulated value of
t
at (
n
2) d.f. for upper
ρ
UV
¼
0.
As usual
α
level of significance, we cannot reject
the null hypothesis; that is, sample correla-
tion coefficient is not significantly different
from zero or the variables are uncorrelated.
Otherwise we reject the null hypothesis, and
sample correlation coefficient is significant
to the population correlation coefficient
=
2
the test statistic is given by
p
ðn
p
1
t ¼ r
uv
2
Þ
r
uv
2
with
ðn
2
Þ
d
:
f
;
where
r
uv
is the sample correlation coeffi-
cient between
u
and
v
.
Example 9.12.
To test whether a specific artificial
hormonal spray has effect on variability of fruit
weight of papaya, initial and final weights from a
sample of eight plants of a particular variety of
papaya were taken at an interval of 15 days of spray.
not
zero and the variables have significant corre-
lation between them.
b
Example
The correlation coefficient
between the age and lactation duration of 20
breeds of Indian cows is found to be 0.850. Test
for the existence of correlation between these
two characters in Indian cows using 1% level of
significance.
9.11.
Initial wt (g) 114 113 119 116 119 116 117 118
Final wt (g) 220 217 226 221 223 221 218 224
Solution.
To work out the significant differences
in variability of fruit weights before and after
