Egieerig 33 Gayheart 5-63 1 Beautif ul Homework 5-63 Suppose t h a t whe t h e ph of a c e r t a i c h e m i c a l compoud i s 5.00, t h e ph measured by a radomly s e l e c t e d begiig chemistry studet i s a radom variable with m e a 5.00 ad stadard deviatio 0.3. A large b a t c h of t h e compoud i s subdivided ad a sample give to e a c h studet i t h e morig l a b ad e a c h studet i a afteroo l a b. Let = t h e average ph a s d e t e r m i e d by t h e morig studets ad = t h e average ph a s d e t e r m i e d by t h e afteroo studets. a. If t h e ph i s a ormal variable ad t h e r e a r e 5 studets i e a c h l a b, comp t e P(-0.1? 0.1.(Hit:? i s a l i e a r combiatio of ormal variables, so i s ormally distribted. Computed µ? ad a?.) b. If t h e r e a r e 36 studets i e a c h l a b, but ph determiatios a r e ot ormal, c a l c u l a t e (a pproximately) P(-0.1? 0.1. Backgroud Iformatio Defiitio(pg. 19) A statistic i s ay quatity whose value c a be c a l c u l a t e d from sample d a t a. Prior to obtaiig d a t a, t h e r e i s ucertaity a s to what value of ay p a r t i c u l a r statistic will result. Therefore, a statistic i s a radom variable ad will be deoted by a uppercase l e t t e r ; a lowercase l e t t e r i s used to represet t h e c a l c u l a t e d or observed value of t h e statistic. By t h i s defiitio - t h e sample m e a of t h e morig class, which i s regarded a s a statistic. - t h e sample m e a of t h e afteroo class, which i s also regarded a s a statistic.? t h e differece b e t e e t h e two sample m e a phs It i s importat to ote t h a t ay statistic, beig a radom variable h a s a probablitiy distributio, e s p e c i a l l y t h e sample meas. For a e x a m p l e, refer to page 19 i t h e t e x t. T h e probability distributio depeds o t h r e e i t e m s: populatio distributio( ormal, logormal), t h e sample size, t h e method of samplig. T h e first two a r e self-explaitory, t h e third r e q u i r e s a defiitio importat to out problem. We a r e cosiderig a (s imple) radom sample of size. Simple radom s a m p l e s must m e e t two c r i t e r i a : 1. T h e i s a r e idepedet rv s. Every i h a s t h e same probability distributio.
Egieerig 33 Gayheart 5-63 T h e ext importat c o c e p t a r e t h e relatioships betwee t h e e x p e c t e d value of a sample m e a ad W ad also t h e variace of a sample m e a ad a, ad. O page 30, t h e followig relatioships a r e proposed. Let 1,... be a radom sample from t h e distributio with a m e a value W ad t h e stadard devaitio a. The, 1. E = W = W. V = a = a ad a = a They also ote t h a t for T o (sample t o t a l ) = 1 +... +, t h e E T o = W, V T o = a, ad a T o = a Cetral Limit Theorem(CLT) Let 1,,... be a radom sample from a distrib tio with m e a W ad variace a. The if i s s fficietly large, h a s approximately a ormal distrib tio with W = W ad a = a, ad also h a s approximately a ormal distrib tio with W t o = W, a T o = a. THE LARGER THE VAL E OF, THE B E T T E R THE APPROIMATION. A good r l e to se CLT, > 30. If t h e size of i s large eo gh we c a se t h e CLT to fid a probability P(a b), the we c a ormailize t h e val e to Z ad se t h e t a b l e s. LINEAR COMBINATIONS R e c a l l from class t h a t : Whether or ot idepedece i s kow: E a 1 1 + a... + a = a 1 E 1 + a E... + a E = a 1 W 1 + a W... + a W If 1... a r e idepedet tha t h e Variace i s: V a 1 1 + a... + a = a 1 V 1 + a V... + a V = a 1 a 1 +... + a a If idepedece i s ukow tha: V a 1 1 + a... + a = >>a i a j Cov i, j Hopefully t h i s summerizes t h e importat p a r t s of t h e readig for t h e sectio. If you eed further e x a m p l e s, r e a d sectios 5.4,5.5 ad 5.6 i t h e t e x t.
Egieerig 33 Gayheart 5-63 3 Part A. Fid P(-0.1? 0.1. From t h e C e t r a l Limit t h e o r e m,s ice we kow t h a t i s ormally distributed ad i s ormally distributed, the we c a say t h a t? i s ormally distributed. T h e hit i d i c a t e s to c o m p u t e W? ad a?. To c o m p u t e W? ad a? use t h e followig equatios Keep i mid t h a t E = W = W ad t h a t V = a = a E 1? = E 1? E V 1? = V 1 + V T h i s i s because t h e variability for subtractio of two radom variables i s t h a t same a s t h e additio of two radom variables. Fid W? ~N W = W = 5.00, a = ~N W = W = 5.00, a = a a = 0.3 5 = 0.3 5 ad the we c a say t h a t? ~N W? = W? W, a? = a + a So t h e solutio i s t h e m e a of t h e morig mius t h e m e a of t h e afteroo class. Sice both c l a s s e s have a m e a of 5.00, W? = 5.00? 5.00 = 0 Fid a?. usig t h e Variace equatio V? = a + a = a + a, where = 5 ( T h e r e a r e 5 studets e a c h testig a sample) ad a =.9 (T h e problem gave a stadard deviatio(a of 0.3), so V? = a + a = 0.09 5 + 0.09 = 5 0.18 5 ad a? = 0.18 = 8. 485 3 10 5? T h e variace c a be added because of t h e defiitio for Liear combiatios o page 38 of t h e t e x t. where t h e variaceis: V a 1 1 + a... + a = a 1 V 1 + a V... + a V = a 1 a 1 +... + a a. Sice a i s squared t h e egative i? i t b e c o m e s positive.
Egieerig 33 Gayheart 5-63 4 Sice? h a s a ormal distributio tha t h e Z t a b l e s c a be used: P x? W? a? x? W?? a? = P? 0.1? 0.1 8. 485 3 10? 8. 485 3 10? = P? 1. 178 5 Z 1. 178 5 = fi? 1.1785? fi 1.1785 = 0.760 Figure 1- T h e ormal distributio for part a. The Normal Distributio 5 4 Percetage 3 1 0-1.5-1 -0.5 0-1 0.5 1 1.5 Radom Variable (-) Part B Here we do t kow t h e distributio of i ad i, but we c a use t h e CLT to fid t h e distributio of?, because 30. V? = a + a = 0.09 36 + 0.09 =.00 5 so a 36? = 0.18 = 7. 071 1 10 36? P x? W? a? x? W?? a? = P? 0.1? 0.1 7.0711 10? 7.0711 10? = P? 1. 41 Z 1. 41 = fi? 1.41? fi 1.41 = 0.8414 Figure 1 T h e ormal distributio for part b.
Egieerig 33 Gayheart 5-63 5 The Normal Distributio 6 5 Percetage 4 3 1 0-1.5-1 -0.5-1 0 0.5 1 1.5 Radom Variable (-)