MGB 0B Homework # 4.4 a α =.05: t = =.05 LSD = α /,n k t.05, 7 t α /,n k MSE + =.05 700 + = 4.8 n i n j 0 0 i =, j = 8.7 0.4 7. i =, j = 8.7.7 5.0 i =, j = 0.4.7. Conclusion: µ differs from µ and µ. b C = ()/ =, α =.05, α = α E / C =.067: t α /,n k = t.008, 7 =.55 (from Excel) LSD = E t α /,n k MSE + =.55 700 + = 0.0 n i n j 0 0 i =, j = 8.7 0.4 7. i =, j = 8.7.7 5.0 i =, j = 0.4.7. Conclusion: µ and µ differ. c q (k, ν = α ) q. 05 (,7).5 MSE ϖ = q α (k, ν) =.5 n g 700 = 9.5 0 i =, j = 8.7 0.4 7. i =, j = 8.7.7 5.0 i =, j = 0.4.7. Conclusion: µ and µ differ. 4.5 a = µ = µ H 0 : µ H : At least two means differ. 0(5.7) + 0(49.7) + 0(44.) x = = 48.96 0 + 0 + 0
SST = n j (x j x) = 0(5.7 48.96) +0(49.7 48.96) + 0(44. 48.96) =,78 j s j SSE = ( n ) = (0 )(94.6) + (0 )(5.6) + (0 )(9.9) =,86 SST,78 Treatments k = SST =,78 = = 589. 0 k SSE,86 Error n k = 87 SSE =,86 = = 59. 0 n k 87 MST MSE = 589.0 =.70 59.0 F =.70, p-value =.086. There is enough evidence to infer that speed of promotion varies between the three sizes of engineering firms. 59.0 b q α (k, ν) = q. 05 (,87).40 ϖ =.40 = 7. 8 0 i =, j = 5.7 49.7.80 i =, j = 5.7 44.44 8.8 i =, j = 49.7 44. 5.04 The means of small and large firms differ. Answer (v) is correct. 4.0 a H 0 : µ = µ = µ = µ 4 = µ 5 4 5 6 7 H : At least two means differ. A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 747.4 4 46.86 4.46 0.007.4 Within Groups 98.7 45 97.89 Total 57. 49 F = 4.46, p-value =.007. There is enough evidence to infer that there are differences in the effect of the new assessment system between the five boroughs.
b Multiple Comparisons LSD Omega Treatment Treatment Difference Alpha = 0.05 Alpha = 0.05 Borough A Borough B -7.4.90 5.40 Borough C -6.8.90 5.40 Borough D -.4.90 5.40 Borough E -.94.90 5.40 Borough B Borough C.4.90 5.40 Borough D 5.00.90 5.40 Borough E.48.90 5.40 Borough C Borough D.76.90 5.40 Borough E.4.90 5.40 Borough D Borough E -.5.90 5.40 The mean assessments in borough A differs from the means in boroughs B and C. c The assessments for each borough are required to be normally distributed with equal variances. d The histograms are approximately bell-shaped and the sample variances are similar. 4.7 Treatments 75 9.67 7.99 Blocks 9 65 69.44 6.05 Error 7 0.48 Total 9,0 > 7 a Rejection region: F Fα,k,n k b+ = F.0,, = 4.60 Conclusion: F = 7.99, p-value =.0006. There is enough evidence to conclude that the treatment means differ. b Rejection region: F > Fα,b,n k b+ = F.0,9, 7 =.5 Conclusion: F = 6.05, p-value =.000. There is enough evidence to conclude that the block means differ. 4.7 a. k =, b = 5, Grand mean = 0.4 SS(Total) = k b i= ij ( x x) = (7 0.4) + (0 0.4) + ( 0.4) + (9 0.4) + ( 0.4)
k + ( 0.4) + (8 0.4) + (6 0.4) + ( 0.4) + (0 0.4) + ( 8 0.4) + (9 0.4) + ( 0.4) + (6 0.4) + ( 0.4) = 99.6 j = SST = b(x[t] x) = 5[(0 0.4) + (.8 0.4) + (9.4 0.4) ] 5. 6 b i = SSB = k(x[b] x) = [(9 0.4) + (9 0.4) + (.7 0.4) + (9. 0.4) + ( 0.4) ] 48. i= SSE = SS(Total) SST SSB = 99.6 5.6 48. = 5.7 b SS(Total) = k b i= ij ( x x) = (7 0.4) + (0 0.4) + ( 0.4) + (9 0.4) + ( 0.4) k + ( 0.4) + (8 0.4) + (6 0.4) + ( 0.4) + (0 0.4) + ( 8 0.4) + (9 0.4) + ( 0.4) + (6 0.4) + ( 0.4) = 99.6 j j = SST = n (x x) = 5(0 0.4) + 5(.8 0.4) + 5(9.4 0.4) 5. 6 SSE = SS(Total) SST = 99.6 5.6 = 84.0 c The variation between all the data is the same for both designs. d The variation between treatments is the same for both designs. e Because the randomized block design divides the sum of squares for error in the one-way analysis of variance into two parts. 4.79 Treatments 0.6 5..86 Blocks 9,00 59.0 6.64 Error 8 6.7 5.97 a = µ = µ H 0 : µ H : At least two means differ. Rejection region: F > Fα,k,n k b+ = F.05,, 8. F =.86, p-value =.4. There is not enough evidence to conclude that there are differences in sales ability between the holders of the three degrees. b H 0 : µ = µ = = µ 0 H : At least two means differ. 4
F = 6.64, p-value = 0. There is sufficient evidence to indicate that there are differences between the blocks of students. The independent samples design would not be recommended. c The commissions for each type of degree are required to be normally distributed with the same variance. d The histograms are bell shaped and the sample variances are similar. 4.89 Factor A 0 67.67.7 Factor B 859 49.5 4.60 Interaction 6 5 85.5.9 Error 84 7845 9.9 Total 95 940 a Rejection region: F > Fα,(a )(b ),n ab F.05,6, 84 =.5 F =.9. There is not enough evidence to conclude that factors A and B interact. b Rejection region: F > Fα,a,n ab F.05,, 84 =.76 F =.7. There is not enough evidence to conclude that differences exist between the levels of factor A. c Rejection region: F > Fα,b,n ab F.05,, 84 =.5 F = 4.60. There is enough evidence to conclude that differences exist between the levels of factor B. 4.95 a Detergents and temperatures b The response variable is the whiteness score. c There are a = 5 detergents and b = temperatures. 9 0 4 5 6 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Sample 97 968.5 7.8 0.0000.06 Columns 967 4 74.9 6.7 0.000.44 Interaction 45 8 06.5.78 0.007.0 Within 490 5 0.4 Total 467 49 d Test for interaction: F =.78, p-value =.007. There is sufficient evidence to conclude that detergents and temperatures interact. The F-tests in Parts e and f are irrelevant. 5
4. H 0 : µ = µ = µ = µ 4 H : At least two means differ. 4 5 6 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups. 0.705 9.7 0.0000.69 Within Groups 7.99 04 0.0769 Total 0. 07 F = 9.7, p-value = 0. There is sufficient evidence to infer that there are differences in changes to the TSE depending on the loss the previous day. 5.5 H 0 : p =.05, p =.07, p =.04, p =.84 4 Cell i H : At least one pi is not equal to its specified value. f i e f e ) i ( i i ( f i ei ) / ei 9 50(.05) =.5 6.5.8 50(.07) = 7.5 5.5.7 4 50(.04) = 0.0 4.0.60 4 94 50(.84) = 0.0-6.0. Total 50 50 χ = 7.9 Rejection region: χ > χα,k = χ.05, = 7.8 χ = 7.9, p-value =.0475. There is enough evidence to infer that the reported side effects of the placebo differ from that of the cold remedy. 5.9 H 0 : The two variables economic option and political affiliation) are independent Cell i H : The two variables are dependent f i e f e ) i ( i i ( f i ei ) / ei 0 444()/000 = 46.96-45.96 4.76 8 444(557)/000 =.99 48.0 9.85 6 444(4)/000 = 6.05 -.05 0.067 4 8 0()/000 = 4.0-5.0 0.588 6
5 67 0(557)/000 = 68.5 -.5 0.0 6 5 0(4)/000 = 8.46 6.54.7 7 50()/000 = 8.75 48.5 8.4 8 88 50(557)/000 =.75-4.75 4.58 9 50(4)/000 = 5.50-4.50 0.570 0 6 76()/000 = 58.6.74 0.9 90 76(557)/000 = 9.75 -.75 0.08 5 76(4)/000 = 4.99 0.0 0.000 Total 000 000 χ = 70.675 Rejection region: α (r )(c.0, 6 χ > χ, = χ = 6.8 χ = 70.675, p-value = 0. There is sufficient evidence to infer that political affiliation affects support for economic options. 5.75 H 0 : The two variables (cold and group) are independent 4 5 6 7 8 9 0 4 5 H : The two variables are dependent χ = 4.9, p-value =.468. There is not enough evidence to infer there are differences between the four groups. A B C D E Contingency Table Group Cold TOTAL 7 8 5 9 8 7 4 6 8 4 TOTAL 54 60 4 chi-squared Stat 4.9 df p-value 0.468 chi-squared Critical 7.847 7