Hypothesis testing Mixed exercise 4

Transcription

1 Hypothesis testing Mixed exercise 4 1 a Let the random variable X denote the number of vehicles passing the point in a 1-minute period. 39 Use the Poisson distribution model X Po, i.e. X Po(.5).5 e.5 P( X = ) = =.1575 (4 d.p.)! b Using the tables P( X8) = 1 P( X 7) = 1.728=.3272 c H :λ=.5 H 1 :λ<.5 Assume H, so that X Po(.5) 2) =.43.43<. 5 There is sufficient evidence at the 5% level to reject H, and conclude that the number of vehicles passing the point in a given period has decreased. 2 Let the random variable X denote the deformed red blood cells found in a 2.5 ml sample of Francesca s blood. H : λ= = 8 H : λ< 8 1 Po(8) 4) =.99.99>.5 There is insufficient evidence at the 5% level to reject H and to suggest that the mean number of deformed red blood cells has decreased. 3 Let the random variable X denote the number of days it takes Peter to complete the first crossword. H : p=.2 H : p<.2 1 Geo(.2) P( 7) = (1.2) =.221 (4 d.p.) X.221>.5 There is insufficient evidence at the 5% level to reject H and to suggest that the crosswords are more difficult. 4 a Let the random variable X denote the number of days until Roisin sees the first fall of snow, so X Geo(.45) 2 P( X 3) = (1.45 = 2 ).3 5 Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 1

2 4 b H :p=.45 H 1 :p<.45 Geo(.45) P( 7) (1.45).277 (4 d.p.) X = =.277 <.5 There is sufficient evidence at the 5% level to reject H, and conclude that the meteorologist is correct. 5 a Let the random variable X denote the number of calls that Scoobie puts through to the wrong extension during a day in which he receives 15 calls, sox B(15,.3). Using a Poisson approximationx Po(15.3), i.e. X Po(4.5) P(X =5)= e =.178 (4 d.p.) 5! b 3) = c Let the random variable Y denote the number of calls that Waldo puts through to the wrong extension during a day in which he receives 3 calls. H : λ= 2 4.5= 9 H : λ< 9 1 Po(9) 4) =.55.55>.5 There is insufficient evidence at the 5% level to reject H and there is therefore no evidence to suggest that the Waldo has decreased the rate at which calls are put through to the wrong extension. a Let the random variable X denote the number of breakdowns in a one-month period, so X Po(1.75). P(X =3)= e =.1552 (4 d.p.) 3! b Let the random variable Y denote the number of breakdowns in a two-month period, so Y Po(3.5). Using the tables P( Y ) = 1 P( Y 5) = 1.857=.1424 c Let the random variable Z denote the number of months in a four-month period in which there are exactly 3 breakdowns. Use the probability from part a to 5 d.p., so Z B(4,.15522). P(Z=2)= 4 (.15522)2 ( ) 2 =.132 (4 d.p.) d Let the random variable M denote the number of breakdowns in a four-month period, so M Po(7). H : λ= 7 H : λ< 7 1 Assume H, so that M Po(7), so require P( Xc) <.5 From the tables P( M 2) =.29 and P( M 3) =.818 P( M 2) <.5 and P( M 3) >.5 so the critical value is 2 Hence the critical region is M 2 Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 2

3 e Actual significance l evel= P( M 2) =.29 7 Let the random variable X denote the number of televisions sold in a two-day period. H : λ= 2 3.5= 7 H : λ> 7 1 Po(7) 11) = 1 P( X1) = 1.915= > 5. There is insufficient evidence at the 5% level to reject H and therefore no evidence to suggest that advert increased the sales. 8 a Let the random variable X denote the rate of visits to the website on a Saturday. H : λ= 8.5 H : λ> Po(8.5) 12) = 1 P( X11) = = >.5 There is insufficient evidence at the 5% level to reject H and therefore no evidence to suggest that the rate of visits is greater on a Saturday. b Requires finding the smallest positive integer c such that P( X c ) <.5 14) = 1 P( X13) = 1.948=.514 and P( X15) = 1 P( X14) = 1.972=.274 P( X14) >.5 and P( X15) <.5, so c= 15 9 Let the random variable X denote the number of workers that are absent for at least one day in the last month, so X B(2,.5). Using a Poisson approximationx Po(2.5), i.e. X Po(1) H : λ= 1 H : λ> 1 1 Po(1) 15) = 1 P( X14) = 1.915= >.5 There is insufficient evidence at the 5% level to reject H and no evidence to suggest that the percentage is higher than the manager thinks. 1 a Let the random variable X denote the number of products tested until the first defective one is found, sox Geo(.15) b P(X =5)=.15(1.15) 4 =.783 (4 d.p.) 2 P( X 3) = (1.15 = 2 ).72 5 Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 1 c H :p=.15 H 1 :p<.15 Require P( Xc) <.5 c 1 So (1.15) <.5 ( c 1)log.85< log.5 log.5 c 1> log.85 Geo(.15) d c> So the critical region is X 2 Actual significance level = P( 2) (1.15) 19 X = =.45 (4 d.p.) 11 a Let the random variable X denote the number of hurricanes in the area in August. Then use a Poisson model and test H :λ=4 H 1 :λ>4 b For the null hypothesis (H :λ=4) to be rejected at 5% level of significance find the smallest positive integer c such that P( X c ) <.5 Assume H, so that X Po(4), so require P( Xc) <.5 8) = 1 P ( X 7) = =.511 and P( X 9) = 1 P( X 8) = 1.978=.214 P( X 8) >.5 and P( X 9) <.5 so the critical value is 9 The critical region is X 9 and the number of hurricanes must increase to 9 for H to be rejected. c As X = 8 is not in the critical region there is insufficient evidence to reject H and therefore the scientist s suggestion should be rejected. 12 a Let the random variable X denote the number of heads recorded in 3 spins of the coin, so X B(3, p). As p is not known to be small, the Poisson approximation cannot be used. H : p=.5 H : p<.5 1 B(3,.5) Significance level 2%, so require P( Xc) <.2 From the binomial cumulative distribution tables P( X 8) =.81 and P( X 9) =.214 P( X 8) <.2 and P ( X 9) >.2 so the critical value is 8 Hence the critical region is X 8 Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 12 b Let the random variable Y denote the number of coin spins until it lands on heads for the first time, so X Geo( p) H : p=.5 H : p<.5 1 Assume H, so that Y Geo(.5) c 1 So (1.5) <.2 Significance level 2% Require P( Yc) <.2 log.2 c 1> log.5 c>.44 So the critical region is Y 7 c The probability that Alison has incorrectly rejected H is P( X 8) =. 81 The probability that Paul has incorrectly rejected H is P( Y 7) =.5 =.15 Challenge a Let N denote the number of wells sunk until the oil company sinks before it strike oil for the third time in the new region, son Negative B(3, p) This assumes that the probability of striking oil remains the same for each well drilled. b H :p=.18 H 1 :p>.18 Assume H, so that N Negative B(3,.18) Require c) <.5 Need to calculate cumulative probability distributions 2 3 3) = = 3) =.18 (1.18) = ) = 3) + = 4) = (1.18) = = ) = 4) + = 5) = (1.18) = = ) = 5) + = ) = (1.18) = =.7587 As 5) <.5 and ) >.5, the critical region is N 5 c Actual significance level= 5) =.437 ( 4 d. p.) Pearson Education Ltd 217. Copying permitted for purchasing institution only. This material is not copyright free. 5