Homework Solution (#4)

Transcription

1 Chapter 0: 4, 9,9,4 and 4 Homework Soluton (#4 Chapter 0. The Analyss of Varance 0.4 (new edton: Page 40; old edton: Page 4 3 ( , so ( ( ] (.08 x Ix SST SSTr 8[ 0. 38, So SSE Then f. 43. Snce F. 05,, 8.95, H : µ µ µ µ s reected at level.05. There are dfferences between at least two average flght tmes for the four treatments. 0.9 (new edton: Page 4; old edton: Page 43 The summary quanttes are x 34. 3, x 39., x , x4 4.9, ( 48.8 x 48.8, ΣΣx 94. 8, so CF 9.5, 4 ( ( 4.9 SST , SSTr , SSE Source Df SS MS F Treatments Error Total 3 4. Snce 3.0 F. 05,3,0 < 3.95 < F.0,3, 0,. 0 < p value <. 05 and H o s reected at level (new edton: Page 437; old edton: Page 49 a : I : x : x 38. x : ΣΣx Thus SST 3.4, SSTr 08.9, SSE 5.45, MSTr.4, MSE.73, f Snce.3 F 4.0, H 0 : µ... µ s reected ,5, 0

2 b The modfed Tukey ntervals are as follows: (The frst number s x x and the second s W MSE MSE Q , I, I ( Par Interval Par Interval Par Interval,.30 ±.,3.03 ±. 3,5 3.3±.*,3. 7 ±.,4. 30 ±. 3, 4.7 ±.*,4.00 ±., ±.04 * 4, ±.*, ±.*, 5.30 ±.* 4, 5.00 ±.*, 4.00 ±. * 3,4. 37 ±. 5,. 9 ±. Astersks dentfy pars of means that are udged sgnfcantly dfferent from one another. c c The 95% t confdence nterval s Σ ± Σ c x t. 05,0 MSE c Σ c x 4 x + 4 x + x + x x5 x 4., Σ. 79, MSE.73, t..08. The resultng nterval s 05,0 (.73( ±.45 ( 4., ±.08. The nterval n the answer secton s a Scheffe nterval, and s substantally wder than the t nterval. 0.9 (new edton: Page 437, old edton: Page 49 E SSTr E Σ X nx Σ n ( ( ( ( Σ [ ( ( ( ] [ ( ( ( ] Var X + nvar X + E X Σ σ σ + µ n + n ( Σ µ ( I σ + Σ [ ] µ + µ Σ α + Σ α Σ ( I σ + Σ α µ n ( I σ + Σ ( [ ( ] µ + α Σ µ + α, from whch E(MSTr s obtaned through dvson by ( I. 0.4 (new edton: Page 439, old edton: Page 43 a. µ true average CFF for the three rs colors. Then the hypotheses are H : µ µ µ 0 3 vs. H : a at least two µ ' s dffer. SST 3,59.7 3, , table follows: ( 04.7 ( 34. ( 9.0 SSTR + + 3, The ANOVA 8 5

3 Source Df SS MS F Treatments Error Total 8.3 Because F.3 < < F. 3,.0 < p-value <.05, so we reect. 05,, 3.0,, H o. There are dfferences n CFF based on rs color. Q. 3.5 b. and W are: 05,3, , so the Modfed Tukey ntervals x x ± W Par (,.33 ±. 7,3.58 ±. 5 *,3.5 ±. 4 Brown Green Blue The CFF s only sgnfcantly dfferent for Brown and Blue rs color. 0.4 (new edton: Page 440, old edton: Page 43 The ordered resduals are.7, -5.7, -4, -.7, -, -, 0, 0, 0,.33,.33,.33,,,.33, 4, 5.33,.33. The correspondng z percentles are.9, -.38, -.09, -.8, -.7, -.5, -.3, -., -.07,.07,.,.3,.5,.7,.8,.09,.38, and.9. The resultng plot of the respectve pars (the Normal Probablty Plot s reasonably straght, and thus there s no reason to doubt the normalty assumpton. Chapter : #, 8, 0, 3, 4, 7 and 8 Chapter. Multfactor Analyss of Varance. (new edton: Page 454, old edton: Page a. MSA 5. 85, MSE 3. 0, f. 83, whch s not sgnfcant at level.05. b. Otherwse extraneous varaton assocated wth houses would tend to nterfere wth our ablty to assess assessor effects. If there really was a dfference between assessors, house varaton mght have hdden such a dfference. Alternatvely, an observed dfference between assessors mght have been due ust to varaton among houses and the manner n whch assessors were allocated to homes.

4 .8 (new edton: Page 455, old edton: Page 447 a. x 4.34, x 4. 43, x , x 7. 30, SST 3. 87, SSTr.458, SSBl , SSE. 887, 3 MSTr.579, MSE. 0938, f.. Snce. F 3. 55, H oa s reected;. 05,, 8 there appears to be a dfference between anesthetcs. b. Q. 05,3,8 3., w.35. x. 434, x. 443, x , so both anesthetc and anesthetc appear to be dfferent from anesthetc 3 but not from one another..0 (new edton: Page 455, old edton: Page 447 Source Df SS MS f Method F Batch Error Total < 8.9 < F 0.39, so.00 < p-value <.0, whch s. 0,,8.00,, 8 sgnfcant. At least two of the curng methods produce dfferng average compressve strengths. (Wth p-value <.00, there are dfferences between batches as well..34 Q. 05,3,8 3.; w ( Method A Method B Method C Methods B and C produce strengths that are not sgnfcantly dfferent, but Method A produces strengths that are dfferent (less than those of both B and C..3 (new edton: Page , old edton: Page 447 a. Wth Y X + d, Y X + d, Y X + d, Y X + d, so all quanttes nsde the parentheses n (.5 reman unchanged when the Y quanttes are substtuted for the correspondng X s (e.g., Y Y X X, etc.. b. Wth Y cx, each sum of squares for Y s the correspondng SS for X multpled by c. However, when F ratos are formed the c factors cancel, so all F ratos computed from Y are dentcal to those computed from X. If Y cx d, the conclusons + reached from usng the Y s wll be dentcal to those reached usng the X s..4 (new edton: Page 45, old edton: Page 448

5 ( ( ( X Σ Σ Σ X E Σ( µ + α + β Σ Σ( µ + α + β I µ + α + Σ β µ Σα Σ β α I I, as desred..7 (new edton: Page 43, old edton: Page 45 a. Source Df SS MS f F.05 Sand Fber, * 4. Sand&Fber Error Total 7 3,05 There appears to be an effect due to carbon fber addton. b. Source Df SS MS f F.05 Sand * 4. Fber * 4. Sand&Fber Error Total There appears to be an effect due to both sand and carbon fber addton to castng hardness. c. Sand% Fber% x The plot below ndcates some effect due to sand and fber addton wth no sgnfcant nteracton. Ths agrees wth the statstcal analyss n part b.

6 mean Sand%.8 (new edton: Page 44, old edton: Page 45 Source Df SS MS f F.05 F.0 Formulaton,53.44, * * Speed * * Formulato n & Speed Error Total 7,574.7 a. There appears to be no nteracton between the two factors. b. Both formulaton and speed appear to have a hghly statstcally sgnfcant effect on yeld. c. Let formulaton Factor A and speed Factor B. The estmatons are: For Factor A: µ µ 4. For Factor B: µ µ µ For Interacton: µ µ 80. µ µ. µ. 03 µ Overall mean: µ α µ µ : α. 9 α. 8 β µ µ : β. 99 β 5. 0 β γ µ µ + α + β : d. ( γ.45 γ. 4 γ. 3 9 γ.45 γ. 39 γ Observed Ftted Resdual Observed Ftted Resdual

7 e From the graphs below, we accept normalty and homoscedastcty of resduals.