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 :

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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

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2

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~/~/

---

-

--

=¾¾¾ ®¾ïþ

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nmY

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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

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210

::

ss

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>

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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

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211

22

210

::

ss

ss

<

³

HH

01,1,1 a---£ nmFV

2/1,1,1 0a---£ nmFV

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211

22

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::

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.

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22

210

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ss

¹

=

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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