# Text&Tests5. Project Maths SUPPLEMENT. Frances O Regan O. D. Morris. Leaving Certificate Higher Level Maths

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1 Project Maths SUPPLEMENT Text&Tests5 Leavig Certificate Higher Level Maths Cotais all the Deferred Material ad Cetral Limit Theorem Fraces O Rega O. D. Morris

2 O.D. Morris, Fraces O Rega, 2014 All rights reserved. No part of this publicatio may be reproduced, stored i a retrieval system or trasmitted i ay form or by ay meas, electroic, mechaical, photocopyig, recordig or otherwise, without the prior writte coset of the copyright holders. First published Jue 2014 by The Celtic Press Groud Floor Block B Liffey Valley Office Campus Dubli 22 ISBN: Prited i Irelad by Turer Prit Group Earl Street Logford

3 Preface This booklet cotais all the Deferred Material from Strad 1 of the Leavig Certificate Higher Level Course. This material was itroduced i September 2013 for examiatio i Jue 2015 ad owards. This chapter is prited as a supplemet for those studets who bought Text & Tests 5 at the begiig of 5th Year i September All future editios of Text & Tests 5 will cotai this ew chapter. Chapter 5 also cotais a sectio o the Cetral Limit Theorem which was belatedly added to the origially published course. O. D. Morris Fraces O Rega Jue 2014

5 I such a large populatio it would be impossible to obtai the weight of each perso ad so the values of ad will ot be kow. However, if we take a radom sample of this populatio, we ca get approximate values for ad. Obviously, the larger the sample, the more accurate we would expect the approximatios to be. If we take a large umber of differet radom samples of size, each sample will have its ow mea, _ x, ad stadard deviatio, _ x. Some of these samples are illustrated o the right. The differet meas of these samples are called the sample meas. If a large umber of samples of the same size are take, you get a correspodigly large umber of meas. These meas form their ow distributio givig us the distributio of sample mea. This distributio is also called the samplig distributio of the mea. The followig example illustrates the shape a distributio might take whe differet samples (of the same size) from a populatio are selected. Example 1 Sample 1 A populatio cosists of five digits 2, 4, 6, 8, 10. (i) Write dow all the possible samples of 2 differet digits that ca occur if radom samples are take. (ii) Fid the mea of each sample ad plot the distributio of the sample meas. (iii) Compare the value of the mea of the sample meas with the value of the populatio mea. x 1 x1 Sample 2 x 2 Sample 3 x 3 Sample 4 Sample 5 Sample 6 x 4 x 5 x 6 (i) The possible samples are: (2, 4), (2, 6), (2, 8), (2, 10), (4, 6), (4, 8), (4, 10), (6, 8), (6, 10), (8, 10) (ii) Their meas are: 3, 4, 5, 6, 5, 6, 7, 7, 8, 9 The distributio of the sample meas is plotted below. f f x Populatio Mea of populatio Sample of meas 60 Mea of sample meas 6 (iii) The mea of the populatio is 6. The mea of the sample meas is also 6. Thus the mea of the sample meas ad the populatio mea are equal. 10 x 5

7 The diagram o the right illustrates how the distributio of the sample mea approximates to a ormal distributio eve whe the uderlyig populatio is skewed. Distributio of sample mea Paret distributio Whe dealig with the samplig distributio of the mea, we covert the give uits to stadard uits usig the formula give o the right. z x x x Example 2 A radom sample of 250 is selected from a populatio havig mea 30 ad stadard deviatio 5. Fid the probability that the sample mea is greater tha Sice 250, the sample mea is ormally distributed sice 30. Chagig to stadard uits we get: _ x z z Now P(x 30.5) P(z 1.581) 1 P(z 1.581) The probability that the mea is greater tha 30.5 is Example 3 A ormal distributio has a mea of 40 ad a stadard deviatio of 4. If 25 items are draw at radom, fid the probability that their mea lies betwee 38 ad

8 Covertig the give uits to stadard uits we get: _ x z For x 38, z For x 40.5, z P(38 x 40.5) P( 2.5 z 0.625) P(z 0.625) P(z 2.5) P(z 0.625) [1 P(z 2.5)] [ ] [ ] P(mea lies betwee 38 ad 40.5) Example 4 A populatio is ormally distributed with mea 12 ad stadard deviatio 3. Fid the sample size such that P( _ x 12.5) 0.05, where _ x is the sample mea. P(z z 1 ) 0.05 P(z z 1 ) 0.95 z _ x z ( ) (1.645)3 0.5 The required sample size is i.e. 98 roud up 8

9 Example 5 A compay istalls ew machies for packig peauts. The compay claims that the machies fill packets with a mea mass of 500 g ad a stadard deviatio of 18 g. To test the compay s claim several samples of size 40 packets are take ad their mea masses, _ x grams, are recorded. (i) Describe the samplig distributio of _ x ad explai your aswer, referrig to the theorem you have used. (ii) Write dow the mea ad stadard deviatio of the distributio of _ x. (iii) Draw a rough sketch of the samplig distributio of _ x. (iv) Fid the probability that the mea of the distributio of _ x is less tha 496. (v) What sample size is required so that P( _ x ) ? (i) The samplig distributio of _ x is approximately ormal as the sample size of 40 is sufficietly large (i.e. 30) to apply The Cetral Limit Theorem. (ii) The mea of the distributio of the sample meas is 18 g, the same as the populatio mea. The stadard deviatio (or stadard error) is (iii) A sketch of the distributio of _ x is show below _ x (iv) Covertig the give uits to z-scores, we use z. For x 496, z P( _ x 496) P(z 1.405) 1 P(z 1.405) The probability that _ x or 7.35%

10 (v) P(z z 1 ) 0.06 P(z z 1 ) z _ x z (1.56) 9.36 (9.36) roud up The sample size required is 88. Exercise Fill i the correct word or symbol to complete the followig statemets: (i) Whe a large umber of samples of size are take from a populatio, the the distributio of _ x, the sample mea, is kow as the of the mea. (ii) As the sample size icreases, the stadard deviatio of the samplig distributio of the sample meas will. (iii) If the mea of the uderlyig populatio is, the mea of the samplig distributio of the meas is. (iv) If the stadard deviatio of a populatio is ad samples of size are take from it, the the stadard deviatio of the distributio of the sample meas is. 2. The diagram o the right shows two curves. Oe of these curves represets a distributio ad the other represets the distributio of the sample meas of size take from this distributio. Which curve represets the distributio of the sample meas? A B 3. Samples of size 36 are take from a populatio with mea 12 ad stadard deviatio 2. The samplig distributio of the meas are plotted i a curve. (i) Describe the shape of this curve amig the theorem you have used to support your descriptio. (ii) Explai why the theorem you have metioed ca be applied whe the shape of the uderlyig populatio is ukow. (iii) Write dow the mea ad stadard deviatio of the samplig distributio of the mea. 10

13 Sectio 5.2 Cofidece iterval for a mea I Sectio 5.1, the Cetral Limit Theorem was used to show that the samplig distributio of the mea approximates to a ormal distributio for large ( 30). I this sectio we itroduce a differet way of presetig iformatio provided by a sample mea to estimate the mea of the populatio from which the sample came. If samples of size are take from a populatio, the meas of the samples will vary. To accommodate this variety, we itroduce the cocept of a cofidece iterval. This iterval will produce a rage of values i which we are quite cofidet the populatio mea lies. The edpoits of this iterval are called cofidece limits. But how do we measure this cofidece? The degree of cofidece is geerally give as a percetage. These percetages are geerally 90%, 95% ad 99%. The most commoly used measure of cofidece is a 95% cofidece level. This meas that there is a 95% probability that the populatio mea lies i the give iterval. I the stadard ormal distributio, we require the values of z such that 95% of the populatio lies i the iterval z 1 z z 1. The work ivolved i fidig the value of z, is show below. We use the stadard ormal tables o pages 36 ad 37 of Formulae ad Tables. From the give diagram, P(z z 1 ) From the tables z z z 1 0 z Thus i the ormal distributio, 95% of the populatio lies withi 1.96 stadard deviatios of the mea. Sice the sample mea is ormally distributed, 95% of the populatio will lie i the iterval _ x 1.96 _ x, where _ x is the stadard error of the mea. _ x If is the populatio mea, the 95% of the sample meas lie i the iterval _ x 1.96 _ x _ x 1.96 _ x where _ x, beig the stadard deviatio of the populatio. This ca be writte as _ x 1.96, which are the ed-poits (or cofidece limits) of the mea. 13

15 (i) The 95% cofidece iterval is give by _ x ( 2 60 ) (0.258) , The 95% cofidece iterval is (ii) A 95% cofidece iterval, meas that o 95 occasios out of 100 the iterval will cotai the true populatio mea. (iii) P(ball bouce lies outside 95% cofidece iterval) Example 3 The heights of people have a stadard deviatio of 11.5 cm. It is required to estimate the mea height of people, with 95% cofidece, to withi 0.4 cm. What sample size should be take i order to achieve this estimate? Let be the mea height of people. The 95% cofidece limits for are _ x stadard error is 0.4 cm 1.96 ( 11.5 ) (1.96) (56.35) Therefore, a sample of at least 3176 should be take. 15

18 12. The weights of pebbles o a beach are distributed with mea 48.6 g ad stadard deviatio 8.5 g. A radom sample of 50 pebbles is chose. (i) Fid the probability that the mea weight will be less tha 49 g. (ii) Fid the limits withi which the cetral 95% of such sample meas would lie. (iii) How large a sample would be eeded i order that the cetral 95% of sample meas would lie i a iterval of width at most 4 g? 13. The 95% cofidece iterval for the mea mark of a group of studets is (54.09, 60.71). This iterval is based o the results from a radom sample of 80 studets. (i) Fid _ x, the mea of the sample. (ii) Fid, the stadard deviatio of the ormal populatio from which the sample is take. Sectio 5.3 Cofidece iterval for a proportio I Sectio 4.4 of Chapter 4 it was show how to fid the cofidece iterval for a populatio proportio usig the margi of error 1, where is the sample size. This cofidece iterval is show agai o the right. The 95% cofidece iterval for a proportio p is p^ 1 p p^ 1 Here is a remider of what a proportio is! If 150 televisio viewers are iterviewed i a sample survey ad 63 say they like a ew situatio comedy, the is the proportio of the sample who like the ew show. 150 This sample proportio, p^, is used as a estimate of the true populatio proportio p of televisio viewers who like the ew show. The otatio p^ is used to deote sample proportio. The otatio p is used to deote populatio proportio. Sice p is geerally ot kow, p^ is used as a estimator for the true populatio proportio, p. If may samples of the same size are take from a populatio, each sample will produce a differet (but similar) proportio. All these proportios form their ow distributio called the samplig distributio of the proportio. The stadard error, p^, of this distributio is give o page 34 of Formulae ad Tables ad is show o the right. p^ p(1 p) 18

19 I this sectio we will use the Stadard Normal Tables ( rather tha the margi of error, 1 ) to get a more accurate cofidece iterval for a populatio proportio. Sice the 95% level of cofidece will be used, the diagram o the right will remid us that 95% of a ormal distributio lies withi 1.96 stadard deviatios of the mea If p^ is the sample proportio ad p is the populatio proportio, the the 95% cofidece iterval for p is give by p^ 1.96 p(1 p) p p^ 1.96 p(1 p) This ca be writte more cocisely as p^ 1.96 p(1 p). The 95% cofidece iterval for a populatio proportio p^ 1.96 p(1 p) Note: A icrease i cofidece levels results i a icrease i the iterval width. Example 1 I a survey carried out i a large city, 170 households out of a radom sample of 250 owed at least oe pet. (i) Fid the stadard error of the samplig distributio of the proportio at the 95% cofidece level. (ii) Fid the 95% cofidece iterval for the proportio of households i the city who ow at least oe pet. (i) The sample proportio p^ Stadard error p^ p(1 p) 0.68(1 0.68) 250 p^

20 (ii) The 95% cofidece iterval is give by p^ 1.96 p(1 p) (0.029) from (i) above (0.6232, ) or about (62%, 74%) Example 2 A radom sample of 250 cars were surveyed passig a certai juctio ad 36 were foud to have K registratios. (i) Determie a 95% cofidece iterval for the proportio of cars i that area that have a K registratio. (ii) What sample size would have to be take i order to estimate the percetage to withi 2%? (i) The sample proportio p^ The 95% cofidece iterval is give by p^ 1.96 p(1 p) ( ) (0.0222) , The 95% cofidece iterval is (0.100, ). (ii) Let be the sample size. We require such that p^ p(1 p) p^ % ( ) 0.02 (0.144)(0.856) (0.02) (0.02) So a sample size of 309 would have to be take. 20

24 To covert the give uits to stadard uits we use _ x the sample mea _ populatio mea z x, where populatio stadard deviatio size of sample For the machie metioed above, _ x z The test statistic is z Come to a coclusio. Sice z does ot lie outside the rage 1.96 z 1.96 it is ot i the critical regio. So we accept the ull hypothesis which states that the mea life of a battery is 120 hours. Note: If, the stadard deviatio of the populatio is ot give, use _ x the stadard deviatio of the sample istead. Example 1 Over the years, a market gardeer foud that the mea yield from his tomato plats was 1.83 kg per plat with a stadard deviatio of 0.35 kg per plat. Oe year he plated 600 of a ew variety ad these yielded 1.87 kg per plat. At the 5% level of sigificace, test whether the mea yield from the ew plats is differet from his ormal variety. 1. H 0 : The mea is H 1 : The mea is ot The level of sigificace is 5%. The critical regio is z 1.96 or z Calculate the test statistic by covertig to stadard uits. x z x ( 600 ) z Sice z ad , we reject the ull hypothesis ad coclude that the ew variety is differet from the ormal variety. 24

25 Usig p-values Suppose we carry out a hypothesis test ad fid the test statistic to be z Sice 2.16 is greater tha 1.96, we reject the ull hypothesis at the 5% level of sigificace ( 0.05). Istead of comparig z 2.16 with z 1.96 (ad z 1.96), we compare the total area of the two coloured regios below with the specific level of sigificace, We use pages 36 ad 37 of Formulae ad Tables to fid the probability that z 2.16 or z P(z 2.16) P(z 2.16) 2P(z 2.16) 2[1 P(z 2.16)] 2[ ] 2[0.0154] The shaded areas above are referred to as the p-value, or probability-value correspodig to the observed value of the test statistic. The value foud above is the p-value that correspods to the test statistic z The p-value is iterpreted as the lowest level of sigificace at which the ull hypothesis could have bee rejected. With a test statistic of z 2.16, we would certaily have rejected the ull hypothesis at the specified level of sigificace ( 0.5). The p-value of gives us a specific or more precise level of sigificace. The smaller the p-value is, the stroger is the evidece agaist H 0 provided by the data. The p-value of a Test Statistic p-value p-value z 1 0 z 1 The p-value is the sum of the two shaded areas. p-value 2 P(z z 1 ), where z 1 is the test statistic. 25

26 Example 2 Calculate the p-value for the sample statistic z Sample statistic is z The sum of the probabilities that z 2.08 ad z 2.08 is the p-value. p-value 2 P(z 2.08 ) [1 P(z 2.08)] 2 [ ] 2(0.0188) p-value Steps ivolved i a Test of Sigificace usig a p-value 1. Write dow the ull hypothesis H 0 ad the alterative hypothesis H State the sigificace level. (O our course 0.5.) 3. Calculate the test statistic. 4. Fid the p-value that correspods to the test statistic. 5. If the p-value 0.05, the result is ot sigificat ad we do ot reject the ull hypothesis H 0. If the p-value 0.05, we reject the ull hypothesis H 0 i favour of the alterative hypothesis H 1. Example 3 A radom sample of 36 observatios is to be take from a distributio with stadard deviatio 10. I the past, the distributio has had a mea of 83, but it is believed that the mea may have chaged. Whe the sample was take it was foud to have a mea of (i) State H 0 ad H 1. (ii) Calculate the value of the test statistic. (iii) Calculate the p-value for the test statistic. (iv) Use the p-value to state if the result is sigificat at the 5% level of sigificace. Explai your coclusio. 26

29 (i) Calculate the sample statistic for this sample. (ii) Calculate the p-value for this sample statistic. (iii) Use the p-value you have foud to ivestigate if the mea score of the sample differs from the mea score of the populatio at the 5% level of sigificace. 11. The security departmet of a warehouse wats to kow whether the average time required by the ight watchma to walk his roud is 12.0 miutes. I a radom sample of 36 rouds, the ight watchma averaged 12.3 miutes with a stadard deviatio of 1.2 miutes. (i) Calculate the test statistic for this sample. (ii) Ca we reject the ull hypothesis that 12.0 miutes at the 5% level of sigificace? (iii) Work out the p-value that correspods to the test statistic foud i (i) above. (iv) If this p-value is used, do you reach the same coclusio with regard to sigificace at the 5% level? 12. The legths of metal bars produced by a particular machie are ormally distributed with mea legth 420 cm ad stadard deviatio 12 cm. The machie is serviced, after which a sample of 100 bars gives a mea legth of 423 cm. (i) Calculate the sample statistic for this sample. (ii) Work out the p-value for this sample statistic. (iii) Use this p-value to determie if there is evidece, at the 5% level, of a chage i the mea legth of the bars produced by the machie, assumig that the stadard deviatio remais the same. 13. A machie is desiged to produce screws with a stated mea legth of 5 mm. A radom sample of 400 screws produced by the machie is foud to have a mea legth of mm ad a stadard deviatio of mm. Estimate the stadard error of the mea, ad obtai a approximate 95% cofidece iterval for the mea of the whole output of this machie. Ivestigate if the mea of the sample differs sigificatly from the stated mea at the 5% level of sigificace. 29

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