TESTS OF HYPOTHESIS I

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1 UNIT 15 STRUCTURE TESTS OF HYPOTHESIS I Tests of Hypothesis I 15.0 Objectives 15.1 Itroductio 15. Poit Estimatio ad Stadard Errors 15.3 Iterval Estimatio 15.4 Cofidece Limits, Cofidece Iterval ad Cofidece Co-efficiet 15.5 Testig Hypothesis Itroductio 15.6 Theory of Testig Hypothesis Level of Sigificace, Type-I ad Type-II Errors ad Power of a Test 15.7 Two-tailed ad Oe-tailed Tests 15.8 Steps to Follow for Testig Hypothesis 15.9 Tests of Sigificace for Populatio Mea Z-test for variables Tests of Sigificace for Populatio Proportio Z-test for Attributes Let Us Sum Up 15.1 Key Words ad Symbols Used Aswers to Self Assessmet Exercises Termial Questios/Exercises Further Readig 15.0 OBJECTIVES After studyig this uit, you should be able to: l l l l l l estimate populatio characteristics (parameters) o the basis of a sample, get familiar with the criteria of a good estimator, differetiate betwee a poit estimator ad a iterval estimator, comprehed the cocept of statistical hypothesis, perform tests of sigificace of populatio mea ad populatio proportio, ad make decisios o the basis of testig hypothesis INTRODUCTION Let us suppose that we have take a radom sample from a populatio with a view to kowig its characteristics, also kow as its parameters. We are the cofroted with the problem of drawig ifereces about the populatio o the basis of the kow sample draw from it. We may look at two differet scearios. I the first case, the populatio is completely ukow ad we would like to throw some light o its parameters with the help of a radom sample draw from the populatio. Thus, if µ deotes the populatio mea, the we ited to make a guess about it o the basis of a radom sample. This is kow as estimatio. For example, oe may be iterested to kow the average icome of people livig i the city of Delhi or the average life i burig hours of a fluorescet tube light produced by Idia Electrical or proportio of people sufferig from T.B. i city B or the percetage of smokers i tow C ad so o. A somewhat differet situatio may arise whe some iformatio about a parameter is either kow or specified ad we would like to verify whether that iformatio holds good for the sample draw from the populatio as well. 5 5

2 Probability ad Hypothesis Testig This is kow as problem of testig of hypothesis. I the previous examples, we may be iterested of i testig whether the average icome i the city of Delhi is, say, Rs.,000 per moth. I the secod example, we may like to verify whether the claims made by Idia Electrical, that their fluorescet lamps would last 5,000 hours, is justified. Some social workers may believe that 0% of the populatio i city B suffers from T.B. We would like to make our commet after a test of hypothesis. I the last example, some huma activists, cocered about the hazards of passive smokig, assert that 30% of the people stayig i tow C are smokers. We may share their opiio oce we have satisfied ourselves after peformig a statistical test of hypothesis. It may be oted that testig of hypothesis plays a vital role i decisio-makig. I the first example, the statisticia may be cocered about whether to bracket Delhi with the top metropolito cities depedig o the average icome based o his/her recommedatios. If o the basis of a statistical test, it is foud that the claim made by the maufacturer of Idia Electrical is justified, the the sales of his lamps would icrease. I the third example if there is evidece, agai o the basis of testig hypothesis, that the social worker is right about his statemet, suitable steps may be udertake to improve the livig coditios of the margialized sectio i the city so that the percetage of people sufferig from T.B. is reduced. Some strict legislatio baig smokig or reducig smokig to a desirable level may be eacted o the basis of a hypothesis tested i the last example. 15. POINT ESTIMATION AND STANDARD ERRORS Estimatio is a itegral part of our daily lives. I order to costruct a ew house or reovate a old house or flat, we demad a estimate of the cost ivolved. A studet estimates his/her chace of success before appearig for a expesive competitive examiatio. 5 6 Now we shall cosider estimatio from the viewpoit of a statisticia. As we discussed i Uit 4: samplig, is the meas to fid the true value of the parameter which ca be correctly obtaied oly through cesus study. I may cases it is ot practicable due to various costraits. Therefore, the alterative approach is to select some items as a sample from the populatio ad collect the data ad aalyse the data, the estimate the chracteristics of the populatio. This is called estimatio. Poit estimate is oe type of estimate. It is a sigle umber which is used as a estimate of the ukow populatio parameter. Let us assume that we have take a radom sample of observatios, x 1, x, x 3 x, from a populatio characterized by a parameter θ (read theta). This symbol θ is used to deote a parameter that could be mea, mode or some measure of variatio etc. Thus θ may be the mea (µ)of a ormal distributio or the probability of success (P) of a biomial distributio with parameters ad p ad so o. I theory of estimatio, we try to fid a statistic (i.e., a fuctio of sample observatios) T which estimates the ukow parameter θ. Thus the sample mea ( x) x i /, x 1, x, x 3 x beig the icome per moth of persos selected at radom from the city of Delhi, may be cosidered to be the estimate of the average icome per moth (µ) of the people of Delhi. This is deoted by µ ˆ x i.e., the estimate of µ is x.

3 To be more precise, x is kow as a poit estimator of µ as we try to estimate the populatio mea (µ) by a sigle value, amely, the sample mea. O the basis of a radom sample of icomes from Delhi, if it is foud that the sample mea is Rs.,000/-, the oe may coclude that the estimate of average icome per moth of the people livig i that city is Rs.,000/-. Tests of Hypothesis I As opposed to a poit estimate, oe may thik of a iterval estimate that is supposed to cotai the average icome of the people of Delhi per moth. This would be discussed i Sectio At this jucture, we must make a distictio betwee the two terms Estimator ad estimate. T is defied to be a estimator of a parameter θ, if T esimtates θ. Thus T is a statistic ad its value may differ from oe sample to aother sample. I other words, T may be cosidered as a radom variable. The probability distributio of T is kow as samplig distributio of T. As already discussed, the sample mea x is a estimator of populatio mea µ. The value of the estimator, as obtaied o the basis of a give sample, is kow as its estimate. Thus x is a estimator of µ, the average icome of Delhi, ad the value of x i.e., Rs.,000/-, as obtaied from the sample, is the estimate of µ. Selectio of the best estimator: Our ext edeavour would be to discuss differet criteria for selectig the best estimator. Ubiasedess ad Miimum Variace: A statistic T is defied to be ubiased for a parameter θ if expectatio of T is θ, i.e., E(T) θ. O the other had if E(T) θ + a (θ), the the differece a (θ) E(T) θ is kow as bias. The bias is kow to be positive if a (θ) > 0 ad egative if a (θ) < 0. Our first priority would be to select a ubiased estimator of θ. However, there may be may ubiased estimators of θ. If x 1, x, x deote sample observatios from a populatio with a ukow parameter θ, the ay of the observatios or ay liear fuctio of them would be a ubiased estimator of θ. I order to choose the best estimator amog these estimators alog with ubiasedess, we itroduce a secod criterio, kow as, miimum variace. A statistic T is defied to be a miimum variace ubiased estimator (MVUE) of θ if T is ubiased for θ ad T has miimum variace amog all the ubiased estimators of θ. We may ote that sample mea ( x ) is a MVUE for µ. We kow that xi x (15.1) ( x E(x) E i ) 1.[ E (xi)] 1 [ µ ] 1. µ [x 1, x, x are take from populatio havig as µ populatio mea] E(x) µ Hece ( x) is a ubiased estimator of µ. 5 7

4 Probability ad Hypothesis Testig Further, variace of ( x) is give by : v (x) v ( x i / ) (where v deotes variace) 1.[ v(x i)] Sice x i1 s are idepedet 1 [where is populatio variace] 1.[ ] (15.) It ca be proved that v ( x ) has the miimum variace amog all the ubiased estimators of µ. Cosistecy: If T is a estimator of θ, the it is obvious that T should be i the eighbourhood of θ. T is kow to be cosistet for θ, if the differece betwee T ad θ ca be made as small as we please by icreasig the sample size sufficietly. We ca further add that T would be a cosistet estimator of θ if i) E (T) θ ad ii) V (T) 0 for a very large i.e., as For example, sample mea ( x) is a cosistet estimator of µ as E ( x) µ Ad V (x) 0 as. It may be oted that if T is a cosistet estimator of θ, the ay fuctio of T is also a cosistet estimator of θ. 5 8 Efficiecy: A statistic T is called as a efficiet estimator of θ if it has the miimum stadard error amog all the estimators of θ for a fixed sample size. Both the sample mea ad sample media are cosistet estimators for µ. But stadard error (a term, to be defied ad explaied i this sectio) of sample mea is less tha that of sample media. Hece sample media is oly a cosistet estimator of µ, whereas sample mea is both cosistet ad efficiet estimator of µ. Sufficiecy: A statistic T is kow to be a sufficiet estimator of θ if T cotais sufficiet iformatio about θ so that we do ot have to look for ay other estimator of θ. Sample mea ( x ) is a sufficiet estimator of µ. Now let us cosider the followig poit estimates that are commoly used. A) Estimatig Populatio Mea: It is obvious that sample mea is the best estimator of populatio mea µ. It is a MVUE. It is both cosistet ad efficiet estimator for µ. Further more, x is a sufficiet estimator for µ. Thus we estimate the average icome of the people of Delhi by the sample mea or the average life of bulbs, maufactured by Idia Electricals, by the correspodig sample mea. B) Estimatig Populatio Proportio: If a discrete radom variable x follows biomial distributio with parameters ad P, the we have µ E(x) P v(x) P (1 p) [ deotig the umber of trials ad P deotig the probability of a success].

5 Hece, it follows that : Tests of Hypothesis I x i P E(p) E P (15.3) V(xi ) ad V (p) V (x i /) P (1 P) P (1 P) (15.4) Thus if we take a radom sample of size from a populatio where the proportio of populatio possessig a certai characteristic is P ad the sample cotais x uits possessig that characteristic, the a estimate of populatio proportio (P) is give by: x Pˆ (15.5) I other words, the estimate of the populatio proportio is give by the correspodig sample estimate i.e., Pˆ p (15.6) From (15.3) E {p} P So p is a ubiased estimator of P. It ca be show that p has the miimum variace amog all the ubiased estimators of p. I other words, p is a MVUE of P. As v (p) P(1 P) 0 as it follows from Eq. (15.4) that p is a cosistet estimator of P. We ca further establish that p is a efficiet as well as a sufficiet estimator of P. Thus we advocate the use of sample proportio to estimate the populatio proportio as p which satisfies all the desirable properties of a estimator. I order to estimate the proportio of people sufferig from T.B. i city B, if we fid the umber of people sufferig from T.B. is x i a radom sample of size, take from city B, the sample estimate p x/ would provide the estimate of the proportio of people i that city sufferig from T.B. Similarly, the percetage of smokers as foud from a radom sample of people of tow C would provide the estimate of the percetage of smokers i tow C. C) Estimatio of Populatio Variace ad Stadard Error: Stadard error of a statistic T, to be deoted by S.E. (T), may be defied as the stadard deviatio of T as obtaied from the samplig distributio of T. I order to compute the stadard error of sample mea, it may be oted that from Eq. (15.) S.E. (x) for simple radom samplig with replacemet (SRSWR). N η S.E. (x) for simple radom samplig without replacemet N 1 (SRSWOR)]. where is the populatio stadard deviatio (S.D.), is sample size, N is N η populatio size ad the factor is kow as fiite populatio corrector N 1 5 9

6 Probability ad Hypothesis Testig (f.p.c.) or fiite populatio multiplier (f.p.m.) which may be igored for a large populatio. I order to fid S.E., it is ecessary to estimate or i case it is ukow. If x 1, x, x deote sample observatios draw from a populatio with mea µ ad variace, the the sample variace: S (x i x) (15.7) may be cosidered to be a estimator of Sice E(x) µ ad V ( z/ ) E (x µ) We have (15.8) s (x x) (from 15-8) i [(xi µ ) (x µ )] [(x µ ) (x µ ) (x µ ) + (x µ ) ] i i (x µ ) (x µ ) (x µ ) + (x µ i i ) (x µ ) (x µ ). (x µ ) + (x µ i ) [sice Σ (x i µ) Σ x i Σµ x µ (x µ )] (x µ ) (x µ ) + (x µ i ) ( x µ ) + (x µ (15.9) i ) As x i is the ith sample observatio from a populatio with µ as mea ad as variace, it follows that : E (x i µ) Ad E (x µ ) v (x) (15.10) From (15-9), E (s ) E (xi µ ).E (x u) 1 E(S ). ( 1) (15.11) Hece S, the sample variace, is a biased estimator of. As 1 E (S ) E 1 s (15.1) 6 0 thus s 1 ( xi x) 1 (s ) is a ubiased estiator of (15.13)

7 so, we use (s ) (x i x) as a estimator of ad 1 Tests of Hypothesis I (x i x) S as a estimator of 1 A estimate of S.E. ( x) is give by: S S.E. (x) for SRSWR > S N for SRSWOR N 1 P (1 P) From (15.4), it follows that v(p) (15.14) P(1 P) S.E.(p) for SRSWR P (1 P) N. for SRSWOR (15.15) N 1 A estimate of stadard error of sample proportio is give by: p (1 p) S.E. (p) for SRSWR > p (1 p). N for SRSWOR (15.16) Let us cosider the followig illustratios to estimate variace from sample ad also estimate the stadard error. Illustratio 1 A sample of 3 fluorescet lights take from Idia Electricals was tested for the lives of the lights i burig hours. The data are preseted below: Table 15.1: The Lives i Hours of 3 Lights Sl. Life Sl. Life Sl. Life No. (Hours) No. (Hours) No. (Hours) Solutio: We are iterested i estimatig the average life of fluorescet lights maufactured by Idia Electricals. As discussed i this sectio, the estimate of the populatio mea (µ) is give by the correspodig sample mea. The 6 1

8 Probability ad Hypothesis Testig µ ˆ x. If we are further iterested i estimatig the stadard error of x, the we are to compute > S.E.(x) s where, s (xi x) η 1 xi x 1 xi ad x, Sample size We igore f.p.c. as the populatio of lights is very large. Table 15.: Computatio of sample mea ad sample S.D. Life i Hours u i x i 5000 u i x i

9 Tests of Hypothesis I Total From the above Table, u 490, u i 374 i ui 490 u As u x 5000 i u x 5000 i Or, X u (approximately) 4985 (s ) (s x ) s (s ) u i u 1 s hece S.E.(x) so the estimate of the average life of lights as maufactured by Idia Electricals is 4985 hours. Estimate of the populatio variace i (hours) ad the stadard error is hours. Illustratio A sample of 350 people from city C cotaied 70 smokers. Fid a estimate of the proportio of smokers i the city. Also fid a estimate of the stadard error of the proportio of smokers i the sample. Solutio: I this case x o. of smokers i the sample 70, 350. x 70 Thus we have p Hece the estimate of the proportios of smokers i the city is 0. or 0%. Further p (1 p) 0. (1 0.) S.E. (p) > The estimate of the stadard error of the proportio of smokers i the sample is

10 Probability ad Hypothesis Testig Self Assessmet Exercise A 1) State, with reasos, whether the followig statemets are true or false. a) Both the statistic ad parameter are fuctios of sample observatios. b) Ay type of samplig would lead to the same iferece about the populatio. c) Statistical iferece is a statistical process to kow about a populatio from the kowledge of a sample draw from it. d) Ay type of estimator ca be used for estimatig a parameter. e) I most cases, decisio-makig depeds o estimatio. f) There may be more tha oe estimator for a parameter. g) Assumptio of ormality is a must for poit estimatio. h) Every cosistet estimator is ecessarily a efficiet estimator. i) A cosistet estimator approaches the parameter with a icrease i sample size. j) Poit estimator is used as a estimate of the ukow populatio parameter. ) Differetiate betwee estimator ad estimate ) I choosig betwee sample mea ad sample media which oe would you prefer? ) The mothly earigs of 0 families, obtaied from a radom sample from a village i West Begal are give below: Sl. Mothly earigs (Rs.) Sl. Mothly earigs (Rs.) No. No Fid a estimate of the average mothly earigs of the village. Also obtai a estimate of the S.E. of the sample estimate

11 Tests of Hypothesis I 5) I a sample of 900 people, 49 people are foud to be cosumers of tea. Estimate the proportio of cosumers of tea i the populatio. Also fid the correspodig stadard error ) Obtai a ubiased estimate of populatio mea ad populatio variace o the basis of the followig sample observatios: 50, 46, 5, 53, 45, 43, 46, 48, INTERVAL ESTIMATION This is aother type of estimatio. As opposed to estimatig a parameter by a sigle value i.e., poit estimatio discussed i the previous sectio, we may thik of a iterval or a rage of values that is supposed to cotai the parameter. A iterval estimate would always be specified by two values i.e., the lower value ad the upper value, withi which the parameter lies. This is kow as Iterval Estimatio. Thus iterval estimatio may be defied as estimatig a iterval to which the ukow parameter θ may belog, i all likelihood. Regardig the estimatio of the average icome of the people of Delhi city, oe may argue that it would be better to provide a iterval which is likely to cotai the populatio mea. Thus, istead of sayig the estimate of the average icome of Delhi is Rs.,000/-, we may suggest that, i all probability, the estimate of the average icome of Delhi would be from Rs. 1,900/- to Rs.,100/-. I the secod example of estimatig the average life of lights produced by Idia Electricals where the estimate came out to be 4985 hours, the poit estimatio may be a boe of cotetio betwee the producer ad the potetial buyer. The buyer may thik that the average life is rather less tha 4985 hours. A iterval estimatio of the life of lights might satisfy both the parties. Figure 15.1 shows some itervals for θ o the basis of differet samples of the same size from a populatio characterized by a parameter θ. A few itervals do ot cotai θ. 6 5

12 Probability ad Hypothesis Testig θ Fig.15.1: Cofidece Itervals to θ 15.4 CONFIDENCE LIMITS, CONFIDENCE INTERVAL AND CONFIDENCE CO-EFFICIENT Let us assume that we have take a radom sample of size from a populatio characterized by a parameter θ. Let us further suppose that based o these sample observatios, it is possible to fid two statistics t 1 ad t such that: P (t 1 < θ) α 1 Ad P (t > θ) α Where α 1 ad α are two small positive umbers. Combiig these two coditios, we may write: P (t 1 θ t ) 1 α (15.17) Where α α 1 + α Equatio (15.17) could be iterpreted as the probability that θ lies betwee t 1 ad t is (1 α), whatever may be the value of θ, satisfyig (15.17). The iterval [t 1, t ], t 1 beig less tha t, that cotais the parameter θ is kow Cofidece Iterval to θ, t 1 beig kow as Lower Cofidece limit ad t as Upper Cofidece Limits. (1 α) is kow to be Cofidece Co-efficiet correspodig to the cofidece iterval [t 1, t ]. Oe may like to kow why the term cofidece comes ito the picture. If we choose α 1 ad α such a way that α 0.01, the the probability that θ would belog to the radom iterval [t 1, t ] is I other words, oe feels 99% cofidet that [t 1, t ] would cotai the ukow parameter θ. Similarly if we select α 0.05, the P [t 1 θ t ] 0.95, thereby implyig that we are 95% cofidet that θ lies betwee t 1 ad t. (15.17) suggests that as α decreases, (1 α) icreases ad the probability that the cofidece iterval [t 1, t ] would iclude the parameter θ also icreases. Hece our edeavour would be to reduce α ad thereby icrease the cofidece co-efficiet (1 α). 6 6

13 Referrig to the estimatio of the average life of lights (θ), if we observe that θ lies betwee 4935 hours ad 5035 hours with probability 0.98, the it would imply that if repeated samples of a fixed size (say 3) are take from the populatio of lights, as maufactured by Idia Electricals, the i 98 per cet of cases, the iterval [4935 hours, 5035 hours] would cotai θ, the average life of lights i the populatio while i per cet of cases, the iterval would ot cotai θ. I this case, the cofidece iterval for θ is [4935 hours, 5035 hours]. Lower Cofidece Limit of θ is 4935 hours, Upper Cofidece Limit of θ is 5035 hours, ad the Cofidece Co-efficiet is 98 per cet. Tests of Hypothesis I Selectio of Cofidece Iterval Our ext task would be to select the basis for estimatig cofidece iterval. Let us assume that we have take a radom sample of size from a ormal populatio characterized by the two parameters µ ad, the populatio mea ad stadard deviatio respectively. Thus, i the case of estimatig a Cofidece Iterval for average icome of people dwellig i Delhi city, we assume that the distributio of icome is ormal ad we have take a radom sample from the city. I aother example cocerig average life of fluorescet lights as produced by Idia Electricals, we assume that the life of a fluorescet light is ormally distributed ad we have take a radom sample from the populatio of fluorescet lights maufactured by Idia Electricals. Figure 15. shows percetage of area uder Normal Curve. It ca be show that if a radom sample of size is draw from a ormal populatio with mea µ ad variace, the ( x), the sample mea also follows ormal distributio with µ as mea ad / as variace. Further as we have observed i Sectio 15.. S.E. (x) From the properties of ormal distributio, it follows that the iterval : [ µ S.E. (x), µ + S.E. (x)] covers 68.7% area. The iterval [ µ S.E. (x), µ + S.E.(x)] covers 95.45% area ad the iterval [ µ 3S.E. (x), µ + 3S.E. (x)] covers 99.73% area. Figure 15. depicts this iformatio % % % 13.59%.14% µ.14% % % 99.73% Fig. 15.: Percetages of Area uder a Normal Curve 6 7

14 Probability ad Hypothesis Testig Now let us cosider a situatio where the assumptio of ormality may ot hold. If the sample size is large eough, the the sample mea x follows approximately, i.e., asymptotically ormal distributio with mea as µ ad stadard error as /, µ ad beig the mea ad S.D. of the populatio uder cosideratio. I case is ukow, we ca replace it by the correspodig sample stadard deviatio. Oe may ask the questio as to how large should be. It is rather difficult to specify a exact value of so that the distributio of x would be asymptotically ormal. Larger the value of, the better. However for practical purposes, if exceeds 30, the we may assume that x is asymptotically ormal. Our ext questio may be what would be the cofidece iterval for µ. Will it be µ ± S.E. (x), or µ ± S.E. (x), or µ ± 3 S.E. (x), or some other iterval? Suppose that the Cofidece Iterval to µ is give by : µ ± u S.E. (x) ad we are to determie u such that : P [x u S.E. (x) µ x + u S.E. (x)] 1 α (15.18) Or, x µ P[ u u] 1 α S.E. (x) Or, P [ u Z u] 1 α [where x µ Z is a stadard ormal variable] S.E. (x) Or. Or, Or. φ (u) φ ( u) 1 α [where φ (K) P(Z K), area uder stadard ormal curve from to K]. φ (u) [ 1 φ (u) 1 α φ (u) α Or. φ ( u) 1 ( α / ) (15.19) Puttig α 0.10 i (15.19), we get φ ( u) or, φ ( u) φ ( ) or. u Thus 100 (1 α) % or 100 (1 0.1)% or 90% cofidece iterval to populatio mea µ is : Give by X 1.645, X Puttig α 0.05, 0.0 ad 0.01 respectively i (15.19) ad proceedig i a similar maer, we get 95% Cofidece Iterval to µ 1.96 x 1.96, x + (15.0) % Cofidece Iterval to µ x.33, x +.33 (15.1)

15 ad 99% Cofidece Iterval to µ x.58, x +.58 (15.) Theoretically we may take ay Cofidece iterval by choosig u accordigly. However i a majority of cases, we prefer 95% or 99% Cofidece Iterval. These are show i Figure 15.3 ad Figure 15.4 below. Tests of Hypothesis I.5% of area uder curve 95% of area uder curve.5% of area uder curve x 1.96 x x Fig. 15.3: 95% Cofidece Iterval for Populatio Mea 0.5% of area uder curve 99% of area uder curve 0.5% of area uder curve x.58 x x +.58 Fig. 15.4: 99% Cofidece Iterval for Populatio Mea Next we cosider Iterval Estimatio i the followig cases: Iterval Estimatio of Populatio Mea As suggested i this sectio uder assumptio of ormality, 95% cofidece iterval to µ, the populatio mea, is give by x 1.96, x If the assumptio of ormality does ot hold but is greater tha 30, the above 95% cofidece iterval still may be used for estimatig populatio mea. I case is ukow, it may be replaced by the correspodig ubiased estimate of, amely S, so log as exceeds 30. However, we may face a difficult situatio i case is ukow ad does ot exceed 30. This problem has bee discussed i the ext uit (Uit-16). Similarly, 99% cofidece iterval to µ is give by : x.58, x +.58 I case is ukow. The 99% cofidece iterval to µ is : S S x.58, x +.58 i case is ukow ad > 30. (15.3) 6 9

16 Probability ad Hypothesis Testig Iterval Estimatio of Ukow Populatio Proportio It ca be assumed that whe is large ad either p or (1 p) is small (oe may specify p 5 ad (1 p) 5), the the sample proportio p is P (1 P) asymptotically ormal with mea as P ad S.E.(p), P beig the ukow populatio proportio i which we are iterested. The estimate of S.E. (p) is give by : p (1 p) S.E.(pˆ ) Hece, 95% cofidece iterval to p is give by : > p ( 1 p) p ( 1 p) p 1. 96, p (15.4) ad 99% cofidece iterval to P is : p.58 p(1 p), p +.58 p(1 p) (15.5) Let us cosider the followig illustratios to uderstad the procedure for iterval estimatio. Illustratio 3 I a radom sample of 1,000 families from the city of Delhi, it was foud that the average of icome as obtaied from the sample is Rs.,000/-, it is further kow that populatio S.D. is Rs. 58. Fid 95% as well as 99% cofidece iterval to populatio mea. Solutio: Let x deote icome of the people of Delhi city. If µ deotes average icome of people dwellig i Delhi, the 95% cofidece iterval to µ is: x 1.96, x ad 99% cofidece iterval to µ is : x.58, x.58 Where x Sample mea; Sample size, ad Populatio stadard deviatio. I our case, x Rs. 000, 1000, Rs x 1.96 Rs Rs x Rs Rs

17 58 x.58 Rs Rs Tests of Hypothesis I ad 58 x +.58 Rs Rs Hece we have 95% cofidece iterval to average icome for the people of Delhi [Rs to Rs ] ad 99% cofidece iterval to average icome for the people of Delhi [Rs to Rs. 01]. Illustratio 4 Calculate the 95% ad 99% cofidece limits to the average life of fluorescet lights produced by Idia Electricals. Solutio: Sice, the populatio stadard deviatio is ukow ad 3 (> 30), we replace by S, the sample S.D. with divisor as ( 1) i our previous example ad get 95% cofidece iterval to µ is: x 1.96 S, x S Similarly, 99% cofidece iterval for µ x.58 S, x +.58 S Where, x Sample mea 4985 hours, Sample size 3; ad S Sample S.D. with ( 1) divisio hours (as computed earlier). x 1.96 S hours 3 Illustratio 5 S x hours 3 S x hours 3 S x hours 3 95% Cofidece Iterval to the average life of lights [ hours, hours]. 99% Cofidece Iterval to the average life of lights [ hours, hours]. While iterviewig 350 people i a city, the umber of smokers was foud to 7 1

18 Probability ad Hypothesis Testig be 70. Obtai 99% lower cofidece limit ad the correspodig upper cofidece limit to the proportio of smokers i the city. Solutio: As discussed i the previous sectio, 99% Lower Cofidece Limit to P, the proportio of smokers i the city is give by: p.58 p (1 p) ad 99% Upper Cofidece Limt to P is: p +.58 p (1 p) provided p 5 ad p (1 p) 5. I this case x o. of smokers 70 o. of people iterviewed 350 x 70 p As p ad (1 p) are rather large, we ca apply the formula for 99% Cofidece Limit as metioed already. 99% Lower Cofidece Limit to P is : 0. (1 0.) % Upper Cofidece Limit to P is : 0. (1 0.) Hece 99% Lower Cofidece Limit ad 99% Upper Cofidece Limit for the proportio of smokers i the city are ad 0.14 respectively. Illustratio 6 I a radom sample of people from a tow, 358 people were foud to be sufferig from T.B. With 95% Cofidece as well as 98% Cofidece, fid the limits betwee which the percetage of the populatio of the tow sufferig from T.B. lies. Solutio: Let x be the umber of people sufferig from T.B. i the sample ad as the umber of people who were examied. The the proportio of people sufferig from T.B. i the sample is give by: x 358 p As p x 358 ad (1 p) p x are both very large umbers, we ca apply the formula for fidig Cofidece Iterval as metioed i the previous sectio. Thus 95% Cofidece Iterval to

19 P, the proportio of the populatio of the tow sufferig from T.B., is give by : Tests of Hypothesis I p 1.96 p (1 p), p p (1 p) ( ), ( ) [ , ] [0.1181, 0.17] I a similar way, 98% Cofidece Iterval to P is give by: p.33 p (1 p), p +.33 p (1 p) ( ), ( ) [0.1150, 0.158] Thereby, we ca say with 95% cofidece that the percetrage of populatio i the tow sufferig from T.B. lies betwee ad 1.7 ad with 98% cofidece that the percetage of populatio sufferig from T.B. lies betwee ad Illustratio 7 A famous shoe compay produces 80,000 pairs of shoes daily. From a sample of 800 pairs, 3% are foud to be of poor quality. Fid the limits for the umber of substadard pair of shoes that ca be expected whe the Cofidece Level is Solutio: Let p be the sample proportio of defective shoes as produced by the shoe compay. I this case sample size () is 800 ad populatio size (N) is 80,000. Sice the populatio is very large, we do ot apply fiite populatio correctio. p 3% 0.03 > p (1 p) 0.03 (1 0.03) S.E. (pˆ ) Thus 99% Lower Cofidece Limit to P, the proportio of defective shoes i the daily productio of the shoe compay is : > p.58 S.E. (pˆ ) similarly 99% Upper Cofidece Limit to p is : P +.58 Ŝ.E. (Pˆ ) Hece, the Lower limit to the umber of substadard i.e., defective pairs of shoes at 99% Level of Cofidece N , (approximately)

20 Probability ad Hypothesis Testig The Upper Limit to the umber of substadard, pairs of shoes at 99% Level of Cofidece is 80, (approximately) 3638 Self Assessmet Exercise B 1) State with reasos, whether the followig statemets are true or false. a) Cofidece Iterval provides a rage of values that may ot cotai the parameter. b) Cofidece Iterval is a fuctio of Cofidece Co-efficiet. c) 95% Cofidece Iterval for populatio mea is x ± 1.96 S.E. (x). d) While computig Cofidece Iterval for populatio mea, if the populatio S.D. is ukow, we ca always replace it by the correspodig sample S.D. p (1 p) e) 99% Upper Cofidece Limit for populatio proportio is p f) Cofidece co-efficiet does ot cotai Lower Cofidece Limit ad Upper Cofidece Limit. g) If p 5 ad p (1 p) 5, oe may apply the formula p ± z α p (1 p) for computig Cofidece Iterval for populatio proportio. h) The iterval µ ± 3 S.E. (x) covers 96% area of the ormal curve. ) Differetiate betwee Poit Estimatio ad Iterval Estimatio ) Distiguish betwee Cofidece Limit ad Cofidece Iterval Out of 5,000 customer s ledger accouts, a sample of 800 accouts was take to test the accuracy of postig ad balacig ad 50 mistakes were foud. Assig limits withi which the umber of wrog postigs ca be expected with 99% cofidece A sample of 0 items is draw at radom from a ormal populatio comprisig 00 items ad havig stadard deviatio as 10. If the sample mea is 40, obtai 95% Iterval Estimate of the populatio mea

21 6) A ew variety of potato grow o 400 plots provided a mea yield of 980 quitals per acre with a S.D. of quitals per acre. Fid 99% Cofidece Limits for the mea yield i the populatio Tests of Hypothesis I 15.5 TESTING HYPOTHESIS INTRODUCTION Referrig to the problem of the status to be give to Delhi City, oe of the criteria for determiig the status would be the average icome of the people of Delhi. Let us suppose that if µ, the average icome of the people is Rs. 3,000 per moth, the Delhi would belog to the group of top cities. I order to estimate µ, we take a radom sample of people livig i that city ad compute x, the sample mea. If x is i the eighbourhood of Rs. 3,000, the we have o hesitatio i declarig the status of Delhi as oe belogig to the top grade. But the most importat questio would be as to what differece betwee the sample mea ad Rs (populatio mea) ca be accepted as the differece due to oly samplig fluctuatios. I order to aswer this questio, let us familiarise ourselves with a few terms associated with the problem. A statemet like The average icome of the people belogig to the city of Delhi is Rs. 3,000 per moth is kow as a ull hypothesis. Thus, a ull hypothesis may be described as a assumptio or a statemet regardig a parameter (populatio mea, µ, i this case) or about the form of a populatio. The term ull is used as we test the hypothesis o the assumptio that there is o differece or, to be more precise, o sigificat differece betwee the value of a parameter ad that of a estimator as obtaied from a radom sample take from the populatio. A hypothesis may be simple or composite. A simple hypothesis is oe that specifies the populatio distributio completely. Thus testig µ 3,000 is a simple hypothesis if the populatio stadard deviatio () is kow. A composite hypothesis is oe that does ot specify the populatio completely. Testig µ 3,000 whe is ukow is a composite hypothesis as it does ot specify the populatio completely. A ull hypothesis is deoted by H 0. Thus we may write : H 0 : µ 3,000 i.e., the ull hypothesis is that the populatio mea is Rs. 3,000. Geerally, we write H 0 : µ µ 0 i.e., the ull hypothesis is that the populatio mea is µ, whereas µ 0 may be ay value as specified i a give situatio. Obviously a ull hypothesis (H 0 ) is to be tested agaist a appropriate alterative hypothesis (H 1 ). Ay hypothesis that cotradicts a ull hypothesis is kow as a alterative hypothesis. If the ull hypothesis is rejected, the alterative hypothesis is accepted. Procedures eablig us to 7 5

22 Probability ad Hypothesis Testig decide whether to accept or reject a hypothesis is kow as test of hypothesis or test of sigificace or decisio rule. Thus, the etire process of hypothesis testig is either to reject or accept H 0 oly. I the preset problem, oe may argue that sice may people of Delhi city are livig i the slums ad eve o the pavemets, the average icome should be less tha Rs So oe alterative hypothesis may be : H 1 : µ < 3,000 i.e., the average icome is less tha Rs. 3,000 or symbolically: or, H 1 : µ < µ 0 i.e., the populatio mea (µ) is less tha µ 0. Agai oe may feel that sice there are may multistoried buildigs ad may ew models of vehicles ru through the streets of the city, the average icome must be more tha Rs. 3,000. So aother alterative hypothesis may be : H : µ > 3000 i.e., the average icome is more tha Rs. 3,000. or, H : µ > µ 0 i.e., the populatio mea is more tha µ o. Lastly, aother group of people may opie that the average icome is sigificatly differet from µ 0. So the third alterative could be : H : µ 3000 i.e., the average icome is aythig but Rs. 3,000. or, H : µ µ 0 i.e., the populatio mea is ot µ THEORY OF TESTING HYPOTHESIS LEVEL OF SIGNIFICANCE, TYPE-I AND TYPE-II ERRORS AND POWER OF A TEST I order to take a decisio about acceptace or rejectio of a ull hypothesis, let us cosider the theory ivolvig testig of hypothesis. Suppose that we have a radom sample of size take from a populatio characterized by a ukow parameter θ. We deote the sample observatios by x (x 1, x, x 3, x ) ad we would like to test H 0 : θ θ 0 agaist H 1 : θ θ If, the x (x 1, x ) ca be represeted as a poit i the -dimesioal plae takig, say x 1, o the horizotal axis ad x o the vertical axis. I a similar way, it is possible to coceive x (x 1, x, x 3, x ) as a poit i the - dimesioal plae. Cosider all the possible samples of a fixed size, i.e., N C i case of SRSWOR ad N i the case of SRSWR, N deotig the populatio size. Next we cosider the sample space formed by all these poits ad let it be deoted by Ω. We divide Ω ito two parts ω ad A Ω ω, the boudary of ω is take withi A. We frame a simple rule which says that if the sample poit x falls o ω, we reject H 0 ad if x falls o A, we accept H 0. ω is kow as the critical regio or rejectio regio ad A, as the acceptace regio. At this jucture, let us make oe poit clear. Acceptace of H 0 does ot mea that H 0 is always true. It just reflects the idea that o the basis of the give data, there is ot eough evidece to support the validity of H 1. I a similar maer rejectio of H 0 idicates the ull hypothesis does ot hold good i the light of the give sample observatios.

23 Type-I ad Type-II Errors Tests of Hypothesis I Now while testig H 0 we are liable to commit two types of errors. I the first case, it may be that H 0 is true but x falls o ω ad as such, we reject H 0. This is kow as type-i error or error of the first kid. Thus type-i error is committed i rejectig a ull hypothesis which is, i fact, true. Secodly, it may be that H 0 is false but x falls o A ad hece we accept H 0. This is kow as type-ii error or error of the secod kid. So type-ii error may be described as the error committed i acceptig a ull hypothesis which is, i fact, false. The two kids of errors are show i Table Table 15.3: Types of Errors i Testig Hypothesis Real Situatio Statistical decisio based o sample H 0 Accepted H 0 Rejected H 0 True Right decisio Type-I error H 0 False Type-II error Right decisio It is obvious that we should take ito accout both types of errors ad must try to reduce them.sice committig these two types of errors may be regarded as radom evets, we may modify our earlier statemet ad suggest that a appropriate test of hypothesis should aim at reducig the probabilities of both types of errors. Let α (read as alpha ) deote the probability of type-i error ad β (read as beta ) the probability of type-ii error. thus by defiitio, we have α The probability of the sample poit fallig o the critical regio whe H 0 is true i.e., the value of θ is θ 0 P (x ω θ 0 ) (15.6) ad β The probability of the sample poit fallig o the critical regio whe H 1 is true, i.e., the value of θ is θ 1 P (x A θ 1 ) (15.7) Surely, our objective would be to reduce both type-i ad type-ii errors. But sice we have take recourse to samplig, it is ot possible to reduce both types of errors simultaeously for a fixed sample size. As we try to reduce α, β icreases ad a reductio i the value of β results i a icrease i the value of α. Thus, we fix α, the probability of type-i error to a give level (say, 5 per cet or 1 per cet) ad subject to that fixatio, we try to reduce β, probability of type-ii error. α is also kow as size of the critical regio. It is further kow as level of sigificace as α costitutes the basis for makig the differece (θ θ 0 ) as sigificat. The selectio of α level of sigificace, depeds o the experimeter. Power of a Test: By defiitio, we have β P (x A θ θ 1 ) [from 15.7] : 1 β 1 P (x A θ θ 1 ) P (x ω θ θ 1 ) [from 15.6] [Sice θ 1 may fall either o ω or A, therefore, P (x ω θ θ 1 ) + P (x A θ θ 1 ) 1 ad we have 1 P (x A θ θ 1 ) P (x ω θ θ 1 )] 7 7

24 Probability ad Hypothesis Testig Now P (x ω θ θ 1 ) is the probability of rejectig H 0 whe H 0 is false ad the alterative hypothesis H 1 is true which should be the desirable property of a appropriate test. It is obvious that a low value of β would esure a high value of (1 β). Hece we try to miimize β, the probability of type-ii error, as the miimizatio of β esures the maximizatio of (1 β). The expressio (1 β) serves as a idicator of the validity of the test as a very high value of (1 β) idicates that the test is doig fie i its edeavour to reject a false hypothesis. Hece (1 β) is kow as power of the test as it tells us how well the test uder cosideratio is performig whe the ull hypothesis is ot true. It is obvious that we should try to make our test as powerful as possible subject to a fixed value of α. Oe may regard power of a test as a fuctio of θ. The fuctio P (θ) 1 β (θ) is kow as the power fuctio of the test. The curve obtaied by plottig P (θ) agaist θ is kow as power curve. Look at the followig figure 15.5 which exhibits a power curve. 1.0 P(θ) 0 θ Fig. 15.5: Power Curve of a Test 15.7 TWO-TAILED AND ONE-TAILED TESTS I order to test the ull hypothesis H 0 : θ θ 0 agaist a plausible alterative hypothesis, let us suppose that we fid a statistic T which is a sufficiet estimator of θ. We assume further that, based o a radom sample take from the populatio characterized by a ukow parameter θ, it is possible to fid a fuctio of T ad θ ad let u u (T, θ) by such a fuctio. T is kow as test statistic for testig H 0 : θ θ 0. Lastly let us assume that whe θ θ 0, u 0 u (T, θ 0 ) i.e., u 0 is the value of u uder H 0 (i.e., assumig the ull hypothesis to be true). Based o the samplig distributio of the test statistic u uder H 0, it may be possible to fid 4 values of u, amely, u α/, u (1-α/), u α ad u (1-α ) for a fixed level of sigificace α, such that : α P (u 0 u α/ ) (15.8) P (u 0 u (1 α/ ) α (15.9) P (u 0 u α ) α (15.30) P (u 0 u (1 α) ) α (15.31) u α may be described as the upper α-poit of the distributio of u ad u (1-α) as the correspodig lower α-poit. Two-tailed test: Addig (15.8) ad (15.9), we get : P (u 0 u α/ ) + P (u 0 u (1-α/) ) α (15.3) 7 8 i.e., the probability that u 0 would exceed u α/ or u 0 is less tha u (1-α/) is α.

25 I order to test H 0 : θ θ 0 agaist H 1 : θ θ 0, if we select a low value of α, say α 0.01, the (15.3) suggests that the probability u 0 is greater tha u a/ or u 0 is less tha u (1-α/) is 0.01 which is pretty low. So o the basis of a radom sample draw from the populatio, if it is foud that u 0 is greater tha u a/ or u 0 is less tha u (1 α/), the we have rather strog evidece that H 0 is ot true. The we reject H 0 : θ θ 0 ad accept the alterative hypothesis H 1 : θ θ 0. As show i the followig Figure 15.6, here the critical regio lies o both tails of the probability distributio of u. Tests of Hypothesis I Critical Regio ω : u 0 u (1-α/) 100 (1 α) % area Acceptace Regio Critical Regio ω : u 0 u α/ 50 α % area u (1- α /) u α / Fig. 15.6: Critical regio of a two-tailed Test 50 α % area If the sample poit x falls o oe of the two tails, we reject H 0 ad accept H 1 : θ θ 0. The statistical test for H 0 : θ θ 0 agaist H 1 : θ θ 0 is kow as both-sided test or two-tailed test as the critical regio, ω lies o both sides of the probability curve, i.e., o the two tails of the curve. The critical regio is ω : u 0 u α/ ad ω : u 0 u (1-α/). It is obvious that a two-tailed test is appropriate whe there are reasos to believe that u differs from θ 0 sigificatly o both the left side ad the right side, i.e., the value of the test statistic u as obtaied from the sample is sigificatly either greater tha or less tha the hypothetical value. For testig the ull hypothesis H 0 : µ 3000, i.e., the average icome of the people of Delhi city is Rs. 3000, oe may thik that the alterative hypothesis would be H 1 : µ 3000 i.e., the average icome is ot Rs ad as such, we may advocate the applicatio of a two-tailed test. Similarly, for testig the ull hypothesis that the average life of lights produced by Idia Electricals is 5,000 hours agaist the alterative hypothesis that the average life is ot 5,000 hours, i.e., for testig H 0 : µ 5,000 agaist H 1 : µ 5,000, we may prescribe a two-tailed test. I the problem cocerig the health of city B, we may be iterested i testig whether 0% of the populatio of city B really suffers from T.B. i.e., testig H 0 : P 0. agaist H 1 : P 0. ad agai a two-tailed test is ecessary ad lastly regardig the harms of smokig, we may like to test H 0 : P 0.3 agaist H 1 : P 0.3. Right-tailed Tests We may thik of testig a ull hypothesis agaist aother pair of alteratives. If we wish to test H 0 : θ θ 0 agaist H 1 : θ > θ 0, the from (15.30) we have P (u 0 u α ) α. This suggests that a low value of α, say α 0.01, implies that the probability that u 0 exceeds u α is So the probability that u 0 exceeds u α is rather small. Thus o the basis of a radom sample draw from this populatio if it is foud that u 0 is greater tha u α, the we have eough evidece to suggest that H 0 is ot true. The we reject H 0 ad accept H 1. This is exhibited i Figure 15.7 as show below: 7 9

26 Probability ad Hypothesis Testig 100 (1 α) % area Acceptace Regio Critical Regio ω : u 0 u α Fig. 15.7: Critical regio of a right-tailed Test u α 100 α % area As show i figure 15.7, the critical regio lies o the right tail of the curve. This is a oe-sided test ad as the critical regio lies o the right tail of the curve, it is kow as right-tailed test or upper-tailed test. We apply a righttailed test whe there is evidece to suggest that the value of the statistic u is sigificatly greater tha the hypothetical value θ 0. I case of testig about the average icome of the citizes of Delhi, if oe has prior iformatio to suggest that the average icome of Delhi is more tha Rs. 3,000, the we would like to test H 0 : µ 3,000 agaist H 1 : µ > 3,000 ad we select the right-tailed test. I a similar maer for testig the hypothesis that the average life of lights by Idia Electricals is more tha 5,000 hours or for testig the hypothesis that more tha 0 per cet suffer from T.B. i city B or for testig the hypothesis that the per cet of smokers i tow C is more tha 30, we apply the righttailed test. Left-tailed test Lastly, we may be iterested to test H 0 : θ θ 0 agaist H : θ < θ 0.From (15.31), we have P (u 0 u 1 α ) α. Choosig α 0.01, this implies that the probability that u 0 would be less tha u α is 0.01, which is surely very low. So, if o the basis of a radom sample take from the populatio, it is foud that u 0 is less tha u 1-α, the we have very serious doubts about the validity of H 0. I this case, we reject H 0 ad accept H : θ < θ 0. This is reflected i Figure 15.8 show below. Critical Regio ω : u 0 u (1-α) 100 (1 α) % area Acceptace Regio 100 α % area u (1- α ) Fig. 15.8: Critical Regio of a Left-tailed Test 8 0 The test for H 0 : θ θ 0 agaist H : θ < θ 0 is aother oe-sided test ad as the critical regio lies o the left tail of the curve, this is kow as a left-

27 tailed test or a lower-tailed test. We apply a left-tailed test whe there is eough idicatio to suggest that the value of the test statistic u is sigificatly less tha the hypothetical value. The for determiig the status of Delhi city, if somebody suggests with evidece that the average icome is less tha Rs. 3,000 ad as such Delhi should ot be regarded as a top grade city, the we are to test H 0 : µ 3000 agaist H 1 : µ < 3000, which is a left-tailed test. We may further ote that we apply left-tailed test whe we would like to test the hypothesis that the average life of lights of Idia Electricals is less tha 5,000 hours or less tha 0 per cet are sufferig from T.B. i city B or less tha 30 per cet are smokers i tow C. Tests of Hypothesis I 15.8 STEPS TO FOLLOW FOR TESTING HYPOTHESIS While testig hypothesis, oe must go through the followig steps. 1) Set up the ull hypothesis H 0 : θ θ 0 ad oe of the alterative hypothesis H : θ θ 0 or H 1 : θ > θ 0 or H : θ < θ 0 depedig upo the problem. Selectig the proper alterative plays a sigificat role i decisio makig i coectio with testig of hypothesis. ) Choose the appropriate test statistic u ad samplig distributio of u uder H 0. I most cases u follows a stadard ormal distributio uder H 0 ad hece Z-test ca be recommeded i such a case. 3) Select α, the level of sigificace of the test if it is ot provided i the give problem. I most cases, we choose α 0.05 ad α 0.01 which are kow as 5% level of sigificace ad 1% level of sigificace. 4) Defie critical regio ω, based o the alterative hypothesis. For testig H 0 : θ θ 0 agaist both-sided alterative H 1 : θ θ 0, the critical regio is give by ω : u 0 u α/ ad ω : u 0 u (1-α/). Similarly, the critical regio for the rightsided alterative is give by ω : u 0 u α ad the critical regio for the left-sided is give by ω : u 0 u 1-α. 5) Obtai the value of u 0 o the basis of the give sample observatios. 6) Reject H 0 if u 0 falls o ω. Otherwise accept H 0. 7) Draw your ow coclusio i very simple laguage which should be uderstood eve by a layma TESTS OF SIGNIFICANCE FOR POPULATION MEAN Z-TEST FOR VARIABLES Let us assume that we have take a radom sample of size from a ormal populatio with mea as µ ad stadard deviatio as. Let the sample observatios be deoted by x 1, x, x 3, x. While testig for the ukow populatio mea µ, we are to cosider the followig cases. Case 1: Whe the stadard deviatio is kow. We wat to test H 0 : µ µ 0 agaist oe of the followig alterative hypothesis. H : µ µ 0 or, H 1 : µ > µ 0 or, H : µ < µ

28 Probability ad Hypothesis Testig As we have discussed i Sectio 15., the best statistic for the parameter µ is x. It has bee proved i that Sectio, E (x) µ. S.E. (x) As such the test statistic : x E(x) z S.E.(x) x µ / is a stadard ormal variable. Uder H 0, i.e., assumig the ull hypothesis to be true, (x µ 0) z0 is a stadard ormal variable. As such, the test is kow as stadard ormal variate test or stadard ormal deviate test or Z-test. I order to fid the critical regio for testig H 0 agaist H from (15.8) ad (15.9), we fid that : ad P (u P (u 0 0 u u ( α /, (1 α /, α ) α ) If we deote the stadard ormal variate by Z, ad the upper α-poit of the stadard ormal distributio by Z α, ad by Z (1 α/) Z α/, (as the stadard ormal distributio is symmetrical about 0), the lower α-poit of the stadard ormal distributio, the the above two equatios are reduced to : α P (Z0 Z α / ) (15.33) α ad P (Z0 Zα / ) (15.34) From (15.33), we have: 1 α P (Z0 < Z( α / ) or or α 1 φ (Zα / ) α φ ( Zα/ ) 1 choosig α.05, φ (Z ) or, φ Z ) (1.96 ) [from Sectio 15.4] ( φ Thus, Z Hece from (15.33) ad (15.34), we have P (Z ) 0.05 ad P (Z )