Quality of tests Mixed exercise 8

Quality of tests Mixed exercise 8 1 a H :p=.35 H 1 :p>.35 Assume H, so that X B(15,.35) Significance level 5%, so require c such that P( Xc) <.5 From the binomial cumulative distribution tables P( X8) = 1 P( X7) = 1.8868=.1132 P( X9) = 1 P( X 8) = 1.9578=.422 P( X8) >.5 and P( X9) <.5 so the critical value is 9 Hence the critical region is X9 b Size= P(Type I error) = P( X 9 p=.35) =.422 c Pow er= P(H is rejected p=.5) = P( X 9 p=.5) = 1 P( X8 p=.5) = 1.6964=.336 2 a H :λ=3.5 H 1 :λ<3.5 Assume H, so that X Po(3.5) Significance level 5%, so require c such that P( Xc) <.5 From the Poisson cumulative distribution tables P( X1) =.1359 and P( X = ) =.32 P( X1) >.5 and P( X = ) <.5 so the critical value is Hence the critical region is X = b Size= P(Type I error)= P(X = λ=.35)=.32 c Power=P(H is rejected λ=3.)= P(X = λ=3.)=.498 3 a H :µ=8 H 1 :µ 8 Standardise the X variable Z= X 8 = 2(X 8) 3 18 9 Assume H, so that X N 8, 18 Significance level 5%, so require 2.5% in each tail From the tables, the critical region for Z is Z>1.96 or Z< 1.96 So the critical values for X are given by 2(X 8)=±1.96 X =6.6141 and X =9.3859 (answers to 4 d.p.) So the critical region for X is X <6.6141 or X >9.3859 (answers to 4 d.p.) b P(Type I error)=significance level=.5 Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 1

3 c Using the normal cumulative distribution function on a calculator: P(Type II error) = P(6.6141X9.3859 µ = 7) = P( X9.3859 µ = 7) P( X6. 6141 µ = 7) =.9996.2926=.77 (4 d.p.) d Power=1 P(Type II error)=1.77=.293 4 a Let the random variable X denote the number of geese observed flying in a migratory pattern in a day, then the null hypothesis H is λ=1 and X Po(1) This hypothesis is only rejected if they observe 3 or fewer geese on the given day and then 2 or fewer geese on the next day; or if they observe 18 or more geese on the given day and then 19 or more geese on the next day. So Size= P(H rejected H true) =P( X3) P( X2) + P( X18) P( X19) =.1 336.2 769 +.14 278.7 187 =. 1312 (7 d.p.) b Power=P(H rejected λ=5) =P( X3) P( X2) + P( X18) P( X19) =.265 26.124 652 +. 5. 1 =.33 (4 d.p.) 5 a H :λ=4.5 H 1 :λ>4.5 Critical region X 8 Size= P( X8 λ= 4.5) = 1 P( X7 λ= 4.5) = 1.9134=.866 b i Power= P( X8 X Po( λ)) = 1 P( X7 X Po ( λ )) λ=4 r=1.9489=.511 λ=6 s=1.744=.256 λ=9 t=1.3239=.6761 Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 2

5 b ii 6 a H :p=.45 H 1 :p<.45 Critical region X 3 Size= P( X 3 X ~ B(15,.45)) =.424 b Power= P( X 3 X ~ B(15, p)) p=.2 s=.6482 p=.4 t=.95 Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 3

6 c 7 a H :λ=2 H 1 :λ>2 (Quality the same) (Quality is poorer) b Assume H, so that X Po(2) Require c such that P( Xc).5 From the Poisson cumulative distribution tables P( X4) = 1 P( X3) = 1.8571=.1429 P( X5) = 1 P( X4) = 1.9473=.527 P( X6) = 1 P( X5) = 1.9834=.166 P( X 5) is closest to.5 so the critical value is 5 Hence the critical region is X5 c Power=P(H is rejected X Po(4)) = P( X5 λ= 4) = 1 P( X4 λ= 4) = 1.6288=.3712 Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 4

8 a H :λ=2 H 1 :λ>2 (as good) (worse) Assume H, so that X Po(2) Require c such that P( Xc).5 b From the Poisson cumulative distribution tables P( X5) = 1 P( X4) = 1.9473=.527 P( X6) = 1 P( X5) = 1.9834=.166 P( X5) is closest to.5 so the critical value is 5 Hence the critical region is X5 c Power= P( H is rejected X Po( 3) ) = P( X5 λ= 3) = 1 P( X4 λ= 3) = 1.8153=.1847 d Let the random variable Y denote the number of faulty garments produced by the machinist over the three days H :λ=6 H 1 :λ>6 Assume H, so that Y Po(6) Require c such that P( Yc).5 From the Poisson cumulative distribution tables P( Y1) = 1 P( Y9) = 1.9161=.839 P( Y11) = 1 P( Y1) = 1.9574=.426 P( Y11) is closest to.5 so the critical value is 11 Hence the critical region is Y11 e Power= P(H is rejected Y Po(9) ) (3 days has mean = 3 3= 9) = P( Y11 λ= 9) = 1 P( X1 λ= 9) = 1.76=.294 f Second test is more powerful as monitors the performance of the machinist over more days. Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 5

9 a H :µ=6 H 1 :µ<6 Critical region is X 2 Size= P(Type I error) = P( X 2 µ = 6) =.62 b Power function= P( X2 X ~ Po( µ )) = P(X = X ~Po(µ))+P(X =1 X ~Po(µ))+P(X =2 X ~Po(µ)) = e µ µ! + e µ µ 1 1! + e µ µ 2 2! =e µ +e µ µ+ 1 2 e µ µ 2 =e µ 1+µ+ µ2 2 = 1 2 e µ (2+2µ+µ 2 ) c s= 1 2 e 2 (2+4+4)=5e 2 =.6767 (4 d.p.) d t= 1 2 e 5 (2+1+25)=18.5e 5 =.1247 (4 d.p.) Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 6

9 e Reading from the graph, at the point that the graph intersects the line Power =.8, µ 1.55 so the power of this test is greater than.8 for µ<1.55 1 a H : p=.1 H 1: p>.1 X B(1, p ) Critical region is X 5 Size= P(Type I error) = P( X5 p=.1) = 1 P( X4 p=.1) = 1.9984=.16 b u= P(H rejected p=.25) = 1 P( X4 p=.25) = 1.9219=.781 =.8 (2 d.p.) H : p=.1 H : p>.1 Y B(5, p ) Critical region is Y 3 Size= P(Type I error) = P( Y3 p=.1) c 1 = 1 P( Y2 p=.1) = 1.9914=.86 d v=p(h rejected p=.35) = 1 P( Y2 p=.35) = 1.7648=.2352 =.24 (2 d.p.) e f i.325 ii With p greater than this value, the technician s test is stronger than the supervisor s. g The test is more powerful for probabilities closer to zero (<.325), and it is quicker to check 5 items for defects than to test 1 items. 11 a H : λ=.3 H 1: λ>.3 X Po(1 λ ) Critical region is X 6 Size= P(Type I error) = P( X6 X Po(3)) = 1 P( X5 X Po(3)) = 1.9161=.839 Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 7

11 b a= P(H rejected X Po(5)) = 1 P( X5 X Po(5) ) = 1.616=.384 =.38 (2 d. p). c H :λ=.3 H 1 :λ>.3 Assume H, so that X Po(15.3), i.e. X Po(4.5) Significance level 5%, so require c such that P( Xc) <.5 From the Poisson cumulative distribution tables P( X8) = 1 P( X7) = 1.9134=.866 P( X9) = 1 P( X8) = 1.9597=.43 P( X8) >.5 and P( X9) <.5 so the critical value is 9 Hence the critical region is X9 d Size= P(Type I error) = P( X 9 X Po(4.5)) = 1 P( X 8 X Po(4.5)) = 1.9597=.43 b= P(H rejected X Po(13.5 )) = 1 P( X8 X Po(13. 5) ) = 1.79=.921 e =.92 (2 d.p. ) f g i.63 ii With λ greater than this value, the manager s test is more powerful. Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 8

Challenge a Let the random variable X be the number of times a pair of ones appears in 12 rolls of the dice, then X B(15, p ) ( ) H 1 1 1 : λ= H 1: λ> X B 12, 36 36 36 Critical region X 2 ( X2 X B 1 1 ( 12, )) 1 P( X1 X B( 1 )) Size= P = 2, 36 36 = 1.9577=.423 (4 d.p.) b Power = P( X2 X B(12, p)) = 1 P( X 1 X B(12, p)) = 1 P( X = X B(12, p)) P( X = 1 X B(12, p)) = 1 (1 p) 12 p(1 p) 12 11 c Size= P(Type I error) = P( X4 X B( 6, 1 )) + P( X3 X B( 6, 1 )) P( X 4 X B( 6, 1 )) 6 6 6 =.872 + (.991298.872 ) =.173 (4 d.p.) d The probability p of a double 1 in this situation is q 1 6 Substituting for p= q 1 in the equation in part b gives 6 12 11 q q q Power = 1 1 12 1 6 6 6 12 11 q q = 1 1 2q 1 6 6 e Power= P( X4 X B 1 1 ( 6, )) + P( X3 X B( 6, )) P X 4 X B( 6, q) ( ) 6 6 4 2 5 6 ( q q q q q ) 4 5 6 5 6 6 ( q q q q q q ) 4 5 6 ( q q q ) =.87+.9913 15 (1 ) + 6 (1 ) + =.87+.9913 15 3 + 15 + 6 6 + =.87+.9913 15 24 + 1 =.87+ 14.8695q 23.7912q + 9.913q 4 5 6 Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 9

Challenge f g As the power of Jane s test is greater than that of Emma s when.1 < q <.4, recommend that Jane s test is used. Pearson Education Ltd 218. Copying permitted for purchasing institution only. This material is not copyright free. 1