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252power 2/8/01 (Open this document in 'Page Layout' view!)

C. POWER FUNCTIONS

1. A one~sided Test.

If .d th bl

{H 0 :P S Po <' . I f

.bar

..we conSI er e pro em: lor a gIVeva ue 0 a, to constructan operatmgc actenshc

HI :P>Po

curve, we must fmd the value of j3 for all possiblevalues of PI above J1.o'k; a practical matter, let

PI equalPo, Po + Yzzaa;, Po +zaa;(=xcv), Po + J;.zaa; and so onuntil,B is almost one.The

power function is a plot of 1- j3 insteadof j3, and crossesthe operating characteristiccurve at the criticalvalue.

{

Ho :ps500Example: Assumethat a = .025andtest whena = 35and n =50, sothat

HI : J1.> 500

ax = c:- = ~ = 4.95 and Xcv = J1.o+ zaa, = 500 + (1.960X4.95) = 509.7 . We thus reject the null.yn .y50

hypothesis if the sample mean is above S09.7. Since the interval between Po and Xcv is about 10, the

interval between values of PI should be about half that, so letpl equal 505, 509.7, 515, and 520. (One

value of PI should be Po, and one should be Xcv .)

Start then, by assuming that PI = 505. Then ,B= pf Accepting' Ho IH 0 is false}

= P{x S 509.7!J1.= SOS}= P{z s Xcv - J1.I

}= P{z s S09.7 - 505}= p{z S O.95}=.5 + .3289 = .8289

. ax 4.95 .

)DO ro,.1

An illustration wouldhelp.SinceourcriticalvalueisS09.7,we accept the null hypothesis only if the samplemean is less than or equal to 509.7. But let us assume that thepopulation mean is actually 505. In all the fol1owin~ diagrams,the 'reiect' region is shaded Southwest to Northeast and theregion representing the probability of wronglv failing to reiectthe null hYPQthesisis shaded Northwest to Southeast.

Then we can draw two Nonnal curves, one centered at 505and one centered at 500. The probability of rejecting the nullhypothesis when it is true is the right shaded area, while theprobability of accepting the null hypothesis when the mean is50S is the left shaded area.

Now, try the secondvalue, 509.7, for PJ'

fJ = pt Accepting H 0IH 0 is false}

=p~ s 509.7/p =509.7}=p{z S XCVa~Pl }

{

S09.7-509.7

}= P Z s = p{z S O}= .5000 Note that j3 at4.95

PI = Xcv is always .5000.

2~b

~roO ""'.1

2

Next, lets try f.J.l= 515.

'~fJ= p{x S;S09.71f.J.= 51S}

=P{z S;509.7-515

}= p{z S;-1.07} = .1423.

4.95

For PI = 520 .

fJ =p{x S; 509.7!p =520}

=P{z S;509.7-520

}= p{z S; -2.08}= .0188

4.95

Finally, to get a limitingvalue, let

PI = 500 . fJ =p{x S;509.71p =500}

{

509.7 - 500

}{ }=P z S; - =P z S;1.96 =.9750. Note that Ii

4.9~

at J.lI =J.lo is always I-a.

We can summarize our results in the table and graph below.

J.lI fJ Power=I- fJ '500 .9750 2.50% .ff505 .8289 17.11%.~509.7 .5000 50.00%515 .1423 85.77%520 .0188 98.12%

.11

0tJ.ol1 ."u.

2. A Two..Sided Test.This is the same as a one sided test except that we must consider values of J11on both sides of J.10 . This is

little additional effort since points at the same distance from J.lo have the same values of fJ .

{

Ho :J.l=165Example: Assume that IX=.05 and test. . when (j =48 and n = 36, so that

HI :p*165

(j 48(j- =-=-=800

x j;; 136 .

Xcv=PoIZ ax = 165+(1.960X8.00)= 165:t15.7~

We thus reject the null hypothesis if the sample mean is below149.3 or above 180.7.

Since the interval between Po and Xcv is actually between 15.6 and 15.7, space Jil at half that, or about 7.8

or 7.9. The points that I picked above 165 were 172.8, 180.7, 188.5,and 196.4. The points below 165 were157.2, 149.3, 141.5 and 133.6. Note that, for example 172.8 and 157.2 are at the same distance from 165.

2 1

3

Now,try 172.8 for J.1J. fJ = P{ Accepting'H 0IH0 is false}

== P~49.3 ~ x:S; 180. 71J.1=17:i.8}

==pr49.3~172.8 <z:s; 180.7;172.8}=p{-2.94:S;Z~0.99}

==.4984 + .3389 = .8373. Nowtry thesame calculation forJ.1]=157.7 , which is the same distance to the left of 165 as

172.8 is to the right.This time

13 =pt Accepting'HolH 0 is false} =P~49.3:S;x ~ 180.71J.1=157.7}

= p{ 149.3; 157.7~ Z ~ 180.7; 157.7} = P{-0.99 ~ Z~ 2.94}= .3389+.4984 = .8373, which is the same

result as for J.1I=172.8. In other wordsthe operatingcharacteristicscurveand power function curve for

values of Pl to the left of Pois the mirror imageof the curve to the right of Po. The table below

summarizes our results and further calculations. xcvLmeans a lower criticalvalue and xcvu means an upper

critical value. These become ZL and Zu . Note that, as above,when PI = Po Power= a, and when

p] ==Xcv Power = 50%.

1"I,r , 'H.,L

IIJ.,. IH" 1,,,.7 , .'.i" ,,,,, ~.1'$).'

3 /')(---

p] I xcvL - p] xcvu - PJ zL Zu 13Power

-=1-13Q"- Q"-x x

165 I 149.3-165 180.7-165 -1.96 1.96 .4750 + .4750 5.00%- =.95008 8

172.8 (Same 149.3-172.8 180.7 -172.8 -2.94 0.99 .4984 + .3389 T16.27%

as 1572) 8 8 = .8373

180.7 (Same 149.3-180.7 180.7 -180.7 -3.93 0.00 .5000 I 5000%

as 149.3) 8 8

188.5 (Same 149.3 -188.5 180.7 -188.5 -4.90 -0.98 .5000-.3365 T 83.65%as 141.5) 8 8 =.1635

196.4 (Same 149.3 -196.4 180.7 -196.4 -5.98 -1. 96 .5000-.4750 I 97.50%

as 133.8) 8 8 = .0250

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