sta108solnsamplefinal

sta108solnsamplefinal - Solution Sample Final Statistics...

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Unformatted text preview: Solution: Sample Final Statistics 108 1. Model: Y : 50 + 314: + 511422 + 4, 4: = X A X, 355(4, 442) = 1,5015. Test H0 : [311 = 0 against H1 21311 75 0, a = .01. Reduced model: Y : fig —|— 613: + E, SSE($) = 8.1815. 4 _ W 1 184815—14015u1 _ F m SS;(m.m2)/(:f3) * 1.5015/(8—6) —22'244- df = (1,5),F(.99; 1,5) = 16.3 Since F* > F(.99;1, 5), we reject H0. Conclusion: linear fit is not adequate. p— —value=shaded area . ‘ From the F—table F(. 99; 1 5) =16.3, F(9951 5)=22..8 21%9 So .005 <p—value< .01. However, p~value should be close to .005. b) When Xh = 5, sch = Xh i X = 5 -— 4.5 : .5, the estimated mean drying time at Xh : 5 is 1% = be + 51mg,” + buzcfi : 4.2154 + (w.2203)(.5) + (.1994)(.5)2 : 4.1551. c) Note that if 2 130 +01$+011$2 = bg+b1(X —X) +011(X *X)2 = to +b1(X — X) + +121“):2 — 2XX +272) 2 53+ng + 24;]X2, where he — be — 51X + 511272: 4. 2154— (— .2203)(4.5) + (.1194)(4.5)2 = 7.2619, b’— _ b] A 2ng1— _ —— 2.203 — (2)(4.5)(.1994)— 42.0149, and b11 4 —b11 = .1994. So the fitted regression in X is Y: 7. 2619 —— 2. 0149X + 1994X2. 2. :1) ANOVA table ——--- Regression 7 = 21. 410 7.1367 19. 346 Error 7?. p = 16 5.903 .3689 27.313 -- Here 33 251% of beta parameters estimated =4. _ _ ‘ 4 . b) Test H; := 131— — 0 against H1 : [31 ¢ 0 at level 0: = .05. .- 13’4”?qu 4271244: 33344559 - 141a: ”"5 5 1 From the t» table 1';(. 975; 16) r: 2.120. Since |t*| < t(.975;16), we cannot reject H0. Conclusion: variable X1 can be dropped from the model. - . __ 2;; pevaluezsum of the shaded areas ‘1'415353 0 144—553 ={2)(area to the right of 1.456) From the t—table, t(.90; 16) = 1.337, t(.95;16) = 1.746. 80 p—value should be between (2)(.05):.1 and (2)(.1)=.2, ie., .1<p—value<.2_ c) In order to calculate T§M3 we need SSE(X2,X3) and SSE(X1,X2, X3). We have SSE(X1,X2,X3). We need to find SSE(X2,X3). From part (b) [3313042, X3) 1 SSE(X1,X2,X3)]/1 . SSE(X1,X2,X3)/(n — 4) ”'6" [SSE(X2, X3) _ 5.903} 5.903/16 ' (1.4559)2 = 4*? z 111* : 2.1196 : Solving the last equation yields SSE(X2, X3) 4 6 6849. So we have T2 SSE X2,X3 l—SSE X1,X21X3! _6. 6849— 5 903_ .1170 Y1 23 — ssmx; X3) _ 6 6849 There is about 11.7% reduction in the residual sum of squares when add variable X1 to the model Y 2 £30 + 16ng + 16ng + 5. Since this reduction is rather small, it seems that there is no need add variable X1 to the model Y t: .60 + ,6ng + 63X 3 + 5 and hence the conclusion in part (b) is consistent with the value of ”$1.23. 3. a) Best one X ~variable and two X —variable models are Y = ,‘30 + ,6ng + e and Y = fie + 51X1+53X3 +5- The FPE criterion is FPEP = ESSEP, where p is the number of beta parameters estimated. Here I‘M'n 4 24. we: 1771? 2 209?:D 160 3 205.71 149 4 208 60 According to the FPE criterion, the most appropriate model seems to be Y: 60 + 51X} + [33X 3 + 5. b) Backward elimination method with F to delete equal to 4.0. Step 1: All variables are in the model The best candidate for deletion is variable X2 and the corresponding value of the F—statistic is [SSE(X1,X3) — SSE(X1, X2, X3)]/1 _ [160 4 149] /1 SSE(X1,X2,X3)/(n _ 4) — W = 1.477. F* = Since this 5“ < F -to—delete, we drop variable X2. Step 2. Variables in the model: X1,X3. Since SSE(X3) is closest to SSE(X1,X3), the best candidate for deletion is variable X1 and the corre sponding value of the F—statistic is [SSE(X3) — SSE(X1,X3)]/1 i [1340 — 160]/1 SSE(X1,X3)/(n 4 3) — W = 220.50. P“ 2 Since this F“ >F~to~de1ete, we cannot drop variable X1. So the most appropriate model seems to be Y 2 60 + 61X 1 + 63X3 + a. A w - _~_._ _ _ 4. a) 31 _ 3;:ng _ 15:33:.11623, 30: b1X=.10317—(.)11623 (25. 833)— 7.31444 The fitted regression line is 17 = 7.31444 + .11623X. Estimated body fat at X— A 30 is Y— _ 3 31444 + (.11623)(30)= 10 3013. b) Estimated correlation r — M - fia— = .6135. Eur ,-,-X)2§j(Y.-,— 1")2 (1010.50)(3s.2650) c) From part (a) we have 1"}, 2 10.8013 at X, = 30. SSE : 294; n Y)2 4 92 23(29- 22—)2— _ 39. 2650— (.11923)2(1010.50) = 22.9137. MSE = SSE/(n — 2): 22. 6137/(18 - 2) _ 1. 4134. 52(Yh) = MSE[l + flh _ (1. 4134)[-,—8 + WP _ .10232. 3(pred) = MMSE + 5209,) = «1.4134 + .10232 : 1.2313. A 95% prediction interval for 1?), it(.975;16)s(pred), 1.9., 10.8013:1:(2.120)(1.2313),i.e., 1030142910, 1.9., (3.191,13.411). (:1) Here we have 9: 4 prediction intervals Bonferroni: B — 15(1— .(é—)Q~W(E’4);n # 2)— - t(. 99375; 16)~ N 2. 823 [From the t— table, t( 9,925 16)— w 2. 724 and t( .99;5 16): 2. 921. 80 t( 99375; 16)~ ~ 2. 823, by interpolation] Scheffe- S: 4F( .95; 4 n— 2) =,/4F( 95; 4, 19) ,/(4)(3. 02) :3 48 [From the F—table, F(.95; 4, 15): 3.06 and F(.95; 4,20) = 2.87. So, F(.95;4, 16) m 3.02, by interpolation] Since B < .5', Bonferroni is the preferred method here. 5. I. When age X : 50, blood pressure is normally distributed with mean=66.8+(1.62)(50)=147.8 and SD2;/2.84 = 1.6852. When age is about 50, proportion of individuals with blood pressure between 145 and 150 is equal to the area under the standard normal curve between W: 1. 66 and 150 1478 _ 1. 6852 —1' 31' From the normal table, the area under the normal curve between —1.66 and 1.31 is equal to (area to the left of 1.31)—(area to the left of —1.66)=.9049—(1—.9515)=.8564. Of those who are about 50 years old, about 85.6% have blood pressure between 145 and 150. 11. Reduced model is r = so 4 61X1+ 922a + 191—3424., + 5 or Y = 90 + mm + X3/2) + 92(X2 + X3/2) + 5. dflSSEmd) = n e 3. For the F—test, the degrees of freedom are (1, n — 4) : (1,21). III. -_-- The most appropriate model seems to be Y2— fig + 51X + 5. ...
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