Agriculture Reference
In-Depth Information
Example 9.2.
Given below are the lengths
(inches) of ten randomly selected panicles of
particular wheat variety. If the length of panicle
assumed to follow a normal distribution with
unknown variance, can we say that the average
length (inches) of panicle for the wheat variety
be 12 in.?
Panicle length (inches): 10, 12, 6, 15, 14, 13,
13, 7, 14, 6
population variance. Against the alternative
hypotheses
2
2
0
2
2
0
H
1
:
ð
1
Þ σ
6¼ σ
; ð
2
Þ σ
> σ
;
and
2
2
ð
3
Þ σ
< σ
0
, the test statistic is
P
n
i¼
1
ðx
i
μÞ
2
2
n
¼
χ
0
σ
with
d.f.
1. For
H
1
: σ
n
0
, the null hypothesis is
rejected if the calculated value of
2
6¼ σ
Solution.
Given that the (1) length of panicle
follows a normal distribution with unknown var-
iance and (2) population hypothetical mean is
12 in.
To test
2
χ
n
is
2
greater
than the table value of
χ
n
at
upper
α
/2 level of significance and at
n
2
χ
n
>
degrees of freedom, that is, cal
tab
H
0
: Population mean
μ ¼
12 against
2
2
χ
or
calculated
χ
n
<
tabulated
α
/2,
n
H
1
:
12. The test statistic under the null
hypothesis is
μ 6¼
χ
2
(1
α
)/2,
n
.
2. For
t ¼
xμ
s=
1) d.f. and the
test is a both-sided test. We have sample mean,
with (
n
p
2
2
0
, the null hypothesis is
rejected if the calculated value of
H
1
: σ
> σ
2
χ
n
is
2
greater
than the table value of
χ
n
at
upper
α
level of significance and at
n
X
n
1
n
1
10
10
x ¼
1
x
i
¼
½
þ
12
þþ
6
¼
11 in
:
2
degrees of freedom, that is, cal
χ
n
>
tab
i¼
χ
2
α
,
n
.
3. For
and
2
2
0
, the null hypothesis is
rejected if the calculated value of
H
1
: σ
< σ
"
#
n
1
X
n
2
9
X
10
1
x
i
2
1
1
χ
n
is
2
2
2
s
¼
i¼
1
x
i
x
ð
Þ
¼
10
:ð
11
Þ
2
at lower
less than the table value of
α
level of significance and at
n
degrees of
freedom, that is, cal
χ
n
22 in
2
¼
12
:
:
2
2
χ
n
<
tab
χ
.
α
n
(1
),
So
Example 9.3.
From the following data on num-
ber of flowers per plant of rose, find out whether
the variance of rose per plant can be taken as 6,
given that the number of flowers per plant of rose
follows a normal distribution with mean number
of flowers per plant as 12.
No. of flowers per plant: 12,11,10,9,13,12,
8,15,16,13
11
12
12
1
t ¼
q
¼
106
¼
0
:
904
:
1
:
22
10
:
at 9 d.f. at 2.5% level of
significance is 2.262 which is greater than the
absolute value of the calculated value of
The table value of
t
t
, that
Solution.
Given that (1) the population mean is
12 (known), (2) the sample has been drawn from
a normal population, (3) sample size is 10.
To test
is,
jj< t
0
:
025
;
9
. So the test is nonsignificant and
the null hypothesis cannot be rejected. That
means that the length of panicle for the given
wheat variety may be taken as 12 in.
3.
2
H
0
:
σ
¼
6 against
H
1
: σ
2
6¼
6.
10
Under the
H
0
,
the test statistics is
χ
¼
Test for Significance for Specified Population
Variance with Known Population Mean from
a Normal Population
If
P
n
2
1
x
i
μ
ð
Þ
¼
½
0
þ
1
þ
4
þ
9
þ
1
þ
0
þ
16
þ
9
þ
16
þ
1
6
i¼
2
σ
x
n
be a random sample drawn
from a normal population with mean
x
1
,
x
2
,
x
3
,
...
,
¼
57
=
6
¼
9
:
5
:
μ
and
From the
table, we have
a value of
2
, that is,
2
) to test
2
variance
σ
N
(
μ; σ
H
0
: σ
¼
2
2
0.975,10
χ
¼
20.483 and
χ
¼
3.25. The
0.025,10
0
where
0
, is any specified value for the
2
, that is, 9.5, lies between
σ
σ
calculated value of
χ
