Table of Contents
Advertisement
Example1 -- Consider two samples drawn from normal populations such that
2
n
= 21, n
= 31, s
= 0.36, and s
1
2
1
2
2
σ
= σ
, at a significance level α = 0.05, against the alternative hypothesis,
1
2
2
2
: σ
≠ σ
H
.
For a two-sided hypothesis, we need to identify s
1
1
2
follows:
2
s
=max(s
M
2
s
=min(s
m
Also,
Therefore, the F test statistics is F
The P-value is
P-value = P(F>F
UTPF(20,30,1.44) = 0.1788...
Since 0.1788... > 0.05, i.e., P-value > α, therefore, we cannot reject the null
: σ
2
= σ
hypothesis that H
1
o
Additional notes on linear regression
In this section we elaborate the ideas of linear regression presented earlier in
the chapter and present a procedure for hypothesis testing of regression
parameters.
The method of least squares
Let x = independent, non-random variable, and Y = dependent, random
variable.
The regression curve of Y on x is defined as the relationship
between x and the mean of the corresponding distribution of the Y's.
Assume that the regression curve of Y on x is linear, i.e., mean distribution of
Y's is given by Α + Βx. Y differs from the mean (Α + Β⋅x) by a value ε, thus
Y = Α + Β⋅x + ε, where ε is a random variable.
To visually check whether the data follows a linear trend, draw a scattergram
or scatter plot.
2
= 0.25. We test the null hypothesis, H
2
2
2
,s
) = max(0.36,0.25) = 0.36 = s
1
2
2
2
,s
) = min (0.36,0.25) = 0.25 = s
1
2
n
= n
= 21,
M
1
n
= n
= 31,
m
2
ν
= n
- 1= 21-1=20,
N
M
ν
= n
-1 = 31-1 =30.
D
m
2
2
= s
/s
=0.36/0.25=1.44
o
M
m
) = P(F>1.44) = UTPF(ν
o
2
.
2
:
o
and s
, as
M
m
2
1
2
2
, ν
,F
) =
N
D
o
Page 18-49
Advertisement
Table of Contents
