MA238 Assignment 4 Solutions (part a)

Transcription

1 (i) Sigle sample tests. Questio. MA38 Assigmet 4 Solutios (part a) (a) (b) (c) H 0 : = 50 sq. ft H A : < 50 sq. ft H 0 : = 3 mpg H A : > 3 mpg H 0 : = 5 mm H A : 5mm Questio. (i) What are the ull ad alterative hypotheses for this study? = 0.05, ad assig the problem solvig team. Recall that the probability, computed assumig that H0 is true (i.e. that the machie is o target), that the test statistic would take a value as extreme or more extreme tha that actually observed, is called the p-value for this test. The smaller the P-value, the stroger the evidece agaist H0 provided by the data. * I this questio the p- value is calculated as twice the P( z.74), which from the Normal tables is estimated as ( ) = Recall that we specified a two-sided test at the oset ad hece we eed to double the probability as the rejectio regio is split across both tails of the distributio. This is iterpreted as a 3% chace of seeig a value of z* as large as we observed if the ull hypothesis is actually true (i.e. 3 i a hudred of gettig a test statistic as large as we did if ideed the ull hypothesis was true). As this is less tha specified of 0.05 (i.e. ay value of z more extreme tha 5 i a hudred will lead us to reject Ho) we reject the ull hypothesis. (ii) H 0 : = 3 H A : 3 I the cotext of this study, iterpret makig a Type I error; iterpret makig a Type II error. A Type I error (i.e. reject the ull hypothesis H 0 whe it was i fact true) i this study would amout to decidig that the process is out of cotrol whe it is i fact workig fie i.e. assig the problem solvig team to work o a problem that is ot actually there. (vi) A 95% C.I. for the true mea diameter is calculated as x z where z. 96 as a 95% C.I. is required. This works out as (iii) A Type II error (i.e. do ot reject the ull hypothesis H 0 whe it was i fact false) i this study would amout to decidig that the process is fie whe i fact it is t fie i.e. do t assig the problem solvig team whe there is actually a problem there for them to fix. This is the cost of a Type I error give the aswer to part (ii) above. =[3.0005, ]. Note that this iterval does ot cotai the value 3mm, the mea value whe the machie is o target as specified i the ull hypothesis. As the iterval estimate for the true mea is ot i agreemet with the ull hypothesis we have further evidece that the ull hypothesis is false. It is worth otig that the machie is ot a lot of target i.e. o average betwee mm to mm. (iv) As the sample size is greater tha 30 the CLT applies. As the level of sigificace was specified as = 0.05, the critical value for this two tailed test is z / = z 0.05 =.96 resultig i a critical regio of z >.96 ad z < (v) z. 74 ad sice z >.96, we reject the ull hypothesis at (vii) z. 30 ad sice -.96 < z <.96, we do ot reject the ull hypothesis at = 0.05, ad we do ot assig the problem solvig team.

2 * I this questio the p- value is calculated as twice the P( z.30), which from the Normal tables is estimated as (-0.903) = 0.9. This is iterpreted as a 9% chace of seeig a value of z* as large as we observed if the ull hypothesis is actually true. As this is greater tha specified of 0.05 we do ot reject the ull hypothesis ad claim that the data we collected are quite cosistet with the ull hypothesis ad the differece i the sample mea compared to the hypothesised value of 3mm is due to atural samplig variatio. Note that as the sample size is large eough we substitute s (the sample stadard deviatio) for (the populatio stadard deviatio). 3. Distributio of TS if H 0 true: As the sample size is large eough, we kow from the Cetral Limit Theorem that if H 0 is true, z has a Normal distributio with mea 0 ad variace i.e. if H 0 is true, we would expect z to be somewhere aroud 0 ad ot expect it to be too far i either directio from 0. Questio 3. The iformatio give i the questio is as follows: (the populatio mea) = 0.5 (the sample size) = 0 x (the sample mea) = 8.9 s (the sample stadard deviatio) =5. ad you are asked to decide, based o the sample statistics provided, whether the true mea age of deliquets is strictly less tha 0.5 or ot. Usig the strategy outlied i the lectures:. State the Null ad Alterative Hypotheses 4. Decide o the Sigificace Level A sigificace level was ot specified by the questio so it is up to you to choose oe! Let s choose a sigificace level of =0.05 (i.e. you have a 5% chace of rejectig the ull hypothesis whe it is i fact true, a so called Type Error ). From your kowledge of the ormal distributio you kow that 5% of all observatios havig a N(0,) distributio are to left of.65 (this is give to you i the formula sheet but you should be able to read this off the Z table) so a suitable decisio rule would be to decide that ay values of z less tha.65 (i.e. to the left of it) are ulikely to occur due to samplig variatio aloe ad therefore represet a extreme result which is ot i keepig with what we would expect if the ull hypothesis was ideed true. Our critical regio therefore (usig =0.05) comprises of ay value of z that is < -.65 (see the figure below).. H 0 : = 0.5. H A : < 0.5 Note that the sociologist is oly iterestig i testig for a mea strictly less tha 0.5 ad therefore we have a oe-sided test. This will become importat whe decidig o a appropriate Critical Regio.. Calculate a appropriate test statistic (TS) As the sample size is greater tha 30 the Cetral Limit Theorem applies ad a suitable test statistic is x - z o which, for this example, works out as z % of z scores -.65 Critical Regio Acceptace Regio 95% of z scores 3 4

3 5. Check whether the value of the TS is i the critical regio ad make a decisio. I this example, z = which is cosiderably less tha.65, ad hece there is covicig evidece that the ull hypothesis is false. * I this questio the p- value is calculated as the P( z 3.37).. Recall that we specified a oe-sided test at the oset ad hece we eed oly cosider the probability i oe tail of the distributio. From the Normal tables this is estimated as ( ) = This is iterpreted as you havig 4 i 0,000 chace of seeig a value of z* as large as we observed if the ull hypothesis is actually true. As this is much smaller tha specified of 0.05 we have very strog evidece agaist reject the ull hypothesis ad claim that the data we collected are ot at all cosistet with the ull hypothesis ad the differece i the sample mea compared to the hypothesised value of 0.5 is due to atural samplig variatio. Coclusio. O the basis of the hypothesis test above, there is strog evidece (at =0.05) that the mea age of bicycle thieves is actually less tha 0.5 years ad ot equal to 0.5 years as stated by the police chief. Questio 4. The iformatio give i the questio is as follows: (the populatio mea) = (the sample size) = 00 x (the sample mea) = 4.7 s (the sample stadard deviatio) = ad you are asked to decide, based o the sample statistics provided, whether the claim that the true average size of a deliquet charge accout is differet from is true or ot. Usig the strategy outlied i the lectures:. State the Null ad Alterative Hypotheses 3. H 0 : = H A : Note that the retail credit associatio is iterestig i testig for a mea deliquet charge accout differet from ad therefore we have a two-sided test. This will become importat whe decidig o a appropriate Critical Regio. 5. Calculate a appropriate test statistic (TS) As the sample size is greater tha 30 the Cetral Limit Theorem applies ad a suitable test statistic is x - z o which, for this example, works out as z Note that as the sample size is large eough we substitute s (the sample stadard deviatio) for (the populatio stadard deviatio). 3. Distributio of TS if H 0 true: If H 0 true the z has a Normal distributio with mea 0 ad variace (i.e. N(0,) ). 4. Decide o the Sigificace Level A sigificace level of =0.05 is specified i the questio (i.e. you have a 5% chace of rejectig the ull hypothesis whe it is i fact true, a so called Type Error ). Remember that you are iterestig i testig for a differece i the populatio mea i both directios (i.e. the true mea could be bigger tha that stated i the ull hypothesis or less tha that stated i the ull hypothesis) ad you wat to have oly a 5% chace of beig wrog. From your kowledge of the ormal distributio you kow that.5% of all observatios havig a N(0,) distributio are to left of.96 ad to the right of +.96 (this is give to you i the formula sheet) so a suitable decisio rule would be to decide that ay values of z less tha.96 (i.e. to the left of it) or greater tha +.96 (i.e. to the right of it) are ulikely to occur due to samplig variatio aloe ad therefore represet a extreme result which is ot i keepig with what we would expect if the ull hypothesis was ideed true. Notice that you have spread the 0.05 over the two tails of the distributio to give you a overall sigificace level of Our critical regio therefore (usig =0.05) comprises of ay value of z that is < -.96 or z > +.96 (see the figure below). ½ % of z scores ½ % of z scores Acceptace Regio % of z scores

4 5. Check whether the value of the TS is i the critical regio ad make a decisio. I this example, z = 6.77 which is cosiderably more tha.96, ad hece there is covicig evidece that the ull hypothesis is false. Coclusio. O the basis of the hypothesis test above, there is strog evidece (at =0.05) that the true average size of a deliquet charge accout is ot as claimed. The secod part of the questio asks you to provide a guess as to what you thik the true average size of a deliquet charge accout is likely to be give that you have just disputed the fact that it is To do this you eed to provide a guess at the true ukow mea usig a cofidece iterval. You are specifically asked to calculate a 95% C.I. which, as the sample size is >30, is calculated as follows (see assigmet ): x z where z. 96 as a 95% C.I. is required. This works out as Usig the strategy outlied i the lectures:. State the Null ad Alterative Hypotheses 5. H 0 : = H A : < 96 Note that the player is iterestig i testig for a mea bowlig score less tha 96 ad therefore we have a oe-sided test. This will become importat whe decidig o a appropriate Critical Regio.. Calculate a appropriate test statistic (TS) As the sample size is greater tha 30 the Cetral Limit Theorem applies ad a suitable test statistic is x - z o which, for this example, works out as z =[.53, 6.0]. Hece, we ca claim that it is quite likely that the true average size of a deliquet charge accout is somewhere betwee.53 ad 6.0. Notice the agreemet betwee the cofidece iterval ad the hypothesis test i that the parameter value specified i the ull hypothesis ( 08.65) is ot cotaied i the iterval that we claim is likely to cotai the true value. Questio 4. The iformatio give i the questio is as follows: (the populatio mea) = 96 pis (the sample size) = 50 x (the sample mea) = 88 pis s (the sample stadard deviatio) = 4.9 ad you are asked to decide, based o the sample statistics provided, whether the claim that the true average umber of pis is less tha 96 or ot (i.e. has his average score worseed). Note that as the sample size is large eough we substitute s (the sample stadard deviatio) for (the populatio stadard deviatio). 3. Distributio of TS if H0 true: If H 0 true the z has a Normal distributio with mea 0 ad variace. 4. Decide o the Sigificace Level A sigificace level of =0.0 specified by the questio (i.e. you have a % chace of rejectig the ull hypothesis whe it is i fact true, a so called Type Error ). Remember that you are iterestig i testig for a differece i populatio mea i oe directio oly (i.e. the true mea is less tha stated i the ull hypothesis). From your kowledge of the ormal distributio you kow that % of all observatios havig a N(0,) distributio are to left of.33 (this is give to you i the formula sheet) so a suitable decisio rule would be to decide that ay values of z less tha.33 (i.e. to the left of it) are ulikely to occur due to samplig variatio aloe ad therefore represet a extreme result which is ot i keepig with what we would expect if the ull hypothesis was ideed true. Our critical regio therefore (usig =0.0) comprises of ay value of z that is < -.33 (see the figure overleaf). 7 8

5 % of z scores 99% of z scores ad you are asked to decide, based o the sample statistics provided, whether there is evidece to suggest a differece i the average weight gai for all patiets o 5mg dose compared to all patiets o 50mg dose (i.e. does it look like oe group gais more weight o average compared to the other group). Lookig at the sample statistics aloe it looks like the 50mg group has a higher average weight gai tha the 5mg group (i.e. the 50mg group look like to have gaied 3 pouds more o average) but this may be just due to samplig variatio ad cosequetly a formal hypothesis test is eeded. Acceptace Regio Usig the strategy outlied i the lectures: 5. Check whether the value of the TS is i the critical regio ad make a decisio. I this example, z = -.7 which is ot less tha.33, ad hece there is o covicig evidece that the ull hypothesis is false. Note that the z score is early i the critical regio so eve though we caot claim that the ull hypothesis is true (while havig oly a % chace of beig wrog), the result is quite worryig i that the z score was quite extreme (although ot extreme eough to reject H 0 ). The bowler should keep accout of a few more scores ad the reaalyze the data ad see how extreme his z score is the. Coclusio. O the basis of the hypothesis test above, there is o evidece (at =0.0) that the bowler s average umber of pis has reduced sigificatly from 96 (i.e. that the ew ball is affectig his performace o average). MA38 Assigmet 5 Solutios (part b) (ii) Two Sample Tests Questio Critical Regio The iformatio give i the questio is as follows: Group (5mg) Group (50mg) Sample size () Sample mea ( x ) 7 0 Sample stadard deviatio (s) 6 7. State the Null ad Alterative Hypotheses. H 0 : = i.e. there is o differece i the average weight gai for patiets o 5mg dose compared to patiets o 50mg dose). H A : i.e. there is a differece i the average weight gai for patiets o 5mg dose compared to patiets o 50mg dose) Note that you are iterested i testig for a mea differece i both directios (i.e. the 5mg group could be bigger, less tha or equal to the 50mg group) ad therefore we have a two-sided test. This will become importat whe decidig o a appropriate Critical Regio.. Calculate a appropriate test statistic (TS) As both the samples are of size greater tha 30 the Cetral Limit Theorem applies ad a suitable test statistic is x x z which, for this example, works out as z Note that as the sample size is large eough we substitute s, s (the sample stadard deviatios) for ad (the populatio stadard deviatios). 3. Distributio of TS if H 0 true: As the sample size is large eough, we kow from the Cetral Limit Theorem that if H0 is true, z has a Normal distributio with mea 0 ad variace i.e. if H 0 is true, we would expect z to be somewhere aroud 0 ad ot expect it to be too far i either directio from

6 4. Decide o the Sigificace Level A sigificace level was ot specified by the questio so it is up to you to choose oe! Let s choose a sigificace level of =0.05 (i.e. you have a 5% chace of rejectig the ull hypothesis whe it is i fact true, a so called Type Error ). From your kowledge of the ormal distributio you kow that.5% of all observatios havig a N(0,) distributio are to left of.96 ad that.5% of all observatios are to the right of.96 (this is give to you i the formula sheet but you should be able to read this off the Z table) so a suitable decisio rule would be to decide that ay values of z less tha.65 or greater tha.96 are ulikely to occur due to samplig variatio aloe ad therefore represet a extreme result which is ot i keepig with what we would expect if the ull hypothesis was ideed true. Our critical regio therefore (usig =0.05) comprises of ay value of z that is < -.96 or greater tha.96 (see the figure below). Acceptace Regio 5. Check whether the value of the TS is i the critical regio ad make a decisio I this example, z = -.30 which is cosiderably less tha.96, ad hece there is covicig evidece that the ull hypothesis is false (i.e. covicig evidece that the 50mg group has i fact a greater average weight gai compared to the 5mg group). Coclusio. 95% of z scores ½ % of z scores ½ % of z scores O the basis of the hypothesis test above, there is strog evidece (at =0.05) that the 50mg group has i fact a greater average weight gai compared to the 5mg group over the week period. As before, a hypothesis test will oly idicate whether there is evidece of a sigificat differece (i.e. a departure from the ull hypothesis that is ot due to samplig variatio aloe) but will ot provide you with a estimate of what the differece is likely to be. I order to do this you eed to calculate a cofidece iterval (usig a required degree of cofidece). We saw i lectures that will provide you with a 00(-)% cofidece iterval for the differece i two populatio meas (e.g. equal to 0.05 will give you a 95% C.I., equal to 0.0 will give you a 99% C.I. etc.). You were asked i this questio to calculate a 95% cofidece iterval (i.e. use = 0.05, cosequetly z =.96 ) which amouts to evaluatig = [ -5.56, -0.44]. We iterpret this iterval as follows: our best guess at the true differece i average weight gais for 5mg group 50mg group (make sure you are clear about the order!) is likely to be betwee 5.56 pouds ad 0.44 pouds (i.e. the 50mg group are likely to gai o average betwee.44 ad 5.56 pouds more compared to the 5mg group over the week period). Notice that this iterval does ot cotai 0 (which would represet o differece i average weight gai betwee the two groups) ad is i agreemet with the hypothesis test. Questio 7. The iformatio give i the questio is as follows: Rocket Rocket Sample size () 8 0 Sample mea ( x ) 36 5 Sample stadard deviatio (s) 5 8 ad you are asked to decide, based o the sample statistics provided, whether there is evidece to suggest that the secod kid of rocket is worse tha the first i terms of it s mea target error (i.e. o the basis of the sample data provided, does it look like Rocket is worse tha Rocket i terms of mea target error). Lookig at the sample statistics aloe it looks like Rocket has a higher mea target error of 6 uits more tha Rocket but this may be just due to samplig variatio ad cosequetly a formal hypothesis test is eeded. Usig the strategy outlied i the lectures:. State the Null ad Alterative Hypotheses x x z H 0 : = H A : < i.e. there is o differece i the mea target error for Rocket compared to Rocket ) i.e. the mea target error for Rocket is strictly less tha that for Rocket )

7 Note that you are iterested i testig for a mea differece i oe directio oly ad therefore we have a oe-sided test. This will become importat whe decidig o a appropriate Critical Regio.. Calculate a appropriate test statistic (TS) As either of the samples are of size greater tha 30, we kow from the Cetral Limit Theorem does ot hold ad we eed to use the t-distributio. Remember that there are two crucial assumptios we have to make i order to use the t-distributio i this cotext ad they are as follows:. Do both samples come from populatios that are ormally distributed?. Are the variaces of both populatios equal?.895 (this is give to you i the t tables by lookig dow the 0.05 colum ad across the 7 df row) so a suitable decisio rule would be to decide that ay values of z less tha.895 are ulikely to occur due to samplig variatio aloe ad therefore represet a extreme result which is ot i keepig with what we would expect if the ull hypothesis was ideed true. Our critical regio therefore (usig =0.05) comprises of ay value of z that is < (see the figure below). 5% of t scores 95% of t scores We ca assume that assumptio is true as it is likely that the mea target error should be ormally distributed give the ature of the measuremet. We caot be sure about the variace assumptio but we do kow (from the lectures) that if the sample sizes are similar the this assumptio is ot too importat Acceptace Regio Give these decisios we ca use t x x s s as a suitable test statistic i order to compare two sample meas to make statemets about two populatio meas, resultig i 3. Distributio of TS if H 0 true: t We kow that if H 0 is true, t has a t-distributio with mi( -, ) degrees of freedom (i.e. if H0 is true, we would expect t to be somewhere aroud 0 ad ot expect it to be too far i either directio from Decide o the Sigificace Level A sigificace level of = 0.05 was specified i the questio (i.e. you have a 5% chace of rejectig the ull hypothesis whe it is i fact true, a so called Type Error. This would amout to you sayig that Rocket was worse tha Rocket whe i fact there was o differece ad you were just aalysig two extreme samples). From your kowledge of the t-distributio tables you kow that 5% of all observatios havig a t- distributio with 7 (i.e. mi(8-, 0 )) degrees of freedom distributio are to left of 5. Check whether the value of the TS is i the critical regio ad make a decisio. I this example, t = -.06 which is less tha.895, ad hece there is covicig evidece that the ull hypothesis is false (i.e. covicig evidece that Rocket has a higher mea target error compared to Rocket ). Coclusio. O the basis of the hypothesis test above, there is strog evidece (at =0.05) that Rocket has a higher mea target error whe compared to Rocket ad Rocket should be used practice as it is more accurate. Note you were ot asked to provide a cofidece iterval i his questio but you ca make a simple guess of the differece i the mea target error by usig the sample meas i.e. the mea target error for Rocket is probably aroud 6 uits more tha that for Rocket. Questio 8. (i). State the Null ad Alterative Hypotheses H 0 : = H A : Critical Regio i.e. there is o differece i the populatio mea talkig time betwee the two batteries) i.e. there is a differece i the populatio mea talkig time betwee the two batteries) 3 4

8 Note that you are iterested i testig for a mea differece i both directios (i.e. the populatio mea for ickel-cadmium battery could be bigger, less tha or equal to that for the ickel-metal hydride battery) ad therefore we have a two-sided test.. Calculate a appropriate test statistic (TS) As either of the samples are of size greater tha 30, the Cetral Limit Theorem does ot hold ad the t-distributio is valid. There are two crucial assumptios to make i order to use the t-distributio i this cotext ad these are as follows:. Do both samples come from populatios that are ormally distributed?. Are the variaces of both populatios equal? Assume that the mea talkig time is ormally distributed for both batteries ad give that the sample sizes ad sample variace are similar assume that the variace assumptio is valid. i fact there was ot ad you were just aalysig two extreme samples). From your kowledge of the t-distributio tables you kow that half a percet (i.e. = 0.005) of all observatios havig a t-distributio with 4 degrees of freedom distributio are to left of.797 ad that 0.5% of all observatios are to the right of.797 (this is give to you i the t tables by lookig dow the colum ad across the 4 df row) so a suitable decisio rule would be to decide that ay values of z less tha.797 or greater tha.797 are ulikely to occur due to samplig variatio aloe ad therefore represet a extreme result which is ot i keepig with what we would expect if the ull hypothesis was ideed true. The critical regio therefore (usig =0.0) comprises of ay value of t that is < or >.797 (see the figure below). t (4 df) 99% of t scores Give these decisios use t x x s s as a suitable test statistic i order to compare two sample meas to make statemets about two populatio meas, which, for this example, works out as ½ % of t scores Acceptace Regio ½ % of t scores 3. Distributio of TS if H 0 true: t t Check whether the value of the TS is i the critical regio ad make a decisio. I this example, t = -.06 which is ot i the critical regio, ad hece there is o covicig evidece (at sigificace level =0.0 ) that the ull hypothesis is false i.e. it is quite plausible that we could get such a differece i sample meas due to samplig variatio aloe if H o was ideed true. Coclusio. We kow that if H 0 is true, t has a t-distributio with mi( -, ) degrees of freedom (i.e. if H 0 is true, we would expect t to be somewhere aroud 0 ad ot expect it to be too far i either directio from 0 uder a t distributio with mi(5-, 5 ) =4 degrees of freedom. 4. Decide o the Sigificace Level A sigificace level of = 0.0 was specified i the questio (i.e. you have a % chace of rejectig the ull hypothesis whe it is i fact true, a so called Type Error. This would amout to you sayig that there was a differece i the average talkig time whe No evidece of a sigificat differece (at =0.0) i the true average talkig time betwee the two battery types. (ii) As either of the samples are of size greater tha 30, the Cetral Limit Theorem does ot hold ad the t-distributio is valid. There are two crucial assumptios to make i order to use the t-distributio i this cotext ad these are as follows:. Do both samples come from populatios that are ormally distributed?. Are the variaces of both populatios equal? 5 6

9 (iii) A approximate 00(-)% cofidece iterval for the true populatio mea differece is calculated as s s xx t where t has mi( -, ) degrees of freedom. Cosequetly, a approximate 99% cofidece iterval for the true populatio mea differece is calculated as ( ) = [ -9.97, 3.006]. Notice that this iterval cotais 0 (which would represet o differece i the true average talkig time betwee the two batteries) ad is therefore i agreemet with the hypothesis test. We iterpret this iterval as follows: our best guess at the true populatio mea differece i average talkig time betwee the two batteries is likely to be betwee 9.97 uits more o average for the ickel-metal hydride battery up to uits more o average for the ickel-cadmium batteries. Note that the iterval is loaded towards beig strictly egative suggestig that ickelmetal hydride batteries may ideed have a higher mea talkig time tha ickel-cadmium batteries which this study may ot have eough power to detect. (iv) A Type II error (i.e. do ot reject the ull hypothesis H 0 whe it was i fact false) i this study would amout to decidig that the mea life time for the two batteries was the same whe i fact they were differet i.e. the ew battery outperforms the old. The likely cosequece is a loss of icome o ot producig a loger life battery ad the time used i productio so far. Questio 9. (i) The boxplots were missig from the assigmet, my apologies! The summary statistics suggest that the mea agular velocity is higher i the Skilled group compared to the Novice group. The boxplots would give a idicatio as to whether the data plausibly arose from a ormal distributio (oe of the assumptios ecessary to carry out the hypothesis test give the small samples). The stadard deviatios are ot equal i each group but are similar. (ii). Histograms of the data for each group suggested that there were o outliers preset ad that the data were reasoably symmetric. This suggests that the mea is a useful measure to use to compare the two samples. If there were outliers preset ad a lack of symmetry you should cosider usig the media. (iii)this is a Observatioal study. We are observig two types of rowers. A experimetal study would be oe where we took a sample of rowers ad radomly assiged them to two traiig methods ad compared the improvemet i fitess across the two methods. (iv). State the Null ad Alterative Hypotheses H 0 : S = N H A : S N i.e. there is o differece i the populatio mea agular velocity betwee the two categories of rowers) i.e. there is a differece i the populatio mea agular velocity betwee the two categories of rowers). Calculate a appropriate test statistic (TS) As either of the samples are of size greater tha 30, the Cetral Limit Theorem does ot hold ad the t-distributio is valid. There are two crucial assumptios to make i order to use the t-distributio i this cotext ad these are as follows: 3. Do both samples come from populatios that are ormally distributed? 4. Are the variaces of both populatios equal? From the evidece provided by the histograms we ca assume that the mea talkig time is ormally distributed for both categories of rower ad give that the sample sizes ad sample variace are similar assume that the variace assumptio is valid. Give these decisios use t x x s s 7 8

10 as a suitable test statistic i order to compare two sample meas to make statemets about two populatio meas, which, for this example, works out as 3. Distributio of TS if H 0 true: t t 5.8. We kow that if H 0 is true, t has a t-distributio with mi( -, ) degrees of freedom (i.e. if H 0 is true, we would expect t to be somewhere aroud 0 ad ot expect it to be too far i either directio from 0 uder a t distributio with mi(0-, 0 ) =9 degrees of freedom. 4. Decide o the Sigificace Level A sigificace level of = 0.05 was specified i the questio (i.e. you have a 5% chace of rejectig the ull hypothesis whe it is i fact true, a so called Type Error. This would amout to you sayig that there was a differece i the average agular velocity whe i fact there was ot ad you were just aalysig two extreme samples). From your kowledge of the t-distributio tables you kow that half a percet (i.e. = 0.05) of all observatios havig a t-distributio with 9 degrees of freedom distributio are to left of.6 ad that.5% of all observatios are to the right of.6 so a suitable decisio rule would be to decide that ay values of t less tha.6 or greater tha.6 are ulikely to occur due to samplig variatio aloe ad therefore represet a extreme result which is ot i keepig with what we would expect if the ull hypothesis was ideed true. The critical regio therefore (usig =0.05) comprises of ay value of t that is <.6 or >.6 (see the figure below). t (9 df) ½ % of t scores 95% of t scores Acceptace Regio 5. Check whether the value of the TS is i the critical regio ad make a decisio..6.6 ½ % of t scores I this example, t = 5.8 which is i the critical regio, ad hece there is covicig evidece (at sigificace level =0.05 ) that the ull hypothesis is false i.e. it is ulikely that we could get such a differece i sample meas due to samplig variatio aloe if H o was ideed true. Coclusio. Evidece of a sigificat differece (at =0.05) i the true average mea agular velocity betwee the two categories of rowers. (v) A approximate 00(-)% cofidece iterval for the true populatio mea differece is calculated as s s xx t where t has mi( -, ) degrees of freedom. Cosequetly, a approximate 95% cofidece iterval for the true populatio mea differece is calculated as ( ) = [ 0.67,.67]. Notice that this iterval does ot cotai 0 (which would represet o differece i the true mea agular velocity betwee the two categories of rowers) ad is therefore i agreemet with the hypothesis test. We iterpret this iterval as follows: our best guess at the true populatio mea differece betwee the two categories of rowers is likely to be betwee 0.67 to.67 uits more o average for Skilled rowers compared to the Novice rowers. It appears that the agular velocity is a importat characteristic i rowig. 9 0

11 Questio 0.. State the Null ad Alterative Hypotheses H 0 : = i.e. there is o differece i the mea amout charged betwee the two plas) distributio are to left of.96 ad that.5% of all observatios are to the right of.96 (this is give to you i the formula sheet but you should be able to read this off the Z table) so a suitable decisio rule would be to decide that ay values of z less tha.65 or greater tha.96 are ulikely to occur due to samplig variatio aloe ad therefore represet a extreme result which is ot i keepig with what we would expect if the ull hypothesis was ideed true. H A : i.e. there is a differece i the mea amout charged betwee the two plas) Our critical regio therefore (usig =0.05) comprises of ay value of z that is < -.96 or greater tha.96. Note that you are iterested i testig for a mea differece i both directios ad therefore we have a two-sided test.. Calculate a appropriate test statistic (TS) As both the samples are of size greater tha 30 the Cetral Limit Theorem applies ad a suitable test statistic is x x z which, for this example, works out as z Note that as the sample size is large eough we substitute s, s (the sample stadard deviatios) for ad (the populatio stadard deviatios). 3. Distributio of TS if H 0 true: As the sample size is large eough, we kow from the Cetral Limit Theorem that if H 0 is true, z has a Normal distributio with mea 0 ad variace i.e. if H 0 is true, we would expect z to be somewhere aroud 0 ad ot expect it to be too far i either directio from Decide o the Sigificace Level A sigificace level was ot specified by the questio so it is up to you to choose oe! Let s choose a sigificace level of =0.05 (i.e. you have a 5% chace of rejectig the ull hypothesis whe it is i fact true, a so called Type Error ). From your kowledge of the ormal distributio you kow that.5% of all observatios followig a N(0,) 5. Check whether the value of the TS is i the critical regio ad make a decisio. I this example, z = -.48, which is i the acceptace regio ad there is o covicig evidece that the ull hypothesis is false. 6. Calculate the P-value. Remember that the P-value is defied as the probability, computed assumig that H0 is true, that the test statistic would take a value as extreme or more extreme tha that actually observed is called the P-value of the test. The smaller the P-value, the stroger the evidece agaist H0 provided by the data. * I this questio the p- value is calculated as twice the P( z.48) = ( ) = 0.4. Recall that we specified a two-sided test at the oset ad hece we eed to double the p- value. We could ot kow the value of the test statistic before we collected the data. This is iterpreted as there is at least a 4% chace of seeig a value of z* as large as we observed if the ull hypothesis is actually true. As this is ot less tha specified of 0.05 we do ot reject the ull hypothesis. Coclusio. O the basis of the hypothesis test above, there is o evidece (at =0.05) of a sigificat differece i the mea amout charged betwee the two plas. The result is ot sigificat there is o clear evidece that oe proposal is a better icetive tha the other. So we ca just go with the oe that is easier ad cheaper to implemet. But if there is o practical differece i cost to the bak, we might choose proposal B, sice the data did lea a bit i that directio. (b) Because the sample sizes are equal ad large, the Cetral Limit Theorem applies ad the test should be reliable i spite of some skewess.