Confidence interval, sample-size formula and test statistic, concerning:

Transcription

1 Cofidece iterval, ample-ize formula ad tet tatitic, cocerig: Populatio mea : (large ample, >30): z ± c z = c E Replace by F if available (rare). µ 0 EXAMPLE: A radom ample of 55 from a pecific populatio yield the mea of 73.5 ad tadard deviatio of 3.. i) Cotruct 90% cofidece iterval for the populatio mea: 73.5 ± = 73.5 ±.93 Awer: (70.57, 76.43) ii) How may more obervatio would we eed to be withi uit of the correct awer (uig the ame level of cofidece)? ( ) = Awer: = 47 iii) Tet H 0 : : = 75 agait H : : < 75 uig " = % Critical value: -.36 Computed value: ( ) = Not eough evidece to reject H 0.

2 Small ample: Need Normal populatio, ue t - itead of z. EXAMPLE: Coider the followig (radom idepedet) ample: 7, 8, 69, 77, 73, from a pecific Normal populatio. i) Cotruct a 80% cofidece iterval for :. G = 37, G = 7764, = 37 5 =74.4 = p$ ± z / c p$ q$ p$ q$ 4 = = ± = 74.4 ± 3.0 or (7.0, 77.60) ii) Tet H 0 : : = 75 agait H : : < 75 uig " = % Critical value: Computed value: ( ) = Not eough evidece to reject H 0. p$ > 5, q$ > 5 Populatio proportio p ( ): z E c $p p p q 0 z c E 0 0 EXAMPLE: I a ample of 7 Brock tudet, 37 ue a bu to get to chool.

3 i) The 95% CI for the proportio of all Brock tudet who ( 7 37) ue bu i: ± 96. = 0.54 ± 0.5 or (39.9%, 6.9%) ii) How may tudet (i total) do we eed to ample to reduce the margi of error to ± 3%? 37 ( 7 37) ( ) = ie iii) Tet H 0 : p = ½ agait H : p ½ uig " = 5% Critical value: ±.96 Computed (oberved) value of the 37 tet tatitic: 7 = (caot reject H 0 ). 4 7 Differece i populatio mea (large ample) ± z c + + EXAMPLE: A ample of 49 adult Caadia male had a average (ample mea) height of 79. cm, with the tadard deviatio of 7.4 cm. For a ample of 55 female, the average height wa 67.8 cm, with the tadard deviatio of 9.4 cm. i) Cotruct a 90% CI for the differece betwee the correpodig populatio mea (male - female):

4 ± (8.7 cm,4. cm) =.4 ±.7 or ii) Tet H 0 : : = : agait H : : > : uig " = % The critical value i.36, the computed value of the tet tatitic: = = Cocluio: Ye, thi repreet a highly igificat evidece that the female populatio mea i maller tha that of the male populatio. Small ample (at leat oe # 30): Both populatio mut be Normal ad have the ame tadard deviatio F. Alo, ue t itead of z. t ( ) + ( ) ± c + + ( ) + ( ) + + EXAMPLE: Same a before, ecept ow our male ample coit of oly 7, 8, 77, 83, 69 cm, ad the female ample i: 6, 58, 74, 7 cm. ad

5 i) Same a before. G = 88, G = 5574, G = 665, G = 075 Thi implie that = = ad ( ) ( ) ( ) ( ) Σ Σ + = Σ + Σ = = So, we have: 05. ± = 05. ± 445. (5.70, 4.60) cm or ii) Same a before. Critical value of t i.998, computed value of the tet tatitic i: 05. = Cocluio: Now, we caot reject the ull hypothei!! Differece i populatio proportio (both ample large i the > 5, > 5 ee). pq $$ pq $$ p$ p$ ± z c + p$ q$ p$ p$ + pq $ $ o o

6 EXAMPLE: Out of a radom ample of 5 Brock female tudet, 8 ue bu to get to chool, wherea for a ample of 4 male tudet the umber i oly 8. i) Cotruct a 80% CI for the populatio differece: 8 8 ± = 0.04 ± 0.35 or ( -.%, 5.9%) ii) Tet, uig 5% level of igificace, whether the two populatio proportio are the ame, agait the p F > p M alterate. Critical value (z c ) i.645, the computed value of the tet tatitic =. i: = Cocluio: We do t have ufficiet evidece to prove that the proportio of female tudet who ue bu i higher tha that of their male couterpart (at 5% level of igificace). Tet for : d (the populatio mea of paired differece). Tet tatitic: EXAMPLE: A radom ample of 7 car wa teted i term of the ditace travelled o oe litre of regular, ad oe litre of highoctae gaolie. The reult are give i the followig table: regular hi-oct d d

7 differece We have to aume that the ditributio of thee differece i (at leat approimately) Normal. H 0 : : d = 0 H : : d 0 " = 5% Critical value: ±.447 (Uig t with 6 df.) To compute the tet tatitic, we firt eed Gd = 0.6 ad Gd = The: / = = Cocluio: No tatitical evidece that either type of gaolie would be more ecoomical tha the other (at ay practical level of igificace). Regreio: Σ SS = Σ Σy SS y = Σy Σ SP y = Σy Σy b = SP y / SS ad a = y b

8 Bet (leat-quare) traight (regreio) lie: y = ba + a Predictio iterval for a ew value of y take at o : Poit etimate: Reidual tadard error: y a+ b p r o SS y b SP y PI: y ± t + ( ) + o SS p c r (Ue t ditributio with - df.) To tet H 0 : $ = 0 agait..., ue b r SS (ame ditributio).

9 Correlatio coefficiet: r SS SP y SS y Coefficiet of determiatio: r (i %)