C291FE2012_Solutions

# Vi it is not sufficient just to say reject h vii

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(vi) It is not sufficient just to say “reject H 0 (vii) Definition of Type I error b) The data are no longer paired; two different sets of cars were used for each Appraiser. So use the two-sample t-test results. c) df = 18; t* = 2.101 (838 – 762) ± 2.101(361)√[1/10 +1/10]) = 76 ± 339 or (-263,415) FE2012 A6 “RVs about RVs” a) D b) C c) (2) & (3) d) 58.48; 0.34 e) 4 f) E g) D Details and Comments: a) 172/500 = 0.344 (Use the overall total as the denominator) b) 54/170 = 0.318 (Use the column total as the denominator) c) (2) = Test of independence, (3) =Test of homogeneity – both are correct here d) E 11 = (172×170)/500 = 58.48; ( Obs Exp ) 2 / Exp =( 54–58.48) 2 /58.48 = 0.34 e) df = (# rows – 1)(# columns – 1) = (3–1)(3–1) = 4 f) From Table X, χ 2 *(4df) at 0.05 = 9.488. Since 6.63 is smaller, the P-value is larger than .05. (In fact, the P-value is larger than 0.01, since χ 2 *(4df) at 0.10 is 7.779.) g) Do not reject H 0 – there is not enough evidence of an association. 4

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FE2012 A7 “Making progress from process” a) = 30.729 – 1.836X b) Each hour of training reduces troubleshooting time by 1.836 min., on average. c) H 0 : β 1 = 0 H a : β 1 ≠ 0 Test stat = –13.35 or 178.10 P-value < 0.001 There is strong evidence of a linear relationship between troubleshooting time and hours of training. d) 93% e) –96% f) (–2.1332, –1.5388) g) Yes; the upper limit is to the left of -1 h) 1.435 i) (16.843,22.583) j) A Details and Comments: a) Read the coefficients from the bottom of the output. Don’t forget the “hat” on the “y”. b) Definition of slope. The slope is negative, so as X goes up, Y goes down. c) Two choices of test statistic: t-stat for slope, or F-stat from ANOVA table d) R 2 = 367.20/394.00 = 0.93 or 93% e) r = –√0.93 = –0.96 or –96%. Don’t forget the minus sign; the slope, and hence the correlation, is negative! f) t*(.025,13df) = 2.160; CI = –1.8360 ± 2.160×0.1376 = –1.8360 ± 0.2972 or (–2.1332, –1.5388) g) Interpretation of CI... h) s e = √MSE = √2.06 = 1.435 i) 6 = 30.729 – 1.836(6) = 19.713; 2 s e = 2(1.435) = 2.870; 19.713 ± 2.870 = (16.843,22.583) j) Horizontal band around x-axis 5
FE2012 A8 “How to ‘excel’ at regression” a) H 0 : β 1 = β 2 = β 3 = β 4 = 0 (The model is not worthwhile.) H a : At least one β i is not zero (The model is worthwhile.) b) 3.06 c) There is evidence that the model is worthwhile d) DEGREES, PREVJOBS e) \$13,052 f) 58% g) It would decrease h) It would not change i) B j ) D Details and Comments: a) Definition of hypotheses in regression ANOVA b) Use Table F, 0.05, with 4 and 15 df c) Since F-stat > F*, or since P-value = 0.0074 (from output), reject the null hypothesis d) P-values of these variables are greater than 0.05, hence likely not significant additions to the model. e) β AGE = 1.3052; 10yrs × 1.3052 = 13.052. Multiply by \$1000 to get \$13,052 f) R 2 = 0.5848 from the output g) Dropping a variable decreases R 2 h) Increasing sample size affects P-values but not the strength of the relationship.

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• Spring '10
• E.Fowler
• Null hypothesis, Statistical hypothesis testing

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