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11
Lecture 22
! Review of Comparing the Means of Two Normal Distributions
! The F Distribution.
! Comparing the Variances of Two Normal Distributions.
17
The F distribution! Consider two independent random variables Y and
Z, such that . Define
Then the distribution of X is called an F distribution with m and n degrees of freedom.
22 ~,~ nm WY cc
mWnY
nWmYX ==//
[ ] 0)()()(
)()( 2/)(
12/
21
21
2/2/21
>+
×GG
+G= +
-
xnmxx
nmnmnmxf nm
mnm
18
m=5,n=20
m=20,n=5
19
Properties of the F distribution! If a random variable X has an F distribution
with m and n degrees of freedom, then 1/Xwill have an F distribution with n and mdegrees of freedom.
! If a random variable X has a t distribution with n degrees of freedom, then will have an F distribution with 1 and n degrees of freedom.
2X
20
Comparing the Variance of Two Normal Distributions
! Suppose that X1,…,Xm form a random sample of mobservations from a normal distribution for which both the mean µ1 and the variance are unknown, suppose also that Y1,…,Yn form a random sample of nobservations from a normal distribution for which both the mean µ2 and the variance are unknown. Suppose the following hypotheses are to be tested at a specified level of significance a0 :
22
211
22
210
::
ss
ss
>
£
HH
21s
22s
21
)1/()1/(
2
2
--
=nSmSV
Y
X
Define
å
å
=
=
-=
-=
n
iniY
m
imiX
YYS
XXS
1
22
1
22
)(
)(
The test statistic we will use is
We will consider the test that rejects H0 when cV ³
22
1,122
2
21
2
21
22
2
21
21
2
~)]1(/[)]1(/[*
~/~/
---
-
--
=¾¾¾ ®¾ïþ
ïýü
nmY
Xtindependen
nY
mX FnSmSV
SS
ss
cs
cs
So when
1,1~* --= nmFVV
22
21 ss =
23
! Note that when
So the size of the test is
! Given a significance level a0, we can choose a constant c such that
0,1,101,1 )Pr( aa ---- =Þ=³ nmnm FccF
)Pr()|Pr(),|Pr(sup 1,122
21
22
21
22
21
cFcVcV nm ³==³=³ --£
ssssss
22
21 ss <
22 2 2 21 2 1 22
2 2 22 21 21 22 2 2
2 1
22
1, 1 21
/( 1)Pr( | ) Pr( | )/( 1)
/( ( 1)) Pr |/( ( 1))
= Pr
X
Y
X
Y
m n
S mV c cS n
S m cS n
F c
s s s s
s s s ss s
ss- -
-³ < = ³ <
-
æ ö-= ³ <ç ÷-è ø
æ ö³ç ÷
è ø
( )1, 1 Pr m nF c- -< ³
24
Example! Suppose that 6 observations X1,…,X6 are selected at
random from a normal distribution for which both the mean µ1 and the variance are unknown, and suppose that it is found that . Suppose also that 21 observations Y1,…,Y21 are selected at random from another normal distribution for which both the mean µ2 and the variance are unknown, and suppose that it is found that
. We shall carry out an F test of the hypotheses
21s
22s
30)(6
12
6 =-å =i i xx
40)(21
12
21 =-å =i i yy
22
211
22
210
::
ss
ss
>
£
HH
25
! The value of V for the given example is
! It is found from the F tables that the 0.05 upper quantile of the F distribution with 5 and 20 degrees of freedom is 2.71.
! Hence, the hypothesis H0 should be rejected at the level of significance a=0.05.
320/405/30==V
26
The Other One-Sided Hypotheses and Two-Sided Hypotheses
! If we want to test the hypotheses
The level a0 F test rejects H0 when
! If we want to test the hypotheses
The level a0 F test rejects H0 whenor
22
211
22
210
::
ss
ss
<
³
HH
01,1,1 a---£ nmFV
2/1,1,1 0a---£ nmFV
22
211
22
210
::
ss
ss
¹
=
HH
2/,1,1 0a--³ nmFV
27
Example! Let X1,…X26 be the log-rainfalls of the 26 seeded
clouds, and let Y1,…,Y26 be the log-rainfalls of 26 unseeded clouds. The observed values of sample statistics are
We want to test the hypotheses
at level of significance a0=0.05.
22
211
22
210
::
ss
ss
¹
=
HH
39.67990.396.63134.5
226
226
==
==
Y
X
SYSX
28
! The value of V for the given example is
! It is found from the F tables that the 0.975 and 0.025 upper quantiles of the F distribution with 25 and 25 degrees of freedom are 0.4484 and 2.2303.
! Hence, the hypothesis H0 should be accepted at the level of significance a=0.05.
9491.025/39.6725/96.63
==V