Biology Reference
In-Depth Information
th
e p
r
obability of finding the result x
B
x
A
0.4 from the data when
x
B
0.1789 is 0.0127. We
find this probability using statistical tables for normal distribution or
software that computes normal distribution. Thus, the p-value of our
one-sided test is p
x
A
is normal with parameters
m ¼
0and
s ¼
2.5
2
¼
0.0127 < 0.05 (see Figure 4-5), and we can reject
the null hypothesis of equal yield from the two corn types (and, thus,
also that the yield from type B is inferior to that from type A).
When either of group variances are unknown, we use a t-test as
exemplified next.
1.5
1
Area to the right of Z:
p
= 0.0127
0.5
0
−
1
−
0.8
−
0.6
−
0.4
−
0.2
0
0.2
0.4
0.6
0.8
1
t-test: As before, we want to test the one-sided alternative
Z = 0.4
H: Average yield of corn B is superior to the average yield of corn A (
m
A
<
m
B
)
FIGURE 4-5.
Relating the p-value and the Z-score. Using a
table, one associates a Z-score of the standard
normal distribution with an area (i.e., a p-value).
For example, Z ¼ 0.4 gives a one-tailed p-value
of 0.0127.
against the null hypothesis:
H
0
: Corn B has the same or inferior average yield to corn A (
m
A
m
B
).
This time, however, we assume the variance in the yield of the corn is
not known. This is, in fact, the more common practical situation. Now
we must rely on empirical estimates of the variances of the two types of
corn, using the formulas in Eq. (4-4) to obtain s
A
¼
0.3169.
The sample means and variances can also be computed using statistical
software, such as MINITAB and SPSS. The output from MINITAB is
given here.
0.4033 and s
B
¼
Descriptive Statistics: A, B
Variable
N
Mean
StDev
Minimum
Maximum
A
10
2.440
0.403
1.900
3.200
B
10
2.840
0.317
2.400
3.400
0.4
The sample distribution for testing the hypothesis H against the
null hypothesis H
0
becomes a t-distribution necessitating use of the
t-test. Because the computation of the t-statistics is quite complex
and is routinely done by computer, we are not going to present the
exact formula here. However, using MINITAB we find the t-statistics
have a value of t
0.35
0.3
0.25
0.2
Area to the right
of
t:
0.15
p
=0.012
0.1
2.466 with 17 degrees of freedom. We again follow
the general paradigm depicted in Figure 4-4(B), this time using Student
t-distribution.
¼
0.05
4
0
0
1
2
3
−
4
−
3
−
2
−
1
t
=2.466
Using appropriate software, we find that the p-value of this one-sided
test is
p
FIGURE 4-6.
p-value calculated from the t-value of the Student's
t-distribution with 17 degrees of freedom.
¼
0.
012 < 0.05, meaning the probability of having an empirical
result x
B
0.4 assuming the null hypothesis holds is 0.012.
Therefore, we can reject the null hypothesis. The MINITAB output from
this test is presented here. Figure 4-6 presents a graphical illustration of
the output.
x
A
