If X = c is the lower boundary of the upper critical region, require P( Xc

Transcription

1 Quality of tests 8A a H :p=.5 H :p>.5, so that X B(,.5), so require suh that P( X) <.5 From the inomial umulative distriution tales P( X5) = P( X4) =.99=.78 P( X6) = P( X 5) =.983=.97 P( X5) >.5 and P( X6) <.5 so the ritial value is 6 Hene the ritial region is X6 P(Type I error) = P( X 6 p=.5) =.97 P(Type II error) = P( X 5 p=.3) =.957 a H :p=.3 H :p<.3, so that X B(,.3) Signifiane level %, so require suh that P( X) <. From the inomial umulative distriution tales P( X) =.355 and P( X) =.76 P( X) >. and P( X) <. so the ritial value is Hene the ritial region is X P(Type I error) = P( X p=.3) =.76 P(Type II error) = P( X p=. 5) = P( X p=.5) =.43= a H :p=.45 H :p.45, so that X B(,.45) If X = is the upper oundary of the lower ritial region, require P( X ) <.5 From the tales P( X) =.33 and P( X) =.996 = and the lower ritial region is X ) <.5 From the tales P( X8) = P( X7) =.976=.74 P( X9) = P( X8) =.9955=.45 = 9 and the upper ritial region is X9 So the ritial region is X or X 9 Pearson Eduation Ltd 8. Copying permitted for purhasing institution only. This material is not opyright free.

2 3 P(Type I error) = P( X p=.45) + P( X p=.45) = =.78 P(Type II error) = P(X8 p=.4) = P( X8 p=.4) P( X p=.4) = = a H :λ=6 H :λ>6, so that X Po(6), so require suh that P( X) <.5 From the Poisson umulative distriution tales P( X) = P( X9) =.96=.839 P( X) = P( X) =.9574=.46 P( X) >.5 and P( X) <.5 so the ritial value is Hene the ritial region is X P(Type I error) = P( X λ = 6) =.46 P(Type II error) = P( X λ = 7) =.95 5 a H :λ=4.5 H :λ<4.5, so that X Po(4.5), so require suh that P( X) <.5 From the Poisson umulative distriution tales P( X) =.6 and P( X = ) =. P( X) >.5 and P( X = ) <.5 so the ritial value is Hene the ritial region is X = P(Type I error)= P(X = λ=4.5)=. P(Type II error) = P( X λ = 3. 5) = P( X = λ = 3.5) 6 a H :λ=9 H :λ 9, so that X =.3=.9698 Po(9) If X = is the upper oundary of the lower ritial region, require P( X ) <.5 From the tales P( X4) =.55 and P( X3) =. = 3 and the lower ritial region is X3 ) <.5 P( X5) = P( X4) =.9585=.45 P( X6) = P( X5) =.978=. = 6 and the upper ritial region is X6 So the ritial region is X3 or X 6 Pearson Eduation Ltd 8. Copying permitted for purhasing institution only. This material is not opyright free.

3 6 P(Type I error) = P( X3 λ = 9) + P( X6 λ = 9) =.+.=.43 P(Type II error) = P(4X5 λ = 8) = P( X5 λ= 8) P( X3 λ= 8) = = a H :p=. H :p<., so that X So (.) <.5 Require P( X) <.5 Geo(.) ( )log.8< log.5 log.5 > log.8 > 4.45 So the ritial value is 5 and the ritial region is X5 5 4 P(Type I error) = P( X 5 p=.) = (.) =.8 =.44 (4 d.p.) P(Type II error) = P( X4 p=.5) = (.5) = = = (4 d.p.) 4 8 a H :p=. H :p<., so that X Signifiane level % Require P( X) <. < So (.). Geo(.) ( )log.98< log. log. > log.98 > So the ritial value is 9 and the ritial region is X9 9 8 P(Type I error) = P( X 9 p=.) = (.) =.98 =. (4 d.p.) P(Type II error) = P( X8 p=.) = (.) = = = (4 d.p.) Pearson Eduation Ltd 8. Copying permitted for purhasing institution only. This material is not opyright free. 3

4 9 a H :p=. H :p., so that X So (.) <.5 Geo(.) ) <.5 log.5 > log.99 > = 369 and the So (.) <.5.99 >.975 upper ritial region is X369 If is the upper oundary of the lower ritial region, require P( X ) <.5 < log.975 log.99 <.59 = and the lower ritial region is X So the ritial region is X or X 369 P(Type I error) = P( X p=.) + P( X 369 p=.) = (.) +(.) 369 = = =.447 (4 d.p.) P(Type II error) = P(3X 368 p=.) = P( X368 p=.) P( X p=.) ( ) = (4 d.p.) = = a i A Type error ours when H is rejeted ut H is in fat true. ii A Type error ours when H is aepted ut H is in fat false. Pearson Eduation Ltd 8. Copying permitted for purhasing institution only. This material is not opyright free. 4

5 H :p=.4 H :p.4, so that X Signifiane level % Geo(.9) ) to e as lose as possile to 5%. First, find P( Xa) <.5 a So (.4) <.5 log.5 a > a> log.996 ould e 748 or 749 X = = X = = P( 748) and P( 749) , and as P( X748) is loser to.5, the upper ritial region is X748 If is the upper oundary of the lower ritial region, require P( X ) to e as lose as possile to 5%. First, find P( X) <.5 So (.4) < >.95 log.95 < <.79 log ould e or 3; P( X) =.996 =.4696 and P( X3) =.996 =.577 and as P( X3) is loser to.5, the lower ritial region is X3 So the ritial region is X3 or X 748 P(Type I error) = P( X3 p=.4) + P( X 748 p=.4) = =.9 (4 d.p.) a H :p=.5 H :p>.5, so that X B(4,.5), so require suh that P( X ) <.5 From the inomial umulative distriution tales P( X4) = P( X3) =.869=.38 P( X5) = P( X 4) =.95=.48 P( X4) >.5 and P( X5) <.5 so the ritial value is 5 Hene the ritial region is X5 P(Type I error) = P( X 5 p=.5) =.48 Pearson Eduation Ltd 8. Copying permitted for purhasing institution only. This material is not opyright free. 5

6 H :p=.5 H :p>.5, so that X Geo(.5) Require P( X).5 So (.5) log.95log.95 log.95 log.95 = So the ritial region is X = d P(Type I error)= P(X = p=.5)=.5 e For David s test X Geo(.588) P(Type II error)= P(X > p=.588)=(.588) =.94 f For Mihael s test X B(4,.588) Find the proaility y using a statistial alulator P(Type II error) = P( X 4 p=.588) =.96 (4 d. p.) Pearson Eduation Ltd 8. Copying permitted for purhasing institution only. This material is not opyright free. 6