Statistical Analysis: Hypothesis Testing and Regression, Exams of Probability and Statistics

Download Statistical Analysis: Hypothesis Testing and Regression and more Exams Probability and Statistics in PDF only on Docsity! 1 Solutions to Even numbered problems in Chapters 5-7 CHAPTER 5 (Sections 5.1-5.8) 5-2. 1) The parameter of interest is the difference in breaking strengths, µ µ1 2− and ∆0 = 10 2) H0 : µ µ1 2 10− = or µ µ1 2= 3) H1 : µ µ1 2 10− > or µ µ1 2> 4) α = 0.05 5) The test statistic is z x x n n 0 1 2 0 1 2 1 2 2 2 = − − + ( ) ∆ σ σ 6) Reject H0 if z0 > zα = 1.645 7) x1 = 162.5 x2 = 155.0 δ = 10 σ1 = 1.0 σ2 = 1.0 n1 = 10 n2 = 12 z0 2 2 162 5 1550 10 10 10 10 12 584= − − + = − ( . . ) ( . ) ( . ) . 8) Since -5.84 < 1.645 do not reject the null hypothesis and conclude there is insufficient evidence to support the use of plastic 1 at α = 0.05. 5-4. x1 = 30.87 x2 = 30.68 σ1 = 0.10 σ2 = 0.15 n1 = 12 n2 = 10 a) 90% two-sided confidence interval: ( ) ( )x x z n n x x z n n1 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 1 2 2 2 − − + ≤ − ≤ − + +α α σ σ µ µ σ σ / / ( )( . . ) . ( . ) ( . ) . . . ( . ) ( . )30 87 30 68 1645 010 12 015 10 30 87 30 68 1645 010 12 015 10 2 2 1 2 2 2 − − + ≤ − ≤ − + +µ µ 0 0987 0 28131 2. .≤ − ≤µ µ We are 90% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between 0.0987 and 0.2813 fl. oz. b) 95% two-sided confidence interval: ( ) ( )x x z n n x x z n n1 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 1 2 2 2 − − + ≤ − ≤ − + +α α σ σ µ µ σ σ / / ( )( . . ) . ( . ) ( . ) . . . ( . ) ( . )3087 30 68 196 010 12 015 10 3087 30 68 196 010 12 015 10 2 2 1 2 2 2 − − + ≤ − ≤ − + +µ µ 0 081 0 2991 2. .≤ − ≤µ µ We are 95% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between 0.081 and 0.299 fl. oz. Comparison of parts a and b: 2 As the level of confidence increases, the interval width also increases (with all other values held constant). c) 95% upper-sided confidence interval: ( )µ µ σ σα1 2 1 2 1 2 1 2 2 2 − ≤ − + +x x z n n ( )µ µ1 2 2 2 3087 30 68 1645 010 12 015 10 − ≤ − + +. . . ( . ) ( . ) µ µ1 2 0 2813− ≤ . With 95% confidence, we bellieve the fill volumne for machine 1 exceeds the fill volume of machine 2 by no more than 0.2813 fl. oz. 5-6. x1 = 89.6 x2 = 92.5 σ1 2 = 1.5 σ2 2 = 1.2 n1 = 15 n2 = 20 a) 95% confidence interval: ( ) ( )x x z n n x x z n n1 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 1 2 2 2 − − + ≤ − ≤ − + +α α σ σ µ µ σ σ / / ( )( . . ) . . . . . . . .89 6 92 5 196 15 15 12 20 89 6 92 5 196 15 15 12 201 2 − − + ≤ − ≤ − + +µ µ − ≤ − ≤ −3684 2 1161 2. .µ µ With 95% confidence, we believe the mean road octane number for formulation 2 exceeds that of formulation 1 by between 2.116 and 3.684. b) 1) The parameter of interest is the difference in mean road octane number, µ µ1 2− and ∆0 = 0 2) H0 : µ µ1 2 0− = or µ µ1 2= 3) H1 : µ µ1 2 0− < or µ µ1 2< 4) α = 0.05 5) The test statistic is z x x n n 0 1 2 0 1 2 1 2 2 2 = − − + ( ) ∆ σ σ 6) Reject H0 if z0 < −zα = −1.645 7) x1 = 89.6 x2 = 92.5 σ1 2 = 1.5 σ2 2 = 1.2 n1 = 15 n2 = 20 z0 2 2 89 6 92 5 0 15 15 12 20 7 254= − − + = − ( . . ) ( . ) ( . ) . 8) Since −7.25 < -1.645 reject the null hypothesis and conclude the mean road octane number for formulation 2 exceeds that of formulation 1 using α = 0.05. c) P-value = P z P z( . ) ( . )≤ − = − ≤ = − =7 25 1 7 25 1 1 0 5-8. 95% level of confidence, E = 1, and z0.025 =1.96 5 P-Value: 0.549 A-Squared: 0.295 Anderson-Darling Normality Test N: 15 StDev: 9.43751 Average: 192.067 205195185175 .999 .99 .95 .80 .50 .20 .05 .01 .001 P ro ba bi lit y type2 Normal Probability Plot 205195185175 type2 210200190180 type1 b) 1) The parameter of interest is the difference in deflection temperature under load, µ µ1 2− 6 2) H0 : µ µ1 2 0− = or µ µ1 2= 3) H1 : µ µ1 2 0− < or µ µ1 2< 4) α = 0.05 5) The test statistic is t x x s n np 0 1 2 0 1 2 1 1 = − − + ( ) ∆ 6) Reject the null hypothesis if t0 < − + −t n nα , 1 2 2 where − t0 05 28. , = −1.701 7) Type 1 Type 2 x1 = 196.4 x2 = 192.067 ∆0 = 0 s n s n s n np = − + − + − ( ) ( )1 1 2 2 2 2 1 2 1 1 2 s1 = 10.48 s2 = 9.44 = + = 14 10 48 14 9 44 28 9 97 2 2( . ) ( . ) . n1 = 15 n2 = 15 t0 196 4 192 067 0 9 97 1 15 1 15 119= − − + = ( . . ) . . 8) Since 1.19 > −1.701 do not reject the null hypothesis and conclude the mean deflection temperature under load for type 2 does not significantly exceed the mean deflection temperature under load for type 1 at the 0.05 level of significance. c) P-value = 2P ( )t > 119. 0.75 < p-value < 0.90 d) ∆ = 5 Use sp as an estimate of σ: d = µ µ2 1 2 5 2 9 97 0 251 − = = sp ( . ) . Using Chart V g) with β = 0.10, d = 0.251 we get n ≅ 100. So, n n1 2 100= = ; Therefore, the sample sizes of 15 are inadequate. 5-18. a) 1) The parameter of interest is the difference in mean impact strength, µ µ1 2− 2) H0 : µ µ1 2 0− = or µ µ1 2= 3) H1 : µ µ1 2 0− < or µ µ1 2< 4) α = 0.05 5) The test statistic is t x x s n s n 0 1 2 0 1 2 1 2 2 2 = − − + ( ) ∆ 6) Reject the null hypothesis if t0 < −tα ν, where t0 05 25. , = −1.708 since ν ν = +             + +       + − = − = ≅ s n s n s n n s n n 1 2 1 2 2 2 2 1 2 1 2 1 2 2 2 21 1 2 27 43 2 25 43 25 . . (truncated) 7) x1 = 290 x2 = 321 7 s1 = 12 s2 = 22 n1 = 10 n2 = 16 t0 2 2 290 321 0 12 10 22 16 4 64= − − + = − ( ) ( ) ( ) . 8) Since −4.64 < −1.708 reject the null hypothesis and conclude that supplier 2 provides gears with higher mean impact strength at the 0.05 level of significance. b) P-value = P(t < −4.64): P-value < 0.0005 c) 1) The parameter of interest is the difference in mean impact strength, µ µ2 1− 2) H0 : µ µ2 1 25− = 3) H1 : µ µ2 1 25− > or µ µ2 1 25> + 4) α = 0.05 5) The test statistic is t x x s n s n 0 2 1 1 2 1 2 2 2 = − − + ( ) δ 6) Reject the null hypothesis if t0 > tα ν, = 1.708 where ν ν = +             + +       + − = − = ≅ s n s n s n n s n n 1 2 1 2 2 2 2 1 2 1 2 1 2 2 2 21 1 2 27 43 2 25 43 25 . . 7) x1 = 290 x2 = 321 ∆0 =25 s1 = 12 s2 = 22 n1 = 10 n2 = 16 t0 2 2 321 290 25 12 10 22 16 0 898= − − + = ( ) ( ) ( ) . 8) Since 0.898 < 1.708, do not reject the null hypothesis and conclude that the mean impact strength from supplier 2 is not at least 25 ft-lb higher that supplier 1 using α = 0.05. 5-20. 1) The parameter of interest is the difference in mean melting point, µ µ1 2− 2) H0 : µ µ1 2 0− = or µ µ1 2= 3) H1 : µ µ1 2 0− ≠ or µ µ1 2≠ 4) α = 0.02 5) The test statistic is t x x s n np 0 1 2 0 1 2 1 1 = − − + ( ) ∆ 6) Reject the null hypothesis if t0 < − + −t n nα / ,2 21 2 where − t0 01 40. , = −2.423 or t0 > t n nα / ,2 21 2+ − where t0 01 40. , = 2.423 7) x1 = 421 x2 = 426 ∆0 = 0 s n s n s n np = − + − + − ( ) ( )1 1 2 2 2 2 1 2 1 1 2 10 2) H0 : µd = 0 3) H1 : µd ≠ 0 4) α = 0.10 5) The test statistic is t d s nd 0 = / 6) Reject the null hpothesis if t0 < −t 0 05 13. , where −t 0 05 13. , = −1.771 or t0 > t 0 05 13. , where t 0 05 13. , = 1.771 7) d = 1.21 sd = 12.68 n = 14 t0 121 12 68 14 0 357= = . . / . 8) Since −1.771 < 0.357 < 1.771 do not reject the null and conclude the data do not support the claim that the two cars have different mean parking times at the 0.10 level of significance. The result is consistent with the confidence interval constructed since 0 is included in the 90% confidence interval. 5-30. d = 868.375 sd = 1290, n = 8 where di = brand 1 - brand 2 99% confidence interval: d t s n d t s n n d d n d−       ≤ ≤ +      − −α αµ/ , / ,2 1 2 1 868 375 3499 1290 8 868 375 3499 1290 8 . . . .−       ≤ ≤ +      µd −727.46 ≤ µd ≤ 2464.21 Since zero is contained within this interval, we are 99% confident there is no significant difference between the two brands of tire. 5-32. 1) The parameter of interest is the difference in blood cholesterol level, µd where di = Before − After. 2) H0 : µd = 0 3) H1 : µd > 0 4) α = 0.05 5) The test statistic is t d s nd 0 = / 6) Reject the null hpothesis if t0 > t 0 05 14. , where t 0 05 14. , = 1.761 7) d = 26.867 sd = 19.04 n = 15 t0 26867 19 04 15 5 465= = . . / . 8) Since 5.465 > 1.761 reject the null and conclude the data support the claim that low the mean difference in cholesterol levels is significanlty less after fat diet and aerobic exercise program at the 0.05 level of significance. 5-34. 1) The parameter of interest is the difference in mean weight, µd where di =Weight Before − Weight After. 2) H0 : µd = 0 3) H1 : µd > 0 4) α = 0.05 5) The test statistic is 11 t d s nd 0 = / 6) Reject the null hpothesis if t0 > t 0 05 9. , where t 0 05 9. , = 1.833 7) d = 17 sd = 6.41 n = 10 t0 17 6 41 10 8 387= = . / . 8) Since 8.387 > 1.833 reject the null and conclude there is evidence to conclude that the mean weight loss is significantly greater than 0; that is, the data support the claim that this particular diet modification program is significantly effective in reducing weight at the 0.05 level of significance. 5-36. 1) The parameter of interest is the difference in mean weight loss, µd where di = Before − After. 2) H0 : µd = 10 3) H1 : µd > 10 4) α = 0.05 5) The test statistic is t d s nd 0 0= − ∆ / 6) Reject the null hpothesis if t0 > t 0 05 9. , where t 0 05 9. , = 1.833 7) d = 17 sd = 6.41 n = 10 t0 17 10 6 41 10 345= − = . / . 8) Since 3.45 > 1.833 reject the null and conclude there is evidence to support the claim that this particular diet modification program is effective in producing a mean weight loss of at least 10 lbs at the 0.05 level of significance. 5-38. a) f0.25,5,10 = 1.59 d) f0.75,5,10 = 1 1 189 0529 0 25 10 5f . , , . .= = b) f0.10,24,9 = 2.28 e) f0.90,24,9 = 1 1 191 0524 0 10 9 24f . , , . .= = c) f0.05,8,15 = 2.64 f) f0.95,8,15 = 1 1 322 0 311 0 05 15 8f . , , . .= = 5-40. 1) The parameters of interest are the variances of concentration, σ σ1 2 2 2, 2) H0 : σ σ1 2 2 2= 3) H1 : σ σ1 2 2 2≠ 4) α = 0.05 5) The test statistic is f s s 0 1 2 2 2 = 6) Reject the null hypothesis if f0 < f0 975 9 15. , , where f0 975 9 15. , , = 0.265 or f0 > f0 025 9 15. , , where f0 025 9 15. , , =3.12 7) n1 = 10 n2 = 16 12 s1 = 4.7 s2 = 5.8 f0 2 2 4 7 58 0 657= = ( . ) ( . ) . 8) Since 0.265 < 0.657 < 3.12 do not reject the null hypothesis and conclude there is insufficient evidence to indicate the two population variances differ significantly at the 0.05 level of significance. 5-42. a) 90% confidence interval for the ratio of variances: s s f s s fn n n n 1 2 2 2 1 2 1 1 1 2 2 2 1 2 2 2 2 1 11 2 1 2       ≤ ≤      − − − − −α α σ σ / , , / , , ( . ) ( . ) . ( . ) ( . ) . 0 35 0 40 0 412 0 35 0 40 2 33 2 2 1 2 2 2 2 2       ≤ ≤       σ σ 0 3605 2 0391 2 2 2 . .≤ ≤ σ σ Since the interval contains 1, we are 90% confident the variances for the rod diameters are not significantly different. b) 95% confidence interval: s s f s s fn n n n 1 2 2 2 1 2 1 1 1 2 2 2 1 2 2 2 2 1 11 2 1 2       ≤ ≤      − − − − −α α σ σ / , , / , , ( . ) ( . ) . ( . ) ( . ) . 0 35 0 40 0 345 0 35 0 40 2 75 2 2 1 2 2 2 2 2       ≤ ≤       σ σ 0 302 2 4061 2 2 2 . .≤ ≤ σ σ We are 95% confident the variances for the rod diameters are not significantly different. The 95% confidence interval is wider than the 90% confidence interval. c) 90% lower-sided confidence interval: s s f n n 1 2 2 2 1 1 1 1 2 2 21 2       ≤− − −α σ σ , , ( . ) ( . ) . 0 35 0 40 0 503 2 2 1 2 2 2       ≤ σ σ 0 440 1 2 2 2 . ≤ σ σ With 90% confidence, we believe the variances for the rod diameters are not significantly different from each other. 5-44. 1) The parameters of interest are the thickness variances, σ σ1 2 2 2, 2) H0 : σ σ1 2 2 2= 3) H1 : σ σ1 2 2 2≠ 4) α = 0.02 5) The test statistic is f s s 0 1 2 2 2 = 6) Reject the null hypothesis if f0 < f0 99 7 7. , , where f0 99 7 7. , , = 0.143 or f0 > f0 01 7 7. , , where f0 01 7 7. , , = 6.99 7) n1 = 8 n2 = 8 15 ( . . ) . . ( . ) . ( . ) ( . . ) . . ( . ) . ( . ) 0 05 0 0267 196 0 05 1 0 05 300 0 0267 1 0 0267 300 0 05 0 0267 196 0 05 1 0 05 300 0 0267 1 0 0267 3001 2 − − − + − ≤ − ≤ − + − + − p p − ≤ − ≤0 0074 0 0541 2. .p p Since this interval contains the value zero, we are 95% confident there is no significant difference in the fraction of defective parts produced by the two machines and that the difference in proportions is between − 0.0074 and 0.054. 5-56. a) One-Way Analysis of Variance ---------------------------------------------------------------------------- Data: Observation Level codes: SCCM Labels: Means plot: LSD Confidence level: 90 Range test: LSD Analysis of variance ------------------------------------------------------------------------------ Source of variation Sum of Squares d.f. Mean square F-ratio Sig. level ------------------------------------------------------------------------------ Between groups 3.6477778 2 1.8238889 3.586 .0534 Within groups 7.6300000 15 .5086667 ------------------------------------------------------------------------------ Total (corrected) 11.277778 17 0 missing value(s) have been excluded. Reject H0 at α = 0.01. C2F6 flow rate does appear to affect etch uniformity. 200160125 5 4 3 FLOW U N IF O R M IT b) Examining the box plots, the 125 and 160 mean levels seem to be the most different. 5-58. a) One-Way Analysis of Variance ------------------------------------------------------------------------------ Data: Density Level codes: Temp Labels: 16 Means plot: LSD Confidence level: 95 Range test: LSD Analysis of variance ------------------------------------------------------------------------------ Source of variation Sum of Squares d.f. Mean square F-ratio Sig. level ------------------------------------------------------------------------------ Between groups .1391104 3 .0463701 2.616 .0827 Within groups .3190714 18 .0177262 ------------------------------------------------------------------------------ Total (corrected) .4581818 21 0 missing value(s) have been excluded. Do not reject H0. There is insignificant evidence to indicate the four firing temperatures affect the density of the brick. b) P-value = 0.0827 5-60. One-Way Analysis of Variance ------------------------------------------------------------------------------ Data: Conductivity Level codes: Coating Labels: Means plot: LSD Confidence level: 95 Range test: LSD Analysis of variance ------------------------------------------------------------------------------ Source of variation Sum of Squares d.f. Mean square F-ratio Sig. level ------------------------------------------------------------------------------ Between groups 1060.5000 4 265.12500 16.349 .0000 Within groups 243.2500 15 16.21667 ------------------------------------------------------------------------------ Total (corrected) 1303.7500 19 0 missing value(s) have been excluded. Reject H0. There appears to be a significant difference among the five coating types in their effect on conductivity. 17 Chapter 6 (Section 6.1-6.4) 6-2. a) y x0 0 1 1= +β β Sxx = − =1432158 339916147820 2 . . Sxy = − =108367 141445 1478 12 75 20. . ( )( . ) $ . . . $ ( . )( ) .. β β 1 0 12 75 20 1478 20 141445 339916 0 0041612 0 0041612 0 3299892 = = = = − = S S xy xx $ . .y x= +0 3299892 0 0041612 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 -50 0 50 100 x y b) $ . . ( ) .y = + =0 3299892 0 0041612 85 0 683689 c) $ . . ( ) .y = + =0 3299892 0 0041612 90 0 7044949 d) $ .β1 0 00416= 6-4. a) Regression Analysis - Linear model: Y = a+bX ------------------------------------------------------------------------------ Dependent variable: Games Independent variable: Yards ------------------------------------------------------------------------------ Standard T Prob. Parameter Estimate Error Value Level ------------------------------------------------------------------------------ Intercept 21.7883 2.69623 8.081 .00000 Slope -7.0251E-3 1.25965E-3 -5.57703 .00001 ------------------------------------------------------------------------------ Analysis of Variance ------------------------------------------------------------------------------ Source Sum of Squares Df Mean Square F-Ratio Prob. Level Model 178.09231 1 178.09231 31.1032 .00001 Residual 148.87197 26 5.72585 ------------------------------------------------------------------------------ Total (Corr.) 326.96429 27 Correlation Coefficient = -0.738027 R-squared = 54.47 percent Stnd. Error of Est. = 2.39287 20 ------------------------------------------------------------------------------ Independent variable coefficient std. error t-value sig.level ------------------------------------------------------------------------------ CONSTANT 47.173999 49.581476 0.9514 0.3555 x1 -9.735202 3.691625 -2.6371 0.0179 x2 0.428287 0.223933 1.9126 0.0739 x3 18.237455 1.311802 13.9026 0.0000 ------------------------------------------------------------------------------ R-SQ. (ADJ.) = 0.9925 SE= 3.479627 MAE= 2.511105 DurbWat= 1.778 Previously: 0.0000 0.000000 0.000000 0.000 20 observations fitted, forecast(s) computed for 0 missing val. of dep. var. If the calculations were to be done by hand, add a cross-product column to the X matrix and use Equation (6-21). a) $ . . . .y x x x= − + +47 174 9 7352 0 4283 18 23751 2 3 b) $ . . ( . ) . ( ) . ( ) .y = − + + =47174 9 7352 14 5 0 4283 220 18 2375 5 9143 6-12. a) 1) The parameter of interest is the regressor variable coefficient, β1 2) H 0 1 0:β = 3) H1 1 0:β ≠ 4) α = 0.05 5) The test statistic is f MS MS SS SS n R E R E 0 1 2 = = − / / ( ) 6) Reject H0 if f0 > fα,1,12 where f0.05,1,12 = 4.75 7) Using results from Exercise 6-1 SS S SS S SS R xy E yy R = = − − = = − = − = $ . ( . ) . . . . β1 2 3298017 59 057143 137 59 159 71429 137 59143 22 123 f0 137 59 22 123 12 74 63= = . . / . 8) Since 74.63 > 4.75 reject H 0 and conclude that compressive strength is a significant in predicting intrinsic permeability of concrete at α = 0.05. We can therefore conclude model specifies a useful linear relationship between these two variables. P value− ≅ 0 000002. b) $ . .σ2 2 22123 12 1844= = − = =MS SS nE E 6-14. a) Refer to ANOVA table of Exercise 6-4. 1) The parameter of interest is the regressor variable coefficient, β1. 21 2) H 0 1 0:β = 3) H1 1 0:β ≠ 4) α = 0.01 5) The test statistic is f MS MS SS SS n R E R E 0 1 2 = = − / / ( ) 6) Reject H0 if f0 > fα,1,26 where f0.01,1,26 = 7.724 7) Using the results of Exercise 6-4 f MS MS R E 0 311032= = . 8) Since 31.1032 > 7.724 reject H0 and conclude the model is useful at α = 0.01. P value− = 0 000007. b) $ .σ2 5 72585= =MSE c) 2) H0 1 0:β = 3) H1 1 0:β ≠ 4) α = 0.01 5) The test statistic is t se 0 1 1 = $ ( $ ) β β 6) Reject H0 if t0 < −tα/2,n-2 where −t0.005,26 = −2.78 or t0 > t0.005,26 = 2.78 7) Using the results from Exercise 6-4 t0 0 0070251 0 00125965 5577= − = − . . . 8) Since −5.577 < −2.78 reject H 0 and conclude the regressor is useful in the model at α = 0.01. 6-16. Refer to ANOVA for Exercise 6-6 a) 1) The parameter of interest is the regressor variable coefficient, β1. 2) H 0 1 0:β = 3) H1 1 0:β ≠ 4) α = 0.01 5) The test statistic is f MS MS SS SS n R E R E 0 1 2 = = − / / ( ) 6) Reject H0 if f0 > fα,1,22 where f0.01,1,10 = 10.049 7) Using the results from Exercise 6-6 f0 28058312 1 37 746089 10 74334 4= = . / . / . 8) Since 74334.4 > 10.049, reject H 0 and conclude the model is useful α = 0.01. P-value < 0.000001 22 b) MSE = SSE/(n − p) = $ .σ2 377461= c) 1) The parameter of interest is the regressor variable coefficient, β1. 2) H 0 1 10:β = 3) H1 1 10:β ≠ 4) α = 0.01 5) The test statistic is t se 0 1 1 0 1 = −$ ( $ ) ,β β β 6) Reject H0 if t0 < −tα/2,n-2 where −t0.005,10 = −3.17 or t0 > t0.005,10 = 3.17 7) Using the results from Exercise 6-6 t0 9 21 10 0 0338 2337= − = − . . . 8) Since −23.37 < −3.17 reject H 0 and conclude the regressor is useful at α = 0.01. P-value = 0. 6-18. a) β1 0 4 82: .t = P-value = 2(4.08 E-5) = 8.16 E-5 β2 0 8 21: .t = P-value = 2(1.91 E-8) = 3.82 E-8 β3 0 0 98: .t = P-value = 2 (0.1689) = 0.3378 b) 2) H 0 3 0:β = 3) H1 3 0:β ≠ 4) α = 0.05 5) Reject H0 if t0 < −tα/2,n-2 where −t0.025,23 = −2.074 or t0 > tα/2,n-2 where t0.025,23 = 2.074 6) t 0 0 98= . 7) Since −2.074 < 0.98 < 2.074 do not reject H0 and conclude that x3 is not useful as a regressor in the model at α = 0.05. 6-20. Analysis of Variance for the Full Regression ------------------------------------------------------------------------------ Source Sum of Squares DF Mean Square F-Ratio P- value ------------------------------------------------------------------------------ Model 12161.6 2 6080.79 9.35303 .0514 Error 1950.42 3 650.141 ------------------------------------------------------------------------------ Total (Corr.) 14112.0 5 R-squared = 0.86179 Stnd. error of est. = 25.4979 R-squared (Adj. for d.f.) = 0.76965 Durbin-Watson statistic = 2.56509 Assume no interaction. a) 1) The parameters of interest are the regression coefficients, βj . 2) H 0 1 2 0:β β= = 3) H1: at least one β j ≠ 0 4) α = 0.05 5) The test statistic is f SS k SS n p R E 0 = − / / ( ) 25 Chapter 7 (Problems 7.1-7.8) 7-1. a) Estimated Effects and Coefficients for Life Term Effect Coef StDev Coef T P Constant 99.38 7.711 12.89 0.000 Material 6.50 3.25 7.711 0.42 0.681 Temp -91.75 -45.87 7.711 -5.95 0.000 Material*Temp -14.50 -7.25 7.711 -0.94 0.366 b) 1-1 150 100 50 Temp Mean Interaction Plot for Life The interaction plot does not indicate a strong interaction between temperature and material. c) The t-ratios are given in the output shown in part a. The t-ratios indicate that temperature is significant, but material and the interaction material*temperature are not at the α = 0.05 level. d) The 95% confidence intervals are given by effect estimate ± 2(s.e.(effect)) where s.e.(effect) = 2[s.e.(coefficient)]. The s.e.(coefficient) is given in the Minitab output of part a. Temperature: s.e.(effect) = 2(7.711) = 15.42 Approximate 95% confidence interval on the the effect of Temperature: 6.5 ± 2(15.42) (−24.34, 37.34) Material: s.e.(effect) = 2(7.711) = 15.42 Approximate 95% confidence interval for the effect of Material: −91.75 ± 2(15.42) (−122.59, −60.91) Material*Temperature: s.e.(effect) = 2(7.711) = 15.42 Approximate 95% confidence interval for the effect of Material*Temperature: −14.50 ± 2(15.42) (−45.34, 16.34) e) The regression equation is Life = 99.4 + 3.25 Material - 45.9 Temp - 7.25 Material*Temp 26 Predictor Coef StDev T P Constant 99.375 7.711 12.89 0.000 Material 3.250 7.711 0.42 0.681 Temp -45.875 7.711 -5.95 0.000 Material*Temp -7.250 7.711 -0.94 0.366 Based on the regression analysis, only temperature appears to be the significant factor. This result is equivalent to that obtained in part c. The final regression analysis and model are Life = 99.4 - 45.9 Temp Predictor Coef StDev T P Constant 99.375 7.448 13.34 0.000 Temp -45.875 7.448 -6.16 0.000 S = 29.79 R-Sq = 73.0% R-Sq(adj) = 71.1% Analysis of Variance Source DF SS MS F P Regression 1 33672 33672 37.94 0.000 Error 14 12426 888 Total 15 46098 The analysis of variance indicates the final regression model is adequate for this set of data. This is evident by p-value ≅ 0.000. P-Value: 0.493 A-Squared: 0.324 Anderson-Darling Normality Test N: 16 StDev: 28.7814 Average: -0.0000000 40200-20-40-60 .999 .99 .95 .80 .50 .20 .05 .01 .001 P ro ba bi lit y RESI1 Normal Probability Plot 27 10-1 50 0 -50 Temp R es id ua l Residuals Versus Temp (response is Life) There does not appear to be any serious departure from normality shown in the normal probability plot of the residuals. The assumption of constant variance does not appear to be violated. The residuals appear to have the same spread for both levels of Temperature. 7-2. a) Estimated Effects and Coefficients for Surface Term Effect Coef StDev Coef T P Constant 73.750 3.955 18.65 0.000 Paint -0.167 -0.083 3.955 -0.02 0.984 Drying 2.833 1.417 3.955 0.36 0.729 Paint*Drying -19.500 -9.750 3.955 -2.47 0.039 b) - 85 75 65 Drying Time Mean Interaction Plot for Surface The interaction plot indicates an interaction between drying time and paint type. c) The t-ratios are given in the output shown in part a. The t-ratios indicate that the interaction of Paint type*Drying Time is significant at the α = 0.05 level. 30 10-1 20 10 0 -10 -20 Interaction Residual Residuals Versus Interaction (response is Surface) There does not appear to be any serious departure from normality shown in the normal probability plot of the residuals. The assumption of constant variance does not appear to be seriously violated. The residuals for the high level of the interaction appear to be slightly more spread out than those for the low level. 7-3. a) Estimated Effects and Coefficients for current Term Effect Coef StDev Coef T P Constant 265.00 2.357 112.43 0.000 Glass -60.00 -30.00 2.357 -12.73 0.000 Phosphor -16.67 -8.33 2.357 -3.54 0.008 Glass*Phosphor -3.33 -1.67 2.357 -0.71 0.500 b) 300 290 280 270 260 250 240 230 Phosphor Mean Interaction Plot for current The interaction plot does not indicate any significant interaction between Glass and Phosphor type. c) The t-ratios are given in the output shown in part a. The t-ratios indicate that both Glass type and Phosphor type are significant at the α = 0.05 level. 31 d) The 95% confidence intervals are given by effect estimate ± 2(s.e.(effect)) where s.e.(effect) = 2[s.e.(coefficient)]. The s.e.(coefficient) is given in the Minitab output of part a. Glass Type: s.e.(effect) = 2(2.357) = 4.714 Approximate 95% confidence interval on the the effect of Glass Type: −60.0 ± 2(4.714) (−69.43, −50.57) Phosphor Type: s.e.(effect) =2(2.357) = 4.714 Approximate 95% confidence interval for the effect of Phosphor Type: −16.67± 2(4.714) (−26.1, −7.242) Glass Type*Phosphor Type: s.e.(effect) =2(2.357) = 4.714 Approximate 95% confidence interval for the effect of Glass Type*Phosphor Type: −3.33 ± 2(4.714) (−12.76, −6.1) e) The regression equation is current = 265 - 30.0 Glass - 8.33 Phosphor - 1.67 Glass*Phosphor Predictor Coef StDev T P Constant 265.000 2.357 112.43 0.000 Glass -30.000 2.357 -12.73 0.000 Phosphor -8.333 2.357 -3.54 0.008 Glass*Ph -1.667 2.357 -0.71 0.500 Based on the regression analysis, both Glass type and Phosphor type appear to be the significant factors. This result is equivalent to that obtained in part c. The regression analysis and final model are: The regression equation is current = 265 - 30.0 Glass - 8.33 Phosphor Predictor Coef StDev T P Constant 265.000 2.291 115.69 0.000 Glass -30.000 2.291 -13.10 0.000 Phosphor -8.333 2.291 -3.64 0.005 S = 7.935 R-Sq = 95.4% R-Sq(adj) = 94.3% Analysis of Variance Source DF SS MS F P Regression 2 11633.3 5816.7 92.38 0.000 Error 9 566.7 63.0 Total 11 12200.0 Source DF Seq SS Glass 1 10800.0 Phosphor 1 833.3 The analysis of variance indicates the final regression model is adequate for this set of data. This is evident by p-value ≅ 0.000. 32 P-Value: 0.292 A-Squared: 0.408 Anderson-Darling Normality Test N: 12 StDev: 7.17741 Average: 0 151050-5 .999 .99 .95 .80 .50 .20 .05 .01 .001 P ro ba bi lit y RESI1 Normal Probability Plot 10-1 20 10 0 -10 Glass R es id ua l Residuals Versus Glass (response is current) 10-1 20 10 0 -10 Phosphor R es id ua l Residuals Versus Phosphor (response is current) 35 10-1 30 20 10 0 -10 -20 -30 -40 -50 -60 position R es id ua l Residuals Versus position (response is density) There does not appear to be any serious departure from normality shown in the normal probability plot of the residuals. The assumption of constant variance does not appear to be violated. The residuals appear to have approximately the same spread for both levels of Temperature and both levels of Position. 7-5. a) Estimated Effects and Coefficients for warping Term Effect Coef StDev Coef T P Constant 19.8750 0.7181 27.68 0.000 temp -1.7500 -0.8750 0.7181 -1.22 0.290 %copper 7.2500 3.6250 0.7181 5.05 0.007 temp*%copper 2.7500 1.3750 0.7181 1.91 0.128 b) 24 1 1 %copper Mean Interaction Plot for warping The interaction plot indicates there may be a slightly significant interaction between %Copper and Temperature. c) The t-ratios are given in the output shown in part a. The t-ratios indicate that only %Copper is significant at the α = 0.05 level. 36 d) The 95% confidence intervals are given by effect estimate ± 2(s.e.(effect)) where s.e.(effect) = 2[s.e.(coefficient)]. The s.e.(coefficient) is given in the Minitab output of part a. Temperature: s.e.(effect) = 2(0.7181) = 1.436 Approximate 95% confidence interval on the the effect of Temperature: −1.75 ± 2(1.436) (−4.622,1.122) %Copper: s.e.(effect) =2(0.7181) = 1.436 Approximate 95% confidence interval for the effect of %Copper: 7.25± 2(1.436) (4.378, 10.122) Temperature*%Copper: s.e.(effect) =2(0.7181) = 1.436 Approximate 95% confidence interval for the effect of Temperature*%Copper: 2.75 ± 2(1.436) (−0.122, 5.622) e) The regression equation is warping = 19.9 - 0.875 temp + 3.63 %copper + 1.38 Temp*%Copper Predictor Coef StDev T P Constant 19.8750 0.7181 27.68 0.000 temp -0.8750 0.7181 -1.22 0.290 %copper 3.6250 0.7181 5.05 0.007 Temp*%Co 1.3750 0.7181 1.91 0.128 Based on the regression analysis, only %Copper appears to be the significant factor. This result is equivalent to that obtained in part c. The regression analysis and final model are The regression equation is warping = 19.9 + 3.62 %copper Predictor Coef StDev T P Constant 19.8750 0.8868 22.41 0.000 %copper 3.6250 0.8868 4.09 0.006 S = 2.508 R-Sq = 73.6% R-Sq(adj) = 69.2% Analysis of Variance Source DF SS MS F P Regression 1 105.13 105.13 16.71 0.006 Error 6 37.75 6.29 Total 7 142.88 The analysis of variance indicates the final regression model is adequate for this set of data. This is evident by p-value ≅ 0.006. 37 P-Value: 0.693 A-Squared: 0.235 Anderson-Darling Normality Test N: 8 StDev: 2.32225 Average: 0.0000000 43210-1-2-3-4 .999 .99 .95 .80 .50 .20 .05 .01 .001 P ro ba bi lit y RESI1 Normal Probability Plot 10-1 4 3 2 1 0 -1 -2 -3 -4 -5 %copper R es id ua l Residuals Versus %copper (response is warping) There does not appear to be any serious departure from normality shown in the normal probability plot of the residuals. The residuals for the low level of %Copper are more spread out than for the high level. The constant variance assumption may be a concern. 7-6. a) Estimated Effects and Coefficients for Fatigue Term Effect Coef StDev Coef T P Constant 2.72187 0.07126 38.20 0.000 Frequenc 1.44125 0.72063 0.07126 10.11 0.000 Envirome -0.14375 -0.07187 0.07126 -1.01 0.333 Frequenc*Envirome 0.03875 0.01937 0.07126 0.27 0.790 b) 40 10-1 0.5 0.4 0.3 0.2 0.1 0.0 -0.1 -0.2 -0.3 -0.4 Frequenc R es id ua l Residuals Versus Frequenc (response is Fatigue) There does not appear to be any serious departure from normality shown in the normal probability plot of the residuals. The residuals for the high level of Frequency are more spread out than for the low level. The constant variance assumption may be a concern. 7-7. a) Estimated Effects and Coefficients for Current Term Effect Coef StDev Coef T P Constant 9.2050 0.07622 120.77 0.000 Polysili -0.4850 -0.2425 0.07622 -3.18 0.033 Anneal 2.0300 1.0150 0.07622 13.32 0.000 Polysili*Anneal 0.3350 0.1675 0.07622 2.20 0.093 b) 1 9 8 Anneal Mean Interaction Plot for Current The interaction plot does not indicate a significant interaction between Polysilicon Doping and Anneal. c) The t-ratios are given in the output shown in part a. The t-ratios indicate that both Polysilicon doping and Anneal are significant at the α = 0.05 level. d) The 95% confidence intervals are given by effect estimate ± 2(s.e.(effect)) where s.e.(effect) = 2[s.e.(coefficient)]. The s.e.(coefficient) is given in the Minitab output of part a. 41 Polysilicon Doping: s.e.(effect) = 2(0.07622) = 0.1524 Approximate 95% confidence interval on the the effect of Polysilicon Doping: −0.485± 2(0.1524) (−0.7898, −0.1802) Anneal: s.e.(effect) =2(0.07622) = 0.1524 Approximate 95% confidence interval for the effect of Anneal: 2.03± 2(0.1524) (1.725, 2.34) Polysilicon doping*Anneal: s.e.(effect) =2(0.07622) = 0.1524 Approximate 95% confidence interval for the effect of Polysilicon doping*Anneal: 0.335 ± 2(0.1524) (0.0302, 0.6398) e) The regression equation is Current = 9.21 - 0.242 Polysilicon + 1.02 Anneal + 0.167 Poly*Anneal Predictor Coef StDev T P Constant 9.20500 0.07622 120.77 0.000 Polysili -0.24250 0.07622 -3.18 0.033 Anneal 1.01500 0.07622 13.32 0.000 Poly*Ann 0.16750 0.07622 2.20 0.093 Based on the regression analysis, both Polysilicon and Anneal appear to be the significant factors. This result is equivalent to that obtained in part c. The regression analysis and final model are The regression equation is Current = 9.21 - 0.242 Polysilicon + 1.02 Anneal Predictor Coef StDev T P Constant 9.2050 0.1013 90.88 0.000 Polysili -0.2425 0.1013 -2.39 0.062 Anneal 1.0150 0.1013 10.02 0.000 S = 0.2865 R-Sq = 95.5% R-Sq(adj) = 93.7% Analysis of Variance Source DF SS MS F P Regression 2 8.7123 4.3561 53.08 0.000 Error 5 0.4103 0.0821 Total 7 9.1226 Source DF Seq SS Polysili 1 0.4704 Anneal 1 8.2418 Unusual Observations Obs Polysili Current Fit StDev Fit Residual St Resid 5 -1.00 8.900 8.432 0.175 0.468 2.06R R denotes an observation with a large standardized residual The analysis of variance indicates the final regression model is adequate for this set of data. This is evident by p-value ≅ 0.000. 42 P-Value: 0.030 A-Squared: 0.747 Anderson-Darling Normality Test N: 8 StDev: 0.242119 Average: 0 0.50.40.30.20.10.0-0.1-0.2 .999 .99 .95 .80 .50 .20 .05 .01 .001 P ro ba bi lit y RESI1 Normal Probability Plot 10-1 0.5 0.4 0.3 0.2 0.1 0.0 -0.1 -0.2 Anneal R es id ua l Residuals Versus Anneal (response is Current)